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ExaminerPrep › Question bank › IB Maths AA HL › Topic 1 Number and algebra

IB Maths Analysis & Approaches · Higher Level · Topic 1

Number and algebra

342 free, high-quality past-paper-style questions on Number and algebra for IB Maths AA HL, each with a full mark scheme. Covers: Geometric sequences and series; Exponents and introduction to logarithms; Counting principles and the extended binomial theorem; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 1 at Standard Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,584 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1AAHL1-q00051.3 Geometric sequences and series Paper 1Easy6 marks

For a geometric sequence with common ratio \(r\neq1\), the sum of the first three terms is 26 and the sum of the first six terms is 728.

(a)

Show that \(r^{3}=27\).

(2)
(b)

Find the first term, \(u_{1}\).

(2)
(c)

Find the least value of \(n\) for which the sum of the first \(n\) terms is greater than \(10^{6}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{S_{6}}{S_{3}}=\dfrac{r^{6}-1}{r^{3}-1}=r^{3}+1\), or \(S_{6}-S_{3}=r^{3}S_{3}\)M1
\(r^{3}+1=\dfrac{728}{26}=28\)A1
\(r^{3}=27\)AG
(b)
\(r=3\) and \(\dfrac{u_{1}(27-1)}{3-1}=26\)(M1)
\(u_{1}=2\)A1
(c)
\(S_{n}=\dfrac{2(3^{n}-1)}{2}=3^{n}-1\gt 10^{6}\)(M1)\(3^{12}=531\,441\), \(3^{13}=1\,594\,323\)
\(n=13\)A1

Answers: (b) \(u_{1}=2\) · (c) \(n=13\)

2AAHL1-q00111.5 Exponents and introduction to logarithms Paper 1Easy7 marks
(a)

Show that \(\dfrac{3^{n+2}-3^{n}}{8\times3^{n-1}}=3\) for all \(n\in\mathbb{Z}\).

(2)
(b)

Find the integers \(a\) and \(b\) for which \(12^{a}\times18^{b}=2^{8}\times3^{7}\).

(3)
(c)

Solve \(5^{2k}\times25^{-1}=125^{k-2}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(3^{n+2}-3^{n}=3^{n-1}(3^{3}-3)\) or \(=3^{n}(9-1)\)M1
\(\dfrac{24\times3^{n-1}}{8\times3^{n-1}}\) or \(\dfrac{8\times3^{n}}{8\times3^{n-1}}\)A1
\(=3\)AG
(b)
\(12^{a}\times18^{b}=2^{2a}3^{a}\times2^{b}3^{2b}=2^{2a+b}\times3^{a+2b}\)M1Writing both bases as products of prime powers.
\(2a+b=8\) and \(a+2b=7\)A1Equating the powers of 2 and of 3.
\(a=3\), \(b=2\)A1
(c)
\(5^{2k-2}=5^{3k-6}\)(M1)
\(k=4\)A1

Answers: (b) \(a=3\), \(b=2\) · (c) \(k=4\)

3AAHL1-q00221.10 Counting principles and the extended binomial theorem Paper 1Easy6 marks

A team of 4 students is chosen from a group of 8 juniors and 6 seniors.

(a)

Find the number of different teams that can be chosen.

(2)
(b)

Find the number of different teams that contain exactly 2 seniors.

(2)
(c)

Find the number of different teams that contain at least one senior.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\({}^{14}C_{4}=\dfrac{14\times13\times12\times11}{24}\)(M1)
\(1001\)A1
(b)
\({}^{8}C_{2}\times{}^{6}C_{2}=28\times15\)(M1)
\(420\)A1
(c)
\(1001-{}^{8}C_{4}=1001-70\)(M1)Or adding the cases with 1, 2, 3 and 4 seniors: 336 + 420 + 160 + 15.
\(931\)A1

Answers: (a) 1001 · (b) 420 · (c) 931

4AAHL1-q00281.11 Partial fractions Paper 1Easy4 marks
(a)

Factorise \(x^{2}-x-6\).

(1)
(b)

Hence express \(\dfrac{7x-1}{x^{2}-x-6}\) in partial fractions.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\((x-3)(x+2)\)A1
(b)
\(7x-1\equiv A(x+2)+B(x-3)\)M1
\(x=3\): \(20=5A\), so \(A=4\); \(x=-2\): \(-15=-5B\), so \(B=3\)A1Award A1 for both values.
\(\dfrac{4}{x-3}+\dfrac{3}{x+2}\)A1

Answers: (a) \((x-3)(x+2)\) · (b) \(\dfrac{4}{x-3}+\dfrac{3}{x+2}\)

5AAHL1-q00311.12 Complex numbers in Cartesian form Paper 1Easy6 marks

Let \(z=3-2i\) and \(w=1+4i\). Give each answer in the form \(a+bi\), where \(a,b\in\mathbb{R}\).

(a)

Find \(z+2w\).

(1)
(b)

Find \(zw\).

(2)
(c)

Find \(\dfrac{z}{w}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(5+6i\)A1
(b)
\(3+12i-2i-8i^{2}\)M1Attempt to expand with \(i^{2}=-1\) used.
\(11+10i\)A1
(c)
multiplying numerator and denominator by \(w^{*}=1-4i\)M1
\(\dfrac{3-12i-2i+8i^{2}}{17}\)A1Denominator 17 must be seen.
\(-\dfrac{5}{17}-\dfrac{14}{17}i\)A1Accept \(\dfrac{-5-14i}{17}\).

Answers: (a) \(5+6i\) · (b) \(11+10i\) · (c) \(-\tfrac{5}{17}-\tfrac{14}{17}i\)

6AAHL1-q00321.12 Complex numbers in Cartesian form Paper 1Easy6 marks

The number \(i\) satisfies \(i^{2}=-1\).

(a)

Write down the value of \(i^{4}\).

(1)
(b)

Find \(i^{2027}\).

(2)
(c)

Find \(\displaystyle\sum_{k=1}^{50}i^{k}\), giving your answer in the form \(a+bi\).

(2)
(d)

Write down the least positive integer \(n\) for which \(\displaystyle\sum_{k=1}^{n}i^{k}=0\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(1\)A1
(b)
\(2027=4\times506+3\), so \(i^{2027}=(i^{4})^{506}\,i^{3}\)M1
\(-i\)A1
(c)
powers repeat in blocks of four with sum \(i-1-i+1=0\), so the sum equals \(i^{49}+i^{50}\)M1Or geometric series \(\dfrac{i(1-i^{50})}{1-i}=\dfrac{2i}{1-i}\).
\(-1+i\)A1
(d)
\(n=4\)A1

Answers: (a) 1 · (b) \(-i\) · (c) \(-1+i\) · (d) 4

7AAHL1-q00331.12 Complex numbers in Cartesian form Paper 1Easy6 marks

The complex number \(z\) satisfies \(2z+iz^{*}=5+4i\).

(a)

Find \(z\) in the form \(a+bi\), where \(a,b\in\mathbb{R}\).

(4)
(b)

Hence find \(\dfrac{z}{z^{*}}\) in the form \(c+di\), where \(c,d\in\mathbb{R}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
substituting \(z=a+bi\), \(z^{*}=a-bi\): \(2a+2bi+ai+b=5+4i\)M1
equating real and imaginary parts: \(2a+b=5\) and \(a+2b=4\)M1
\(a=2\)A1
\(b=1\), so \(z=2+i\)A1
(b)
\(\dfrac{2+i}{2-i}\times\dfrac{2+i}{2+i}\)M1
\(\dfrac{3}{5}+\dfrac{4}{5}i\)A1FT from their \(z\).

Answers: (a) \(z=2+i\) · (b) \(\tfrac{3}{5}+\tfrac{4}{5}i\)

8AAHL1-q00351.12 Complex numbers in Cartesian form Paper 1Easy8 marks

The following Argand diagram, drawn on a unit grid, shows the points A and B, which represent the complex numbers \(z_{1}\) and \(z_{2}\) respectively.

ReImO1234123AB
(a)

Write down \(z_{1}\) and \(z_{2}\) in the form \(a+bi\).

(1)
(b)

Find \(|z_{1}|\), \(|z_{2}|\) and \(|z_{2}-z_{1}|\).

(3)
(c)

Show that triangle OAB is right-angled and isosceles.

(2)
(d)

Write down the size of angle AOB.

(1)
(e)

Write down the complex number represented by the reflection of B in the real axis.

(1)
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WorkingMarkNotes
(a)
\(z_{1}=1+3i,\ z_{2}=4+2i\)A1Both required.
(b)
\(|z_{1}|=\sqrt{10}\)A1
\(|z_{2}|=\sqrt{20}\;(=2\sqrt{5})\)A1
\(z_{2}-z_{1}=3-i\), so \(|z_{2}-z_{1}|=\sqrt{10}\)A1
(c)
\(\text{OA}=\text{AB}=\sqrt{10}\), so the triangle is isoscelesR1
\(\text{OA}^{2}+\text{AB}^{2}=10+10=20=\text{OB}^{2}\), so (by the converse of Pythagoras) angle OAB is a right angleR1
triangle OAB is right-angled and isoscelesAG
(d)
\(\dfrac{\pi}{4}\)A1Accept \(45^{\circ}\).
(e)
\(z_{2}^{*}=4-2i\)A1

Answers: (a) \(1+3i,\ 4+2i\) · (b) \(\sqrt{10},\ 2\sqrt5,\ \sqrt{10}\) · (d) \(\pi/4\) · (e) \(4-2i\)

9AAHL1-q00361.12 Complex numbers in Cartesian form Paper 1Easy5 marks

Let \(w=\dfrac{k+3i}{2-i}\), where \(k\in\mathbb{R}\).

(a)

Given that \(w\) is purely imaginary, find the value of \(k\).

(3)
(b)

For this value of \(k\), find \(|w|\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(w=\dfrac{(k+3i)(2+i)}{5}\)M1Multiplying by the conjugate of the denominator.
\(w=\dfrac{(2k-3)+(k+6)i}{5}\)A1
\(\operatorname{Re}(w)=0\Rightarrow k=\dfrac{3}{2}\)A1
(b)
\(w=\dfrac{7.5}{5}i=1.5i\)M1Or \(|w|=\dfrac{|k+3i|}{|2-i|}=\dfrac{\sqrt{2.25+9}}{\sqrt5}\).
\(|w|=\dfrac{3}{2}\)A1

Answers: (a) \(k=\tfrac32\) · (b) \(|w|=\tfrac32\)

10AAHL1-q00401.13 Polar and Euler forms Paper 1Easy6 marks
(a)

Express \(-1-\sqrt3\,i\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
(b)

Express \(4\operatorname{cis}\dfrac{5\pi}{6}\) in the form \(a+bi\).

(2)
(c)

Express \(-3\operatorname{cis}\dfrac{\pi}{5}\) in the form \(r\operatorname{cis}\theta\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
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WorkingMarkNotes
(a)
\(r=2\)A1
\(2\mathrm{e}^{-\frac{2\pi}{3}i}\)A1Award A0 for \(\theta=\frac{\pi}{3}\) or \(\frac{4\pi}{3}\).
(b)
\(4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)=4\left(-\dfrac{\sqrt3}{2}+\dfrac{1}{2}i\right)\)M1
\(-2\sqrt3+2i\)A1
(c)
\(-1=\operatorname{cis}\pi\), so \(-3\operatorname{cis}\dfrac{\pi}{5}=3\operatorname{cis}\dfrac{6\pi}{5}\)M1
\(3\operatorname{cis}\left(-\dfrac{4\pi}{5}\right)\)A1

Answers: (a) \(2\mathrm{e}^{-2\pi i/3}\) · (b) \(-2\sqrt3+2i\) · (c) \(3\operatorname{cis}(-4\pi/5)\)

11AAHL1-q00501.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy5 marks

Consider the equation \(z^{3}-7z^{2}+17z-15=0\), where \(z\in\mathbb{C}\). One root of the equation is \(2-i\).

(a)

Write down a second root, giving a reason for your answer.

(1)
(b)

Find a quadratic factor of \(z^{3}-7z^{2}+17z-15\) with real coefficients.

(2)
(c)

Find the third root.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(2+i\), since the coefficients are real, non-real roots occur in conjugate pairsA1Reason required.
(b)
\(\big(z-(2-i)\big)\big(z-(2+i)\big)=(z-2)^{2}+1\)M1
\(z^{2}-4z+5\)A1
(c)
\(z^{3}-7z^{2}+17z-15=(z^{2}-4z+5)(z-c)\), comparing constant terms \(-5c=-15\)M1Or sum of roots \(2+i+2-i+c=7\).
\(z=3\)A1

Answers: (a) \(2+i\) · (b) \(z^{2}-4z+5\) · (c) \(3\)

12AAHL1-q00521.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

Let \(z=\sqrt3-i\).

(a)

Express \(z\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
(b)

Use De Moivre’s theorem to find \(z^{9}\) in the form \(a+bi\).

(2)
(c)

Find the least positive integer \(n\) for which \(z^{n}\) is real and positive.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(r=2\)A1
\(2\mathrm{e}^{-\frac{\pi}{6}i}\)A1
(b)
\(z^{9}=2^{9}\mathrm{e}^{-\frac{9\pi}{6}i}=512\mathrm{e}^{-\frac{3\pi}{2}i}\)M1
\(512i\)A1
(c)
\(-\dfrac{n\pi}{6}=-2k\pi\), \(k\in\mathbb{Z}^{+}\)M1Award M1 for \(\frac{n\pi}{6}\) a multiple of \(2\pi\).
\(n=12\)A1Award A0 for \(n=6\) (\(z^{6}=-64\) is negative).

Answers: (a) \(2\mathrm{e}^{-\pi i/6}\) · (b) \(512i\) · (c) 12

13AAHL1-q00581.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

Consider the equation \(z^{3}+8=0\), where \(z\in\mathbb{C}\).

(a)

Find the three roots of the equation, giving your answers in the form \(a+bi\).

(4)
(b)

State why the two non-real roots are complex conjugates of each other.

(1)
(c)

Write down the sum of the three roots.

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(z^{3}=8\mathrm{e}^{\pi i}\Rightarrow z=2\mathrm{e}^{\frac{\pi+2k\pi}{3}i}\)M1Or factorizing \((z+2)(z^{2}-2z+4)=0\).
\(z=-2\)A1
\(2\mathrm{e}^{\pm\frac{\pi}{3}i}\)A1Or \(z=\dfrac{2\pm\sqrt{-12}}{2}\).
\(1+\sqrt3\,i,\ 1-\sqrt3\,i\)A1
(b)
the polynomial \(z^{3}+8\) has real coefficientsR1
(c)
\(0\)A1

Answers: (a) \(-2,\ 1+\sqrt3 i,\ 1-\sqrt3 i\) · (c) 0

14AAHL1-q00661.15 Proof by induction, contradiction and counterexample Paper 1Easy7 marks

A sequence is defined by \(u_{1}=1\) and \(u_{n+1}=3u_{n}+2^{n}\), for \(n\in\mathbb{Z}^{+}\).

(a)

Find \(u_{2}\), \(u_{3}\) and \(u_{4}\).

(2)
(b)

Prove by mathematical induction that \(u_{n}=3^{n}-2^{n}\), for \(n\in\mathbb{Z}^{+}\).

(5)
Show mark scheme
WorkingMarkNotes
(a)
\(u_{2}=3(1)+2=5\)(M1)
\(u_{3}=19\), \(u_{4}=65\)A1Both values required.
(b)
\(n=1\): \(3-2=1=u_{1}\), so true for \(n=1\)A1
assume true for \(n=k\): \(u_{k}=3^{k}-2^{k}\)M1
\(u_{k+1}=3\left(3^{k}-2^{k}\right)+2^{k}\)M1
\(=3^{k+1}-2\times2^{k}=3^{k+1}-2^{k+1}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least three of the previous marks have been awarded.

Answers: (a) \(u_{2}=5,\ u_{3}=19,\ u_{4}=65\)

15AAHL1-q00681.15 Proof by induction, contradiction and counterexample Paper 1Easy7 marks
(a)

Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n}r\times r!=(n+1)!-1\), for \(n\in\mathbb{Z}^{+}\).

(5)
(b)

Hence find the value of \(n\) for which \(\displaystyle\sum_{r=1}^{n}r\times r!=719\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(n=1\): LHS \(=1\), RHS \(=2!-1=1\), so true for \(n=1\)A1
assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}r\times r!=(k+1)!-1\)M1
\(\displaystyle\sum_{r=1}^{k+1}r\times r!=(k+1)!-1+(k+1)(k+1)!\)M1
\(=(k+1)!\,(k+2)-1=(k+2)!-1\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least three of the previous marks have been awarded.
(b)
\((n+1)!=720=6!\)(M1)
\(n=5\)A1

Answers: (b) \(n=5\)

16AAHL1-q00731.15 Proof by induction, contradiction and counterexample Paper 1Easy7 marks

Let \(a\) be a rational number and \(b\) be an irrational number.

(a)

Prove by contradiction that \(a+b\) is irrational.

(3)
(b)

Given also that \(a\neq0\), prove by contradiction that \(ab\) is irrational.

(2)
(c)

Use a counterexample to show that the sum of two irrational numbers is not always irrational.

(2)
Show mark scheme
WorkingMarkNotes
(a)
assume \(a+b=c\), where \(c\) is rationalM1
\(b=c-a\), and the difference of two rational numbers is rational: \(\dfrac{p}{q}-\dfrac{s}{t}=\dfrac{pt-qs}{qt}\)A1
so \(b\) is rational, a contradiction; hence \(a+b\) is irrationalR1
(b)
assume \(ab=c\in\mathbb{Q}\); then \(b=\dfrac{c}{a}\) (valid since \(a\neq0\))M1
\(\dfrac{c}{a}\) is a quotient of rationals so is rational, a contradiction; hence \(ab\) is irrationalR1
(c)
eg \(\sqrt{2}\) and \(-\sqrt{2}\) (or \(1+\sqrt{2}\) and \(1-\sqrt{2}\)), which are both irrationalA1
their sum is \(0\) (or \(2\)), which is rational, so the statement is falseR1Must explain why the example is a counterexample.

Answers: (c) eg \(\sqrt{2}+(-\sqrt{2})=0\)

17AAHL1-q00771.15 Proof by induction, contradiction and counterexample Paper 1Easy6 marks

For each of the following statements, use a counterexample to show that the statement is not always true.

(a)

If \(f'(a)=0\), then the graph of \(y=f(x)\) has a local maximum or a local minimum point at \(x=a\).

(2)
(b)

For all functions \(f\) and \(g\) with domain \(\mathbb{R}\), \((f\circ g)(x)=(g\circ f)(x)\) for every \(x\in\mathbb{R}\).

(2)
(c)

For all \(a,b\gt0\), \(\ln(a+b)\neq\ln a+\ln b\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
eg \(f(x)=x^{3}\), \(a=0\): \(f'(0)=0\)A1
\(f'(x)=3x^{2}\ge0\) on both sides of \(0\) (\(f\) is increasing), so \((0,0)\) is a point of inflexion, not a maximum or minimumR1
(b)
eg \(f(x)=x+1\), \(g(x)=2x\): \((f\circ g)(x)=2x+1\) and \((g\circ f)(x)=2x+2\)A1Any pair of functions that do not commute, with both composites found.
at \(x=0\), \((f\circ g)(0)=1\neq2=(g\circ f)(0)\), so the statement is falseR1Must evaluate at a specific \(x\) or show the composites differ for all \(x\).
(c)
\(a=b=2\)A1Any \(a,b\) with \(a+b=ab\), eg \(a=3,\ b=\tfrac{3}{2}\).
\(\ln(2+2)=\ln4\) and \(\ln2+\ln2=\ln4\), so they are equalR1

Answers: (a) eg \(f(x)=x^{3}\) at \(x=0\) · (b) eg \(f(x)=x+1,\ g(x)=2x\) · (c) eg \(a=b=2\)

18AAHL1-q00821.16 Systems of linear equations Paper 1Easy5 marks

Consider the system of equations

\(\begin{aligned}x+2y-z&=3\\2x-y+3z&=6\\3x+y+z&=8\end{aligned}\)

(a)

Use row reduction to solve the system.

(4)
(b)

Give a geometric interpretation of your answer.

(1)
Show mark scheme
WorkingMarkNotes
(a)
attempt to eliminate \(x\): \(R_{2}-2R_{1}\): \(-5y+5z=0\); \(R_{3}-3R_{1}\): \(-5y+4z=-1\)M1Any valid elimination method.
two correct equations in two variables, eg \(y=z\) and \(-5y+4z=-1\)A1
\(z=1,\ y=1\)A1
\(x=2\)A1
(b)
three planes intersecting at the single point \((2,1,1)\)A1FT from their unique solution.

Answers: (a) \(x=2,\ y=1,\ z=1\) · (b) three planes meeting at \((2,1,1)\)

19IBMA2H-045-06-021.9 The binomial theorem Paper 1Easy6 marks

No calculator

In the expansion of \((3+ax)^6\), where \(a>0\), the coefficient of \(x^2\) is 540.

(a)

Find the value of \(a\).

(4)
(b)

Hence find the coefficient of \(x^3\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 identifies the \(x^2\) term as \(\binom{6}{2}3^4(ax)^2\) ; A1 coefficient \(1215a^2\) ; M1 sets \(1215a^2=540\) and solves ; A1 \(a=\frac{2}{3}\)✓ 4
Part (b)
M1 \(\binom{6}{3}3^3a^3=20\times27\times\frac{8}{27}\) ; A1 160✓ 2

Common error: Squaring only \(x\) and not \(a\) in \((ax)^2\) gives \(15\times81a=540\) and a wrong value of \(a\). The whole bracket \((ax)\) is raised to the power.

20IBMA2H-045-06-041.6 Deductive proof Paper 1Easy7 marks

No calculator

The sum of the first \(n\) terms of an arithmetic sequence is given by \(S_n=3n^2+5n\).

(a)

Find the first term and the common difference.

(3)
(b)

Find an expression for \(u_n\) in terms of \(n\).

(2)
(c)

Show that 500 is a term of the sequence and state which term it is.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses \(u_1=S_1\) and \(u_2=S_2-S_1\), or compares \(S_n\) with \(\frac{d}{2}n^2+\left(u_1-\frac{d}{2}\right)n\) ; A1 \(u_1=8\) ; A1 \(d=6\) (from \(u_2=14\) or \(\frac{d}{2}=3\))✓ 3
Part (b)
M1 \(u_n=8+6(n-1)\) or \(u_n=S_n-S_{n-1}\) ; A1 \(u_n=6n+2\)✓ 2
Part (c)
M1 sets \(6n+2=500\) ; A1 \(n=83\), which is a positive integer, so 500 is the 83rd term✓ 2

Common error: \(S_2=22\) is the sum of the first two terms, not the second term. The second term is \(S_2-S_1=14\).

21IBMA2H-045-03-011.11 Partial fractions Paper 1Easy4 marks

No calculator

Solve the equation \(4^{x+1}=8^{2x-3}\), giving your answer as an exact fraction.

(a)

Find the exact value of \(x\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes both sides as powers of 2✓ 1
A1 \(2^{2x+2}=2^{6x-9}\)✓ 1
M1 equates the exponents: \(2x+2=6x-9\)✓ 1
A1 \(x=\frac{11}{4}\)✓ 1

Exam tip: Rewrite every term with the same prime base before equating powers. Then the problem becomes a linear equation and no logarithms are needed.

22IBMA2H-045-03-041.5 Exponents and introduction to logarithms Paper 1Easy4 marks

No calculator

Answer the following, giving exact values.

(a)

Find \(\log_5 125\).

(1)
(b)

Find \(\log_4 2\).

(1)
(c)

Solve \(\log_x 81=4\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 3✓ 1
Part (b)
A1 \(\frac{1}{2}\)✓ 1
Part (c)
M1 converts to exponential form: \(x^4=81\)✓ 1
A1 \(x=3\); reject \(-3\) because a logarithm base must be positive✓ 1

Common error: \(\log_4 2\) asks for the power of 4 that gives 2. It is \(\frac{1}{2}\), not 2.

23IBMA2H-045-04-011.12 Complex numbers in Cartesian form Paper 1Easy5 marks

No calculator

In an alternating-current circuit, the impedances of two components are represented by the complex numbers \(Z_1 = 3+4i\) ohms and \(Z_2 = 2-i\) ohms. When the components are connected in series, the total impedance is \(Z_1+Z_2\). When they are connected in parallel, the total impedance is \(\frac{Z_1Z_2}{Z_1+Z_2}\).

(a)

Write down the total impedance when the components are connected in series.

(1)
(b)

Find \(Z_1Z_2\) in the form \(a+bi\).

(2)
(c)

Find the total impedance when the components are connected in parallel. Give your answer in the form \(a+bi\), where \(a, b \in \mathbb{Q}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(5+3i\)✓ 1
Part (b)
M1 expands \((3+4i)(2-i)\) to \(6-3i+8i-4i^2\) and uses \(i^2=-1\)✓ 1
A1 \(10+5i\)✓ 1
Part (c)
M1 multiplies \(\frac{10+5i}{5+3i}\) by \(\frac{5-3i}{5-3i}\), giving denominator \(34\)✓ 1
A1 \(\frac{65}{34}-\frac{5}{34}i\)✓ 1

Exam tip: To divide by a complex number, multiply numerator and denominator by the conjugate of the denominator. The denominator then becomes the real number \(a^2+b^2\).

24IBMA2H-045-04-021.12 Complex numbers in Cartesian form Paper 1Easy4 marks

No calculator

The complex number \(z\) satisfies \(2z + iz^* = 7-4i\), where \(z^*\) is the complex conjugate of \(z\). Find \(z\) in the form \(x+iy\), where \(x, y \in \mathbb{R}\).

(a)

Find \(z\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes \(z=x+iy\) and \(z^*=x-iy\): \(2x+2iy+ix+y = 7-4i\)✓ 1
M1 equates real and imaginary parts: \(2x+y=7\) and \(x+2y=-4\)✓ 1
A1 \(x=6\)✓ 1
A1 \(y=-5\), so \(z=6-5i\)✓ 1

Common error: \(i(x-iy) = ix + y\), not \(ix - y\). Keep track of the sign when \(i \times (-iy)\) gives \(-i^2y = +y\).

25IBMA2H-045-04-041.9 The binomial theorem Paper 1Easy5 marks

No calculator

Let \(k\) be a positive integer.

(a)

Show that \(\frac{1}{k}-\frac{1}{k+1} = \frac{1}{k(k+1)}\).

(2)
(b)

Hence find the exact value of \(\sum_{k=1}^{99} \frac{1}{k(k+1)}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes the left-hand side over the common denominator \(k(k+1)\)✓ 1
A1 \(\frac{(k+1)-k}{k(k+1)} = \frac{1}{k(k+1)}\)✓ 1
Part (b)
M1 rewrites each term as \(\frac{1}{k}-\frac{1}{k+1}\)✓ 1
M1 writes out terms to show cancellation: \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right)\)✓ 1
A1 \(1-\frac{1}{100}=\frac{99}{100}\)✓ 1

Exam tip: Write out the first two and the last term of a telescoping sum. This shows which terms survive the cancellation.

26IBMA2H-045-04-091.6 Deductive proof Paper 1Easy5 marks

No calculator

This question is about proof and counterexample.

(a)

Prove that the difference between the squares of any two consecutive odd integers is a multiple of 8.

(3)
(b)

Show that the statement “\(n^2+n+41\) is prime for every positive integer \(n\)” is false.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 represents consecutive odd integers as \(2n-1\) and \(2n+1\), \(n\in\mathbb{Z}\)✓ 1
A1 \((2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n\)✓ 1
R1 \(n\) is an integer, so \(8n\) is a multiple of 8✓ 1
Part (b)
M1 chooses a suitable value, e.g. \(n=40\)✓ 1
A1 \(40^2+40+41=1681=41^2\), which is not prime✓ 1

Exam tip: A single counterexample is enough to disprove a general claim. Checking many cases that work never proves one.

27IBMA2H-045-00-011.15 Proof by induction, contradiction and counterexample Paper 1Easy4 marks

No calculator

A student testing a computer program claims that both of the following statements are true. Each statement is false.

(a)

Statement 1: “n2 − n + 11 is prime for every positive integer n.” Use a counterexample to show that Statement 1 is false.

(2)
(b)

Statement 2: “For all real numbers a and b, if a2 > b2 then a > b.” Use a counterexample to show that Statement 2 is false.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 chooses a suitable value, e.g. n = 11, and substitutes: 121 − 11 + 11 = 121✓ 1
A1 121 = 11 × 11, so it is not prime and Statement 1 is false✓ 1
(Also accept n = 12, which gives 143 = 11 × 13.)—
Part (b)
M1 chooses a negative a with |a| > |b|, e.g. a = −3, b = 1✓ 1
A1 shows a2 = 9 > 1 = b2 but −3 < 1, so Statement 2 is false✓ 1

Exam tip: A single counterexample earns full marks, but only if you show the numerical check and state clearly why it breaks the statement.

28IBMA2H-045-00-091.15 Proof by induction, contradiction and counterexample Paper 1Easy6 marks

No calculator

This question uses proof by contradiction.

(a)

Prove that there are no integers a and b such that 6a + 9b = 1.

(3)
(b)

Prove that if x is rational and y is irrational, then x + y is irrational.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 assumes integers a, b exist with 6a + 9b = 1✓ 1
M1 6a + 9b = 3(2a + 3b), a multiple of 3✓ 1
R1 1 is not a multiple of 3: contradiction, so no such integers exist✓ 1
Part (b)
M1 assumes x + y = p/q is rational, with x = r/s (p, q, r, s integers, q, s ≠ 0)✓ 1
M1 y = p/q − r/s = (ps − qr)/(qs)✓ 1
R1 this is a ratio of integers with non-zero denominator, so y is rational: contradiction, hence x + y is irrational✓ 1

Exam tip: Always begin a proof by contradiction by stating exactly what you are assuming, and finish by naming the contradiction.

29IBMA2H-045-02-011.11 Partial fractions Paper 1Easy4 marks

No calculator

In control engineering, the response of a motor controller is described by the transfer function \(G(s)=\frac{7s-1}{s^2-s-6}\).
Express \(G(s)\) in partial fractions.

(a)

Express \(G(s)\) in the form \(\frac{A}{s-3}+\frac{B}{s+2}\), where \(A,B\in\mathbb{Z}\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 factorises the denominator as \((s-3)(s+2)\)✓ 1
M1 forms the identity \(7s-1\equiv A(s+2)+B(s-3)\) and substitutes values or compares coefficients✓ 1
A1 \(A=4\)✓ 1
A1 \(B=3\), so \(G(s)=\frac{4}{s-3}+\frac{3}{s+2}\)✓ 1

Exam tip: Substitute the root of each linear factor, here s = 3 and s = −2, into the identity. Each substitution removes one unknown, so you never need to solve simultaneous equations.

30IBMA2H-045-02-021.13 Polar and Euler forms Paper 1Easy6 marks

No calculator

The tip of a robotic arm, viewed from above, is represented in the complex plane by \(z=-2\sqrt{3}+2i\), with the base of the arm at the origin.

OReImOtip (z)π/3
Anticlockwise rotation about the base O. Not to scale
(a)

Write \(z\) in the form \(re^{i\theta}\), where \(r>0\) and \(-\pi<\theta\le\pi\).

(3)
(b)

The arm rotates anticlockwise about its base through an angle of \(\frac{\pi}{3}\). Find the new position of the tip, giving your answer in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\), and in the form \(a+bi\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=\sqrt{12+4}=4\)✓ 1
M1 finds the argument and places it in the second quadrant, e.g. \(\pi-\arctan\frac{1}{\sqrt3}\)✓ 1
A1 \(z=4e^{\frac{5\pi i}{6}}\)✓ 1
Part (b)
M1 multiplies by \(e^{\frac{\pi i}{3}}\), giving \(4e^{\frac{7\pi i}{6}}\)✓ 1
A1 \(4e^{-\frac{5\pi i}{6}}\)✓ 1
A1 \(-2\sqrt3-2i\)✓ 1

Common error: Reading arctan(2/(−2√3)) straight from the formula gives −π/6, which lies in the wrong quadrant. Locate the point first: negative real part and positive imaginary part means the second quadrant.

31IBMA2H-045-02-041.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

No calculator

In a digital signal processor, a sample is repeatedly multiplied by the complex gain \(w=1+i\).

(a)

Write \(w\) in the form \(re^{i\theta}\).

(2)
(b)

Hence find \(w^8\).

(2)
(c)

Find the smallest positive integer \(n\) for which \(w^n\) is a negative real number, and state the value of \(w^n\) for this \(n\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=\sqrt2\)✓ 1
A1 \(\theta=\frac{\pi}{4}\), so \(w=\sqrt2e^{\frac{\pi i}{4}}\)✓ 1
Part (b)
M1 \((\sqrt2)^8e^{2\pi i}\)✓ 1
A1 \(w^8=16\)✓ 1
Part (c)
M1 requires \(\frac{n\pi}{4}\) to be an odd multiple of \(\pi\)✓ 1
A1 \(n=4\) and \(w^4=(\sqrt2)^4e^{\pi i}=-4\)✓ 1

Common error: An argument of 2π gives a positive real number, not a negative one. A negative real number needs an argument of π, 3π, 5π and so on.

32IBMA2H-045-01-011.16 Systems of linear equations Paper 1Easy5 marks

No calculator

A simple model in veterinary medicine estimates the daily resting energy requirement \(B\) (in kcal per day) of a mammal of mass \(M\) kg using \(B = 70M^{\frac{3}{4}}\).

(a)

Find the daily resting energy requirement of a dog of mass 81 kg.

(2)
(b)

A second animal has a daily resting energy requirement of 560 kcal. Find its mass.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(81^{\frac{3}{4}} = (81^{\frac{1}{4}})^3 = 3^3 = 27\)✓ 1
A1 \(B = 1890\) kcal per day✓ 1
Part (b)
A1 \(M^{\frac{3}{4}} = 8\)✓ 1
M1 \(M = 8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4\)✓ 1
A1 \(M = 16\) kg✓ 1

Exam tip: To undo a power of \(\frac{3}{4}\), raise both sides to the power \(\frac{4}{3}\). Take the root first, then the power, to keep the numbers small without a calculator.

33IBMA2H-045-01-021.7 Rational exponents and laws of logarithms Paper 1Easy5 marks

No calculator

Consider the equation \(\log_3 x + \log_9 x = 6\), where \(x > 0\).

(a)

Show that \(\log_9 x = \frac{1}{2}\log_3 x\).

(2)
(b)

Hence solve the equation.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 change of base: \(\log_9 x = \frac{\log_3 x}{\log_3 9}\)✓ 1
A1 \(\log_3 9 = 2\), so \(\log_9 x = \frac{1}{2}\log_3 x\)✓ 1
Part (b)
M1 \(\frac{3}{2}\log_3 x = 6\)✓ 1
A1 \(\log_3 x = 4\)✓ 1
A1 \(x = 81\)✓ 1

Common error: Logs with different bases cannot be added directly. Convert to a common base first; \(\log_3 x + \log_9 x\) is not \(\log_{12} x\) or \(\log_{27} x\).

34IBMA2H-045-01-031.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

No calculator

The behaviour of an electrical circuit is described by the characteristic equation \(z^3 + az^2 + bz - 20 = 0\), where \(a, b \in \mathbb{R}\). One root of the equation is \(1 + 3i\).

(a)

Write down a second complex root.

(1)
(b)

Find the quadratic factor of \(z^3 + az^2 + bz - 20\) with these two roots.

(2)
(c)

Find the real root, and the values of \(a\) and \(b\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1 - 3i\)✓ 1
Part (b)
M1 \((z - (1+3i))(z - (1-3i)) = (z-1)^2 + 9\)✓ 1
A1 \(z^2 - 2z + 10\)✓ 1
Part (c)
M1 \((z^2 - 2z + 10)(z - c)\) has constant term \(-10c = -20\), so the real root is \(c = 2\)✓ 1
A1 \(a = -(2 + 2) = -4\)✓ 1
A1 \(b = 10 + 2(2) = 14\)✓ 1

Exam tip: The conjugate root theorem needs real coefficients. Say this when you use it, because the reasoning can carry a mark.

35IBMA2H-045-01-041.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

No calculator

In an AC circuit model, a voltage phasor is represented by the complex number \(z = \sqrt{3} + i\).

(a)

Write \(z\) in the form \(r\,\text{cis}\,\theta\), where \(r > 0\) and \(-\pi < \theta \le \pi\).

(2)
(b)

Use De Moivre's theorem to find \(z^6\).

(2)
(c)

Find the least positive integer \(n\) for which \(z^n\) is a positive real number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = 2\)✓ 1
A1 \(\theta = \frac{\pi}{6}\), so \(z = 2\,\text{cis}\,\frac{\pi}{6}\)✓ 1
Part (b)
M1 \(z^6 = 2^6\,\text{cis}\,\pi\)✓ 1
A1 \(z^6 = -64\)✓ 1
Part (c)
M1 requires \(\frac{n\pi}{6} = 2k\pi\), \(k \in \mathbb{Z}\)✓ 1
A1 \(n = 12\)✓ 1

Common error: A real result is not always a positive one. \(z^6 = -64\) is real but negative, so the argument must be a multiple of \(2\pi\), not just a multiple of \(\pi\).

36IBMA2H-045-05-011.10 Counting principles and the extended binomial theorem Paper 1Easy4 marks

No calculator

A launch vehicle carries eight different satellites, which are released one at a time. Two of the satellites, X and Y, share a docking frame.

(a)

Find the number of possible release orders in which X and Y are released consecutively, in either order.

(2)
(b)

Find the number of possible release orders in which X and Y are not released consecutively.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 treats X and Y as a single unit, giving 7! arrangements of the units, multiplied by 2 for the internal order (or equivalent: 7 adjacent position pairs × 2 × 6!)✓ 1
A1 2 × 7! = 10080✓ 1
Part (b)
M1 subtracts from the total number of orders: 8! − 10080 (or equivalent: arranges the other six in 6! ways, then places X and Y in two of the 7 gaps in 7 × 6 ways)✓ 1
A1 40320 − 10080 = 30240✓ 1

Exam tip: For 'not together' conditions, the complement (total minus together) is usually the fastest route. Remember the factor of 2 for the internal order of the block.

37IBMA2H-045-05-041.8 Infinite geometric series Paper 1Easy4 marks

No calculator

A motion sensor records the arc length of each swing of a damped pendulum. The first swing has arc length 12 cm, and each swing is a constant fraction r of the previous one. If the pendulum could swing indefinitely, the total arc length travelled would be 48 cm.

(a)

Find the value of r.

(2)
(b)

Find the arc length of the fourth swing.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{12}{1-r}=48\)✓ 1
A1 \(r=\frac{3}{4}\)✓ 1
Part (b)
M1 \(12\left(\frac{3}{4}\right)^3\)✓ 1
A1 \(\frac{81}{16}\) cm (= 5.0625 cm)✓ 1

Common error: Using r⁴ for the fourth term. The n-th term is u₁rⁿ⁻¹, so the fourth swing uses r³.

38IBMA2H-045-07-011.1 Standard form Paper 1Easy4 marks

No calculator

A cloud-storage company's data centre holds a total of 4.8×1017 bytes of customer data. Each hard drive in the centre stores 1.6×1013 bytes, and every drive is completely full.

(a)

Find the number of hard drives in the data centre. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(b)

Each drive uses 6.5 W of power. Find the total power used by all the drives, giving your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to divide: \(\frac{4.8\times10^{17}}{1.6\times10^{13}}\)✓ 1
A1 3×104 drives✓ 1
Part (b)
M1 multiplies: 3×104 × 6.5 = 19.5×104✓ 1
A1 1.95×105 W✓ 1

Common error: 19.5×10⁴ is correct in value but is not in standard form, because 19.5 is not between 1 and 10. Adjust the power of ten to give 1.95×10⁵.

39IBMA2H-045-07-041.8 Infinite geometric series Paper 1Easy5 marks

No calculator

In a geometric sequence, the second term is −6 and the fifth term is 48.

(a)

Find the common ratio.

(2)
(b)

Find the first term.

(1)
(c)

Find the sum of the first eight terms.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{u_1r^4}{u_1r}=r^3=\frac{48}{-6}=-8\)✓ 1
A1 \(r=-2\)✓ 1
Part (b)
A1 \(u_1=\frac{-6}{-2}=3\)✓ 1
Part (c)
M1 \(S_8=\frac{3(1-(-2)^8)}{1-(-2)}\)✓ 1
A1 −255✓ 1

Common error: A negative common ratio is allowed. \(r^3=-8\) has exactly one real solution, \(r=-2\). Use brackets so that \((-2)^8=256\) is evaluated correctly.

40IBMA3H-1-18r5-091.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy5 marks

No calculator


Let w = 1 + i√3.

(a)

Write w in the form r(cos θ + i sin θ), where r > 0 and 0 ≤ θ < 2π.

(2)
(b)

Use De Moivre's theorem to find w9 in the form a + bi, where a and b are integers.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 r = 2✓ 1
A1 θ = π/3✓ 1
Part (b)
M1 w9 = 29(cos 3π + i sin 3π)✓ 1
A1 512(−1 + 0i)✓ 1
A1 -512 + 0i✓ 1

Common error: Raising only the argument to the power. De Moivre's theorem also raises the modulus: 2⁹, not 2.

41IBMA3H-1-18r5-111.10 Counting principles and the extended binomial theorem Paper 1Easy5 marks

No calculator


A hospital has 6 surgeons and 4 anaesthetists. An operating team consists of 3 surgeons and 2 anaesthetists.

(a)

Find the number of different teams that can be chosen.

(2)
(b)

Two of the surgeons, Dr A and Dr B, cannot work in the same team. Find the number of possible teams.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 C(6, 3) × C(4, 2)✓ 1
A1 20 × 6 = 120✓ 1
Part (b)
M1 subtract the number of teams that contain both Dr A and Dr B✓ 1
A1 4 × C(4, 2) = 24 teams containing both✓ 1
A1 120 − 24 = 96✓ 1

Exam tip: For "cannot be together" conditions, it is often quicker to count all arrangements and subtract those that break the rule.

42IBMA3H-1-18r6-081.5 Exponents and introduction to logarithms Paper 1Easy4 marks

No calculator

(a)

Express \(\dfrac{5x+1}{(x-1)(x+2)}\) in the form \(\dfrac{A}{x-1}+\dfrac{B}{x+2}\), where \(A\) and \(B\) are constants.

(3)
(b)

Hence find the exact value of \(\dfrac{5x+1}{(x-1)(x+2)}\) when \(x=2\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(5x+1=A(x+2)+B(x-1)\)✓ 1
A1 \(A=2\)✓ 1
A1 \(B=3\)✓ 1
Part (b)
A1 \(\frac21+\frac34=\frac{11}{4}\) (= 2.75)✓ 1

Exam tip: Substituting the root of each linear factor into the identity removes the other constant at once.

43IBMA3H-1-18r6-091.12 Complex numbers in Cartesian form Paper 1Easy5 marks

No calculator

Let \(z=3-2\mathrm{i}\) and \(w=1+4\mathrm{i}\).

(a)

Find \(zw\) in the form \(a+b\mathrm{i}\).

(2)
(b)

Find \(\dfrac zw\) in the form \(a+b\mathrm{i}\), where \(a,b\in\mathbb{Q}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((3-2\mathrm{i})(1+4\mathrm{i})\) using \(\mathrm{i}^2=-1\)✓ 1
A1 \(11+10\mathrm{i}\)✓ 1
Part (b)
M1 multiply numerator and denominator by \(1-4\mathrm{i}\)✓ 1
A1 numerator \(-5-14\mathrm{i}\) and denominator 17✓ 1
A1 \(-\frac5{17}-\frac{14}{17}\mathrm{i}\)✓ 1

Common error: When dividing, multiply by the conjugate of the denominator, not of the numerator.

44IBMA3H-1-18r10-071.13 Polar and Euler forms Paper 1Easy4 marks

No calculator

The position of the tip of a robotic arm, in metres, is represented by the complex number \(z=-1+\sqrt3\,i\) on an Argand diagram.

OReImOz = −1 + √3i
Not to scale
(a)

Find the modulus of \(z\) and the argument of \(z\), in radians, with \(-\pi<\arg z\le\pi\).

(3)
(b)

Write \(z\) in the form \(re^{i\theta}\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(|z|=\sqrt{1+3}=2\)✓ 1
identifies that \(z\) is in the second quadrant (or uses \(\tan^{-1}\sqrt3=\frac\pi3\) with \(\pi-\frac\pi3\))✓ 1
\(\arg z=\frac{2\pi}{3}\)✓ 1
Part (b)
\(z=2e^{\frac{2\pi}{3}i}\)✓ 1

Common error: Sketch the point first: \(\tan^{-1}\) alone gives \(-\frac{\pi}{3}\), which lies in the wrong quadrant.

45IBMA3H-1-18r10-081.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy4 marks

No calculator

A robot's turn command multiplies its position by \(w=\sqrt3-i\) each time it is applied. Starting from the position \(1\), it applies the command 6 times.

(a)

Write \(w\) in the form \(r(\cos\theta+i\sin\theta)\) with \(r>0\).

(2)
(b)

Use de Moivre's theorem to find \(w^6\) in the form \(a+bi\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(r=2\)✓ 1
\(\theta=-\frac\pi6\)✓ 1
Part (b)
\(w^6=2^6\left(\cos(-\pi)+i\sin(-\pi)\right)\)✓ 1
\(w^6=-64\)✓ 1

Exam tip: Raise the modulus to the power and multiply the argument by the power, then evaluate the cosine and sine exactly.

46IBMA3H-1-19r11-021.3 Geometric sequences and series Paper 1Easy4 marks

No calculator

A pump delivers water to a reservoir on successive days. The volumes delivered, in megalitres (ML), form a geometric sequence with first term 18 ML. If the pump ran for ever, the total volume delivered would be 54 ML.

(a)

Find the common ratio of the sequence.

(2)
(b)

Find the total volume delivered from day 3 onwards, if the pump ran for ever.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{18}{1-r}=54\)✓ 1
A1 \(r=\frac{2}{3}\)✓ 1
Part (b)
M1 \(u_3=18\left(\frac{2}{3}\right)^2=8\) and use of \(\frac{u_3}{1-r}\) (or \(54-18-12\))✓ 1
A1 24 ML✓ 1

Common error: Starting the sum from the wrong term: the series from day 3 has first term \(u_3\), not \(u_1\).

47IBMA3H-1-19r11-081.15 Proof by induction, contradiction and counterexample Paper 1Easy4 marks

No calculator

A pharmacy labels production batches using the expression \(n^2+n+11\), where \(n=1,2,3,\ldots\). A technician claims that every label produced in this way is a prime number.

(a)

Show that the claim is true for \(n=1\), \(n=2\) and \(n=3\).

(2)
(b)

Show that the claim is false by finding a suitable value of \(n\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes the three values of \(n\)✓ 1
A1 labels 13, 17 and 23, each of which is prime✓ 1
Part (b)
M1 tries values until a counterexample is found, e.g. \(n=10\): \(100+10+11\)✓ 1
A1 \(121=11^2\) is not prime, so the claim is false✓ 1

Exam tip: One counterexample is enough to disprove a statement; many confirming cases never prove it.

48IBMA3H-1-19r11-141.3 Geometric sequences and series Paper 1Easy4 marks

No calculator

In an online game, the probability that a player wins a round is \(p=0.363636\ldots\), where the digits 36 recur.

(a)

Write \(p\) as the sum of an infinite geometric series and state its first term and common ratio.

(2)
(b)

Hence show that \(p=\frac{4}{11}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(p=0.36+0.0036+0.000036+\ldots\) with first term \(0.36\)✓ 1
A1 common ratio \(0.01\)✓ 1
Part (b)
M1 \(S_\infty=\frac{0.36}{1-0.01}\)✓ 1
A1 \(\frac{36}{99}=\frac{4}{11}\) (= 0.363636)✓ 1

Exam tip: A recurring decimal with a repeating block of two digits has common ratio \(0.01\).

49IBMA3H-1-18r13-011.10 Counting principles and the extended binomial theorem Paper 1Easy5 marks

No calculator

Six different robots queue to use a charging station. Robots \(R_1\) and \(R_2\) are two of the six.

(a)

Find the number of different orders in which the six robots can queue.

(1)
(b)

Find the number of orders in which \(R_1\) and \(R_2\) are next to each other.

(2)
(c)

Find the number of orders in which \(R_1\) and \(R_2\) are not next to each other.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(6! = 720\)✓ 1
Part (b)
M1 treat \(R_1R_2\) as one block, giving \(5!\) orders, and multiply by 2 for the order within the block✓ 1
A1 240✓ 1
Part (c)
M1 \(720 - 240\)✓ 1
A1 480✓ 1

Common error: Forgetting that the two robots in the block can swap places, which doubles the count.

50IBMA3H-1-18r13-131.3 Geometric sequences and series Paper 1Easy5 marks

No calculator

The lengths of successive extension movements of a robot arm form a geometric sequence. The 2nd movement is 6 mm long and the 5th movement is 162 mm long.

(a)

Find the common ratio.

(2)
(b)

Find the length of the 1st movement.

(1)
(c)

Find the total length of the first 8 movements.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r^3 = \frac{162}{6} = 27\)✓ 1
A1 \(r = 3\)✓ 1
Part (b)
A1 2 mm✓ 1
Part (c)
M1 \(S_8 = \frac{2(3^8 - 1)}{3-1}\)✓ 1
A1 6560 mm✓ 1

Exam tip: To find \(r\) from two non-adjacent terms, divide them and take the root that matches the gap in positions.

51IBMA3H-1-18r14-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy5 marks

No calculator

The characteristic equation of a robot's balance controller is \(z^3-7z^2+19z-13=0\). The controller is stable only if every root has a negative real part. One root is \(3+2i\).

(a)

Write down another root.

(1)
(b)

Find the third root.

(3)
(c)

State, with a reason, whether the controller is stable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(3-2i\) (complex conjugate, since the coefficients are real)✓ 1
Part (b)
M1 using the quadratic factor \((z-(3+2i))(z-(3-2i))\) or the sum of the roots being 7✓ 1
A1 \(z^2-6z+13\) (or sum of the two complex roots \(=6\))✓ 1
A1 \(z=1\)✓ 1
Part (c)
A1 not stable, since the real parts 3, 3 and 1 are positive (ECF)✓ 1

Exam tip: For a polynomial with real coefficients, non-real roots come in conjugate pairs; the sum of all roots equals minus the coefficient of \(z^{n-1}\).

52IBMA3H-1-18r16-021.9 The binomial theorem Paper 1Easy4 marks

No calculator

The coefficient of \(x^2\) in the expansion of \((1 + kx)^8\) is 252, where \(k > 0\).

(a)

Find the value of k.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(\binom{8}{2}(kx)^2\)✓ 1
A1 \(28k^2\)✓ 1
M1 \(28k^2 = 252\) so \(k^2 = 9\)✓ 1
A1 \(k = 3\) (since \(k > 0\))✓ 1

Common error: The coefficient includes \(k^2\), not just \(k\): the whole factor \((kx)\) is raised to the power 2.

53IBMA3H-1-18r24-031.2 Arithmetic sequences and series Paper 1Easy5 marks

No calculator

A sports scientist plans an athlete's weekly running distance as an arithmetic sequence. In week 1 the athlete runs 12 km and in week 5 the athlete runs 20 km.

(a)

Find the common difference.

(2)
(b)

Find the distance run in week 10.

(1)
(c)

Find the total distance run in the first 10 weeks.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 12 + 4d = 20✓ 1
A1 d = 2 km✓ 1
Part (b)
A1 12 + 9 × 2 = 30 km (ECF from (a))✓ 1
Part (c)
M1 S10 = (10/2)(12 + 30) (or (10/2)(2 × 12 + 9 × 2))✓ 1
A1 210 km (ECF)✓ 1

Common error: The nth term uses (n − 1) differences, not n: week 10 is nine steps after week 1.

54IBMA3H-1-18r24-041.10 Counting principles and the extended binomial theorem Paper 1Easy6 marks

No calculator

A rail company sells adult, child and senior day tickets at prices €a, €c and €s. Group 1 has 2 adults, 1 child and 1 senior and pays €40. Group 2 has 1 adult, 2 children and 1 senior and pays €35. Group 3 has 1 adult, 1 child and 2 seniors and pays €37.

(a)

Write down three linear equations for a, c and s.

(2)
(b)

Solve the system to find a, c and s.

(3)
(c)

Find the total price for 2 adults, 3 children and 1 senior.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms at least one correct equation, for example 2a + c + s = 40✓ 1
A1 a + 2c + s = 35 and a + c + 2s = 37 (all three correct)✓ 1
Part (b)
M1 adds the three equations: 4(a + c + s) = 112 (or any valid elimination)✓ 1
A1 a + c + s = 28✓ 1
A1 a = 12, c = 7, s = 9 (accept any valid method)✓ 1
Part (c)
A1 2(12) + 3(7) + 9 = €54 (ECF from (b))✓ 1

Exam tip: With a symmetric system, adding all the equations first often gives the total a + c + s and makes the rest quick.

55IBMA3H-1-20r29-011.13 Polar and Euler forms Paper 1Easy6 marks

No calculator

A robot's displacement from its charging dock is modelled by the complex number \(z=-1+\sqrt3\,i\), in metres, in a plane.

OReImOz = −1 + √3i
Not to scale
(a)

Find the modulus and the argument of \(z\), where \(-\pi<\arg z\le\pi\).

(3)
(b)

Write \(z\) in the form \(re^{i\theta}\).

(1)
(c)

Find \(z^3\) in its simplest form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(|z|=\sqrt{1+3}\)✓ 1
A1 \(|z|=2\)✓ 1
A1 \(\arg z=\frac{2\pi}{3}\)✓ 1
Part (b)
A1 \(z=2e^{2\pi i/3}\)✓ 1
Part (c)
M1 \(z^3=2^3e^{2\pi i}\)✓ 1
A1 \(z^3=8\)✓ 1

Common error: Sketch the point first: \(-1+\sqrt3\,i\) lies in the second quadrant, so \(\arctan(-\sqrt3)=-\frac\pi3\) must be adjusted by adding π.

56IBMA3H-1-20r29-021.13 Polar and Euler forms Paper 1Easy5 marks

No calculator

Two complex numbers are \(z_1=2\,\text{cis}\frac{\pi}{3}\) and \(z_2=3\,\text{cis}\frac{\pi}{6}\), where \(\text{cis}\,\theta=\cos\theta+i\sin\theta\). A fashion design program applies a scaling and a rotation by multiplying coordinates by these numbers.

(a)

Find \(z_1z_2\), giving your answer in modulus-argument form.

(2)
(b)

Find \(\dfrac{z_1}{z_2}\) in modulus-argument form.

(2)
(c)

Write \(z_1z_2\) in the form \(a+bi\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply the moduli and add the arguments✓ 1
A1 \(6\,\text{cis}\frac{\pi}{2}\)✓ 1
Part (b)
M1 divide the moduli and subtract the arguments✓ 1
A1 \(\frac23\,\text{cis}\frac{\pi}{6}\)✓ 1
Part (c)
A1 \(6i\)✓ 1

Exam tip: To multiply in polar form: multiply moduli, add arguments. To divide: divide moduli, subtract arguments.

57IBMA3H-1-20r35-021.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy4 marks

No calculator

Simplify and solve the following, showing your working.

(a)

Evaluate \(27^{\frac23}\times16^{-\frac34}\), giving your answer as a fraction.

(2)
(b)

Solve \(4^{x+1}=32\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(27^{\frac23}=9\) and \(16^{-\frac34}=\dfrac18\)✓ 1
A1 \(\dfrac98\)✓ 1
Part (b)
M1 write both sides as powers of 2: \(2^{2x+2}=2^5\)✓ 1
A1 \(x=\dfrac32\)✓ 1

Exam tip: For a fractional power, take the root first and then the power: it keeps the numbers small enough to do without a calculator.

58IBMA3H-1-20r35-041.2 Arithmetic sequences and series Paper 1Easy6 marks

No calculator

A square courtyard is paved with identical square tiles, with \(n\) tiles along each side, where \(n\) is an odd positive integer. The courtyard is then extended into a larger square with \(n+2\) tiles along each side.

n + 2n × n tilesn + 2
Shaded: the original n × n courtyard (n = 5 shown). Not to scale
(a)

Prove that the number of extra tiles needed is always a multiple of 8.

(4)
(b)

Find the number of extra tiles needed when \(n=31\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 let \(n=2k-1\) for a positive integer \(k\), so the new side is \(2k+1\) tiles✓ 1
M1 extra tiles \(=(2k+1)^2-(2k-1)^2=(4k^2+4k+1)-(4k^2-4k+1)\)✓ 1
A1 extra tiles \(=8k\)✓ 1
R1 \(k\) is an integer, so \(8k\) is a multiple of 8✓ 1
Part (b)
M1 \(33^2-31^2\) or \(8k\) with \(k=16\)✓ 1
A1 128 tiles✓ 1

Exam tip: In a proof, define the general case with a letter (here 2k − 1 for an odd number) and finish with a sentence saying why the result follows.

59IBMA3H-1-19r39-081.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Easy6 marks

No calculator

A designer for a spiral facade places the nth panel anchor at the point \(z^{n}\) in an Argand diagram, where \(z=\sqrt{3}+i\).

(a)

Write \(z\) in the form \(r(\cos\theta+i\sin\theta)\), where \(r>0\) and \(-\pi<\theta\le\pi\).

(3)
(b)

Use de Moivre's theorem to find the position of the fifth anchor, \(z^{5}\), in the form \(a+bi\) where \(a\) and \(b\) are exact.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to find the modulus or the argument✓ 1
A1 \(r=2\)✓ 1
A1 \(\theta=\dfrac{\pi}{6}\)✓ 1
Part (b)
M1 \(z^5=2^5\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\)✓ 1
A1 \(32\left(-\dfrac{\sqrt3}{2}+\dfrac{1}{2}i\right)\)✓ 1
A1 \(-16\sqrt{3}+16i\)✓ 1

Common error: Raise the modulus to the power \(n\) as well as multiplying the argument by \(n\); a frequent slip is to leave the modulus as 2.

60IBMA3H-1-19r39-111.3 Geometric sequences and series Paper 1Easy5 marks

No calculator

A coral colony has 3×105 polyps. Each polyp deposits 4×10−9 g of calcium carbonate every second.

(a)

Find the total mass deposited by the colony each second. Give your answer in the form a×10k, where 1 ≤ a < 10 and k is an integer.

(3)
(b)

Find the time, in seconds, for the colony to deposit 3 kg of calcium carbonate. Give your answer in standard form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies 3×105 by 4×10−9✓ 1
A1 12×10−4 g✓ 1
A1 1.2×10−3 g✓ 1
Part (b)
M1 divides 3×103 g by 1.2×10−3 g s−1✓ 1
A1 2.5×106 s✓ 1

Common error: Change 3 kg to 3×10³ g before dividing; mixing kilograms and grams changes the answer by a factor of 1000.

61IBMA3H-1-18r47-091.15 Proof by induction, contradiction and counterexample Paper 1Easy5 marks

No calculator

Prove by mathematical induction that \(7^n - 1\) is divisible by 6 for all positive integers \(n\).

(a)

Write a complete proof.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 when \(n = 1\), \(7^1 - 1 = 6\), which is divisible by 6, so the statement is true for \(n = 1\)✓ 1
M1 assume true for \(n = k\): \(7^k - 1 = 6m\) for some integer \(m\)✓ 1
M1 consider \(7^{k+1} - 1 = 7(7^k - 1) + 6\)✓ 1
A1 \(= 42m + 6 = 6(7m + 1)\), which is divisible by 6✓ 1
R1 true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), so true for all positive integers \(n\) by induction✓ 1

Exam tip: The final sentence of an induction proof must state all three ingredients: the base case, the assumption, and the conclusion for all positive integers.

62IBMA3H-1-18r47-101.15 Proof by induction, contradiction and counterexample Paper 1Easy6 marks

No calculator

An archaeologist tests two claims about a collection of clay tablets.

(a)

Claim 1: "For every positive integer \(n\), the number \(n^2 + n + 41\) is prime." Show that this claim is false.

(2)
(b)

Claim 2: "Exactly 100 tablets can be packed into boxes, each box holding either 6 or 9 tablets, with every box full." Use proof by contradiction to show that this is impossible.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute a suitable value, e.g. \(n = 40\) (or \(n = 41\))✓ 1
A1 \(40^2 + 40 + 41 = 1681 = 41^2\) (not prime) (or \(41^2 + 41 + 41 = 41 \times 43\)); counterexample found✓ 1
Part (b)
M1 assume it is possible: \(6a + 9b = 100\) with \(a\) and \(b\) non-negative integers✓ 1
M1 factorise: \(3(2a + 3b) = 100\)✓ 1
R1 \(2a + 3b\) is an integer, so 100 would be divisible by 3✓ 1
R1 but \(100 = 3 \times 33 + 1\) is not divisible by 3, a contradiction, so the assumption is false and the packing is impossible✓ 1

Common error: A single counterexample is enough to disprove a statement; many examples that work do not prove it.

63AAHL1-q00021.2 Arithmetic sequences and series Paper 1Medium7 marks

Consider the positive integers from 1 to 500 inclusive.

(a)

Find the sum of all the multiples of 3 in this range.

(2)
(b)

Find the sum of all the multiples of 15 in this range.

(2)
(c)

Hence find the sum of all the integers from 1 to 500 inclusive that are divisible by neither 3 nor 5.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(3+6+\cdots+498\) with 166 terms: \(\dfrac{166}{2}(3+498)\)(M1)Award for identifying \(n=166\) and \(u_{n}=498\) and substituting.
\(41\,583\)A1
(b)
\(15+30+\cdots+495\) with 33 terms: \(\dfrac{33}{2}(15+495)\)(M1)
\(8415\)A1
(c)
sum of 1 to 500 is \(125\,250\); sum of multiples of 5 is \(\dfrac{100}{2}(5+500)=25\,250\)(A1)
\(125\,250-41\,583-25\,250+8415\)M1Must add back the multiples of 15 (counted twice).
\(66\,832\)A1

Answers: (a) \(41\,583\) · (b) \(8415\) · (c) \(66\,832\)

64AAHL1-q00041.2 Arithmetic sequences and series Paper 1Medium8 marks

The sum of the interior angles of a convex polygon with \(n\) sides is \(180(n-2)\) degrees, and each interior angle of a convex polygon is less than \(180^{\circ}\).

The interior angles of a particular convex polygon with \(n\) sides form an arithmetic sequence with first term \(100^{\circ}\) and common difference \(10^{\circ}\).

(a)

Show that \(n^{2}-17n+72=0\).

(3)
(b)

Find the value of \(n\), justifying your answer.

(3)
(c)

Show that there is no convex polygon whose interior angles form an arithmetic sequence with first term \(100^{\circ}\) and common difference \(20^{\circ}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{n}{2}\big(2(100)+10(n-1)\big)=180(n-2)\)M1Equating the arithmetic series to the angle sum.
\(5n^{2}+95n=180n-360\)A1Or \(n(5n+95)=180n-360\).
\(5n^{2}-85n+360=0\), and dividing by 5A1
\(n^{2}-17n+72=0\)AG
(b)
\((n-8)(n-9)=0\), so \(n=8\) or \(n=9\)A1
if \(n=9\), the largest angle is \(100+8(10)=180^{\circ}\), which is not possible for a convex polygonR1
\(n=8\)A1Largest angle \(170^{\circ}\). Award A0 for \(n=8\) without rejecting \(n=9\).
(c)
\(\dfrac{n}{2}\big(200+20(n-1)\big)=180(n-2)\Rightarrow n^{2}-9n+36=0\)M1
discriminant \(=81-144=-63\lt0\), so there is no real (and hence no integer) value of \(n\)R1

Answers: (b) \(n=8\)

65AAHL1-q00061.3 Geometric sequences and series Paper 1Medium8 marks

Three consecutive terms of a geometric sequence have a product of 216 and a sum of 21. Let the terms be \(\dfrac{a}{r}\), \(a\) and \(ar\), where \(r\gt 0\).

(a)

Show that \(a=6\).

(1)
(b)

Find the two possible values of \(r\).

(3)
(c)

For the value of \(r\) with \(r\lt 1\), the three terms are the first three terms of an infinite geometric series. Find the sum to infinity.

(2)
(d)

The squares of the terms of this infinite series form another infinite geometric series. Find the sum to infinity of this new series.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{a}{r}\times a\times ar=a^{3}=216\)A1
\(a=6\)AG
(b)
\(\dfrac{6}{r}+6+6r=21\)M1
\(6r^{2}-15r+6=0\) or \(2r^{2}-5r+2=0\)A1
\(r=2\) or \(r=\dfrac{1}{2}\)A1
(c)
terms \(12,\ 6,\ 3\); \(S_{\infty}=\dfrac{12}{1-\frac{1}{2}}\)(M1)
\(24\)A1
(d)
\(144+36+9+\cdots\): first term \(144\), common ratio \(\left(\tfrac12\right)^{2}=\tfrac14\); \(\dfrac{144}{1-\frac14}\)(M1)
\(192\)A1

Answers: (b) \(r=2\) or \(r=\tfrac12\) · (c) \(24\) · (d) \(192\)

66AAHL1-q00131.6 Deductive proof Paper 1Medium6 marks

Let \(n\in\mathbb{Z}\).

(a)

Show that \(n(n+3)=n^{2}+3n\) and \((n+1)(n+2)=n^{2}+3n+2\).

(1)
(b)

Hence show that \(n(n+1)(n+2)(n+3)+1\equiv\left(n^{2}+3n+1\right)^{2}\).

(3)
(c)

Hence write \(2024\times2025\times2026\times2027+1\) in the form \(m^{2}\), where \(m\in\mathbb{Z}^{+}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(n(n+3)=n^{2}+3n\) and \((n+1)(n+2)=n^{2}+2n+n+2=n^{2}+3n+2\)A1
(b)
LHS \(=\big[n(n+3)\big]\big[(n+1)(n+2)\big]+1=\left(n^{2}+3n\right)\left(n^{2}+3n+2\right)+1\)M1Must start from the LHS.
let \(u=n^{2}+3n\): \(u(u+2)+1=u^{2}+2u+1\)A1Or full expansion to \(n^{4}+6n^{3}+11n^{2}+6n+1\).
\(=(u+1)^{2}\)A1Or showing that \(\left(n^{2}+3n+1\right)^{2}\) also expands to \(n^{4}+6n^{3}+11n^{2}+6n+1\).
\(\equiv\left(n^{2}+3n+1\right)^{2}\)AG
(c)
\(n=2024\): \(m=2024^{2}+3(2024)+1\)M1
\(4\,102\,649^{2}\)A1\(2024^{2}=4\,096\,576\)

Answers: (c) \(4\,102\,649^{2}\)

67AAHL1-q00161.7 Rational exponents and laws of logarithms Paper 1Medium8 marks
(a)

Solve \(x^{\log_{10}x}=100x\), for \(x\gt0\).

(4)
(b)(i)

Show that \(4^{\log_{2}x}=x^{2}\), for \(x\gt0\).

(2)
(b)(ii)

Hence solve \(4^{\log_{2}x}+2x=15\), for \(x\gt0\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
taking \(\log_{10}\) of both sides: \(\log_{10}x\times\log_{10}x=\log_{10}100+\log_{10}x\)M1
\(\left(\log_{10}x\right)^{2}-\log_{10}x-2=0\)A1
\(\left(\log_{10}x-2\right)\left(\log_{10}x+1\right)=0\), so \(\log_{10}x=2\) or \(\log_{10}x=-1\)A1
\(x=100\) or \(x=\dfrac{1}{10}\)A1
(b)(i)
\(4^{\log_{2}x}=\left(2^{2}\right)^{\log_{2}x}=2^{2\log_{2}x}\)M1
\(=2^{\log_{2}\left(x^{2}\right)}=x^{2}\)A1Or \(\log_{2}\left(4^{\log_{2}x}\right)=\log_{2}x\times\log_{2}4=2\log_{2}x=\log_{2}x^{2}\).
\(4^{\log_{2}x}=x^{2}\)AG
(b)(ii)
\(x^{2}+2x-15=0\)M1
\(x=3\) onlyA1Must reject \(x=-5\) (\(x\gt0\)).

Answers: (a) \(x=100\) or \(x=\tfrac{1}{10}\) · (b)(ii) \(x=3\)

68AAHL1-q00171.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

Two populations of insects are modelled by \(P(t)=200\mathrm{e}^{0.35t}\) and \(Q(t)=50\times3^{t/2}\), where \(t\ge0\) is the time in weeks.

(a)

Write \(Q(t)\) in the form \(50\mathrm{e}^{kt}\), giving the exact value of \(k\).

(2)
(b)

Show that the two populations are equal when \(t=\dfrac{4\ln2}{\ln3-0.7}\).

(3)
(c)

Explain why \(Q(t)\gt P(t)\) for all values of \(t\) greater than the value in part (b).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(3^{t/2}=\mathrm{e}^{\frac{t}{2}\ln3}\)(M1)Using \(a^{x}=\mathrm{e}^{x\ln a}\).
\(k=\dfrac{\ln3}{2}\)A1Accept \(\ln\sqrt3\).
(b)
\(200\mathrm{e}^{0.35t}=50\mathrm{e}^{\frac{t}{2}\ln3}\Rightarrow4=\mathrm{e}^{t\left(\frac{\ln3}{2}-0.35\right)}\)M1
\(\ln4=t\left(\dfrac{\ln3}{2}-0.35\right)\)A1
\(t=\dfrac{2\ln4}{\ln3-0.7}\) and \(2\ln4=4\ln2\)A1
\(t=\dfrac{4\ln2}{\ln3-0.7}\)AG
(c)
\(\dfrac{Q(t)}{P(t)}=\dfrac{1}{4}\mathrm{e}^{t\left(\frac{\ln3}{2}-0.35\right)}\) and \(\ln3\gt\ln\mathrm{e}=1\gt0.7\), so \(\dfrac{\ln3}{2}-0.35\gt0\)R1
so \(\dfrac{Q(t)}{P(t)}\) is increasing; it equals 1 at the time in (b), so it is greater than 1 afterwardsR1

Answers: (a) \(k=\dfrac{\ln3}{2}\)

69AAHL1-q00181.8 Infinite geometric series Paper 1Medium8 marks

An infinite geometric series has first term \(u_{1}\) and common ratio \(r\), where \(0\lt|r|\lt1\). Its sum to infinity is 12, and the sum of the terms in the odd positions, \(u_{1}+u_{3}+u_{5}+\cdots\), is 9.

(a)

Show that \(u_{1}+u_{3}+u_{5}+\cdots=\dfrac{u_{1}}{1-r^{2}}\).

(2)
(b)

Find the value of \(r\) and the value of \(u_{1}\).

(3)
(c)

Write down the sum of the terms in the even positions, \(u_{2}+u_{4}+u_{6}+\cdots\).

(1)
(d)

Show that, for every infinite geometric series with \(u_{1}\gt0\) and \(0\lt|r|\lt1\), the sum of the terms in the odd positions is more than half of the sum to infinity.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(u_{1}+u_{1}r^{2}+u_{1}r^{4}+\cdots\) is geometric with first term \(u_{1}\) and common ratio \(r^{2}\)M1
\(0\lt r^{2}\lt1\), so the series convergesR1
\(\dfrac{u_{1}}{1-r^{2}}\)AG
(b)
\(\dfrac{u_{1}}{1-r}\div\dfrac{u_{1}}{1-r^{2}}=\dfrac{12}{9}\), so \(1+r=\dfrac{4}{3}\)M1Or substituting \(u_{1}=12(1-r)\) into \(\dfrac{u_{1}}{1-r^{2}}=9\).
\(r=\dfrac{1}{3}\)A1
\(u_{1}=12\left(1-\dfrac13\right)=8\)A1
(c)
\(12-9=3\)A1
(d)
\(\dfrac{\text{odd-position sum}}{S_{\infty}}=\dfrac{u_{1}/(1-r^{2})}{u_{1}/(1-r)}=\dfrac{1}{1+r}\)M1
\(-1\lt r\lt1\Rightarrow0\lt1+r\lt2\), so \(\dfrac{1}{1+r}\gt\dfrac{1}{2}\); as \(S_{\infty}\gt0\), the odd-position sum is more than \(\tfrac12S_{\infty}\)R1
odd-position sum \(\gt\dfrac{1}{2}S_{\infty}\)AG

Answers: (b) \(r=\tfrac13\), \(u_{1}=8\) · (c) \(3\)

70AAHL1-q00201.9 The binomial theorem Paper 1Medium8 marks

Let \(n\in\mathbb{Z}^{+}\), \(n\ge3\).

(a)

Write down the first four terms, in ascending powers of \(x\), of the expansion of \((1+x)^{n}\).

(2)
(b)

Hence show that \(1.01^{100}\gt2.6\).

(3)
(c)

Find the coefficient of \(x^{2}\) and the coefficient of \(x^{3}\) in the expansion of \((1+x)^{100}(1-x)^{100}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(1+nx+\dfrac{n(n-1)}{2}x^{2}\)A1
\(+\dfrac{n(n-1)(n-2)}{6}x^{3}\)A1Accept \({}^{n}C_{2}x^{2}+{}^{n}C_{3}x^{3}\).
(b)
\(n=100\), \(x=0.01\): \(1+100(0.01)+4950(0.0001)+161\,700(0.000001)\)M1
\(=1+1+0.495+0.1617=2.6567\)A1
all the remaining terms of the expansion are positive, so \(1.01^{100}\gt2.6567\gt2.6\)R1
\(1.01^{100}\gt2.6\)AG
(c)
\((1+x)^{100}(1-x)^{100}=\left(1-x^{2}\right)^{100}\)M1Or multiplying the two expansions: \(2\times4950-100\times100\).
coefficient of \(x^{2}\): \(-100\)A1
coefficient of \(x^{3}\): \(0\)A1Only even powers of \(x\) appear.

Answers: (a) \(1+nx+\tfrac{n(n-1)}{2}x^{2}+\tfrac{n(n-1)(n-2)}{6}x^{3}\) · (c) \(-100\) and \(0\)

71AAHL1-q00231.10 Counting principles and the extended binomial theorem Paper 1Medium6 marks

Eight different people, including Ana, Ben, Chloe, Dev and Eva, are to sit in a row of eight seats.

(a)

Find the number of possible arrangements.

(1)
(b)

Find the number of arrangements in which Ana and Ben sit next to each other.

(2)
(c)

Find the number of arrangements in which no two of Chloe, Dev and Eva sit next to each other.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(8!=40\,320\)A1
(b)
treating Ana and Ben as one unit: \(7!\times2!\)(M1)
\(10\,080\)A1
(c)
arrange the other 5 people: \(5!\), creating 6 gaps (including the ends)(M1)
place Chloe, Dev and Eva in 3 different gaps: \({}^{6}P_{3}=120\)(M1)
\(120\times120=14\,400\)A1

Answers: (a) 40 320 · (b) 10 080 · (c) 14 400

72AAHL1-q00241.10 Counting principles and the extended binomial theorem Paper 1Medium7 marks

A four-digit code is formed using four different digits chosen from 1, 2, 3, 4, 5, 6, 7, 8.

(a)

Use the formula for \({}^{n}P_{r}\) to find the number of possible codes.

(2)
(b)

Find the number of codes that are even.

(2)
(c)

Find the number of codes that are even and greater than 6000.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\({}^{8}P_{4}=\dfrac{8!}{4!}=8\times7\times6\times5\)(M1)
\(1680\)A1
(b)
last digit from \(\{2,4,6,8\}\): \(4\times{}^{7}P_{3}=4\times210\)(M1)
\(840\)A1
(c)
first digit 6 or 8: \(2\times3\times{}^{6}P_{2}=180\)(M1)First digit even uses one of the even digits, leaving 3 for the last.
first digit 7: \(4\times{}^{6}P_{2}=120\)(M1)
\(300\)A1

Answers: (a) 1680 · (b) 840 · (c) 300

73AAHL1-q00251.10 Counting principles and the extended binomial theorem Paper 1Medium8 marks
(a)

Find the first four terms, in ascending powers of \(x\), of the expansion of \((1-4x)^{-\frac{1}{2}}\).

(3)
(b)

Write down the set of values of \(x\) for which the expansion is valid.

(1)
(c)

Show that, when \(x=0.01\), \((1-4x)^{-\frac12}=\dfrac{5\sqrt6}{12}\).

(2)
(d)

Hence find an approximation for \(\sqrt6\), giving your answer to six decimal places.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(1+\left(-\tfrac12\right)(-4x)+\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2!}(-4x)^{2}+\dfrac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{3!}(-4x)^{3}\)(M1)
\(1+2x+6x^{2}\)A1
\(+20x^{3}\)A1
(b)
\(|x|\lt\dfrac{1}{4}\)A1Accept \(-\tfrac14\lt x\lt\tfrac14\).
(c)
\((0.96)^{-\frac12}=\dfrac{1}{\sqrt{0.96}}=\dfrac{10}{\sqrt{96}}\)M1
\(=\dfrac{10}{4\sqrt6}=\dfrac{5}{2\sqrt6}=\dfrac{5\sqrt6}{12}\)A1
\(\dfrac{5\sqrt6}{12}\)AG
(d)
\(\dfrac{5\sqrt6}{12}\approx1+0.02+0.0006+0.00002=1.02062\)M1
\(\sqrt6\approx\dfrac{12}{5}\times1.02062=2.449488\)A1

Answers: (a) \(1+2x+6x^{2}+20x^{3}\) · (b) \(|x|\lt\tfrac14\) · (d) \(2.449488\)

74AAHL1-q00261.10 Counting principles and the extended binomial theorem Paper 1Medium8 marks
(a)

Show that the first three terms of the expansion of \((4-x)^{-\frac12}\), in ascending powers of \(x\), are \(\dfrac{1}{2}+\dfrac{x}{16}+\dfrac{3x^{2}}{256}\).

(3)
(b)

State the set of values of \(x\) for which this expansion is valid.

(1)
(c)

Hence find the first three terms of the expansion of \(\dfrac{2+x}{\sqrt{4-x}}\).

(2)
(d)

Find the value of \(k\) for which the coefficient of \(x^{2}\) in the expansion of \(\dfrac{k+x}{\sqrt{4-x}}\) is zero.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\((4-x)^{-\frac12}=4^{-\frac12}\left(1-\dfrac{x}{4}\right)^{-\frac12}=\dfrac{1}{2}\left(1-\dfrac{x}{4}\right)^{-\frac12}\)M1
\(\left(1-\dfrac{x}{4}\right)^{-\frac12}=1+\left(-\tfrac12\right)\left(-\tfrac{x}{4}\right)+\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2}\left(-\tfrac{x}{4}\right)^{2}+\cdots\)A1
\(=1+\dfrac{x}{8}+\dfrac{3x^{2}}{128}+\cdots\)A1
\(\dfrac{1}{2}+\dfrac{x}{16}+\dfrac{3x^{2}}{256}\)AG
(b)
\(|x|\lt4\)A1
(c)
\((2+x)\left(\dfrac12+\dfrac{x}{16}+\dfrac{3x^{2}}{256}\right)\)(M1)
\(1+\dfrac{5x}{8}+\dfrac{11x^{2}}{128}\)A1
(d)
\(\dfrac{3k}{256}+\dfrac{1}{16}=0\)(M1)
\(k=-\dfrac{16}{3}\)A1

Answers: (b) \(|x|\lt4\) · (c) \(1+\tfrac{5}{8}x+\tfrac{11}{128}x^{2}\) · (d) \(k=-\tfrac{16}{3}\)

75AAHL1-q00291.11 Partial fractions Paper 1Medium7 marks

Let \(k\in\mathbb{Z}^{+}\).

(a)

Express \(\dfrac{4}{(2k-1)(2k+3)}\) in partial fractions.

(3)
(b)

Hence show that \(\displaystyle\sum_{k=1}^{n}\dfrac{4}{(2k-1)(2k+3)}=\dfrac{4}{3}-\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\).

(3)
(c)

Write down the value of \(\displaystyle\sum_{k=1}^{\infty}\dfrac{4}{(2k-1)(2k+3)}\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(4\equiv A(2k+3)+B(2k-1)\)M1
\(A=1\), \(B=-1\)A1From \(k=\tfrac12\) and \(k=-\tfrac32\).
\(\dfrac{1}{2k-1}-\dfrac{1}{2k+3}\)A1
(b)
writing out terms: \(\left(1-\tfrac15\right)+\left(\tfrac13-\tfrac17\right)+\left(\tfrac15-\tfrac19\right)+\cdots+\left(\tfrac{1}{2n-1}-\tfrac{1}{2n+3}\right)\)M1At least the first three and the last term.
all terms cancel except \(1+\dfrac13\) and \(-\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\)R1
\(1+\dfrac{1}{3}=\dfrac{4}{3}\)A1
\(\dfrac{4}{3}-\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\)AG
(c)
\(\dfrac{4}{3}\)A1

Answers: (a) \(\dfrac{1}{2k-1}-\dfrac{1}{2k+3}\) · (c) \(\tfrac43\)

76AAHL1-q00341.12 Complex numbers in Cartesian form Paper 1Medium8 marks

The complex number \(a+bi\), where \(a,b\in\mathbb{R}\), satisfies \((a+bi)^{2}=-5+12i\).

(a)

Find the two possible pairs of values of \(a\) and \(b\).

(5)
(b)

Hence solve the equation \(z^{2}+2z+6-12i=0\), giving your answers in the form \(p+qi\), where \(p,q\in\mathbb{Z}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(a^{2}-b^{2}+2abi=-5+12i\)M1Attempt to expand and equate parts.
\(a^{2}-b^{2}=-5\) and \(ab=6\)A1
eliminating one variable, e.g. \(a^{2}-\dfrac{36}{a^{2}}=-5\Rightarrow a^{4}+5a^{2}-36=0\)M1
\((a^{2}+9)(a^{2}-4)=0\Rightarrow a^{2}=4\)A1\(a^{2}=-9\) must be rejected or not seen, since \(a\) is real.
\(a=2,\ b=3\) or \(a=-2,\ b=-3\)A1
(b)
completing the square: \((z+1)^{2}=-5+12i\)M1
\(z+1=\pm(2+3i)\)A1FT from (a).
\(z=1+3i\) or \(z=-3-3i\)A1

Answers: (a) \(a=2,b=3\) or \(a=-2,b=-3\) · (b) \(1+3i,\ -3-3i\)

77AAHL1-q00381.12 Complex numbers in Cartesian form Paper 1Medium6 marks

Let \(z=a+bi\) and \(w=c+di\), where \(a,b,c,d\in\mathbb{R}\), and let Z and W be the points representing \(z\) and \(w\) on an Argand diagram with origin O.

(a)

Show that \(zw^{*}+z^{*}w=2(ac+bd)\).

(2)
(b)

Hence explain why OZ is perpendicular to OW if and only if \(zw^{*}\) is purely imaginary, for non-zero \(z\) and \(w\).

(2)
(c)

Given that \(z=3+2i\) and \(w=k-6i\), where \(k\in\mathbb{R}\), and that OZ is perpendicular to OW, find \(k\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\((a+bi)(c-di)+(a-bi)(c+di)\)M1Attempt to expand both products.
\(ac-adi+bci+bd+ac+adi-bci+bd\)A1
\(=2(ac+bd)\)AG
(b)
\(ac+bd\) is the scalar product of \(\begin{pmatrix}a\\b\end{pmatrix}\) and \(\begin{pmatrix}c\\d\end{pmatrix}\), so OZ \(\perp\) OW \(\iff ac+bd=0\)R1
\(z^{*}w=(zw^{*})^{*}\), so \(zw^{*}+z^{*}w=2\operatorname{Re}(zw^{*})\); hence OZ \(\perp\) OW \(\iff\operatorname{Re}(zw^{*})=0\)R1
(c)
\(3k+2(-6)=0\)M1
\(k=4\)A1

Answers: (c) \(k=4\)

78AAHL1-q00411.13 Polar and Euler forms Paper 1Medium7 marks

Let \(z_{1}=-2+2\sqrt3\,i\) and \(z_{2}=-\sqrt3-i\).

OReImOz₁z₂
Not to scale
(a)

Find the modulus and argument of \(z_{1}\) and of \(z_{2}\), giving each argument in the interval \(-\pi\lt\theta\le\pi\).

(3)
(b)

Find \(z_{1}z_{2}\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
(c)

Find \(\dfrac{z_{1}}{z_{2}}\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\). Hence write down \(\dfrac{z_{1}}{z_{2}}\) in the form \(a+bi\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(|z_{1}|=4,\ |z_{2}|=2\)A1
\(\arg z_{1}=\dfrac{2\pi}{3}\)A1
\(\arg z_{2}=-\dfrac{5\pi}{6}\)A1Award A0 for \(\frac{\pi}{6}\) (wrong quadrant).
(b)
\(4\times2=8\) and \(\dfrac{2\pi}{3}-\dfrac{5\pi}{6}\)M1
\(8\mathrm{e}^{-\frac{\pi}{6}i}\)A1
(c)
\(\dfrac{4}{2}=2\) and \(\dfrac{2\pi}{3}+\dfrac{5\pi}{6}=\dfrac{3\pi}{2}\), adjusted by \(-2\pi\)M1
\(2\mathrm{e}^{-\frac{\pi}{2}i}=-2i\)A1Both forms required. Award A0 for \(2\mathrm{e}^{\frac{3\pi}{2}i}\).

Answers: (a) \(4,\ 2\pi/3\); \(2,\ -5\pi/6\) · (b) \(8\mathrm{e}^{-\pi i/6}\) · (c) \(2\mathrm{e}^{-\pi i/2}=-2i\)

79AAHL1-q00421.13 Polar and Euler forms Paper 1Medium6 marks

The following Argand diagram shows a square OABC, with vertices labelled anticlockwise. The point A represents the complex number \(3+i\).

ReImOA (3 + i)BC
Diagram not to scale
(a)

By considering a rotation about O, find the complex number represented by C.

(2)
(b)

Write down the complex number represented by B.

(1)
(c)

The square is rotated anticlockwise about O through an angle of \(\dfrac{\pi}{3}\) and then enlarged with centre O and scale factor 2. Find, in the form \(a+bi\), the complex number represented by the image of B.

(3)
Show mark scheme
WorkingMarkNotes
(a)
multiplying by \(i\) (rotation of \(\frac{\pi}{2}\) anticlockwise): \(i(3+i)\)M1
\(-1+3i\)A1
(b)
\((3+i)+(-1+3i)=2+4i\)A1FT from (a).
(c)
multiplying by \(2\mathrm{e}^{\frac{\pi}{3}i}\)M1
\(2\mathrm{e}^{\frac{\pi}{3}i}=1+\sqrt3\,i\)A1
\((2+4i)(1+\sqrt3\,i)=(2-4\sqrt3)+(4+2\sqrt3)i\)A1FT from (b).

Answers: (a) \(-1+3i\) · (b) \(2+4i\) · (c) \((2-4\sqrt3)+(4+2\sqrt3)i\)

80AAHL1-q00441.13 Polar and Euler forms Paper 1Medium6 marks

Let \(z=1+\mathrm{e}^{i\theta}\), where \(0\lt\theta\lt 2\pi\) and \(\theta\neq\pi\).

(a)

Show that \(z=2\cos\left(\dfrac{\theta}{2}\right)\mathrm{e}^{\frac{\theta}{2}i}\).

(2)
(b)

For \(0\lt\theta\lt\pi\), write down \(|z|\) and \(\arg z\) in terms of \(\theta\).

(2)
(c)

For \(\pi\lt\theta\lt 2\pi\), find \(|z|\) and \(\arg z\) in terms of \(\theta\), where \(-\pi\lt\arg z\le\pi\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(z=\mathrm{e}^{\frac{\theta}{2}i}\left(\mathrm{e}^{-\frac{\theta}{2}i}+\mathrm{e}^{\frac{\theta}{2}i}\right)\)M1
\(\mathrm{e}^{-\frac{\theta}{2}i}+\mathrm{e}^{\frac{\theta}{2}i}=2\cos\dfrac{\theta}{2}\)A1
\(z=2\cos\left(\dfrac{\theta}{2}\right)\mathrm{e}^{\frac{\theta}{2}i}\)AG
(b)
\(|z|=2\cos\dfrac{\theta}{2}\)A1
\(\arg z=\dfrac{\theta}{2}\)A1
(c)
\(\cos\frac{\theta}{2}\lt 0\), so \(|z|=-2\cos\dfrac{\theta}{2}\)A1
\(z=-2\cos\frac{\theta}{2}\,\mathrm{e}^{(\frac{\theta}{2}-\pi)i}\), so \(\arg z=\dfrac{\theta}{2}-\pi\)A1

Answers: (b) \(2\cos\frac{\theta}{2},\ \frac{\theta}{2}\) · (c) \(-2\cos\frac{\theta}{2},\ \frac{\theta}{2}-\pi\)

81AAHL1-q00451.13 Polar and Euler forms Paper 1Medium7 marks

Let \(u=-1+i\) and \(v=1+\sqrt3\,i\).

(a)

Find \(uv\) in the form \(a+bi\).

(1)
(b)

By writing \(u\) and \(v\) in Euler form, find \(uv\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(3)
(c)

Hence find the exact values of \(\cos\dfrac{11\pi}{12}\) and \(\sin\dfrac{11\pi}{12}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\((-1-\sqrt3)+(1-\sqrt3)i\)A1
(b)
\(u=\sqrt2\,\mathrm{e}^{\frac{3\pi}{4}i}\) and \(v=2\mathrm{e}^{\frac{\pi}{3}i}\)A1Both required.
\(uv=2\sqrt2\,\mathrm{e}^{\frac{13\pi}{12}i}\), argument adjusted by \(-2\pi\)M1Multiplying moduli and adding arguments.
\(uv=2\sqrt2\,\mathrm{e}^{-\frac{11\pi}{12}i}\)A1Award A0 for \(\theta=\frac{13\pi}{12}\) (outside the interval).
(c)
equating real and imaginary parts: \(2\sqrt2\cos\left(-\dfrac{11\pi}{12}\right)=-1-\sqrt3\) and \(2\sqrt2\sin\left(-\dfrac{11\pi}{12}\right)=1-\sqrt3\)M1
\(\cos\dfrac{11\pi}{12}=-\dfrac{1+\sqrt3}{2\sqrt2}=-\dfrac{\sqrt2+\sqrt6}{4}\)A1Accept any equivalent exact form.
\(\sin\dfrac{11\pi}{12}=\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac{\sqrt6-\sqrt2}{4}\)A1Accept any equivalent exact form. Award A0 for a negative value (sign of \(\sin(-\theta)\) not handled).

Answers: (a) \((-1-\sqrt3)+(1-\sqrt3)i\) · (b) \(2\sqrt2\,\mathrm{e}^{-11\pi i/12}\) · (c) \(\cos\frac{11\pi}{12}=-\frac{\sqrt2+\sqrt6}{4}\), \(\sin\frac{11\pi}{12}=\frac{\sqrt6-\sqrt2}{4}\)

82AAHL1-q00511.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

The polynomial \(p(z)=z^{4}+az^{3}+bz^{2}+cz+d\) has real coefficients. Two of the roots of \(p(z)=0\) are \(1+i\) and \(2-3i\).

(a)

Find the values of \(a\), \(b\), \(c\) and \(d\).

(5)
(b)

Explain why \(p(x)\gt 0\) for all \(x\in\mathbb{R}\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(1-i\) and \(2+3i\) are also roots (conjugate pairs)M1
\((z-1-i)(z-1+i)=z^{2}-2z+2\)A1
\((z-2+3i)(z-2-3i)=z^{2}-4z+13\)A1
\(p(z)=(z^{2}-2z+2)(z^{2}-4z+13)\), attempt to expandM1
\(a=-6,\ b=23,\ c=-34,\ d=26\)A1All four required.
(b)
the four roots are all non-real, so the graph of \(y=p(x)\) does not meet the \(x\)-axisR1
\(p\) is continuous and \(p(0)=26\gt 0\) (or each quadratic factor has positive leading coefficient and negative discriminant), so \(p(x)\gt 0\) for all real \(x\)R1

Answers: (a) \(a=-6,\ b=23,\ c=-34,\ d=26\)

83AAHL1-q00531.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium8 marks

Consider the equation \(z^{4}=-8+8\sqrt3\,i\).

(a)

Express \(-8+8\sqrt3\,i\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
(b)

Find the four roots of the equation, giving your answers in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(4)
(c)

Write down the root that lies in the first quadrant in the form \(a+bi\), and hence write down the other three roots in the form \(a+bi\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(r=16\)A1
\(16\mathrm{e}^{\frac{2\pi}{3}i}\)A1
(b)
modulus of each root \(=16^{\frac14}=2\)A1
one root using De Moivre: \(2\mathrm{e}^{\frac{\pi}{6}i}\)M1
adding multiples of \(\dfrac{2\pi}{4}=\dfrac{\pi}{2}\) to the argumentM1
\(2\mathrm{e}^{\frac{\pi}{6}i},\ 2\mathrm{e}^{\frac{2\pi}{3}i},\ 2\mathrm{e}^{-\frac{\pi}{3}i},\ 2\mathrm{e}^{-\frac{5\pi}{6}i}\)A1All four with arguments in \((-\pi,\pi]\).
(c)
\(\sqrt3+i\)A1
\(-1+\sqrt3\,i,\ -\sqrt3-i,\ 1-\sqrt3\,i\)A1Obtained by repeated multiplication by \(i\), or otherwise.

Answers: (a) \(16\mathrm{e}^{2\pi i/3}\) · (c) \(\sqrt3+i,\ -1+\sqrt3 i,\ -\sqrt3-i,\ 1-\sqrt3 i\)

84AAHL1-q00541.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium9 marks

Let \(\operatorname{cis}\theta=\cos\theta+i\sin\theta\).

(a)

Prove by mathematical induction that \((\operatorname{cis}\theta)^{n}=\operatorname{cis}(n\theta)\) for all \(n\in\mathbb{Z}^{+}\).

(6)
(b)

Show that \((\operatorname{cis}\theta)^{-1}=\operatorname{cis}(-\theta)\), and hence show that \((\operatorname{cis}\theta)^{-m}=\operatorname{cis}(-m\theta)\) for all \(m\in\mathbb{Z}^{+}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(n=1\): \((\operatorname{cis}\theta)^{1}=\operatorname{cis}\theta\), so true for \(n=1\)A1
assume true for \(n=k\): \((\operatorname{cis}\theta)^{k}=\cos k\theta+i\sin k\theta\)M1Do not award for “let \(n=k\)”.
\((\operatorname{cis}\theta)^{k+1}=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)\)M1
\(=\cos k\theta\cos\theta-\sin k\theta\sin\theta+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)\)A1
\(=\cos(k+1)\theta+i\sin(k+1)\theta\)A1Compound angle identities used.
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); hence true for all \(n\in\mathbb{Z}^{+}\) by mathematical inductionR1Award only if at least three of the previous marks have been awarded.
(b)
\(\dfrac{1}{\cos\theta+i\sin\theta}\times\dfrac{\cos\theta-i\sin\theta}{\cos\theta-i\sin\theta}=\dfrac{\cos\theta-i\sin\theta}{\cos^{2}\theta+\sin^{2}\theta}\)M1
\(=\cos(-\theta)+i\sin(-\theta)=\operatorname{cis}(-\theta)\)A1
\((\operatorname{cis}\theta)^{-m}=\big(\operatorname{cis}(-\theta)\big)^{m}=\operatorname{cis}(-m\theta)\), using part (a) with \(-\theta\) in place of \(\theta\)R1

Answers: Proof

85AAHL1-q00551.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium9 marks

Let \(z=\cos\theta+i\sin\theta\).

(a)

Use De Moivre’s theorem to show that \(z^{n}+z^{-n}=2\cos n\theta\), for \(n\in\mathbb{Z}^{+}\).

(2)
(b)

Expand \(\left(z+\dfrac1z\right)^{4}\).

(2)
(c)

Hence show that \(\cos^{4}\theta=\dfrac18\left(\cos4\theta+4\cos2\theta+3\right)\).

(2)
(d)

Hence find the exact value of \(\displaystyle\int_{0}^{\frac{\pi}{4}}\cos^{4}\theta\,\mathrm{d}\theta\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(z^{n}=\cos n\theta+i\sin n\theta\) and \(z^{-n}=\cos(-n\theta)+i\sin(-n\theta)=\cos n\theta-i\sin n\theta\)M1
adding: imaginary parts cancelA1
\(z^{n}+z^{-n}=2\cos n\theta\)AG
(b)
binomial expansion with coefficients \(1,4,6,4,1\)M1
\(z^{4}+4z^{2}+6+4z^{-2}+z^{-4}\)A1
(c)
\((2\cos\theta)^{4}=(z^{4}+z^{-4})+4(z^{2}+z^{-2})+6\)M1
\(16\cos^{4}\theta=2\cos4\theta+8\cos2\theta+6\)A1
\(\cos^{4}\theta=\dfrac18(\cos4\theta+4\cos2\theta+3)\)AG
(d)
integrating the result of (c) term by termM1
\(\dfrac18\left[\dfrac{\sin4\theta}{4}+2\sin2\theta+3\theta\right]_{0}^{\frac{\pi}{4}}\)A1
\(\dfrac18\left(0+2+\dfrac{3\pi}{4}\right)=\dfrac{8+3\pi}{32}\)A1Accept \(\dfrac14+\dfrac{3\pi}{32}\).

Answers: (b) \(z^{4}+4z^{2}+6+4z^{-2}+z^{-4}\) · (d) \(\dfrac{8+3\pi}{32}\)

86AAHL1-q00611.15 Proof by induction, contradiction and counterexample Paper 1Medium10 marks
(a)

Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n}r\,2^{r}=(n-1)2^{n+1}+2\), for \(n\in\mathbb{Z}^{+}\).

(7)
(b)

Hence find \(\displaystyle\sum_{r=11}^{20}r\,2^{r}\), giving your answer in the form \(p\times2^{q}\), where \(p\) is an odd integer and \(q\in\mathbb{Z}^{+}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(n=1\): LHS \(=1\times2=2\), RHS \(=0\times4+2=2\), so true for \(n=1\)A1Must see both sides evaluated.
assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}r\,2^{r}=(k-1)2^{k+1}+2\)M1Do not award for “let \(n=k\)” only.
consider \(n=k+1\): \(\displaystyle\sum_{r=1}^{k+1}r\,2^{r}=(k-1)2^{k+1}+2+(k+1)2^{k+1}\)M1Adding the correct \((k+1)\)th term to the assumed sum.
\(=2k\times2^{k+1}+2\)A1
\(=k\times2^{k+2}+2\)A1
\(=\big((k+1)-1\big)2^{(k+1)+1}+2\)A1Shows the result in the required form for \(n=k+1\).
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(b)
\(\displaystyle\sum_{r=1}^{20}-\sum_{r=1}^{10}=\big(19\times2^{21}+2\big)-\big(9\times2^{11}+2\big)\)M1Award for subtracting the sum to 10, not to 11.
\(=2^{11}\left(19\times2^{10}-9\right)=2^{11}(19\,456-9)\)A1
\(19\,447\times2^{11}\)A1Award A0 for \(39\,827\,456\) if not written in the required form \(p\times2^{q}\).

Answers: (b) \(19\,447\times2^{11}\) \((=39\,827\,456)\)

87AAHL1-q00621.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks

Prove by mathematical induction that \(5^{2n}+3n-1\) is divisible by 9 for all \(n\in\mathbb{Z}^{+}\).

(6)
Show mark scheme
WorkingMarkNotes
\(n=1\): \(25+3-1=27=9\times3\), so true for \(n=1\)A1
assume true for \(n=k\): \(5^{2k}+3k-1=9m\), \(m\in\mathbb{Z}\)M1
consider \(n=k+1\): \(5^{2k+2}+3(k+1)-1=25\times5^{2k}+3k+2\)M1
\(=25(9m-3k+1)+3k+2\)A1Or \(=25\big(5^{2k}+3k-1\big)-72k+27\).
\(=225m-72k+27=9(25m-8k+3)\), which is divisible by 9A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.

Answers: Proof: \(5^{2(k+1)}+3(k+1)-1=9(25m-8k+3)\).

88AAHL1-q00631.15 Proof by induction, contradiction and counterexample Paper 1Medium10 marks

Let \(f(x)=x\,\mathrm{e}^{2x}\), for \(x\in\mathbb{R}\). The \(n\)th derivative of \(f\) is denoted by \(f^{(n)}(x)\).

(a)

Find \(f'(x)\) and \(f''(x)\).

(3)
(b)

Prove by mathematical induction that \(f^{(n)}(x)=2^{n-1}(2x+n)\,\mathrm{e}^{2x}\), for \(n\in\mathbb{Z}^{+}\).

(6)
(c)

Hence write down, in terms of \(n\), the solution of the equation \(f^{(n)}(x)=0\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
attempt at the product ruleM1
\(f'(x)=\mathrm{e}^{2x}+2x\,\mathrm{e}^{2x}\;\big(=(2x+1)\mathrm{e}^{2x}\big)\)A1
\(f''(x)=2\mathrm{e}^{2x}+2(2x+1)\mathrm{e}^{2x}=(4x+4)\mathrm{e}^{2x}\)A1Accept \(4(x+1)\mathrm{e}^{2x}\).
(b)
\(n=1\): \(2^{0}(2x+1)\mathrm{e}^{2x}=f'(x)\), so true for \(n=1\)A1May refer to part (a).
assume true for \(n=k\): \(f^{(k)}(x)=2^{k-1}(2x+k)\mathrm{e}^{2x}\)M1
\(f^{(k+1)}(x)=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2^{k-1}(2x+k)\mathrm{e}^{2x}\right)\)M1Attempt to differentiate the assumed expression.
\(=2^{k-1}\left(2\mathrm{e}^{2x}+2(2x+k)\mathrm{e}^{2x}\right)\)A1
\(=2^{k}\big(2x+(k+1)\big)\mathrm{e}^{2x}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(c)
\(x=-\dfrac{n}{2}\)A1Since \(2^{n-1}\mathrm{e}^{2x}\gt 0\).

Answers: (a) \(f'(x)=(2x+1)\mathrm{e}^{2x}\), \(f''(x)=4(x+1)\mathrm{e}^{2x}\) · (c) \(x=-\tfrac{n}{2}\)

89AAHL1-q00651.15 Proof by induction, contradiction and counterexample Paper 1Medium9 marks

Consider the statement \(P(n)\): \(3^{n}\gt 2^{n}+5n^{2}\).

(a)

Show that \(P(4)\) is false.

(1)
(b)

Prove by mathematical induction that \(P(n)\) is true for all integers \(n\ge5\).

(7)
(c)

The argument used in the inductive step of part (b) is valid for every \(k\ge2\). Explain why this does not mean that \(P(n)\) is true for all \(n\in\mathbb{Z}^{+}\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(3^{4}=81\) and \(2^{4}+5(16)=96\); \(81\lt96\), so \(P(4)\) is falseA1Both values required.
(b)
\(n=5\): \(3^{5}=243\gt2^{5}+5(25)=157\), so \(P(5)\) is trueA1
assume \(P(k)\) true for some \(k\ge5\): \(3^{k}\gt2^{k}+5k^{2}\)M1
\(3^{k+1}=3\times3^{k}\gt3\times2^{k}+15k^{2}\)M1
\(=2^{k+1}+2^{k}+15k^{2}\)A1
need to show \(2^{k}+15k^{2}\ge5(k+1)^{2}\), i.e. \(2^{k}+10k^{2}-10k-5\ge0\)M1Award for comparing with the required RHS.
\(10k^{2}-10k-5=10k(k-1)-5\gt0\) for \(k\ge5\) (and \(2^{k}\gt0\)), so \(3^{k+1}\gt2^{k+1}+5(k+1)^{2}\)A1Any valid justification for \(k\ge5\).
\(P(5)\) is true, and \(P(k)\) true implies \(P(k+1)\) true, so \(P(n)\) is true for all \(n\ge5\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(c)
induction needs a true starting case; \(P(1)\), \(P(2)\), \(P(3)\) and \(P(4)\) are all false (eg \(P(4)\) from part (a)), so the chain of implications cannot start before \(n=5\)R1Accept “there is no valid base case below 5”.

Answers: (a) \(81\lt96\)

90AAHL1-q00671.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks
(a)

Prove by mathematical induction that \(\displaystyle\prod_{r=2}^{n}\left(1-\frac{1}{r^{2}}\right)=\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\cdots\left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2n}\), for all integers \(n\ge2\).

(6)
(b)

Find the least value of \(n\) for which the product is less than \(0.501\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(n=2\): LHS \(=\dfrac{3}{4}\), RHS \(=\dfrac{3}{4}\), so true for \(n=2\)A1
assume true for \(n=k\): product to \(k\) \(=\dfrac{k+1}{2k}\)M1
product to \(k+1\) \(=\dfrac{k+1}{2k}\left(1-\dfrac{1}{(k+1)^{2}}\right)\)M1
\(=\dfrac{k+1}{2k}\times\dfrac{k(k+2)}{(k+1)^{2}}\)A1Must see \((k+1)^{2}-1=k(k+2)\) or equivalent.
\(=\dfrac{k+2}{2(k+1)}\)A1
true for \(n=2\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\ge2\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(b)
\(\dfrac{n+1}{2n}\lt0.501\Rightarrow n+1\lt1.002n\Rightarrow n\gt500\)M1
\(n=501\)A1Award A0 for \(n=500\) (product exactly \(0.501\)).

Answers: (b) \(n=501\)

91AAHL1-q00691.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks

In this question you may use the product rule and the quotient rule for differentiation, and the result \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x)=1\). You must not use the result \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{n}\right)=nx^{n-1}\).

(a)

Prove by mathematical induction that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{n}\right)=nx^{n-1}\), for \(n\in\mathbb{Z}^{+}\).

(5)
(b)

Hence, by writing \(x^{-m}=\dfrac{1}{x^{m}}\), show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{-m}\right)=-mx^{-m-1}\), for \(m\in\mathbb{Z}^{+}\) and \(x\neq0\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(n=1\): \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x)=1=1\times x^{0}\), so true for \(n=1\)A1
assume true for \(n=k\): \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{k}\right)=kx^{k-1}\)M1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{k+1}\right)=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x\times x^{k}\right)=1\times x^{k}+x\times kx^{k-1}\)M1Product rule applied to \(x\times x^{k}\).
\(=x^{k}+kx^{k}=(k+1)x^{k}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least three of the previous marks have been awarded.
(b)
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{x^{m}}\right)=\dfrac{x^{m}\times0-1\times mx^{m-1}}{x^{2m}}\)M1Quotient rule using part (a).
\(=\dfrac{-mx^{m-1}}{x^{2m}}\)A1
\(=-mx^{m-1-2m}\)A1
\(=-mx^{-m-1}\)AG

Answers: Proofs.

92AAHL1-q00701.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks

A student claims that \(\displaystyle\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}+1\) for all \(n\in\mathbb{Z}^{+}\).

(a)

Show that, if the claim is true for \(n=k\), then it is true for \(n=k+1\).

(3)
(b)

Show that the claim is false for \(n=1\).

(1)
(c)

Explain why part (a) does not prove the claim.

(1)
(d)

Write down the correct expression for \(\displaystyle\sum_{r=1}^{n}r(r+1)\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{k(k+1)(k+2)}{3}+1+(k+1)(k+2)\)M1
\(=\dfrac{(k+1)(k+2)(k+3)}{3}+1\)A1Must see \((k+1)(k+2)\) factorized out, eg \(\dfrac{(k+1)(k+2)}{3}(k+3)\).
which is the claim with \(n=k+1\)R1
(b)
LHS \(=1\times2=2\); RHS \(=\dfrac{1\times2\times3}{3}+1=3\neq2\)A1
(c)
there is no (true) base case: the implication in (a) only shows that if one case is true then the next is, but the first case is falseR1
(d)
\(\dfrac{n(n+1)(n+2)}{3}\)A1

Answers: (b) LHS = 2, RHS = 3 · (d) \(\tfrac{n(n+1)(n+2)}{3}\)

93AAHL1-q00711.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks
(a)

Prove by contradiction that \(\log_{2}3\) is irrational.

(4)
(b)

Hence prove that \(\log_{2}12\) is irrational.

(2)
Show mark scheme
WorkingMarkNotes
(a)
assume \(\log_{2}3\) is rational: \(\log_{2}3=\dfrac{p}{q}\), \(p,q\in\mathbb{Z}^{+}\)M1Must state the assumption; \(p,q\gt0\) since \(\log_{2}3\gt0\).
\(2^{p/q}=3\Rightarrow2^{p}=3^{q}\)A1
\(2^{p}\) is even (as \(p\ge1\)) and \(3^{q}\) is oddR1
this is a contradiction, so \(\log_{2}3\) is irrationalR1
(b)
\(\log_{2}12=\log_{2}4+\log_{2}3=2+\log_{2}3\)M1
if \(\log_{2}12=r\) were rational, then \(\log_{2}3=r-2\) would be rational, contradicting part (a)R1

Answers: Proofs.

94AAHL1-q00741.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks
(a)

Prove by contradiction that there are no integers \(a\) and \(b\) such that \(a^{2}-4b=2\).

(5)
(b)

Hence prove that there is no quadratic equation \(x^{2}+px+q=0\), with \(p,q\in\mathbb{Z}\), whose two roots differ by \(\sqrt{2}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
assume integers \(a,b\) exist with \(a^{2}=4b+2\)M1
\(4b+2\) is even, so \(a^{2}\) is even and hence \(a\) is evenA1
\(a=2m\Rightarrow4m^{2}=4b+2\)A1
LHS is a multiple of 4 but \(4b+2\) leaves remainder 2 on division by 4 (or \(2m^{2}=2b+1\): even = odd)R1
contradiction, so no such integers existR1
(b)
roots \(\dfrac{-p\pm\sqrt{p^{2}-4q}}{2}\), so they differ by \(\sqrt{p^{2}-4q}\)M1
\(\sqrt{p^{2}-4q}=\sqrt{2}\Rightarrow p^{2}-4q=2\)A1
by part (a) no integers \(p,q\) satisfy this, so no such equation existsR1

Answers: Proofs.

95AAHL1-q00751.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks

Let \(f(x)=x^{3}+x+1\), for \(x\in\mathbb{R}\).

(a)

Show that the equation \(f(x)=0\) has exactly one real root, and that this root lies between \(-1\) and \(0\).

(3)
(b)

Prove by contradiction that this root is irrational.

(5)
Show mark scheme
WorkingMarkNotes
(a)
\(f'(x)=3x^{2}+1\gt0\) for all \(x\), so \(f\) is increasingA1
an increasing function can cross the \(x\)-axis at most onceR1
\(f(-1)=-1\lt0\) and \(f(0)=1\gt0\), and \(f\) is continuous, so the root lies between \(-1\) and \(0\)A1
(b)
assume the root is \(\dfrac{p}{q}\), with \(p,q\in\mathbb{Z}\), \(q\neq0\), having no common factorM1
\(\dfrac{p^{3}}{q^{3}}+\dfrac{p}{q}+1=0\Rightarrow p^{3}+pq^{2}+q^{3}=0\)A1
considering parity: \(p\) and \(q\) cannot both be even, so the cases are both odd, \(p\) even and \(q\) odd, or \(p\) odd and \(q\) evenM1
both odd: odd + odd + odd is odd; \(p\) even, \(q\) odd: even + even + odd is odd; \(p\) odd, \(q\) even: odd + even + even is oddR1All three cases needed.
in every case \(p^{3}+pq^{2}+q^{3}\) is odd so cannot equal 0: a contradiction; hence the root is irrationalR1

Answers: (a) \(f'(x)=3x^{2}+1\gt0\); \(f(-1)=-1,\ f(0)=1\)

96AAHL1-q00831.16 Systems of linear equations Paper 1Medium9 marks

Consider the system of equations, where \(a,b\in\mathbb{R}\):

\(\begin{aligned}x+y+z&=2\\x+2y+3z&=5\\2x+3y+az&=b\end{aligned}\)

(a)

Show that the system can be reduced to \(x+y+z=2\), \(y+2z=3\), \((a-4)z=b-7\).

(3)
(b)

Write down the value of \(a\) for which the system does not have a unique solution.

(1)
(c)(i)

For this value of \(a\),

find the value of \(b\) for which the system has infinitely many solutions;

(1)
(c)(ii)

find the general solution of the system in this case.

(3)
(d)

For the value of \(a\) found in part (b) and \(b\neq7\), describe the geometric arrangement of the three planes represented by the equations.

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(R_{2}-R_{1}\): \(y+2z=3\)M1
\(R_{3}-2R_{1}\): \(y+(a-2)z=b-4\)A1
subtracting: \((a-4)z=b-7\)A1
system as givenAG
(b)
\(a=4\)A1
(c)(i)
\(b=7\)A1
(c)(ii)
letting \(z=\lambda\)M1Any parameter.
\(y=3-2\lambda\)A1
\(x=\lambda-1\)A1Accept equivalent forms, eg \(x=t,\ y=1-2t,\ z=t+1\); or \(\mathbf{r}=\begin{pmatrix}-1\\3\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\1\end{pmatrix}\).
(d)
the planes form a triangular prism: each pair meets in a line, the three lines are parallel, and there is no common pointA1Accept “no two planes are parallel, and there is no common point of intersection”.

Answers: (b) \(a=4\) · (c)(i) \(b=7\) · (c)(ii) \(x=\lambda-1,\ y=3-2\lambda,\ z=\lambda\) · (d) triangular prism

97AAHL1-q00841.16 Systems of linear equations Paper 1Medium8 marks

A circle has equation \(x^{2}+y^{2}+Dx+Ey+F=0\), where \(D,E,F\in\mathbb{R}\). The circle passes through the points \((0,3)\), \((7,2)\) and \((-2,-1)\).

The circle with centre \((h,k)\) and radius \(r\) has equation \((x-h)^{2}+(y-k)^{2}=r^{2}\).

(a)

Show that \(3E+F=-9\), and find two further linear equations in \(D\), \(E\) and \(F\).

(2)
(b)

Solve the system of equations to find \(D\), \(E\) and \(F\).

(4)
(c)

Hence find the coordinates of the centre and the radius of the circle.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\((0,3)\): \(9+3E+F=0\Rightarrow3E+F=-9\)A1
\(7D+2E+F=-53\) and \(-2D-E+F=-5\)A1Both required.
(b)
eliminating \(F\), eg \(7D-E=-44\) and \(-2D-4E=4\)M1
one correct equation in two variablesA1
\(D=-6,\ E=2\)A1
\(F=-15\)A1
(c)
completing the square: \((x-3)^{2}+(y+1)^{2}=15+9+1\)M1
centre \((3,-1)\), radius \(5\)A1Both required.

Answers: (b) \(D=-6,\ E=2,\ F=-15\) · (c) centre \((3,-1)\), radius 5

98AAHL1-q00851.16 Systems of linear equations Paper 1Medium7 marks

Consider the system of equations, where \(k\in\mathbb{R}\):

\(\begin{aligned}x-y+2z&=5\\3x+y-z&=2\\x+2y+z&=k\end{aligned}\)

For a particular value of \(k\), the solution \(x,\,y,\,z\), in that order, forms an arithmetic sequence.

(a)

By writing \(x=y-d\) and \(z=y+d\), use the first two equations to find the value of \(y\) and the value of \(d\).

(4)
(b)

Hence find the value of \(k\).

(1)
(c)

Show that the system has a unique solution for every value of \(k\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
substituting into the first two equationsM1
\(2y+d=5\) and \(3y-4d=2\)A1
solving simultaneously(M1)
\(y=2,\ d=1\)A1
(b)
\(x=1,\ z=3\Rightarrow k=1+4+3=8\)A1
(c)
eliminating \(x\): \(R_{2}-3R_{1}\): \(4y-7z=-13\); \(R_{3}-R_{1}\): \(3y-z=k-5\)M1
these two equations are not multiples of each other (\(\tfrac{4}{3}\neq\tfrac{-7}{-1}\)), so \(y\), \(z\) and then \(x\) are uniquely determined for every \(k\)R1Accept a correct general solution \(x=\tfrac{25-k}{17},\ y=\tfrac{7k-22}{17},\ z=\tfrac{4k+19}{17}\).

Answers: (a) \(y=2,\ d=1\) · (b) \(k=8\)

99IBMA2H-045-06-051.11 Partial fractions Paper 1Medium7 marks

No calculator

Consider the expansion of \(\left(x^2-\dfrac{2}{x}\right)^6\).

(a)

Find the term independent of \(x\).

(4)
(b)

Hence find the term independent of \(x\) in the expansion of \((1+x^3)\left(x^2-\dfrac{2}{x}\right)^6\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 general term \(\binom{6}{r}(x^2)^{6-r}(-2x^{-1})^r\) ; M1 power of \(x\) is \(12-3r\), set equal to 0 ; A1 \(r=4\) ; A1 \(\binom{6}{4}(-2)^4=240\)✓ 4
Part (b)
R1 constant term comes from \(1\times\)(constant term) plus \(x^3\times\)(\(x^{-3}\) term) ; A1 \(x^{-3}\) term: \(r=5\), coefficient \(\binom{6}{5}(-2)^5=-192\) ; A1 \(240-192=48\)✓ 3

Exam tip: Write the general term with all powers of \(x\) collected into a single index such as \(x^{12-3r}\). Any required power can then be found by solving one linear equation.

100IBMA2H-045-06-071.10 Counting principles and the extended binomial theorem Paper 1Medium8 marks

No calculator

Let \(f(x)=(1-x)^{-\frac{1}{2}}\).

(a)

Find the first four terms of the binomial expansion of \(f(x)\) in ascending powers of \(x\).

(4)
(b)

State the set of values of \(x\) for which the expansion is valid.

(1)
(c)

By substituting \(x=\frac{1}{10}\), find an approximation for \(\sqrt{10}\), giving your answer as a single fraction.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses \(1+n(-x)+\frac{n(n-1)}{2!}(-x)^2+\frac{n(n-1)(n-2)}{3!}(-x)^3\) with \(n=-\frac{1}{2}\) ; A1 \(\frac{1}{2}x\) ; A1 \(\frac{3}{8}x^2\) ; A1 \(\frac{5}{16}x^3\), giving \(1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3\)✓ 4
Part (b)
A1 \(|x|<1\)✓ 1
Part (c)
M1 recognises \((0.9)^{-\frac{1}{2}}=\sqrt{\frac{10}{9}}=\frac{\sqrt{10}}{3}\) ; M1 \(\sqrt{10}\approx3\left(1+\frac{1}{20}+\frac{3}{800}+\frac{5}{16000}\right)\) ; A1 \(\frac{10119}{3200}\)✓ 3

Common error: The term inside the bracket is \(-x\), not \(x\). Forgetting the sign makes the odd-power terms negative. All terms here are positive because both \(n\) and \(-x\) are negative.

101IBMA2H-045-06-081.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

Let \(z=2+\mathrm{i}\).

(a)

Use the binomial theorem to express \(z^5\) in the form \(a+b\mathrm{i}\), where \(a,b\in\mathbb{Z}\).

(4)
(b)

Hence write down \((2-\mathrm{i})^5\) and find \((2+\mathrm{i})^5+(2-\mathrm{i})^5\).

(2)
(c)

By considering \((2+\mathrm{i})^5(2-\mathrm{i})^5\), show that \(38^2+41^2=5^5\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expansion \(\sum_{r=0}^{5}\binom{5}{r}2^{5-r}\mathrm{i}^r\) with six terms ; A1 correct powers \(\mathrm{i}^2=-1,\ \mathrm{i}^3=-\mathrm{i},\ \mathrm{i}^4=1,\ \mathrm{i}^5=\mathrm{i}\) in \(32+80\mathrm{i}-80-40\mathrm{i}+10+\mathrm{i}\) ; A1 real part \(-38\) ; A1 imaginary part 41, so \(z^5=-38+41\mathrm{i}\)✓ 4
Part (b)
R1 \((2-\mathrm{i})^5=-38-41\mathrm{i}\), since replacing \(\mathrm{i}\) by \(-\mathrm{i}\) changes the sign of the odd-power terms only ; A1 sum \(=-76\)✓ 2
Part (c)
R1 \((2+\mathrm{i})^5(2-\mathrm{i})^5=((2+\mathrm{i})(2-\mathrm{i}))^5=5^5\) and also \(=(-38+41\mathrm{i})(-38-41\mathrm{i})=38^2+41^2\)✓ 1

Exam tip: Cycle the powers of i as 1, i, −1, −i before you expand, and keep the real and imaginary terms in separate columns. Most lost marks come from a single sign slip.

102IBMA2H-045-06-101.11 Partial fractions Paper 1Medium8 marks

No calculator

The binomial expansion of \((1+ax)^n\), where \(n\in\mathbb{Q}\) and \(a\in\mathbb{R}\), begins \(1-6x+27x^2+\dots\)

(a)

Find the value of \(n\) and the value of \(a\).

(5)
(b)

Find the coefficient of \(x^3\).

(2)
(c)

State the values of \(x\) for which the expansion is valid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(na=-6\) ; M1 \(\frac{n(n-1)}{2}a^2=27\) ; M1 eliminates \(a\), e.g. \(\frac{n(n-1)a^2}{n^2a^2}=\frac{54}{36}\) ; A1 \(n=-2\) ; A1 \(a=3\)✓ 5
Part (b)
M1 \(\frac{(-2)(-3)(-4)}{3!}\times3^3\) ; A1 \(-108\)✓ 2
Part (c)
A1 \(|x|<\frac{1}{3}\)✓ 1

Common error: Validity needs \(|ax|<1\), which here is \(|3x|<1\). Writing \(|x|<1\) ignores the coefficient of \(x\) inside the bracket.

103IBMA2H-045-03-051.16 Systems of linear equations Paper 1Medium7 marks

No calculator

Consider the system of equations
\(x+y+z=6\)
\(x+2y+3z=14\)
\(x+4y+kz=30\)
where \(k\in\mathbb{R}\).

(a)

Show that, for \(k\neq7\), the system has a unique solution, and find this solution. Show that it does not depend on \(k\).

(4)
(b)

For \(k=7\), explain why the system has infinitely many solutions, and find the general solution.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 eliminates \(x\), e.g. R2 − R1 and R3 − R1✓ 1
A1 \(y+2z=8\) and \(3y+(k-1)z=24\)✓ 1
M1 eliminates \(y\) to obtain \((k-7)z=0\)✓ 1
A1 since \(k\neq7\), \(z=0\), giving \(y=8\), \(x=-2\), which is independent of \(k\)✓ 1
Part (b)
R1 when \(k=7\) the reduced equation becomes \(0=0\), leaving only two independent equations✓ 1
M1 lets \(z=\lambda\) and solves for \(y\) and \(x\)✓ 1
A1 \(x=-2+\lambda\), \(y=8-2\lambda\), \(z=\lambda\)✓ 1

Exam tip: Keep the parameter in your working until the final step. The single equation \((k-7)z=0\) shows both cases at once.

104IBMA2H-045-03-081.5 Exponents and introduction to logarithms Paper 1Medium5 marks

No calculator

Consider the equation \(9^{x}-4\times3^{x+1}+27=0\).

(a)

Show that, using the substitution \(u=3^{x}\), the equation becomes \(u^{2}-12u+27=0\).

(2)
(b)

Hence solve the equation for \(x\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(9^{x}=(3^{2})^{x}=(3^{x})^{2}=u^{2}\)✓ 1
A1 \(4\times3^{x+1}=4\times3\times3^{x}=12u\), giving \(u^2-12u+27=0\)✓ 1
Part (b)
M1 factorises: \((u-3)(u-9)=0\)✓ 1
A1 \(u=3\) or \(u=9\)✓ 1
A1 \(x=1\) or \(x=2\)✓ 1

Common error: \(3^{x+1}\) is \(3\times3^{x}\), not \(3^{x}+1\). Splitting the power correctly is what makes the quadratic appear.

105IBMA2H-045-03-091.16 Systems of linear equations Paper 1Medium9 marks

No calculator

An engineer models the shape of a guide cable using \(y=ax^{3}+bx^{2}+cx+d\), where \(y\) metres is the vertical position of the cable relative to a reference line and \(x\) metres is the horizontal distance from a support. The cable passes through the points \((-1,-1)\), \((0,2)\), \((1,3)\) and \((2,8)\).

(a)

Write down the value of \(d\).

(1)
(b)

Show that \(a\), \(b\) and \(c\) satisfy \(-a+b-c=-3\), \(a+b+c=1\) and \(4a+2b+c=3\).

(2)
(c)

Solve the system to find \(a\), \(b\) and \(c\).

(4)
(d)

Find the vertical position of the cable when \(x=1.5\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(d=2\)✓ 1
Part (b)
M1 substitutes the three remaining points with \(d=2\), e.g. \(8a+4b+2c+2=8\)✓ 1
A1 simplifies all three to the given equations✓ 1
Part (c)
M1 adds the first two equations to eliminate \(a\) and \(c\)✓ 1
A1 \(b=-1\)✓ 1
M1 substitutes to obtain \(a+c=2\) and \(4a+c=5\), and solves✓ 1
A1 \(a=1\), \(c=1\)✓ 1
Part (d)
M1 substitutes into \(y=x^3-x^2+x+2\)✓ 1
A1 \(y=\frac{37}{8}\) m (4.625 m)✓ 1

Exam tip: Use the point with \(x=0\) first, because it gives a constant immediately. Look for two equations that add to cancel several unknowns at once.

106IBMA2H-045-04-031.6 Deductive proof Paper 1Medium6 marks

No calculator

This question is about the squares of consecutive integers.

(a)

Prove that the sum of the squares of any three consecutive integers is always 2 more than a multiple of 3.

(4)
(b)

Hence prove that the sum of the squares of three consecutive integers can never be a perfect square.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 represents the integers algebraically, e.g. as \(n-1\), \(n\), \(n+1\) with \(n \in \mathbb{Z}\) (accept any equivalent, e.g. \(n\), \(n+1\), \(n+2\))✓ 1
A1 correct expansion, e.g. \((n^2-2n+1)+n^2+(n^2+2n+1)\)✓ 1
A1 \(=3n^2+2\) (or equivalent, e.g. \(3n^2+6n+5=3(n^2+2n+1)+2\))✓ 1
R1 since \(n\) is an integer, the bracketed term or \(n^2\) is an integer, so the sum is a multiple of 3 plus 2✓ 1
Part (b)
M1 considers any integer \(m\) as \(3k\) or \(3k\pm1\), giving \(m^2=9k^2\) or \(m^2=3(3k^2\pm2k)+1\)✓ 1
R1 every perfect square is a multiple of 3 or 1 more than a multiple of 3, never 2 more, so the sum cannot be a perfect square✓ 1

Exam tip: Centring the consecutive integers on \(n\) (as \(n-1\), \(n\), \(n+1\)) makes the linear terms cancel, so the algebra is much shorter.

107IBMA2H-045-04-051.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium6 marks

No calculator

The stability of a feedback control system is studied using its characteristic equation
\(z^3+az^2+bz+10=0\), where \(a, b \in \mathbb{R}\).
One root of the equation is \(1+i\).

(a)

Write down the other non-real root of the equation.

(1)
(b)

Find the real root.

(2)
(c)

Find the value of \(a\) and the value of \(b\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-i\)✓ 1
Part (b)
M1 uses product of roots \(=-10\): \((1+i)(1-i)r=-10\), or factorises using \(z^2-2z+2\)✓ 1
A1 \(2r=-10\), so \(r=-5\)✓ 1
Part (c)
M1 expands \((z^2-2z+2)(z+5)\), or uses sums and products of roots✓ 1
A1 \(a=3\)✓ 1
A1 \(b=-8\)✓ 1

Exam tip: For a polynomial with real coefficients, non-real roots come in conjugate pairs. Then \((z-(1+i))(z-(1-i)) = z^2-2z+2\) is a real quadratic factor.

108IBMA2H-045-04-061.12 Complex numbers in Cartesian form Paper 1Medium7 marks

No calculator

Let \(w = x+iy\), where \(x, y \in \mathbb{R}\).

(a)

Given that \(w^2=-21-20i\), find the two possible values of \(w\).

(4)
(b)

Hence solve \(z^2+5iz-1+5i=0\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands \((x+iy)^2\) and equates parts: \(x^2-y^2=-21\), \(2xy=-20\)✓ 1
M1 eliminates a variable, e.g. \(y=-\frac{10}{x}\) giving \(x^4+21x^2-100=0\)✓ 1
A1 \((x^2+25)(x^2-4)=0\), so \(x=\pm2\)✓ 1
A1 \(w=2-5i\) or \(w=-2+5i\)✓ 1
Part (b)
M1 finds the discriminant \((5i)^2-4(-1+5i)=-21-20i\)✓ 1
M1 uses \(z=\frac{-5i\pm(2-5i)}{2}\)✓ 1
A1 \(z=1-5i\) or \(z=-1\)✓ 1

Common error: Reject \(x^2=-25\), because \(x\) is real. Once \(x\) is found, use \(2xy=-20\) to fix the sign of \(y\). Do not pair every sign of \(x\) with every sign of \(y\).

109IBMA2H-045-04-071.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks

No calculator

This question is about inequalities involving real numbers.

(a)

Prove that \(p^2+q^2 \ge 2pq\) for all \(p, q \in \mathbb{R}\).

(2)
(b)

Hence prove that \(a+b \ge 2\sqrt{ab}\) for all \(a, b > 0\).

(2)
(c)

Hence prove that \((a+b)(b+c)(c+a) \ge 8abc\) for all \(a, b, c > 0\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 starts from \((p-q)^2 \ge 0\) for all real \(p, q\)✓ 1
A1 expands to \(p^2-2pq+q^2 \ge 0\) and rearranges to \(p^2+q^2\ge 2pq\)✓ 1
Part (b)
M1 lets \(p=\sqrt{a}\) and \(q=\sqrt{b}\), which are real because \(a, b>0\)✓ 1
A1 substitutes into the previous result to give \(a+b\ge 2\sqrt{a}\sqrt{b}=2\sqrt{ab}\)✓ 1
Part (c)
M1 applies the previous result to each factor: \(a+b\ge2\sqrt{ab}\), \(b+c\ge2\sqrt{bc}\), \(c+a\ge2\sqrt{ca}\)✓ 1
R1 multiplies the inequalities, valid because both sides of each are positive, giving \(\ge 8\sqrt{a^2b^2c^2}=8abc\) since \(abc>0\)✓ 1

Common error: Do not start from the result you are trying to prove and work towards a true statement. Begin with a known fact such as \((p-q)^2\ge0\) and deduce forwards.

110IBMA2H-045-04-101.12 Complex numbers in Cartesian form Paper 1Medium12 marks

No calculator

Let \(z=x+iy\), where \(x, y\in\mathbb{R}\) and \(z\ne 2\). The complex number \(w\) is defined by \(w=\frac{z+i}{z-2}\).

(a)

Find \(w\) in the form \(a+bi\) when \(z=3+i\).

(2)
(b)

Show that \(\text{Re}(w)=\frac{x^2+y^2-2x+y}{(x-2)^2+y^2}\).

(3)
(c)

Given that \(w\) is purely imaginary, show that \(z\) lies on a circle. State the centre and radius of this circle.

(4)
(d)

Find the real value of \(z\) for which \(w\) is purely imaginary and non-zero.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes to get \(\frac{3+2i}{1+i}\) and multiplies by \(\frac{1-i}{1-i}\)✓ 1
A1 \(w=\frac{5}{2}-\frac{1}{2}i\)✓ 1
Part (b)
M1 multiplies numerator and denominator by \((x-2)-iy\)✓ 1
A1 denominator \((x-2)^2+y^2\)✓ 1
A1 real part of numerator \(x(x-2)+y(y+1)=x^2+y^2-2x+y\)✓ 1
Part (c)
M1 sets \(x^2+y^2-2x+y=0\)✓ 1
M1 completes the square: \((x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{5}{4}\)✓ 1
A1 centre \(\left(1,-\frac{1}{2}\right)\)✓ 1
A1 radius \(\frac{\sqrt5}{2}\)✓ 1
Part (d)
M1 substitutes \(y=0\): \(x^2-2x=0\), so \(x=0\) or \(x=2\)✓ 1
R1 rejects \(x=2\) because the denominator \(z-2\) is then zero✓ 1
A1 \(z=0\), which gives \(w=-\frac{1}{2}i\ne0\)✓ 1

Common error: A point on a locus may still be excluded. Check that \(z\ne2\) and that \(w\ne0\), since \(0\) is not purely imaginary.

111IBMA2H-045-00-031.15 Proof by induction, contradiction and counterexample Paper 1Medium7 marks

No calculator

Prove by mathematical induction that 32n + 3 × 2n+1 is divisible by 7 for all n ∈ ℤ+.

(a)

Write a complete proof.

(7)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 n = 1: 9 + 12 = 21 = 7 × 3, so true for n = 1✓ 1
M1 assume true for n = k: 9k + 6 × 2k = 7m, m ∈ ℤ✓ 1
M1 considers n = k + 1: 9k+1 + 6 × 2k+1 = 9 × 9k + 12 × 2k✓ 1
M1 uses the assumption, e.g. 9(9k + 6 × 2k) − 54 × 2k + 12 × 2k✓ 1
A1 = 63m − 42 × 2k✓ 1
A1 = 7(9m − 6 × 2k), which is divisible by 7✓ 1
R1 true for n = 1 and true for n = k implies true for n = k + 1, so true for all n ∈ ℤ+ by induction✓ 1

Exam tip: Rewrite the expression as 9ⁿ + 6 × 2ⁿ first; it makes the inductive step far easier to organise.

112IBMA2H-045-00-041.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks

No calculator

This question uses proof by contradiction.

(a)

Prove that there are no positive integers x and y such that x2 − y2 = 10.

(4)
(b)

Prove that log310 is irrational.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 assumes positive integers x, y exist with x2 − y2 = 10✓ 1
M1 factorises: (x − y)(x + y) = 10 with both factors positive integers✓ 1
R1 (x + y) + (x − y) = 2x is even, so the two factors have the same parity✓ 1
R1 the only factor pairs of 10 are 1 × 10 and 2 × 5, each of opposite parity: contradiction, so no such integers exist✓ 1
Part (b)
M1 assumes log310 = p/q with p, q positive integers (positive since log310 > 0)✓ 1
M1 obtains 3p/q = 10, so 3p = 10q✓ 1
R1 3p is odd but 10q is even✓ 1
R1 an odd number cannot equal an even number: contradiction, so log310 is irrational✓ 1

Common error: In (b), state that p and q are positive integers; without this, 10q need not be an even integer and the parity argument breaks down.

113IBMA2H-045-00-111.15 Proof by induction, contradiction and counterexample Paper 1Medium10 marks

No calculator

A file-synchronisation program performs un operations on its nth pass, where u1 = 1 and un+1 = 3un + 2n for n ∈ ℤ+.

(a)

Find u2, u3 and u4.

(2)
(b)

Prove by mathematical induction that un = 3n − 2n for all n ∈ ℤ+.

(6)
(c)

Hence prove that un is odd for all n ∈ ℤ+.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 u2 = 5, u3 = 19✓ 1
A1 u4 = 65✓ 1
Part (b)
A1 n = 1: 3 − 2 = 1 = u1, so true for n = 1✓ 1
M1 assumes uk = 3k − 2k✓ 1
M1 substitutes into the recurrence: uk+1 = 3(3k − 2k) + 2k✓ 1
A1 = 3k+1 − 2 × 2k✓ 1
A1 = 3k+1 − 2k+1✓ 1
R1 true for n = 1 and true for n = k implies true for n = k + 1, so true for all n ∈ ℤ+✓ 1
Part (c)
R1 3n is a product of odd numbers so is odd; 2n is even for n ≥ 1✓ 1
R1 odd − even = odd, so un is odd✓ 1

Exam tip: Use the recurrence relation in the inductive step: substitute the assumed formula for uk straight into uk+₁ = 3uk + 2k.

114IBMA2H-045-00-121.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks

No calculator

This question compares 4n with quadratic expressions in n.

(a)

Show by counterexample that the statement “4n ≥ 3n2 + 2 for all n ∈ ℤ+” is false.

(1)
(b)

Prove by mathematical induction that 4n ≥ 3n2 + 1 for all n ∈ ℤ+.

(7)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 n = 1: 4 < 5, so the statement is false✓ 1
Part (b)
A1 n = 1: 4 ≥ 4, so true for n = 1✓ 1
M1 assumes 4k ≥ 3k2 + 1✓ 1
M1 4k+1 = 4 × 4k ≥ 12k2 + 4✓ 1
M1 compares with the target 3(k + 1)2 + 1 = 3k2 + 6k + 4✓ 1
A1 (12k2 + 4) − (3k2 + 6k + 4) = 9k2 − 6k = 3k(3k − 2)✓ 1
R1 3k(3k − 2) > 0 for k ≥ 1, so 4k+1 ≥ 3(k + 1)2 + 1✓ 1
R1 true for n = 1 and true for n = k implies true for n = k + 1, so true for all n ∈ ℤ+✓ 1

Common error: Starting from the target inequality and working towards something true is not a valid proof; begin from 4k⁺¹ = 4 × 4k and build a chain of inequalities.

115IBMA2H-045-02-051.8 Infinite geometric series Paper 1Medium11 marks

No calculator

A telescopic radio antenna extends in stages. At stage \(r\), the extra length added is \(u_r=\frac{3}{(3r-1)(3r+2)}\) metres, for \(r=1,2,3,\dots\) Let \(S_n=\sum_{r=1}^{n}u_r\) be the total extension after \(n\) stages.

(a)

Express \(u_r\) in partial fractions.

(3)
(b)

Hence show that \(S_n=\frac12-\frac{1}{3n+2}\).

(3)
(c)

Find the limiting value of the total extension as \(n\to\infty\). Explain why the antenna never actually reaches this length.

(2)
(d)

The antenna locks once the total extension reaches exactly \(0.48\) m. Find the number of stages needed.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms the identity \(3\equiv A(3r+2)+B(3r-1)\), or uses an equivalent method such as the cover-up rule✓ 1
A1 \(A=1\)✓ 1
A1 \(B=-1\), so \(u_r=\frac{1}{3r-1}-\frac{1}{3r+2}\)✓ 1
Part (b)
M1 writes out at least the first two terms and the last term: \(\left(\frac12-\frac15\right)+\left(\frac15-\frac18\right)+\dots+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)✓ 1
M1 shows that the intermediate terms cancel in pairs✓ 1
A1 \(S_n=\frac12-\frac{1}{3n+2}\) (AG)✓ 1
Part (c)
A1 as \(n\to\infty\), \(\frac{1}{3n+2}\to0\), so \(S_n\to\frac12\) m✓ 1
R1 \(\frac{1}{3n+2}>0\) for every \(n\), so \(S_n<\frac12\) for all \(n\) and the limit is never attained✓ 1
Part (d)
M1 sets \(\frac12-\frac{1}{3n+2}=0.48\)✓ 1
A1 \(\frac{1}{3n+2}=\frac{1}{50}\), so \(3n+2=50\)✓ 1
A1 \(n=16\)✓ 1

Exam tip: In a 'show that' telescoping sum, write out the first two terms and the last term explicitly so that the cancellation can be seen. Stating the result without this earns little credit.

116IBMA2H-045-02-061.13 Polar and Euler forms Paper 1Medium11 marks

No calculator

In a simplified model of a three-phase generator, the three voltage phasors are the roots of the equation \(z^3=27i\).

(a)

Solve \(z^3=27i\), giving your answers in the form \(re^{i\theta}\), where \(r>0\) and \(-\pi<\theta\le\pi\).

(4)
(b)

Show that the sum of the three phasors is zero.

(2)
(c)

Find the product of the three phasors, giving your answer in the form \(a+bi\).

(2)
(d)

The three phasors are plotted as points in the complex plane. Find the exact area of the triangle formed.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes \(27i=27e^{i\left(\frac{\pi}{2}+2k\pi\right)}\)✓ 1
A1 \(r=3\) and \(\theta=\frac{\pi}{6}+\frac{2k\pi}{3}\)✓ 1
A1 \(z_1=3e^{\frac{\pi i}{6}}\)✓ 1
A1 \(z_2=3e^{\frac{5\pi i}{6}}\) and \(z_3=3e^{-\frac{\pi i}{2}}\)✓ 1
Part (b)
M1 converts to Cartesian form: \(\frac{3\sqrt3}{2}+\frac32i\), \(-\frac{3\sqrt3}{2}+\frac32i\), \(-3i\), or uses the fact that the coefficient of \(z^2\) in \(z^3-27i=0\) is zero✓ 1
A1 shows that the real parts sum to 0 and the imaginary parts sum to 0 (AG)✓ 1
Part (c)
M1 multiplies the moduli and adds the arguments: \(27e^{i\left(\frac{\pi}{6}+\frac{5\pi}{6}-\frac{\pi}{2}\right)}\), or uses the product of the roots of \(z^3-27i=0\)✓ 1
A1 \(27i\)✓ 1
Part (d)
M1 uses a valid method for the area, e.g. notes the points are equally spaced on a circle of radius 3, so the triangle is equilateral, and finds a side \(|z_1-z_3|=\sqrt{\frac{27}{4}+\frac{81}{4}}\); or splits the triangle into three triangles with apex at the origin, each with area \(\frac12\times3\times3\sin\frac{2\pi}{3}\)✓ 1
A1 side length \(3\sqrt3\), or one sub-triangle area \(\frac{9\sqrt3}{4}\)✓ 1
A1 area \(=\frac{27\sqrt3}{4}\)✓ 1

Common error: Dividing only θ = π/2 by 3 gives just one root. Include the +2kπ in the argument before dividing by 3 to obtain all three roots.

117IBMA2H-045-02-081.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

No calculator

A slow-release patch delivers a medication into the bloodstream at a rate of \(R(t)=\frac{10}{(t+1)(t+3)}\) mg per hour, where \(t\) is the time in hours after the patch is applied.

(a)

Express \(R(t)\) in partial fractions.

(3)
(b)

Hence find the exact amount of medication delivered in the first 3 hours, giving your answer in the form \(k\ln2\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms the identity \(10\equiv A(t+3)+B(t+1)\)✓ 1
A1 \(A=5\)✓ 1
A1 \(B=-5\), so \(R(t)=\frac{5}{t+1}-\frac{5}{t+3}\)✓ 1
Part (b)
M1 integrates to logarithms✓ 1
A1 \(\left[5\ln(t+1)-5\ln(t+3)\right]_0^3\)✓ 1
M1 substitutes the limits: \(5(\ln4-\ln6)-5(\ln1-\ln3)\)✓ 1
A1 \(5\ln2\) mg✓ 1

Common error: The lower limit t = 0 does not give zero here, because ln 3 ≠ 0. Always substitute both limits fully before combining the logarithms.

118IBMA2H-045-02-091.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium10 marks

No calculator

When a pure audio tone \(\cos\theta\) passes through an overdriven amplifier, the output contains a \(\cos^3\theta\) term, which creates harmonics. Let \(z=e^{i\theta}\).

(a)

Show that \(z^n+z^{-n}=2\cos n\theta\) for \(n\in\mathbb{Z}^+\).

(2)
(b)

By expanding \(\left(z+z^{-1}\right)^3\), show that \(\cos^3\theta=\frac14\cos3\theta+\frac34\cos\theta\).

(4)
(c)

State the amplitude of the third-harmonic component \(\cos3\theta\) in the output.

(1)
(d)

Hence find the exact value of \(\int_0^{\pi/3}\cos^3\theta\,d\theta\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses de Moivre's theorem, \(z^n=\cos n\theta+i\sin n\theta\) and \(z^{-n}=\cos n\theta-i\sin n\theta\)✓ 1
A1 adds the two to obtain \(2\cos n\theta\) (AG)✓ 1
Part (b)
M1 applies the binomial expansion✓ 1
A1 \(z^3+3z+3z^{-1}+z^{-3}\)✓ 1
M1 groups the terms as \((z^3+z^{-3})+3(z+z^{-1})=2\cos3\theta+6\cos\theta\)✓ 1
A1 uses \((z+z^{-1})^3=(2\cos\theta)^3=8\cos^3\theta\) and divides by 8 to obtain the result (AG)✓ 1
Part (c)
A1 \(\frac14\)✓ 1
Part (d)
M1 integrates \(\frac14\cos3\theta+\frac34\cos\theta\) term by term✓ 1
A1 \(\left[\frac{1}{12}\sin3\theta+\frac34\sin\theta\right]_0^{\pi/3}\)✓ 1
A1 \(\frac{1}{12}\sin\pi+\frac34\cdot\frac{\sqrt3}{2}-0=\frac{3\sqrt3}{8}\)✓ 1

Exam tip: After expanding, pair each \(z^n\) with \(z^{-n}\). Each pair converts directly to \(2\cos n\theta\), so the whole expansion becomes cosines without any further trigonometric identities.

119IBMA2H-045-01-051.13 Polar and Euler forms Paper 1Medium7 marks

No calculator

In a simplified model, the positions of three communication satellites in a circular orbit are represented on an Argand diagram by the three solutions of \(z^3 = 8i\).

(a)

Find the three solutions, giving each in the form \(x + iy\).

(4)
(b)

The three satellites form the vertices of a triangle. Find the exact area of this triangle.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(8i = 8\,\text{cis}\,\frac{\pi}{2}\)✓ 1
M1 \(z = 2\,\text{cis}\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)\), \(k = 0, 1, 2\) (arguments \(\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\) or equivalent)✓ 1
A1 \(\sqrt{3} + i\)✓ 1
A1 \(-\sqrt{3} + i\) and \(-2i\)✓ 1
Part (b)
M1 a correct side length, e.g. \(|(\sqrt{3}+i) - (-\sqrt{3}+i)| = 2\sqrt{3}\)✓ 1
M1 a correct area method, e.g. \(\frac{1}{2}(2\sqrt{3})^2\sin 60^\circ\) for the equilateral triangle, or \(\frac{1}{2} \times 2\sqrt{3} \times 3\) using base \(2\sqrt{3}\) and perpendicular height \(1-(-2) = 3\)✓ 1
A1 area \(= 3\sqrt{3}\) square units✓ 1

Exam tip: The \(n\)th roots of a complex number are equally spaced around a circle. Here they are \(\frac{2\pi}{3}\) apart, so the triangle must be equilateral, which shortens the area calculation.

120IBMA2H-045-01-071.1 Standard form Paper 1Medium6 marks

No calculator

Let \(\log_a 2 = p\) and \(\log_a 3 = q\), where \(a > 0\), \(a \ne 1\). Give each answer in terms of \(p\) and \(q\).

(a)

Find \(\log_a 72\).

(2)
(b)

Find \(\log_a\left(\frac{\sqrt{a}}{18}\right)\).

(2)
(c)

Find \(\log_6 a\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(72 = 2^3 \times 3^2\), so \(\log_a 72 = 3\log_a 2 + 2\log_a 3\)✓ 1
A1 \(3p + 2q\)✓ 1
Part (b)
M1 \(\frac{1}{2}\log_a a - \log_a 18\)✓ 1
A1 \(\frac{1}{2} - p - 2q\)✓ 1
Part (c)
M1 \(\log_6 a = \frac{1}{\log_a 6}\)✓ 1
A1 \(\frac{1}{p + q}\)✓ 1

Common error: \(\log_a \sqrt{a}\) is \(\frac{1}{2}\), not \(\sqrt{a}\) and not 1. Write the root as a power, \(a^{\frac{1}{2}}\), before applying the power law.

121IBMA2H-045-01-081.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium8 marks

No calculator

In control engineering, the poles of a system are the zeros of a polynomial. For one system the polynomial is \(p(z) = z^4 - 6z^3 + 18z^2 - 30z + 25\). It is given that \(2 - i\) is a zero of \(p(z)\).

(a)

Explain why \(2 + i\) is also a zero, and find the corresponding real quadratic factor of \(p(z)\).

(3)
(b)

Find the other two zeros of \(p(z)\).

(3)
(c)

The four zeros are plotted on an Argand diagram and joined to form a quadrilateral. Find its area.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 the coefficients of \(p(z)\) are all real, so the non-real zeros occur in conjugate pairs✓ 1
M1 \((z - 2 + i)(z - 2 - i) = (z-2)^2 + 1\)✓ 1
A1 \(z^2 - 4z + 5\)✓ 1
Part (b)
M1 write \(p(z) = (z^2 - 4z + 5)(z^2 + bz + c)\) and compare coefficients, e.g. \(b - 4 = -6\) and \(5c = 25\), or use polynomial division✓ 1
A1 \(z^2 - 2z + 5\)✓ 1
A1 \(z = 1 \pm 2i\)✓ 1
Part (c)
M1 identifies a trapezium with parallel sides of lengths 4 (on \(x = 1\)) and 2 (on \(x = 2\)), a distance 1 apart, or uses another valid method such as splitting into triangles✓ 1
A1 area \(= \frac{1}{2}(4 + 2) \times 1 = 3\) square units✓ 1

Exam tip: When you compare coefficients, use the \(z^3\) term and the constant term first. Each one gives a single unknown directly, so you do not need simultaneous equations. Then check your factor against the \(z^2\) and \(z\) terms.

122IBMA2H-045-01-091.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium10 marks

No calculator

This question uses De Moivre's theorem to solve a cubic equation.

(a)

Use De Moivre's theorem to show that \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\).

(4)
(b)

By letting \(x = \cos\theta\), where \(0 \le \theta \le \pi\), find the three solutions of \(8x^3 - 6x - 1 = 0\). Give each in the form \(\cos k\pi\).

(4)
(c)

Hence find the exact value of \(\cos\frac{\pi}{9}\cos\frac{5\pi}{9}\cos\frac{7\pi}{9}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta\)✓ 1
M1 binomial expansion, equating real parts: \(\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta\)✓ 1
M1 substitute \(\sin^2\theta = 1 - \cos^2\theta\)✓ 1
A1 \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\)✓ 1
Part (b)
M1 \(2(4x^3 - 3x) = 1\) gives \(\cos 3\theta = \frac{1}{2}\)✓ 1
M1 \(3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}\) for \(0 \le 3\theta \le 3\pi\)✓ 1
A1 \(\theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}\)✓ 1
A1 \(x = \cos\frac{\pi}{9}, \cos\frac{5\pi}{9}, \cos\frac{7\pi}{9}\). These are three distinct roots, because cosine is one-to-one on \([0, \pi]\)✓ 1
Part (c)
M1 the product of the roots of \(8x^3 - 6x - 1 = 0\) is \(-\frac{d}{a} = -\frac{-1}{8}\)✓ 1
A1 \(\frac{1}{8}\)✓ 1

Exam tip: "Hence" in the final part points to the product of the roots of the cubic. You do not need to evaluate any cosines: the product comes straight from the coefficients, \(-\frac{d}{a}\).

123IBMA2H-045-05-031.11 Partial fractions Paper 1Medium7 marks

No calculator

Let \(f(x)=(4+x)^{-\frac{1}{2}}\).

(a)

Find the first three terms in the expansion of \(f(x)\) in ascending powers of x.

(4)
(b)

State the set of values of x for which the expansion is valid.

(1)
(c)

Use your expansion to find an approximation for \(\frac{1}{\sqrt{4.2}}\), giving your answer as a single fraction.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes \(f(x)\) in the form \(4^{-\frac{1}{2}}\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}=\frac{1}{2}\left(1+\frac{x}{4}\right)^{-\frac{1}{2}}\). Alternatively, uses the general term \(\binom{n}{r}4^{n-r}x^r\) with \(n=-\frac{1}{2}\), so the constant term is \(4^{-\frac{1}{2}}=\frac{1}{2}\)✓ 1
M1 applies the extended binomial theorem with \(n=-\frac{1}{2}\), including a correct \(x^2\) coefficient structure: \(1+\left(-\frac{1}{2}\right)\frac{x}{4}+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}\left(\frac{x}{4}\right)^2\)✓ 1
A1 constant term \(\frac{1}{2}\) and \(x\) term \(-\frac{x}{16}\) both correct✓ 1
A1 \(x^2\) term \(\frac{3x^2}{256}\), giving \(\frac{1}{2}-\frac{x}{16}+\frac{3x^2}{256}\)✓ 1
Part (b)
A1 \(|x|<4\) (accept \(-4<x<4\))✓ 1
Part (c)
M1 substitutes \(x=0.2=\frac{1}{5}\) into the expansion: \(\frac{1}{2}-\frac{1}{80}+\frac{3}{6400}\)✓ 1
A1 \(\frac{3123}{6400}\)✓ 1

Common error: A common error is forgetting to take out the factor 4⁻¹/² before expanding. The extended binomial theorem needs the form (1 + u)ⁿ, so the constant 4 must also be raised to the power n.

124IBMA2H-045-05-061.11 Partial fractions Paper 1Medium8 marks

No calculator

The expansion of \((1+ax)^n\), where \(n\in\mathbb{Z}\), \(n<0\) and \(a\in\mathbb{R}\), in ascending powers of x begins \(1-6x+27x^2+\dots\)

(a)

Find the value of n and the value of a.

(5)
(b)

Find the coefficient of \(x^3\).

(2)
(c)

State the range of values of x for which the expansion is valid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(na=-6\)✓ 1
M1 \(\frac{n(n-1)}{2}a^2=27\)✓ 1
M1 eliminates a, e.g. \(\frac{n(n-1)a^2}{(na)^2}=\frac{54}{36}\) so \(\frac{n-1}{n}=\frac{3}{2}\)✓ 1
A1 \(n=-2\)✓ 1
A1 \(a=3\)✓ 1
Part (b)
M1 \(\frac{n(n-1)(n-2)}{3!}a^3\) with their values✓ 1
A1 \(\frac{(-2)(-3)(-4)}{6}\times27=-108\)✓ 1
Part (c)
A1 \(|x|<\frac{1}{3}\)✓ 1

Exam tip: Dividing one equation by the square of the other removes a cleanly and avoids messy substitution.

125IBMA2H-045-05-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium6 marks

No calculator

Consider the infinite series \(S=1+\cos^2\theta+\cos^4\theta+\cos^6\theta+\dots\), where \(0<\theta<\frac{\pi}{2}\).

(a)

Explain why the series has a finite sum.

(2)
(b)

Show that \(S=\frac{1}{\sin^2\theta}\).

(2)
(c)

Given that \(S=4\), find the exact value of θ.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 the series is geometric with common ratio \(\cos^2\theta\)✓ 1
R1 for \(0<\theta<\frac{\pi}{2}\), \(0<\cos^2\theta<1\), so \(|r|<1\) and the series converges✓ 1
Part (b)
M1 \(S=\frac{1}{1-\cos^2\theta}\)✓ 1
A1 uses \(\sin^2\theta+\cos^2\theta=1\) to obtain \(\frac{1}{\sin^2\theta}\) (AG)✓ 1
Part (c)
M1 \(\sin^2\theta=\frac{1}{4}\) so \(\sin\theta=\frac{1}{2}\) (positive in the given interval)✓ 1
A1 \(\theta=\frac{\pi}{6}\)✓ 1

Exam tip: When asked to explain convergence, state both the common ratio and why its modulus is less than 1 on the given interval; each piece of reasoning earns credit.

126IBMA2H-045-07-071.15 Proof by induction, contradiction and counterexample Paper 1Medium7 marks

No calculator

This question is about the series \(\displaystyle\sum_{r=1}^{n}\frac{r}{2^r}\).

(a)

Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n}\frac{r}{2^r}=2-\frac{n+2}{2^n}\) for all \(n\in\mathbb{Z}^+\).

(6)
(b)

Hence write down the value of \(\displaystyle\sum_{r=1}^{\infty}\frac{r}{2^r}\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 when \(n=1\): LHS \(=\frac{1}{2}\) and RHS \(=2-\frac{3}{2}=\frac{1}{2}\), so the result is true for \(n=1\)✓ 1
M1 assumes the result is true for \(n=k\): \(\sum_{r=1}^{k}\frac{r}{2^r}=2-\frac{k+2}{2^k}\)✓ 1
M1 adds the \((k+1)\)th term: \(2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}\)✓ 1
M1 uses the common denominator \(2^{k+1}\): \(2-\frac{2k+4-k-1}{2^{k+1}}\)✓ 1
A1 \(=2-\frac{k+3}{2^{k+1}}=2-\frac{(k+1)+2}{2^{k+1}}\)✓ 1
R1 the result is true for \(n=1\), and if true for \(n=k\) it is true for \(n=k+1\); hence it is true for all \(n\in\mathbb{Z}^+\)✓ 1
Part (b)
A1 2, since \(\frac{n+2}{2^n}\to0\) as \(n\to\infty\)✓ 1

Exam tip: The reasoning mark needs a full concluding sentence: state the base case, the inductive step and the conclusion. In the algebra, aim to write the result explicitly in terms of \(k+1\).

127IBMA2H-045-07-101.8 Infinite geometric series Paper 1Medium7 marks

No calculator

An infinite geometric series has first term \(a\) and common ratio \(r\), where \(|r|<1\). The sum to infinity of the series is 27. The squares of the terms of the series form a new infinite series, whose sum to infinity is 243.

(a)

Show that \(\dfrac{a}{1+r}=9\).

(2)
(b)

Find the value of \(a\) and the value of \(r\).

(3)
(c)

Find the exact value of the sum to infinity of the cubes of the terms of the original series.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 the squares form a geometric series with first term \(a^2\) and ratio \(r^2\), so \(\frac{a^2}{1-r^2}=243\)✓ 1
M1 writes \(1-r^2=(1-r)(1+r)\) and divides by \(\frac{a}{1-r}=27\) to obtain \(\frac{a}{1+r}=9\)✓ 1
Part (b)
M1 \(a=27(1-r)\) and \(a=9(1+r)\)✓ 1
M1 \(27-27r=9+9r\)✓ 1
A1 \(r=\frac{1}{2}\) and \(a=\frac{27}{2}\)✓ 1
Part (c)
M1 \(\frac{a^3}{1-r^3}=\frac{19683/8}{7/8}\)✓ 1
A1 \(\frac{19683}{7}\)✓ 1

Exam tip: Squaring or cubing every term of a geometric series gives another geometric series, with ratio \(r^2\) or \(r^3\). Dividing one sum-to-infinity equation by the other removes the difference of two squares in a single step.

128IBMA2H-045-07-121.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

Consider the series \(\displaystyle\sum_{k=1}^{8}(1+i)^k\).

(a)

Write \(1+i\) in modulus–argument form. Hence show that \((1+i)^8=16\).

(2)
(b)

Find \(\displaystyle\sum_{k=1}^{8}(1+i)^k\), giving your answer in the form \(x+iy\), where \(x,y\in\mathbb{Z}\).

(4)
(c)

Find the exact modulus of the sum \(\displaystyle\sum_{k=1}^{8}(1+i)^k\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1+i=\sqrt2\,\mathrm{cis}\frac{\pi}{4}\)✓ 1
A1 by de Moivre, \((1+i)^8=(\sqrt2)^8\,\mathrm{cis}\,2\pi=16(\cos2\pi+i\sin2\pi)=16\)✓ 1
Part (b)
M1 recognises a geometric series with first term \(1+i\) and common ratio \(1+i\)✓ 1
M1 substitutes into the sum formula: \(S_8=\frac{(1+i)((1+i)^8-1)}{(1+i)-1}\)✓ 1
A1 \(=\frac{15(1+i)}{i}\)✓ 1
A1 uses \(\frac{1}{i}=-i\) to obtain \(15(1+i)(-i)=15-15i\)✓ 1
Part (c)
A1 \(\sqrt{15^2+(-15)^2}=15\sqrt2\)✓ 1

Common error: The denominator is \(r-1=(1+i)-1=i\), not \(1+i\). To divide by \(i\), multiply the numerator by \(-i\), since \(\frac{1}{i}=-i\).

129IBMA3H-1-18r5-051.11 Partial fractions Paper 1Medium5 marks

No calculator


Find the constant term in the expansion of (2x + 3/x)6.

(a)

Find the constant term.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 general term C(6, r)(2x)6−r(3/x)r✓ 1
A1 power of x is 6 − 2r✓ 1
M1 set 6 − 2r = 0, so r = 3✓ 1
A1 C(6, 3) × 23 × 33 = 20 × 8 × 27✓ 1
A1 4320✓ 1

Exam tip: Find the power of x in the general term first. Setting it equal to zero tells you which value of r to use.

130IBMA3H-1-18r5-061.9 The binomial theorem Paper 1Medium7 marks

No calculator


The expansion of (3 + kx)5, where k is a non-zero constant, is written in ascending powers of x.

(a)

Show that the coefficients of x2 and x3 are 270k2 and 90k3 respectively.

(2)
(b)

The coefficient of x3 is twice the coefficient of x2. Find the value of k.

(3)
(c)

Using your value of k, find the coefficient of x4.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 C(5, 2) × 33 × k2 or C(5, 3) × 32 × k3✓ 1
A1 10 × 27 = 270 and 10 × 9 = 90, giving 270k2 and 90k3 (AG)✓ 1
Part (b)
M1 90k3 = 2 × 270k2✓ 1
R1 divide by k2 since k ≠ 0✓ 1
A1 k = 6✓ 1
Part (c)
M1 C(5, 4) × 3 × 64 = 15 × 64 (ECF from (b))✓ 1
A1 19440✓ 1

Common error: Dividing both sides by k² without saying k ≠ 0 can lose the reasoning mark. State why the division is allowed.

131IBMA3H-1-18r5-081.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator


The polynomial p(z) = z3 + az2 + bz − 13, where a and b are real, has a root 2 + 3i.

(a)

Write down another complex root of p(z).

(1)
(b)

Find the value of a, the value of b, and the third root of p(z).

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 2 − 3i (conjugate root, since the coefficients are real)✓ 1
Part (b)
M1 form the quadratic factor (z − (2 + 3i))(z − (2 − 3i))✓ 1
A1 z2 − 4z + 13✓ 1
M1 third root r from the product of roots: 13r = 13, or by division✓ 1
A1 third root = 1✓ 1
A1 a = −5✓ 1
A1 b = 17✓ 1

Common error: Stating the conjugate root without justification. Mention that the coefficients are real.

132IBMA3H-1-18r5-121.9 The binomial theorem Paper 1Medium7 marks

No calculator


Let f(x) = (1 + 2x)−3.

(a)

Find the first four terms of the expansion of f(x) in ascending powers of x.

(4)
(b)

State the range of values of x for which the expansion is valid.

(1)
(c)

Use x = 0.01 in your expansion to estimate 1/1.023. Give your answer to 5 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the binomial expansion with n = −3 and 2x in place of x✓ 1
A1 1 − 6x✓ 1
A1 + 24x2✓ 1
A1 − 80x3✓ 1
Part (b)
A1 |x| < 1/2✓ 1
Part (c)
M1 substitute x = 0.01: 1 − 0.06 + 0.0024 − 0.00008 (ECF from (a))✓ 1
A1 0.94232✓ 1

Common error: Forgetting to raise the 2 to a power: the coefficient of x² comes from (2x)² = 4x², not 2x².

133IBMA3H-1-18r5-141.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks

No calculator


This question is about whole numbers.

(a)

A student claims that n2 + n + 41 is prime for every positive integer n. Show, by counterexample, that the claim is false.

(2)
(b)

A beekeeper packs honey jars into small boxes holding 4 jars and large boxes holding 6 jars. Use proof by contradiction to show that it is impossible to pack exactly 13 jars using whole boxes only.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 n = 40 gives 402 + 40 + 41 = 1681✓ 1
R1 1681 = 412, which is not prime, so the claim is false✓ 1
Part (b)
M1 assume there are non-negative integers x and y with 4x + 6y = 13✓ 1
A1 LHS = 2(2x + 3y), which is even✓ 1
A1 RHS = 13, which is odd✓ 1
R1 contradiction, so the assumption is false and the packing is impossible✓ 1

Exam tip: A counterexample needs a specific value and a check that it breaks the claim. One case is enough to disprove a "for all" statement.

134IBMA3H-1-18r6-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium6 marks

No calculator

The cubic equation \(z^3-5z^2+11z-15=0\) has real coefficients, and \(z=1+2\mathrm{i}\) is one root.

(a)

Write down another root of the equation.

(1)
(b)

Find a quadratic factor with real coefficients.

(2)
(c)

Hence find the third root of the equation.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-2\mathrm{i}\)✓ 1
Part (b)
M1 \((z-(1+2\mathrm{i}))(z-(1-2\mathrm{i}))\) or sum 2 and product 5 of the roots✓ 1
A1 \(z^2-2z+5\)✓ 1
Part (c)
M1 divide the cubic by \(z^2-2z+5\) or compare coefficients✓ 1
A1 \((z^2-2z+5)(z-3)\)✓ 1
A1 \(z=3\)✓ 1

Exam tip: Complex roots of a real polynomial come in conjugate pairs; the product of all roots can provide a quick check on the final root.

135IBMA3H-1-18r6-111.11 Partial fractions Paper 1Medium7 marks

No calculator

The function \(f\) is defined by \(f(x)=\dfrac{3x+7}{(x+1)(x+3)}\) for \(x\in\mathbb{R}\), \(x\ne-1\), \(x\ne-3\).

(a)

Express \(f(x)\) in the form \(\dfrac{A}{x+1}+\dfrac{B}{x+3}\).

(3)
(b)

Write down the equations of the vertical asymptotes of the graph of \(y=f(x)\).

(1)
(c)

Solve \(f(x)=1\). Give the larger solution in exact form.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(3x+7=A(x+3)+B(x+1)\)✓ 1
A1 \(A=2\)✓ 1
A1 \(B=1\)✓ 1
Part (b)
A1 \(x=-1\) and \(x=-3\)✓ 1
Part (c)
M1 \(3x+7=(x+1)(x+3)\)✓ 1
A1 \(x^2+x-4=0\)✓ 1
A1 \(x=\frac{-1+\sqrt{17}}{2}\) (≈ 1.56), other solution \(\frac{-1-\sqrt{17}}2\)✓ 1

Common error: After clearing denominators, check that each solution is not one of the excluded values.

136IBMA3H-1-18r6-121.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium6 marks

No calculator

Find the two square roots of \(5-12\mathrm{i}\). Give each root in the form \(a+b\mathrm{i}\) with \(a,b\in\mathbb{R}\).

(a)

Find the two square roots.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 let \((a+b\mathrm{i})^2=5-12\mathrm{i}\) and equate real and imaginary parts✓ 1
A1 \(a^2-b^2=5\) and \(2ab=-12\)✓ 1
M1 eliminate \(b\) using \(b=-\frac6a\)✓ 1
A1 \(a^4-5a^2-36=0\) so \((a^2-9)(a^2+4)=0\)✓ 1
R1 reject \(a^2=-4\) since \(a\) is real✓ 1
A1 \(a=3,b=-2\) or \(a=-3,b=2\): roots \(3-2\mathrm{i}\) and \(-3+2\mathrm{i}\) (real part magnitude 3)✓ 1

Exam tip: Equate real and imaginary parts to get two real equations, then eliminate one unknown.

137IBMA3H-1-18r10-101.16 Systems of linear equations Paper 1Medium7 marks

No calculator

The currents \(x\), \(y\) and \(z\) (in amperes) in three motors of a robot satisfy the system
\(x+2y-z=3\)
\(2x-y+3z=1\)
\(3x+y+2z=k\)
where \(k\) is a constant.

(a)

Find the value of \(k\) for which the system has at least one solution, and explain why there is no solution for any other value of \(k\).

(3)
(b)

For this value of \(k\), find the general solution in terms of a parameter \(t\), where \(z=t\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
adding the first two equations gives \(3x+y+2z=4\)✓ 1
\(k=4\)✓ 1
if \(k\ne4\) the third equation contradicts \(3x+y+2z=4\), so no solution✓ 1
Part (b)
substitute \(z=t\) and eliminate to find \(y\) in terms of \(t\)✓ 1
\(y=1+t\)✓ 1
\(x=1-t\)✓ 1
general solution \((x,y,z)=(1-t,\,1+t,\,t)\)✓ 1

Common error: When two rows combine to give the third row's left side, check whether the right sides also agree; if not, there is no solution.

138IBMA3H-1-18r10-111.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium6 marks

No calculator

A robot's stability depends on the polynomial equation \(z^3-6z^2+pz+q=0\), where \(p\) and \(q\) are real constants. One root is \(2-i\).

(a)

Write down another root of the equation.

(1)
(b)

Find a quadratic factor with real coefficients having both of these roots.

(2)
(c)

Find the third root, and the values of \(p\) and \(q\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(2+i\)✓ 1
Part (b)
sum of roots \(=4\) and product \(=(2-i)(2+i)=5\)✓ 1
\(z^2-4z+5\)✓ 1
Part (c)
third root: sum of roots is 6, so \(6-4=2\)✓ 1
\((z^2-4z+5)(z-2)\) expanded✓ 1
\(p=13\) and \(q=-10\)✓ 1

Exam tip: For a polynomial with real coefficients, non-real roots come in conjugate pairs; the sum of roots is minus the \(z^2\) coefficient.

139IBMA3H-1-18r10-121.13 Polar and Euler forms Paper 1Medium7 marks

No calculator

The four vertices of a square fashion motif are drawn on an Argand diagram at the solutions of \(z^4=-16\), where one unit represents 10 cm.

(a)

Write \(-16\) in modulus-argument form and hence find the modulus and the four arguments of the solutions, with \(0<\arg z<2\pi\).

(3)
(b)

Write the four solutions in the form \(a+bi\).

(2)
(c)

Find the area of the square, in square units.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(-16=16(\cos\pi+i\sin\pi)\)✓ 1
modulus \(=16^{1/4}=2\) and argument \(\frac{\pi+2k\pi}{4}\)✓ 1
arguments \(\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4\)✓ 1
Part (b)
\(\sqrt2+\sqrt2\,i\) and \(-\sqrt2+\sqrt2\,i\)✓ 1
\(-\sqrt2-\sqrt2\,i\) and \(\sqrt2-\sqrt2\,i\)✓ 1
Part (c)
side length \(=2\sqrt2\)✓ 1
area \(=8\)✓ 1

Common error: Use the argument \(\pi+2k\pi\) before dividing by 4; dividing \(\pi\) alone gives only one of the four roots.

140IBMA3H-1-19r11-061.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

No calculator

Water flows into a storage tank at a rate \(R(t)=\frac{5t+7}{(t+1)(t+3)}\) m3 per hour, where \(t\) is the time in hours and \(0\le t\le2\).

(a)

Find constants \(A\) and \(B\) such that \(R(t)=\frac{A}{t+1}+\frac{B}{t+3}\).

(3)
(b)

Hence find the exact volume of water that flows into the tank in the first 2 hours, giving your answer in the form \(p\ln5+q\ln3\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(5t+7=A(t+3)+B(t+1)\)✓ 1
A1 \(A=1\)✓ 1
A1 \(B=4\)✓ 1
Part (b)
M1 integrates both terms using \(\int\frac{1}{t+a}\,dt=\ln|t+a|\)✓ 1
A1 \(\left[\ln(t+1)+4\ln(t+3)\right]_0^2\)✓ 1
M1 substitutes limits: \(\ln3+4\ln5-(\ln1+4\ln3)\)✓ 1
A1 \(4\ln5-3\ln3\) m3 (which is 3.1419 m3)✓ 1

Exam tip: Check partial fractions by substituting a convenient value, such as \(t=0\): \(\frac{7}{3}=1+\frac{4}{3}\).

141IBMA3H-1-19r11-091.6 Deductive proof Paper 1Medium6 marks

No calculator

In an online game, tokens are sold in small packs of 12 tokens and large packs of 18 tokens. A player wants to buy exactly 100 tokens by buying \(m\) small packs and \(n\) large packs, where \(m\) and \(n\) are non-negative integers.

(a)

Use proof by contradiction to show that it is impossible to buy exactly 100 tokens.

(4)
(b)

Find one combination of \(m\) and \(n\) that gives exactly 102 tokens.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 assumes there are integers \(m,n\ge0\) with \(12m+18n=100\)✓ 1
M1 factorises: \(6(2m+3n)=100\)✓ 1
A1 \(2m+3n=\frac{50}{3}\)✓ 1
R1 \(2m+3n\) is an integer but \(\frac{50}{3}\) is not, a contradiction, so the assumption is false and 100 tokens cannot be bought✓ 1
Part (b)
M1 \(12m+18n=102\Rightarrow2m+3n=17\)✓ 1
A1 any valid pair, e.g. \(m=4\), \(n=3\) (also \(m=1,n=5\) or \(m=7,n=1\))✓ 1

Common error: A proof by contradiction must end by stating that the assumption is false and so the original statement is true.

142IBMA3H-1-19r11-101.15 Proof by induction, contradiction and counterexample Paper 1Medium7 marks

No calculator

In a game, completing level \(r\) awards \(r\times2^r\) points. Let \(S_n\) be the total points earned from completing levels 1 to \(n\), so \(S_n=\sum_{r=1}^{n}r\times2^r\).

(a)

Calculate \(S_3\).

(1)
(b)

Prove by mathematical induction that \(S_n=(n-1)2^{n+1}+2\) for all \(n\in\mathbb{Z}^+\).

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2+8+24=34\)✓ 1
Part (b)
A1 \(n=1\): LHS \(=2\), RHS \(=0\times4+2=2\), so true for \(n=1\)✓ 1
M1 assumes true for \(n=k\): \(S_k=(k-1)2^{k+1}+2\)✓ 1
M1 \(S_{k+1}=S_k+(k+1)2^{k+1}\)✓ 1
A1 \(=(k-1)2^{k+1}+2+(k+1)2^{k+1}=2k\times2^{k+1}+2\)✓ 1
A1 \(=k\times2^{k+2}+2=((k+1)-1)2^{(k+1)+1}+2\)✓ 1
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{Z}^+\)✓ 1

Exam tip: In the inductive step, write out what you are aiming for (the statement with \(k+1\) in place of \(n\)) before you start.

143IBMA3H-1-18r13-041.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

The equation \(z^3 + az^2 + bz + 52 = 0\), where \(a, b\in\mathbb{R}\), has a root \(z = 2 - 3i\).

(a)

Write down another complex root of the equation.

(1)
(b)

Find the real root of the equation.

(3)
(c)

Find the values of \(a\) and \(b\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2 + 3i\)✓ 1
Part (b)
M1 quadratic factor from the pair: \(z^2 - 4z + 13\) (sum 4, product 13)✓ 1
M1 product of roots \(= -52\), so \(13 r = -52\), or divide the cubic by the quadratic✓ 1
A1 \(r = -4\)✓ 1
Part (c)
M1 expand \((z^2-4z+13)(z+4)\), or use sum and pairwise products of roots✓ 1
A1 \(a = 0\)✓ 1
A1 \(b = -3\)✓ 1

Exam tip: A polynomial with real coefficients always has non-real roots in conjugate pairs. The pair gives a real quadratic factor.

144IBMA3H-1-18r13-051.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

Let \(z = \sqrt{3} - i\).

(a)

Write \(z\) in the form \(r\,\mathrm{cis}\,\theta\), where \(r>0\) and \(-\pi<\theta\leq\pi\).

(2)
(b)

Use de Moivre's theorem to show that \(z^9 = 512i\).

(3)
(c)

Find the smallest positive integer \(n\) for which \(z^n\) is a positive real number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = \sqrt{3+1} = 2\)✓ 1
A1 \(\theta = -\frac{\pi}{6}\)✓ 1
Part (b)
M1 \(z^9 = 2^9\,\mathrm{cis}\left(-\frac{9\pi}{6}\right)\)✓ 1
A1 \(512\,\mathrm{cis}\left(-\frac{3\pi}{2}\right)\)✓ 1
A1 \(= 512(0 + i) = 512i\) (AG)✓ 1
Part (c)
M1 \(-\frac{n\pi}{6} = -2\pi k\) for a positive integer \(k\)✓ 1
A1 \(n = 12\)✓ 1

Common error: Giving the argument as \(+\frac{\pi}{6}\). The point \((\sqrt3, -1)\) is in the fourth quadrant, so the argument is negative.

145IBMA3H-1-18r14-081.16 Systems of linear equations Paper 1Medium7 marks

No calculator

A robot has three motors with settings \(x\), \(y\) and \(z\). A calibration gives the system
\(x+2y+z=4\)
\(2x+y-z=5\)
\(3x+3y+kz=m\)
where \(k\) and \(m\) are constants.

(a)

By subtracting the sum of the first two equations from the third, show that \(kz=m-9\).

(2)
(b)

When \(k=0\) and \(m=9\), the system has infinitely many solutions. Find the general solution, in terms of a parameter.

(3)
(c)

When \(k=0\) and \(m=7\), state the number of solutions and give a geometric interpretation.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 subtracting \((x+2y+z)+(2x+y-z)=3x+3y+0z=9\) from the third equation✓ 1
A1 \(kz=m-9\) (AG)✓ 1
Part (b)
M1 eliminating one variable from the first two equations✓ 1
A1 \(x+y=3\) (with \(x=3-t\), \(y=t\))✓ 1
A1 \((x,y,z)=(3-t,\ t,\ 1-t)\), \(t\in\mathbb{R}\)✓ 1
Part (c)
A1 no solutions (the equation becomes \(0=-2\))✓ 1
R1 the three planes have no common point (they meet pairwise in parallel lines, forming a triangular prism)✓ 1

Exam tip: If row operations lead to \(0=c\) with \(c\ne0\) the system is inconsistent; if they lead to \(0=0\) there are infinitely many solutions.

146IBMA3H-1-18r16-011.11 Partial fractions Paper 1Medium5 marks

No calculator

Consider the expansion of \(\left(x^2 + \frac{2}{x}\right)^6\).

(a)

Find the term that is independent of x.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 general term \(\binom{6}{r}(x^2)^{6-r}\left(\frac{2}{x}\right)^r\)✓ 1
A1 power of x is \(12 - 3r\)✓ 1
M1 \(12 - 3r = 0\)✓ 1
A1 \(r = 4\)✓ 1
A1 \(\binom{6}{4}\times 2^4 = 240\)✓ 1

Exam tip: For a term independent of x, write the general term, combine the powers of x and set the exponent equal to zero.

147IBMA3H-1-18r24-071.8 Infinite geometric series Paper 1Medium7 marks

No calculator

A rail crew lays track so that the length laid each day forms an arithmetic sequence. On day 1 the crew lays 2.0 km and on day 9 the crew lays 6.0 km.

(a)

Find the common difference.

(2)
(b)

Find the total length of track laid in the first 9 days.

(2)
(c)

The whole project is 135 km long and is completed at the end of day n, with all 135 km laid in exactly n days. Find n.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2.0 + 8d = 6.0✓ 1
A1 d = 0.5 km✓ 1
Part (b)
M1 S9 = (9/2)(2.0 + 6.0)✓ 1
A1 36 km (ECF from (a) if used)✓ 1
Part (c)
M1 (n/2)(4 + 0.5(n − 1)) = 135✓ 1
M1 simplifies to n2 + 7n − 540 = 0 and solves, (n − 20)(n + 27) = 0✓ 1
A1 n = 20 (reject n = −27)✓ 1

Exam tip: For a sum, choose whichever form of the series formula uses the information you already have: (n/2)(first + last) or (n/2)(2u₁ + (n − 1)d).

148IBMA3H-1-20r29-041.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

Let \(z=1-\sqrt3\,i\).

(a)

Write \(z\) in modulus-argument form \(r\,\text{cis}\,\theta\).

(2)
(b)

Use de Moivre's theorem to show that \(z^5=16+16\sqrt3\,i\).

(3)
(c)

Find the smallest positive integer \(n\) for which \(z^n\) is a positive real number, and state the value of \(z^n\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=2\)✓ 1
A1 \(\theta=-\frac{\pi}{3}\)✓ 1
Part (b)
M1 \(z^5=2^5\text{cis}\left(-\frac{5\pi}{3}\right)\)✓ 1
A1 \(=32\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)✓ 1
A1 \(=32\left(\frac12+\frac{\sqrt3}{2}i\right)=16+16\sqrt3\,i\)✓ 1
Part (c)
M1 \(-\frac{n\pi}{3}\) is a multiple of \(2\pi\), so \(n=6\)✓ 1
A1 \(z^6=2^6=64\)✓ 1

Common error: \(z^n\) is real when \(n\theta\) is a multiple of π, but it is a positive real only when \(n\theta\) is a multiple of 2π. Check the sign.

149IBMA3H-1-20r29-051.13 Polar and Euler forms Paper 1Medium8 marks

No calculator

The vertices of a triangular logo are placed in the complex plane at the three solutions of \(z^3=8i\).

(a)

Find the three solutions in the form \(r\,\text{cis}\,\theta\) with \(-\pi<\theta\le\pi\).

(4)
(b)

Write the three solutions in the form \(a+bi\).

(2)
(c)

Find the exact area of the triangular logo.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(8i=8\,\text{cis}\left(\frac{\pi}{2}+2k\pi\right)\)✓ 1
M1 \(z=2\,\text{cis}\left(\frac{\pi}{6}+\frac{2k\pi}{3}\right)\)✓ 1
A1 \(2\,\text{cis}\frac{\pi}{6}\) and \(2\,\text{cis}\frac{5\pi}{6}\)✓ 1
A1 \(2\,\text{cis}\left(-\frac{\pi}{2}\right)\)✓ 1
Part (b)
A1 \(\sqrt3+i\) and \(-\sqrt3+i\)✓ 1
A1 \(-2i\)✓ 1
Part (c)
M1 base \(2\sqrt3\) and height 3 (or equilateral triangle of side \(2\sqrt3\))✓ 1
A1 \(3\sqrt3\)✓ 1

Exam tip: The n roots of \(z^n=w\) are equally spaced by \(\frac{2\pi}{n}\) on a circle: find one, then rotate.

150IBMA3H-1-20r29-061.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

The polynomial \(p(z)=z^3+az^2+bz+13\) has real coefficients \(a\) and \(b\), and one root is \(2+3i\).

(a)

Write down another root of \(p(z)=0\).

(1)
(b)

Find the third root.

(3)
(c)

Find the value of \(a\) and the value of \(b\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2-3i\)✓ 1
Part (b)
M1 product of the roots \(=-13\)✓ 1
M1 \((2+3i)(2-3i)=13\)✓ 1
A1 third root is \(-1\)✓ 1
Part (c)
M1 sum of roots \(=-a\)✓ 1
A1 \(a=-3\)✓ 1
A1 \(b=13+(-1)(4)=9\)✓ 1

Exam tip: Complex roots of a real polynomial come in conjugate pairs. The pair gives a real quadratic factor \(z^2-4z+13\).

151IBMA3H-1-20r29-071.15 Proof by induction, contradiction and counterexample Paper 1Medium6 marks

No calculator

Use mathematical induction to prove that \(3^{2n+1}+2^{n+2}\) is divisible by 7 for all \(n\in\mathbb{Z}^+\).

(a)

Prove the statement.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(n=1\): \(3^3+2^3=35=7\times5\), so the statement is true for \(n=1\)✓ 1
M1 assume true for \(n=k\): \(3^{2k+1}+2^{k+2}=7m\) for an integer \(m\)✓ 1
M1 consider \(n=k+1\): \(3^{2k+3}+2^{k+3}\)✓ 1
M1 rewrite as \(9\times3^{2k+1}+2\times2^{k+2}\)✓ 1
A1 \(=2\left(3^{2k+1}+2^{k+2}\right)+7\times3^{2k+1}=14m+7\times3^{2k+1}\), a multiple of 7✓ 1
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{Z}^+\)✓ 1

Common error: The conclusion must state both parts: the base case holds, and truth at \(k\) implies truth at \(k+1\).

152IBMA3H-1-20r29-091.15 Proof by induction, contradiction and counterexample Paper 1Medium7 marks

No calculator

This question is about disproving and proving statements.

(a)

A student claims that if \(n\) is prime then \(2^n-1\) is prime. Use a counterexample to show that the claim is false.

(2)
(b)

Prove by contradiction that \(\log_23\) is irrational.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(n=11\) (prime) gives \(2^{11}-1=2047\)✓ 1
A1 \(2047=23\times89\), which is not prime✓ 1
Part (b)
M1 assume \(\log_23=\frac pq\) with positive integers \(p,q\)✓ 1
M1 \(2^{p/q}=3\), so \(2^p=3^q\)✓ 1
A1 \(2^p\) is even since \(p\ge1\)✓ 1
A1 \(3^q\) is odd✓ 1
R1 contradiction, so \(\log_23\) is irrational✓ 1

Common error: A proof by contradiction must begin by assuming the opposite of what you want to prove, and end by naming the contradiction.

153IBMA3H-1-20r29-121.11 Partial fractions Paper 1Medium6 marks

No calculator

This question is about exponent laws.

(a)

Simplify \(\left(27a^{6}b^{3}\right)^{-2/3}\), giving your answer without negative exponents.

(3)
(b)

Solve \(8^{x}\times4^{x-1}=\dfrac{1}{32}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 apply the power \(-\frac23\) to each factor✓ 1
A1 \(27^{-2/3}=\frac19\)✓ 1
A1 \(\frac{1}{9a^4b^2}\)✓ 1
Part (b)
M1 write each term as a power of 2: \(2^{3x}\times2^{2x-2}=2^{-5}\)✓ 1
M1 \(5x-2=-5\)✓ 1
A1 \(x=-\frac35\)✓ 1

Common error: \(8^x\times4^{x-1}\) is \(2^{3x}\times2^{2(x-1)}\). The power of 4 needs the bracket multiplied by 2, not just \(x\).

154IBMA3H-1-20r35-081.11 Partial fractions Paper 1Medium7 marks

No calculator

Let \(p=\log_3x\) and \(q=\log_3y\), where \(x>0\) and \(y>0\).

(a)

Express \(\log_3\!\left(\dfrac{9x}{y^2}\right)\) in terms of \(p\) and \(q\).

(2)
(b)

Express \(\log_9(xy)\) in terms of \(p\) and \(q\).

(2)
(c)

Given that \(\log_3\!\left(\dfrac{9x}{y^2}\right)=3\) and \(\log_9(xy)=2\), find the value of \(x\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of laws of logarithms: \(\log_39+\log_3x-2\log_3y\)✓ 1
A1 \(2+p-2q\)✓ 1
Part (b)
M1 change of base: \(\dfrac{\log_3(xy)}{\log_39}\)✓ 1
A1 \(\dfrac{p+q}{2}\)✓ 1
Part (c)
M1 form simultaneous equations: \(p-2q=1\) and \(p+q=4\)✓ 1
A1 \(q=1\) and \(p=3\)✓ 1
A1 \(x=3^3=27\) (and \(y=3\))✓ 1

Exam tip: To change the base of a logarithm, divide by the logarithm of the old base to the new base; log₉ of anything becomes half of log₃ of it.

155IBMA3H-1-20r35-101.3 Geometric sequences and series Paper 1Medium8 marks

No calculator

An infinite geometric series has first term 4 and common ratio \(r=\dfrac{2x-1}{5}\).

(a)

Find the set of values of \(x\) for which the series converges.

(3)
(b)

Find the sum to infinity when \(x=1\).

(2)
(c)

Find the value of \(x\) for which the sum to infinity is 10.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 convergence requires \(|r|<1\)✓ 1
A1 \(-5<2x-1<5\)✓ 1
A1 \(-2<x<3\)✓ 1
Part (b)
M1 \(r=\dfrac15\) and \(S_\infty=\dfrac{4}{1-r}\)✓ 1
A1 5✓ 1
Part (c)
M1 \(\dfrac{4}{1-r}=10\)✓ 1
A1 \(r=0.6\)✓ 1
A1 \(\dfrac{2x-1}{5}=0.6\), so \(x=2\) (this lies in the interval from (a))✓ 1

Common error: Always check a value found from S∞ against the convergence condition |r| < 1; a value outside the interval cannot be correct.

156IBMA3H-1-20r35-111.15 Proof by induction, contradiction and counterexample Paper 1Medium8 marks

No calculator

In a warehouse, crates are stacked in rows so that row \(r\) contains \(r(r+1)\) crates, for \(r=1,2,3,\dots\)

(a)

Prove by induction that \(\displaystyle\sum_{r=1}^{n}r(r+1)=\dfrac{n(n+1)(n+2)}{3}\) for all \(n\in\mathbb{Z}^+\).

(6)
(b)

Hence find the total number of crates in the first 12 rows.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 for \(n=1\): LHS \(=1\times2=2\) and RHS \(=\dfrac{1\times2\times3}{3}=2\), so true for \(n=1\)✓ 1
M1 assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}r(r+1)=\dfrac{k(k+1)(k+2)}{3}\)✓ 1
M1 for \(n=k+1\) the sum is \(\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)\)✓ 1
A1 \(=\dfrac{(k+1)(k+2)(k+3)}{3}\) after taking out the common factor \((k+1)(k+2)\)✓ 1
A1 this is the formula with \(n=k+1\)✓ 1
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{Z}^+\) by induction✓ 1
Part (b)
M1 \(\dfrac{12\times13\times14}{3}\)✓ 1
A1 728 crates✓ 1

Exam tip: In the inductive step, write down exactly what you are trying to reach (the formula with k + 1 in place of n) so that you know what to factorise towards.

157IBMA3H-1-20r35-141.15 Proof by induction, contradiction and counterexample Paper 1Medium7 marks

No calculator

A depot packs items into crates that each hold either 12 items or 18 items. An order can be shipped only if it is made up of whole crates.

(a)

Prove by contradiction that it is impossible to ship exactly 1000 items.

(4)
(b)

Prove by contradiction that the number of items in a non-empty shipment of whole crates is never a prime number.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 assume \(12a+18b=1000\) for non-negative integers \(a\) and \(b\)✓ 1
M1 factorise: \(6(2a+3b)=1000\)✓ 1
A1 so \(2a+3b=\dfrac{1000}{6}=\dfrac{500}{3}\)✓ 1
R1 \(2a+3b\) is an integer but \(\dfrac{500}{3}\) is not: contradiction, so the order cannot be shipped✓ 1
Part (b)
M1 assume the total \(N=12a+18b\) is prime, with \(a,b\) non-negative integers not both zero, and factorise: \(N=6(2a+3b)\)✓ 1
A1 \(N\ge12\) and \(N\) has the factors 2 and 3, so \(N\) has a factor other than 1 and itself✓ 1
R1 this contradicts \(N\) being prime, so the total is never prime✓ 1

Common error: In a proof by contradiction, 'not possible' must come from a statement that is false for every integer, such as 'an integer equals a fraction', not from trying a few cases.

158IBMA3H-1-19r39-031.11 Partial fractions Paper 1Medium5 marks

No calculator

Find the constant term in the expansion of \(\left(x^{2}+\dfrac{2}{x}\right)^{9}\).

(a)

Find the constant term. Show your working.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 general term \(\binom{9}{r}(x^2)^{9-r}\left(\dfrac{2}{x}\right)^{r}\)✓ 1
A1 power of \(x\) is \(18-3r\)✓ 1
M1 sets \(18-3r=0\) to obtain \(r=6\)✓ 1
A1 \(\binom{9}{6}=84\) and \(2^6=64\)✓ 1
A1 constant term = 5376✓ 1

Common error: Remember that \((2/x)^r\) contributes \(2^r\) to the coefficient as well as \(x^{-r}\) to the power; forgetting \(2^r\) is a frequent slip.

159IBMA3H-1-19r39-041.11 Partial fractions Paper 1Medium8 marks

No calculator

Let \(f(x)=\dfrac{7x+4}{(x+1)(x-2)}\) for \(x>2\).

(a)

Express \(f(x)\) in the form \(\dfrac{A}{x+1}+\dfrac{B}{x-2}\), where \(A\) and \(B\) are constants to be found.

(4)
(b)

Hence find the exact value of \(\displaystyle\int_3^5 f(x)\,dx\), giving your answer in the form \(\ln k\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(7x+4\equiv A(x-2)+B(x+1)\)✓ 1
M1 substitutes \(x=-1\) and \(x=2\) (or equates coefficients)✓ 1
A1 \(A=1\)✓ 1
A1 \(B=6\)✓ 1
Part (b)
M1 integrates to obtain logarithmic terms✓ 1
A1 \(\ln(x+1)+6\ln(x-2)\)✓ 1
M1 substitutes limits: \((\ln 6+6\ln 3)-(\ln 4+6\ln 1)\)✓ 1
A1 \(\ln\dfrac{2187}{2}\)✓ 1

Exam tip: After integrating, combine logarithms using \(a\ln b=\ln b^a\) and \(\ln a-\ln b=\ln\frac{a}{b}\) so the answer takes the single-logarithm form requested.

160IBMA3H-1-19r39-061.11 Partial fractions Paper 1Medium6 marks

No calculator

Let \(f(x)=\dfrac{2x^{2}-3x+7}{(x-1)(x+2)}\).

(a)

Write \(f(x)\) in the form \(A+\dfrac{B}{x-1}+\dfrac{C}{x+2}\), where \(A\), \(B\) and \(C\) are constants.

(5)
(b)

Write down the equation of the horizontal asymptote of the graph of \(y=f(x)\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 recognises the fraction is improper (equal degrees) and writes it in the given form✓ 1
A1 \(A=2\) (by division or by letting \(x\to\infty\))✓ 1
M1 correct method for \(B\) or \(C\) (e.g. cover-up or substituting \(x=1\), \(x=-2\))✓ 1
A1 \(B=2\)✓ 1
A1 \(C=-7\)✓ 1
Part (b)
A1 \(y=2\)✓ 1

Exam tip: When the numerator and denominator have the same degree, divide first so that the remaining fraction is proper before splitting it into partial fractions.

161IBMA3H-1-19r39-071.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium8 marks

No calculator

The cubic equation \(z^{3}+pz^{2}+qz-13=0\), where \(p,q\in\mathbb{R}\), has a root \(z=2+3i\).

(a)

Write down another root of the equation.

(1)
(b)

Find the third root of the equation.

(3)
(c)

Find the values of \(p\) and \(q\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2-3i\)✓ 1
Part (b)
M1 uses that the product of the three roots is \(13\) (or divides by the factor \(z^2-4z+13\))✓ 1
M1 \((2+3i)(2-3i)=13\)✓ 1
A1 third root is \(1\)✓ 1
Part (c)
M1 uses sum of roots \(=-p\)✓ 1
A1 \(p=-5\)✓ 1
M1 uses sum of products of pairs of roots \(=q\) (or expands \((z-1)(z^2-4z+13)\))✓ 1
A1 \(q=17\)✓ 1

Exam tip: Complex roots of a polynomial with real coefficients come in conjugate pairs, so a cubic with one non-real root must have exactly one real root.

162IBMA3H-1-19r39-131.9 The binomial theorem Paper 1Medium5 marks

No calculator

Consider the expansion of \((2-x)^{6}\).

(a)

Find the first four terms of the expansion of \((2-x)^6\) in ascending powers of \(x\).

(3)
(b)

Hence find the coefficient of \(x^{3}\) in the expansion of \((1+x)(2-x)^{6}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the binomial theorem with terms \(\binom{6}{r}2^{6-r}(-x)^r\)✓ 1
A1 \(64-192x\)✓ 1
A1 \(+\,240x^2-160x^3\)✓ 1
Part (b)
M1 combines the two contributing products: \(1\times(-160)\) and \(1\times240\) (ECF from (a))✓ 1
A1 80✓ 1

Common error: When a bracket such as \((1+x)\) multiplies an expansion, each term of the bracket contributes to the required power; adding only one product loses marks.

163IBMA3H-1-19r39-141.9 The binomial theorem Paper 1Medium7 marks

No calculator

In the expansion of \((1+kx)^{n}\), where \(n\) is a positive integer and \(k>0\), the coefficient of \(x\) is 15 and the coefficient of \(x^{2}\) is 90.

(a)

Find the values of \(n\) and \(k\).

(5)
(b)

Find the coefficient of \(x^{3}\) in the expansion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(nk=15\)✓ 1
M1 \(\binom{n}{2}k^2=90\)✓ 1
M1 eliminates \(k\): \(\dfrac{n(n-1)}{2}\times\dfrac{225}{n^2}=90\)✓ 1
A1 \(n=5\)✓ 1
A1 \(k=3\)✓ 1
Part (b)
M1 \(\binom{5}{3}\times3^3\) (ECF from (a))✓ 1
A1 270✓ 1

Exam tip: Write each coefficient as an equation using \(\binom{n}{r}k^r\) and eliminate one unknown by substitution.

164IBMA3H-1-18r47-131.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Medium7 marks

No calculator

Let \(z = \dfrac{11 + 2i}{2 - i}\).

(a)

Write \(z\) in the form \(a + bi\), where \(a, b \in \mathbb{Z}\).

(3)
(b)

The number \(z\) is a root of the equation \(w^2 + bw + c = 0\), where \(b\) and \(c\) are real. Find the value of \(b\) and the value of \(c\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply numerator and denominator by \(2 + i\)✓ 1
A1 \(\frac{(11+2i)(2+i)}{5} = \frac{20 + 15i}{5}\)✓ 1
A1 \(z = 4 + 3i\)✓ 1
Part (b)
R1 the coefficients are real, so the complex conjugate \(4 - 3i\) is also a root✓ 1
M1 use sum and product of roots, or expand \((w - (4 + 3i))(w - (4 - 3i))\)✓ 1
A1 \(b = -8\)✓ 1
A1 \(c = 4^2 + 3^2 = 25\)✓ 1

Common error: Real coefficients guarantee that non-real roots come in conjugate pairs; the product of a root and its conjugate is the real number \(a^2 + b^2\).

165IBMA3H-1-18r47-141.12 Complex numbers in Cartesian form Paper 1Medium7 marks

No calculator

The complex number \(z = a + bi\), where \(a\) and \(b\) are real, satisfies \((2 + i)z + 3z^* = 17 + 5i\), where \(z^*\) is the complex conjugate of \(z\).

(a)

Show that \(5a - b = 17\) and \(a - b = 5\).

(3)
(b)

Hence find \(z\).

(2)
(c)

Find the exact value of \(|z|\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute \(z = a + bi\) and \(z^* = a - bi\) and expand \((2 + i)(a + bi)\)✓ 1
A1 real part: \(2a - b + 3a = 17\), so \(5a - b = 17\)✓ 1
A1 imaginary part: \(a + 2b - 3b = 5\), so \(a - b = 5\)✓ 1
Part (b)
M1 solve simultaneously, e.g. subtract to get \(4a = 12\)✓ 1
A1 \(z = 3 - 2i\)✓ 1
Part (c)
M1 \(\sqrt{3^2 + (-2)^2}\) (ECF from (b))✓ 1
A1 \(\sqrt{13}\)✓ 1

Exam tip: When an equation involves both \(z\) and \(z^*\), write \(z = a + bi\) and equate real and imaginary parts to get two real equations.

166AAHL1-q00031.2 Arithmetic sequences and series Paper 1Hard8 marks

A sum of \(n\) consecutive odd positive integers has first term \(2k+1\), where \(k\in\mathbb{N}\) and \(n\in\mathbb{Z}^{+}\), \(n\ge2\).

(a)

Show that the sum is equal to \(n(n+2k)\).

(2)
(b)

Explain why, if the sum is 120, both \(n\) and \(n+2k\) must be even.

(2)
(c)

Hence find all the ways in which 120 can be written as a sum of at least two consecutive odd positive integers.

(4)
Show mark scheme
WorkingMarkNotes
(a)
arithmetic series with \(u_{1}=2k+1\), \(d=2\): \(\dfrac{n}{2}\big(2(2k+1)+2(n-1)\big)\)M1
\(\dfrac{n}{2}(4k+2n)\) or \(n(2k+1+n-1)\)A1
\(=n(n+2k)\)AG
(b)
\(n\) and \(n+2k\) differ by the even number \(2k\), so they are both odd or both evenR1
if both were odd their product would be odd, but 120 is even; so both are evenR1
(c)
factor pairs of 120 with both factors even and \(n\le n+2k\): \(2\times60,\ 4\times30,\ 6\times20,\ 10\times12\)M1Award for at least two correct pairs and no incorrect ones.
\(n=2,\ k=29\): \(59+61\) and \(n=4,\ k=13\): \(27+29+31+33\)A1
\(n=6,\ k=7\): \(15+17+19+21+23+25\)A1
\(n=10,\ k=1\): \(3+5+7+\cdots+21\); and these are the only four waysA1Award the final A1 only if exactly these four sums are given (no others, e.g. not \(n=1\) or \(n=12\) with \(k\lt 0\)).

Answers: (c) \(59+61\); \(27+29+31+33\); \(15+17+\cdots+25\); \(3+5+\cdots+21\)

167AAHL1-q00141.6 Deductive proof Paper 1Hard6 marks

Let \(n\in\mathbb{Z}\). You may use the fact that a perfect square that is a multiple of 5 is always a multiple of 25.

(a)

Show that \((n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}\equiv5\left(n^{2}+2\right)\).

(2)
(b)

By considering the possible final digits of \(n^{2}\), show that \(n^{2}+2\) is never a multiple of 5.

(2)
(c)

Hence prove that the sum of the squares of five consecutive integers is never a perfect square.

(2)
Show mark scheme
WorkingMarkNotes
(a)
expanding: \(n^{2}-4n+4+n^{2}-2n+1+n^{2}+n^{2}+2n+1+n^{2}+4n+4\)M1At least two non-trivial brackets expanded correctly.
\(5n^{2}+10\)A1
\(\equiv5\left(n^{2}+2\right)\)AG
(b)
the final digit of \(n^{2}\) is one of 0, 1, 4, 5, 6, 9R1
so the final digit of \(n^{2}+2\) is one of 2, 3, 6, 7, 8, 1, never 0 or 5; hence \(n^{2}+2\) is not a multiple of 5R1
(c)
the sum \(5\left(n^{2}+2\right)\) is a multiple of 5, so if it were a perfect square it would be a multiple of 25R1
this would need \(n^{2}+2\) to be a multiple of 5, which contradicts (b); so the sum is never a perfect squareR1

Answers: (a)–(c) proofs

168AAHL1-q00151.7 Rational exponents and laws of logarithms Paper 1Hard6 marks

Let \(\log_{2}3=p\) and \(\log_{3}5=q\).

(a)

Show that \(\log_{2}5=pq\).

(2)
(b)

Hence express \(\log_{12}45\) in terms of \(p\) and \(q\).

(4)
Show mark scheme
WorkingMarkNotes
(a)
\(\log_{3}5=\dfrac{\log_{2}5}{\log_{2}3}\)M1Change of base to base 2.
\(q=\dfrac{\log_{2}5}{p}\)A1
\(\log_{2}5=pq\)AG
(b)
\(\log_{12}45=\dfrac{\log_{2}45}{\log_{2}12}\)M1
\(\log_{2}45=\log_{2}9+\log_{2}5=2p+pq\)A1
\(\log_{2}12=\log_{2}4+\log_{2}3=2+p\)A1
\(\dfrac{2p+pq}{2+p}\)A1Accept \(\dfrac{p(2+q)}{p+2}\).

Answers: (b) \(\log_{12}45=\dfrac{p(2+q)}{p+2}\)

169AAHL1-q00191.8 Infinite geometric series Paper 1Hard13 marks

A sequence of shapes is formed as follows. Stage 0 is an equilateral triangle with sides of length 9 cm. To form each new stage, every side of the previous stage is divided into three equal parts, and an equilateral triangle is drawn outwards on the middle part, which is then removed. The first three stages are shown.

Stage 0Stage 1Stage 2
Diagram not to scale

The area of an equilateral triangle with side \(s\) is \(\dfrac{\sqrt3}{4}s^{2}\).

(a)(i)

For stage \(n\):

write down an expression for the number of sides;

(1)
(a)(ii)

show that the perimeter is \(P_{n}=27\left(\dfrac{4}{3}\right)^{n}\) cm.

(2)
(b)

Find the first stage at which the perimeter is greater than 100 cm.

(3)
(c)

Show that the areas added at stages \(1, 2, 3, \ldots\) form a geometric sequence with first term \(\dfrac{27\sqrt3}{4}\) cm2 and common ratio \(\dfrac{4}{9}\).

(3)
(d)

Find the exact area that the shapes approach as \(n\to\infty\).

(3)
(e)

Comment on the perimeter and the area of the shape as \(n\to\infty\).

(1)
Show mark scheme
WorkingMarkNotes
(a)(i)
\(3\times4^{n}\)A1
(a)(ii)
each side has length \(\dfrac{9}{3^{n}}\)M1
\(P_{n}=3\times4^{n}\times\dfrac{9}{3^{n}}=27\times\dfrac{4^{n}}{3^{n}}\)A1
\(P_{n}=27\left(\dfrac{4}{3}\right)^{n}\)AG
(b)
\(27\left(\dfrac{4}{3}\right)^{n}\gt100\)(M1)
\(P_{4}=\dfrac{256}{3}\approx85.3\) and \(P_{5}=\dfrac{1024}{9}\approx113.8\)(A1)
stage 5A1
(c)
at stage \(n\), one triangle is added on each of the \(3\times4^{n-1}\) sides of stage \(n-1\), each of side \(\dfrac{9}{3^{n}}\)M1
area added \(=3\times4^{n-1}\times\dfrac{\sqrt3}{4}\times\dfrac{81}{9^{n}}\); stage 1: \(3\times\dfrac{\sqrt3}{4}\times9=\dfrac{27\sqrt3}{4}\)A1
ratio of consecutive added areas \(=\dfrac{4}{9}\) (4 times as many triangles, each \(\tfrac19\) of the area)A1
geometric with \(u_{1}=\dfrac{27\sqrt3}{4}\), \(r=\dfrac{4}{9}\)AG
(d)
\(S_{\infty}=\dfrac{\frac{27\sqrt3}{4}}{1-\frac{4}{9}}=\dfrac{243\sqrt3}{20}\)(M1)
adding the area of stage 0: \(\dfrac{81\sqrt3}{4}+\dfrac{243\sqrt3}{20}\)(M1)
\(\dfrac{162\sqrt3}{5}\) cm2A1Accept \(\dfrac{648\sqrt3}{20}\).
(e)
the perimeter increases without bound (\(\tfrac43\gt1\)) but the area approaches a finite limitR1

Answers: (a)(i) \(3\times4^{n}\) · (b) stage 5 · (d) \(\dfrac{162\sqrt3}{5}\) cm2

170AAHL1-q00211.9 The binomial theorem Paper 1Hard7 marks

Let \(n,r\in\mathbb{Z}^{+}\), with \(1\le r\le n\).

(a)

Show that \(\dfrac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\dfrac{n-r+1}{r}\).

(2)
(b)

In the expansion of \((1+x)^{n}\), the coefficients of \(x^{r-1}\), \(x^{r}\) and \(x^{r+1}\) are in the ratio \(1:3:6\). Find the value of \(n\) and the value of \(r\).

(5)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{n!}{r!\,(n-r)!}\times\dfrac{(r-1)!\,(n-r+1)!}{n!}\)M1
\(\dfrac{(r-1)!}{r!}=\dfrac{1}{r}\) and \(\dfrac{(n-r+1)!}{(n-r)!}=n-r+1\)A1
\(\dfrac{n-r+1}{r}\)AG
(b)
\(\dfrac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=3\Rightarrow\dfrac{n-r+1}{r}=3\)M1
\(n=4r-1\)A1
\(\dfrac{{}^{n}C_{r+1}}{{}^{n}C_{r}}=2\Rightarrow\dfrac{n-r}{r+1}=2\)M1Using (a) with \(r\) replaced by \(r+1\).
\(n=3r+2\)A1
\(r=3\), \(n=11\)A1Check: \(55:165:330=1:3:6\).

Answers: (b) \(n=11\), \(r=3\)

171AAHL1-q00301.11 Partial fractions Paper 1Hard14 marks

Let \(f(x)=\dfrac{3}{(1-2x)(1+x)}\).

(a)

Express \(f(x)\) in partial fractions.

(3)
(b)

Hence find the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^{3}\).

(4)
(c)

Show that the coefficient of \(x^{n}\) in the expansion of \(f(x)\) is \(2^{n+1}+(-1)^{n}\).

(2)
(d)

State the set of values of \(x\) for which the expansion of \(f(x)\) is valid, justifying your answer.

(2)
(e)

Find the value of \(n\) for which the coefficient of \(x^{n}\) is 2049.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(3\equiv A(1+x)+B(1-2x)\)M1
\(A=2\) and \(B=1\)A1From \(x=\tfrac12\) and \(x=-1\).
\(f(x)=\dfrac{2}{1-2x}+\dfrac{1}{1+x}\)A1
(b)
using \((1-2x)^{-1}\) and \((1+x)^{-1}\) with the extended binomial theoremM1
\(2(1-2x)^{-1}=2+4x+8x^{2}+16x^{3}+\cdots\)A1
\((1+x)^{-1}=1-x+x^{2}-x^{3}+\cdots\)A1
\(f(x)=3+3x+9x^{2}+15x^{3}+\cdots\)A1
(c)
coefficient of \(x^{n}\) in \(2(1-2x)^{-1}\) is \(2\times2^{n}\); in \((1+x)^{-1}\) it is \((-1)^{n}\)M1
\(2^{n+1}+(-1)^{n}\)A1
\(2^{n+1}+(-1)^{n}\)AG
(d)
\((1-2x)^{-1}\) needs \(|x|\lt\tfrac12\) and \((1+x)^{-1}\) needs \(|x|\lt1\); both must holdR1
\(|x|\lt\dfrac{1}{2}\)A1
(e)
\(2^{n+1}+(-1)^{n}=2049\)M1
\(n\) odd gives \(2^{n+1}=2050\), not a power of 2, so \(n\) is evenR1
\(2^{n+1}=2048=2^{11}\), so \(n=10\)A1

Answers: (a) \(\dfrac{2}{1-2x}+\dfrac{1}{1+x}\) · (b) \(3+3x+9x^{2}+15x^{3}\) · (d) \(|x|\lt\tfrac12\) · (e) \(n=10\)

172AAHL1-q00371.12 Complex numbers in Cartesian form Paper 1Hard7 marks

Consider the equation \(z^{2}=z^{*}\), where \(z\in\mathbb{C}\).

(a)

By taking the modulus of both sides, show that \(|z|=0\) or \(|z|=1\).

(2)
(b)

Find all the solutions of the equation, giving your answers in the form \(a+bi\), where \(a,b\in\mathbb{R}\).

(5)
Show mark scheme
WorkingMarkNotes
(a)
\(|z^{2}|=|z|^{2}\) and \(|z^{*}|=|z|\), so \(|z|^{2}=|z|\)M1
\(|z|\big(|z|-1\big)=0\)A1
\(|z|=0\) or \(|z|=1\)AG
(b)
substituting \(z=a+bi\): \(a^{2}-b^{2}+2abi=a-bi\)M1
\(a^{2}-b^{2}=a\) and \(2ab=-b\)A1
\(b(2a+1)=0\Rightarrow b=0\) or \(a=-\tfrac12\)M1Considering both cases.
\(b=0\): \(a^{2}=a\Rightarrow z=0\) or \(z=1\)A1
\(a=-\tfrac12\): \(b^{2}=\tfrac34\Rightarrow z=-\dfrac{1}{2}\pm\dfrac{\sqrt3}{2}i\)A1Both required.

Answers: (b) \(0,\ 1,\ -\tfrac12+\tfrac{\sqrt3}{2}i,\ -\tfrac12-\tfrac{\sqrt3}{2}i\)

173AAHL1-q00431.13 Polar and Euler forms Paper 1Hard8 marks

On an Argand diagram, the points P and Q represent the complex numbers \(1+2i\) and \(5\) respectively. The point R is such that triangle PQR is equilateral, and P, Q, R are in anticlockwise order, as shown in the following diagram.

ReImOPQR
Diagram not to scale
(a)

Write down the complex number represented by \(\overrightarrow{\text{PQ}}\).

(1)
(b)

Explain why the complex number represented by \(\overrightarrow{\text{PR}}\) is \((4-2i)\,\mathrm{e}^{\frac{\pi}{3}i}\).

(1)
(c)

Find the complex number represented by R, giving your answer in the form \(a+bi\), where \(a,b\) are exact.

(4)
(d)

Find the exact area of triangle PQR.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(5-(1+2i)=4-2i\)A1
(b)
\(\overrightarrow{\text{PR}}\) has the same length as \(\overrightarrow{\text{PQ}}\) and is obtained by rotating it anticlockwise through \(\frac{\pi}{3}\); multiplying by \(\mathrm{e}^{\frac{\pi}{3}i}\) (modulus 1, argument \(\frac{\pi}{3}\)) does exactly thisR1
(c)
\(\mathrm{e}^{\frac{\pi}{3}i}=\dfrac12+\dfrac{\sqrt3}{2}i\)A1
\((4-2i)\left(\dfrac12+\dfrac{\sqrt3}{2}i\right)=2+2\sqrt3\,i-i+\sqrt3\)M1
adding \(1+2i\) to \(\overrightarrow{\text{PR}}\)M1
\((3+\sqrt3)+(1+2\sqrt3)i\)A1
(d)
\(\dfrac12|4-2i|^{2}\sin\dfrac{\pi}{3}=\dfrac12(20)\dfrac{\sqrt3}{2}\)M1
\(5\sqrt3\)A1

Answers: (a) \(4-2i\) · (c) \((3+\sqrt3)+(1+2\sqrt3)i\) · (d) \(5\sqrt3\)

174AAHL1-q00491.13 Polar and Euler forms Paper 1Hard15 marks

Let \(w=1+i\). For \(n\in\mathbb{N}\), the point \(\text{P}_{n}\) on an Argand diagram represents the complex number \(w^{n}\). The following diagram shows the points \(\text{P}_{0}\) to \(\text{P}_{5}\), joined in order; triangle \(\text{OP}_{3}\text{P}_{4}\) is shaded.

ReImO−4−22−4−22P₀P₁P₂P₃P₄P₅
(a)

Write \(w\) in the form \(r\mathrm{e}^{i\theta}\).

(2)
(b)

Show that \(w^{n+1}-w^{n}=i\,w^{n}\).

(2)
(c)

Hence explain why, for every \(n\), triangle \(\text{OP}_{n}\text{P}_{n+1}\) is right-angled and isosceles.

(3)
(d)

Show that the area of triangle \(\text{OP}_{n}\text{P}_{n+1}\) is \(2^{n-1}\).

(2)
(e)

Find the least value of \(N\) for which the total area of the triangles \(\text{OP}_{0}\text{P}_{1},\ \text{OP}_{1}\text{P}_{2},\ \ldots,\ \text{OP}_{N-1}\text{P}_{N}\) exceeds 1000.

(4)
(f)

Find the least positive integer \(n\) for which \(\text{P}_{n}\) lies on the positive real axis, and write down the distance \(\text{OP}_{n}\) for this \(n\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(r=\sqrt2\)A1
\(\sqrt2\,\mathrm{e}^{\frac{\pi}{4}i}\)A1
(b)
\(w^{n+1}-w^{n}=w^{n}(w-1)\)M1
\(w-1=i\)A1
\(w^{n+1}-w^{n}=i\,w^{n}\)AG
(c)
\(\overrightarrow{\text{P}_{n}\text{P}_{n+1}}\) represents \(iw^{n}\) and \(|iw^{n}|=|w^{n}|\), so \(\text{P}_{n}\text{P}_{n+1}=\text{OP}_{n}\)R1
multiplying by \(i\) is a rotation through \(\frac{\pi}{2}\), so \(\text{P}_{n}\text{P}_{n+1}\) is perpendicular to \(\text{OP}_{n}\)R1
the right angle is at \(\text{P}_{n}\), and the two sides meeting there are equalA1
(d)
area \(=\dfrac12|w^{n}|^{2}=\dfrac12\left(\sqrt2\right)^{2n}\)M1
\(=\dfrac12\times2^{n}\)A1
\(=2^{n-1}\)AG
(e)
\(\displaystyle\sum_{n=0}^{N-1}2^{n-1}=\dfrac12\cdot\dfrac{2^{N}-1}{2-1}\)M1Recognizing a geometric series.
\(\dfrac{2^{N}-1}{2}\)A1
\(\dfrac{2^{N}-1}{2}\gt 1000\Rightarrow 2^{N}\gt 2001\)M1
\(N=11\)A1\(2^{10}=1024,\ 2^{11}=2048\).
(f)
\(\arg w^{n}=\dfrac{n\pi}{4}=2k\pi\Rightarrow n=8\)A1
\(\text{OP}_{8}=\left(\sqrt2\right)^{8}=16\)A1

Answers: (a) \(\sqrt2\,\mathrm{e}^{\pi i/4}\) · (e) \(N=11\) · (f) \(n=8\), distance 16

175AAHL1-q00561.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard8 marks

Let \(\omega=\mathrm{e}^{\frac{2\pi}{5}i}\).

(a)

Show that \(1+\omega+\omega^{2}+\omega^{3}+\omega^{4}=0\).

(2)
(b)

Hence show that \(\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}=-\dfrac12\).

(3)
(c)

Hence find the exact value of \(\cos\dfrac{2\pi}{5}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(1+\omega+\omega^{2}+\omega^{3}+\omega^{4}=\dfrac{\omega^{5}-1}{\omega-1}\)M1Geometric series, or factorizing \(\omega^{5}-1\).
\(\omega^{5}=\mathrm{e}^{2\pi i}=1\) and \(\omega\neq1\)R1
sum \(=0\)AG
(b)
taking real parts: \(1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}=0\)M1
\(\cos\frac{8\pi}{5}=\cos\frac{2\pi}{5}\) and \(\cos\frac{6\pi}{5}=\cos\frac{4\pi}{5}\)A1
\(1+2\left(\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}\right)=0\)A1
\(\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}=-\dfrac12\)AG
(c)
with \(c=\cos\frac{2\pi}{5}\): \(\cos\frac{4\pi}{5}=2c^{2}-1\), so \(c+2c^{2}-1=-\frac12\)M1
\(4c^{2}+2c-1=0\)A1
\(c=\dfrac{-1+\sqrt5}{4}\), the positive root since \(0\lt\frac{2\pi}{5}\lt\frac{\pi}{2}\)A1Reason for rejecting the negative root required.

Answers: (c) \(\dfrac{\sqrt5-1}{4}\)

176AAHL1-q00571.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard6 marks

Let \(w=\dfrac{1+\sqrt3\,i}{1-i}\).

(a)

Express \(w\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(3)
(b)

Find the least positive integer \(n\) for which \(w^{n}\) is real, and find the value of \(w^{n}\) for this \(n\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(1+\sqrt3\,i=2\mathrm{e}^{\frac{\pi}{3}i}\)A1
\(1-i=\sqrt2\,\mathrm{e}^{-\frac{\pi}{4}i}\)A1
\(w=\sqrt2\,\mathrm{e}^{\frac{7\pi}{12}i}\)A1
(b)
\(w^{n}=2^{\frac n2}\mathrm{e}^{\frac{7n\pi}{12}i}\) is real \(\iff\dfrac{7n}{12}\in\mathbb{Z}\); since 7 and 12 have no common factor, \(n\) must be a multiple of 12R1
\(n=12\)A1
\(w^{12}=2^{6}\mathrm{e}^{7\pi i}=-64\)A1

Answers: (a) \(\sqrt2\,\mathrm{e}^{7\pi i/12}\) · (b) \(n=12,\ w^{12}=-64\)

177AAHL1-q00601.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard14 marks

Let \(c=\cos\theta\) and \(s=\sin\theta\).

(a)

Use De Moivre’s theorem to show that \(\cos4\theta=8c^{4}-8c^{2}+1\).

(4)
(b)

Hence show that the solutions of the equation \(8x^{4}-8x^{2}+1=0\) are \(x=\cos\dfrac{k\pi}{8}\) for \(k=1,3,5,7\).

(4)
(c)

By solving \(8x^{4}-8x^{2}+1=0\) as a quadratic equation in \(x^{2}\), find the exact value of \(\cos^{2}\dfrac{\pi}{8}\).

(3)
(d)

By considering the product of the roots of \(8x^{4}-8x^{2}+1=0\), find the exact value of \(\cos\dfrac{\pi}{8}\cos\dfrac{3\pi}{8}\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(\cos4\theta+i\sin4\theta=(c+is)^{4}\)M1
expanding and taking the real part: \(\cos4\theta=c^{4}-6c^{2}s^{2}+s^{4}\)M1
\(=c^{4}-6c^{2}(1-c^{2})+(1-c^{2})^{2}\)A1
\(=c^{4}-6c^{2}+6c^{4}+1-2c^{2}+c^{4}\)A1
\(=8c^{4}-8c^{2}+1\)AG
(b)
letting \(x=\cos\theta\): the equation becomes \(\cos4\theta=0\)M1
\(4\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},\dfrac{7\pi}{2}\Rightarrow\theta=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{5\pi}{8},\dfrac{7\pi}{8}\)A1
\(\cos\) is one-to-one on \([0,\pi]\), so these give four distinct values of \(x\)R1
a quartic has at most four roots, so these are all the solutionsR1
\(x=\cos\dfrac{k\pi}{8},\ k=1,3,5,7\)AG
(c)
\(x^{2}=\dfrac{8\pm\sqrt{64-32}}{16}\)M1
\(x^{2}=\dfrac{2\pm\sqrt2}{4}\)A1
\(\cos^{2}\dfrac{\pi}{8}=\dfrac{2+\sqrt2}{4}\), the larger value since \(\cos\frac{\pi}{8}\gt\cos\frac{3\pi}{8}\gt 0\)A1Reason required for the choice.
(d)
product of roots \(=\dfrac{1}{8}\)M1
\(\cos\frac{5\pi}{8}=-\cos\frac{3\pi}{8}\) and \(\cos\frac{7\pi}{8}=-\cos\frac{\pi}{8}\), so product \(=\left(\cos\frac{\pi}{8}\cos\frac{3\pi}{8}\right)^{2}\)A1
\(\cos\dfrac{\pi}{8}\cos\dfrac{3\pi}{8}=\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}\)A1Must reject the negative square root (both factors positive).

Answers: (c) \(\dfrac{2+\sqrt2}{4}\) · (d) \(\dfrac{\sqrt2}{4}\)

178AAHL1-q00641.15 Proof by induction, contradiction and counterexample Paper 1Hard10 marks

Let \(x\in\mathbb{R}\) with \(\sin x\neq0\).

(a)

Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n}\sin\big((2r-1)x\big)=\dfrac{\sin^{2}(nx)}{\sin x}\), for \(n\in\mathbb{Z}^{+}\).

(7)
(b)

Hence solve the equation \(\sin x+\sin3x+\sin5x+\sin7x=0\), for \(0\lt x\lt\pi\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(n=1\): LHS \(=\sin x\), RHS \(=\dfrac{\sin^{2}x}{\sin x}=\sin x\), so true for \(n=1\)A1
assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}\sin\big((2r-1)x\big)=\dfrac{\sin^{2}(kx)}{\sin x}\)M1
consider \(n=k+1\): \(\dfrac{\sin^{2}(kx)}{\sin x}+\sin\big((2k+1)x\big)=\dfrac{\sin^{2}(kx)+\sin x\sin\big((2k+1)x\big)}{\sin x}\)M1
\(\sin x\sin\big((2k+1)x\big)=\tfrac{1}{2}\big(\cos(2kx)-\cos\big((2k+2)x\big)\big)\)M1From \(\cos(A-B)-\cos(A+B)=2\sin A\sin B\) with \(A=(2k+1)x,\ B=x\); or by expanding \(\sin(2kx+x)\).
\(=\tfrac{1}{2}\big(1-2\sin^{2}(kx)\big)-\tfrac{1}{2}\big(1-2\sin^{2}\big((k+1)x\big)\big)=\sin^{2}\big((k+1)x\big)-\sin^{2}(kx)\)A1Use of the double angle identity for cosine.
so the sum \(=\dfrac{\sin^{2}\big((k+1)x\big)}{\sin x}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(b)
\(n=4\): \(\dfrac{\sin^{2}(4x)}{\sin x}=0\Rightarrow\sin4x=0\)M1
\(4x=\pi,\,2\pi,\,3\pi\) (as \(0\lt4x\lt4\pi\))(M1)
\(x=\dfrac{\pi}{4},\ \dfrac{\pi}{2},\ \dfrac{3\pi}{4}\)A1All three values and no others.

Answers: (b) \(x=\tfrac{\pi}{4},\ \tfrac{\pi}{2},\ \tfrac{3\pi}{4}\)

179AAHL1-q00721.15 Proof by induction, contradiction and counterexample Paper 1Hard10 marks
(a)

Prove by contradiction that \(\sqrt{6}\) is irrational.

(5)
(b)

Hence prove by contradiction that \(\sqrt{2}+\sqrt{3}\) is irrational.

(3)
(c)

By considering \(\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\), deduce that \(\sqrt{3}-\sqrt{2}\) is irrational.

(2)
Show mark scheme
WorkingMarkNotes
(a)
assume \(\sqrt{6}=\dfrac{p}{q}\), \(p,q\in\mathbb{Z}^{+}\) with no common factorM1
\(p^{2}=6q^{2}\), so \(p^{2}\) is even and hence \(p\) is evenA1
\(p=2m\Rightarrow4m^{2}=6q^{2}\Rightarrow2m^{2}=3q^{2}\)A1
\(3q^{2}\) is even, so \(q^{2}\), and hence \(q\), is evenR1
\(p\) and \(q\) have common factor 2, a contradiction; so \(\sqrt{6}\) is irrationalR1
(b)
assume \(\sqrt{2}+\sqrt{3}=r\in\mathbb{Q}\); squaring: \(5+2\sqrt{6}=r^{2}\)M1
\(\sqrt{6}=\dfrac{r^{2}-5}{2}\)A1
\(\dfrac{r^{2}-5}{2}\) is rational, contradicting part (a); so \(\sqrt{2}+\sqrt{3}\) is irrationalR1
(c)
\(\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\), so \(\sqrt{3}-\sqrt{2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}\)M1
if \(\sqrt{3}-\sqrt{2}\) (which is nonzero) were rational, its reciprocal \(\sqrt{3}+\sqrt{2}\) would be rational, contradicting part (b)R1

Answers: Proofs.

180AAHL1-q00761.15 Proof by induction, contradiction and counterexample Paper 1Hard9 marks

A student conjectures that every positive integer can be written as \(a^{2}+b^{2}+c^{2}\), where \(a,b,c\) are integers.

(a)

Show that 7 cannot be written in this form, and hence explain why the conjecture is false.

(2)
(b)

Show that, for every integer \(k\), \(k^{2}\) leaves a remainder of 0, 1 or 4 when divided by 8.

(3)
(c)

Hence prove that no integer of the form \(8n+7\), \(n\in\mathbb{Z}\), can be written as the sum of three integer squares.

(2)
(d)

Show that 101 and 102 can each be written as the sum of three integer squares, and hence find the least integer greater than 100 that cannot.

(2)
Show mark scheme
WorkingMarkNotes
(a)
only squares \(\le7\) are \(0,1,4\); possible sums of three of these that are \(\le7\): \(0,1,2,3,4,5,6\)M1Or a systematic check of cases.
7 is not obtained, so 7 is a counterexample and the conjecture is falseR1
(b)
considering \(k\) even and \(k\) oddM1
\(k=2m\): \(k^{2}=4m^{2}\), which leaves remainder 0 (\(m\) even) or 4 (\(m\) odd)A1
\(k=2m+1\): \(k^{2}=4m(m+1)+1\); \(m(m+1)\) is even, so remainder 1A1
(c)
sums of three remainders from \(\{0,1,4\}\): \(0,1,2,3,4,5,6,8,9,12\)M1Remainders mod 8: \(0,1,2,3,4,5,6,0,1,4\).
none gives remainder 7, so \(8n+7\) is never a sum of three squaresR1
(d)
\(101=10^{2}+1^{2}+0^{2}\), \(102=10^{2}+1^{2}+1^{2}\)A1
\(103\ (=8\times12+7)\)A1

Answers: (d) \(103\)

181AAHL1-q00791.15 Proof by induction, contradiction and counterexample Paper 1Hard13 marks

Every odd integer can be written in the form \(4k+1\) or \(4k+3\), where \(k\in\mathbb{Z}\).

(a)

Show that the product of two integers of the form \(4k+1\) is also of the form \(4k+1\).

(2)
(b)

Hence explain why every positive integer of the form \(4k+3\) has at least one prime factor of the form \(4k+3\).

(2)
(c)(i)

Assume that there are only finitely many primes of the form \(4k+3\), and that they are \(3,\,p_{2},\,p_{3},\,\ldots,\,p_{m}\). Let \(N=4\left(3p_{2}p_{3}\cdots p_{m}\right)-1\).

Show that \(N\) is of the form \(4k+3\).

(1)
(c)(ii)

Explain why \(N\) is not divisible by any of \(3,\,p_{2},\,\ldots,\,p_{m}\).

(2)
(c)(iii)

Hence complete a proof by contradiction that there are infinitely many primes of the form \(4k+3\).

(2)
(d)(i)

A student suggests that the number \(N\) constructed from any finite list of such primes is itself prime.

Find \(N\) for the list \(3,\,7\), and show that it is prime.

(2)
(d)(ii)

Find \(N\) for the list \(3,\,7,\,11\), and hence show that the student’s suggestion is false.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\((4a+1)(4b+1)=16ab+4a+4b+1\)M1
\(=4(4ab+a+b)+1\), which is of the form \(4k+1\)A1
(b)
an integer of the form \(4k+3\) is odd, so all its prime factors are odd, ie of the form \(4k+1\) or \(4k+3\)R1
if all were of the form \(4k+1\), by (repeated use of) part (a) the product would be of the form \(4k+1\), not \(4k+3\); so at least one prime factor is of the form \(4k+3\)R1
(c)(i)
\(N=4P-1=4(P-1)+3\), where \(P=3p_{2}\cdots p_{m}\)A1
(c)(ii)
each \(p_{i}\) divides \(4P\)M1
if \(p_{i}\) also divided \(N\), it would divide \(4P-N=1\), which is impossibleR1
(c)(iii)
by (b), \(N\) has a prime factor of the form \(4k+3\), and by (c)(ii) this prime is not in the listR1
this contradicts the assumption that the list contains all such primes; so there are infinitely many primes of the form \(4k+3\)R1
(d)(i)
\(N=4\times21-1=83\)A1
83 is not divisible by 2, 3, 5 or 7 (and \(\sqrt{83}\lt10\)), so it is primeR1Must check primes up to \(\sqrt{83}\).
(d)(ii)
\(N=4\times231-1=923\)A1
\(923=13\times71\), so \(N\) is not prime; the suggestion is falseR1Must show a factorization.

Answers: (d)(i) \(N=83\) (prime) · (d)(ii) \(N=923=13\times71\)

182AAHL1-q00891.16 Systems of linear equations Paper 1Hard16 marks

Consider the system of equations, where \(a,b\in\mathbb{R}\):

\(\begin{aligned}x+y+az&=1\\x+ay+z&=1\\ax+y+z&=b\end{aligned}\)

(a)

Show that the system reduces to \(x+y+az=1\), \((a-1)(y-z)=0\), \((1-a)(a+2)z=b-a\).

(4)
(b)

For \(a\neq1\) and \(a\neq-2\), find the unique solution, giving \(x\), \(y\) and \(z\) in terms of \(a\) and \(b\).

(3)
(c)(i)

Consider the case \(a=1\).

State the condition on \(b\) for the system to be consistent, and describe the solution set in this case.

(2)
(c)(ii)

Describe the geometric arrangement of the three planes when \(b\neq1\).

(1)
(d)(i)

Consider the case \(a=-2\).

Find the value of \(b\) for which the system is consistent.

(2)
(d)(ii)

For this value of \(b\), find the general solution of the system.

(3)
(e)

Hence find the solution when \(a=3\) and \(b=5\).

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(R_{2}-R_{1}\): \((a-1)y+(1-a)z=0\), ie \((a-1)(y-z)=0\)M1
\(R_{3}-aR_{1}\): \((1-a)y+(1-a^{2})z=b-a\)A1
adding the two new equations to eliminate \(y\)M1
\(\big((1-a)+(1-a^{2})\big)z=(1-a)(2+a)z=b-a\)A1
\((1-a)(a+2)z=b-a\)AG
(b)
\(z=\dfrac{b-a}{(1-a)(a+2)}\)A1
\(y=z=\dfrac{b-a}{(1-a)(a+2)}\)A1Since \(a\neq1\), \(y-z=0\).
\(x=1-y-az=1-\dfrac{(1+a)(b-a)}{(1-a)(a+2)}\)A1Accept equivalent forms, eg \(x=\dfrac{ab+b-2}{(a-1)(a+2)}\).
(c)(i)
all three equations become \(x+y+z=1,\ x+y+z=1,\ x+y+z=b\); consistent only if \(b=1\)A1
then all three planes coincide: infinitely many solutions forming a plane, eg \(x=1-\lambda-\mu,\ y=\lambda,\ z=\mu\)A1
(c)(ii)
two coincident planes, parallel to (and distinct from) the third planeA1
(d)(i)
\(a=-2\): third reduced equation becomes \(0=b+2\)M1Or adding the three original equations: \(0=2+b\).
\(b=-2\)A1
(d)(ii)
\(y=z=\lambda\) from the second reduced equationM1
\(x=1-y+2z=1+\lambda\)A1
\(x=1+\lambda,\ y=\lambda,\ z=\lambda\)A1Accept \(\mathbf{r}=\begin{pmatrix}1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\1\end{pmatrix}\) or equivalent.
(e)
\(x=\dfrac{9}{5},\ y=-\dfrac{1}{5},\ z=-\dfrac{1}{5}\)A1FT from (b).

Answers: (b) \(y=z=\dfrac{b-a}{(1-a)(a+2)}\), \(x=1-(1+a)z\) · (c)(i) \(b=1\) · (d)(i) \(b=-2\) · (d)(ii) \(x=1+\lambda,\ y=z=\lambda\) · (e) \(\left(\tfrac{9}{5},-\tfrac{1}{5},-\tfrac{1}{5}\right)\)

183IBMA2H-045-06-111.11 Partial fractions Paper 1Hard13 marks

No calculator

In this question \(n\in\mathbb{Z}^+\) and \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).

(a)

Show that \(r\binom{n}{r}=n\binom{n-1}{r-1}\) for \(1\le r\le n\).

(3)
(b)

Hence show that \(\displaystyle\sum_{r=1}^{n}r\binom{n}{r}=n\,2^{n-1}\).

(3)
(c)

An arithmetic sequence is defined by \(u_r=a+rd\) for \(r=0,1,2,\dots\) Show that \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}u_r=2^{n-1}(2a+nd)\).

(3)
(d)

For a particular arithmetic sequence of this form, \(u_0=5\) and \(\displaystyle\sum_{r=0}^{10}\binom{10}{r}u_r=30720\). Find the common difference \(d\).

(2)
(e)

Show that, for any such sequence, \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}u_r\) equals \(2^n\) multiplied by the mean of \(u_0\) and \(u_n\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes \(r\binom{n}{r}=\frac{r\,n!}{r!(n-r)!}\) ; M1 cancels to \(\frac{n!}{(r-1)!(n-r)!}\) ; A1 writes as \(\frac{n(n-1)!}{(r-1)!\,((n-1)-(r-1))!}=n\binom{n-1}{r-1}\)✓ 3
Part (b)
M1 applies part (a) to give \(n\sum_{r=1}^{n}\binom{n-1}{r-1}\) ; M1 re-indexes with \(s=r-1\) to obtain \(n\sum_{s=0}^{n-1}\binom{n-1}{s}\) ; R1 uses \((1+1)^{n-1}=\sum_{s=0}^{n-1}\binom{n-1}{s}=2^{n-1}\)✓ 3
Part (c)
M1 splits as \(a\sum_{r=0}^{n}\binom{n}{r}+d\sum_{r=0}^{n}r\binom{n}{r}\) ; A1 \(a\,2^n+d\,n\,2^{n-1}\), using \((1+1)^n\) and part (b) ; A1 factorises to \(2^{n-1}(2a+nd)\)✓ 3
Part (d)
M1 \(2^9(10+10d)=30720\) ; A1 \(d=5\)✓ 2
Part (e)
M1 \(u_n=a+nd\), so the mean of \(u_0\) and \(u_n\) is \(\frac{2a+nd}{2}\) ; R1 \(2^{n-1}(2a+nd)=2^n\cdot\frac{2a+nd}{2}=2^n\cdot\frac{u_0+u_n}{2}\)✓ 2

Exam tip: In 'show that' parts, every line of algebra must be visible. Write the re-indexing step and name the identity \((1+1)^m=2^m\) explicitly; a bare statement of the result earns little.

184IBMA2H-045-03-111.16 Systems of linear equations Paper 1Hard14 marks

No calculator

Consider the system of equations
\(2^{x}\times4^{y}\times8^{z}=32\)
\(\dfrac{9^{x}\times3^{z}}{3^{y}}=243\)
\(e^{x}\times e^{3y}\times e^{kz}=e^{c}\)
where \(k,c\in\mathbb{R}\).

(a)

Show that the first two equations are equivalent to \(x+2y+3z=5\) and \(2x-y+z=5\).

(3)
(b)

Find the value of \(k\) for which the system does not have a unique solution.

(3)
(c)

For this value of \(k\), find the value of \(c\) for which the system has infinitely many solutions.

(2)
(d)

For these values of \(k\) and \(c\), find the general solution of the system.

(4)
(e)

Find the particular solution for which \(x+y+z=0\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes the first equation with base 2: \(2^{x+2y+3z}=2^{5}\)✓ 1
A1 equates exponents to give \(x+2y+3z=5\)✓ 1
A1 writes \(9^{x}=3^{2x}\) and \(243=3^5\), giving \(3^{2x-y+z}=3^{5}\), so \(2x-y+z=5\)✓ 1
Part (b)
M1 writes the third equation as \(x+3y+kz=c\) and attempts a determinant or elimination✓ 1
A1 determinant \(=-5k+20\), or an equivalent reduced row✓ 1
A1 \(k=4\)✓ 1
Part (c)
M1 expresses the third row as a combination: \(\frac{7}{5}R_1-\frac{1}{5}R_2\), or eliminates to obtain a condition on \(c\)✓ 1
A1 \(c=6\)✓ 1
Part (d)
M1 lets \(z=\lambda\)✓ 1
M1 solves \(x+2y=5-3\lambda\) and \(2x-y=5-\lambda\) simultaneously✓ 1
A1 \(x=3-\lambda\)✓ 1
A1 \(y=1-\lambda\), \(z=\lambda\)✓ 1
Part (e)
M1 \((3-\lambda)+(1-\lambda)+\lambda=0\), so \(\lambda=4\)✓ 1
A1 \(x=-1\), \(y=-3\), \(z=4\)✓ 1

Exam tip: Convert exponential equations to a common base and equate powers to get linear equations. After that, the usual methods for systems apply.

185IBMA2H-045-04-111.12 Complex numbers in Cartesian form Paper 1Hard13 marks

No calculator

Let \(z=a+bi\) and \(w=c+di\), where \(a, b, c, d\in\mathbb{Z}\).

(a)

Find \(zw\) in Cartesian form. Hence prove that \((a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\).

(4)
(b)

Deduce that if each of two positive integers \(m\) and \(n\) can be written as a sum of two squares of integers, then so can \(mn\).

(2)
(c)

By considering \((1+2i)(2+3i)\) and \((1+2i)(2-3i)\), write 65 as a sum of two squares in two different ways.

(4)
(d)

Using \(2=1^2+1^2\), express 130 as a sum of the squares of two positive integers.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands \((a+bi)(c+di)\) using \(i^2=-1\)✓ 1
A1 \(zw=(ac-bd)+(ad+bc)i\)✓ 1
M1 uses \(|zw|^2=|z|^2|w|^2\), i.e. \((ac-bd)^2+(ad+bc)^2=(a^2+b^2)(c^2+d^2)\) [or expands $(ac-bd)^2+(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2$]✓ 1
A1 completes the argument correctly, e.g. states \(|z|^2=a^2+b^2\) and \(|w|^2=c^2+d^2\) [or factorises to \(a^2(c^2+d^2)+b^2(c^2+d^2)=(a^2+b^2)(c^2+d^2)\)]✓ 1
Part (b)
R1 writes \(m=a^2+b^2\) and \(n=c^2+d^2\) with integers \(a, b, c, d\)✓ 1
R1 then \(mn=(ac-bd)^2+(ad+bc)^2\), where \(ac-bd\) and \(ad+bc\) are integers because integers are closed under addition, subtraction and multiplication✓ 1
Part (c)
M1 expands \((1+2i)(2+3i)\) using \(i^2=-1\)✓ 1
A1 \(-4+7i\)✓ 1
A1 \((1+2i)(2-3i)=8+i\)✓ 1
A1 \(65=5\times13\), so \(65=4^2+7^2=8^2+1^2\)✓ 1
Part (d)
M1 multiplies \(1+i\) by one of the products found in the previous part✓ 1
A1 e.g. \((1+i)(8+i)=7+9i\) [or \((1+i)(-4+7i)=-11+3i\)]✓ 1
A1 \(130=7^2+9^2\) [or \(11^2+3^2\)]✓ 1

Exam tip: \(|zw|^2=|z|^2|w|^2\) connects complex multiplication to sums of two squares. Multiplying by a conjugate instead gives a second, different representation.

186IBMA2H-045-04-121.13 Polar and Euler forms Paper 1Hard10 marks

No calculator

Let \(z=x+iy\), where \(x, y\in\mathbb{R}\). Consider the equation \(z^2=2iz^*\), where \(z^*\) is the complex conjugate of \(z\).

(a)

Find all solutions of the equation, giving your answers in the form \(x+iy\).

(6)
(b)

The three non-zero solutions are represented by points on an Argand diagram.
(i) Show that these points are the vertices of an equilateral triangle.
(ii) Find the exact area of this triangle.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes \(z=x+iy\), \(z^*=x-iy\): \(x^2-y^2+2xyi=2y+2xi\)✓ 1
A1 real parts \(x^2-y^2=2y\)✓ 1
A1 imaginary parts \(2xy=2x\)✓ 1
R1 factorises to \(x(y-1)=0\) and considers both cases rather than dividing by \(x\)✓ 1
A1 \(x=0\): \(-y^2=2y\), so \(z=0\) or \(z=-2i\)✓ 1
A1 \(y=1\): \(x^2=3\), so \(z=\sqrt3+i\) or \(z=-\sqrt3+i\)✓ 1
Part (b)
M1 attempts the distances between pairs of points, using \(|z_1-z_2|\)✓ 1
A1 \(|(\sqrt3+i)-(-\sqrt3+i)|=2\sqrt3\)✓ 1
A1 \(|(\sqrt3+i)-(-2i)|=|\sqrt3+3i|=\sqrt{12}=2\sqrt3\), and by symmetry \(|(-\sqrt3+i)-(-2i)|=2\sqrt3\), so all three sides are equal✓ 1
A1 area \(=\frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3\)✓ 1

Common error: Dividing \(2xy=2x\) by \(x\) loses the solutions with \(x=0\), including \(z=-2i\). Factorise instead.

187IBMA2H-045-00-071.15 Proof by induction, contradiction and counterexample Paper 1Hard15 marks

No calculator

Let f(x) = xe2x, x ∈ ℝ. The nth derivative of f is written f(n)(x).

(a)

Find f ′(x) and show that f ″(x) = 2(2x + 2)e2x.

(3)
(b)

Prove by mathematical induction that f(n)(x) = 2n−1(2x + n)e2x for all n ∈ ℤ+.

(7)
(c)

Find the least value of n for which f(n)(0) > 1000.

(3)
(d)

For each n, the graph of y = f(n)(x) meets the x-axis at exactly one point. Find the value of n for which this point is (−12, 0).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the product rule✓ 1
A1 f ′(x) = e2x + 2xe2x = (2x + 1)e2x✓ 1
A1 f ″(x) = 2e2x + 2(2x + 1)e2x = (4x + 4)e2x = 2(2x + 2)e2x✓ 1
Part (b)
A1 n = 1: 20(2x + 1)e2x = f ′(x), so true for n = 1✓ 1
M1 assumes true for n = k: f(k)(x) = 2k−1(2x + k)e2x✓ 1
M1 differentiates the assumed expression to obtain f(k+1)(x)✓ 1
A1 = 2k−1[2e2x + 2(2x + k)e2x]✓ 1
A1 = 2k(2x + k + 1)e2x✓ 1
R1 this is the formula with n = k + 1, i.e. 2(k+1)−1(2x + (k + 1))e2x✓ 1
R1 true for n = 1 and true for n = k implies true for n = k + 1, so true for all n ∈ ℤ+✓ 1
Part (c)
M1 f(n)(0) = n × 2n−1✓ 1
M1 evaluates consecutive values, e.g. n = 7 gives 448, n = 8 gives 1024✓ 1
A1 n = 8✓ 1
Part (d)
M1 since 2n−1 ≠ 0 and e2x > 0, 2x + n = 0 so x = −n/2✓ 1
A1 n = 24✓ 1

Exam tip: In the inductive step, factor out 2ke²x explicitly and then write the target expression for n = k + 1 beside it so the match is obvious.

188IBMA2H-045-02-111.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard11 marks

No calculator

A phased-array radar uses \(n\) identical antennas. The combined signal is modelled by \(S=\sum_{k=0}^{n-1}e^{ik\theta}\), where \(\theta\) is the phase difference between adjacent antennas and \(\theta\) is not a multiple of \(2\pi\).

(a)

Show that \(e^{i\theta}-1=2i\sin\left(\frac{\theta}{2}\right)e^{\frac{i\theta}{2}}\).

(3)
(b)

Hence show that \(S=e^{\frac{i(n-1)\theta}{2}}\,\frac{\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\).

(4)
(c)

Find the exact value of \(|S|\) when \(n=4\) and \(\theta=\frac{\pi}{3}\).

(2)
(d)

For an array of \(n=6\) antennas, find the smallest positive value of \(\theta\) for which the combined signal is zero.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 factorises out \(e^{\frac{i\theta}{2}}\): \(e^{\frac{i\theta}{2}}\left(e^{\frac{i\theta}{2}}-e^{-\frac{i\theta}{2}}\right)\)✓ 1
M1 uses \(e^{ix}-e^{-ix}=2i\sin x\)✓ 1
A1 obtains \(2i\sin\left(\frac{\theta}{2}\right)e^{\frac{i\theta}{2}}\) (AG)✓ 1
Part (b)
M1 sums the geometric series: \(S=\frac{e^{in\theta}-1}{e^{i\theta}-1}\)✓ 1
M1 applies the result of part (a) to both numerator and denominator✓ 1
A1 \(\frac{2i\sin\left(\frac{n\theta}{2}\right)e^{\frac{in\theta}{2}}}{2i\sin\left(\frac{\theta}{2}\right)e^{\frac{i\theta}{2}}}\)✓ 1
A1 simplifies the exponentials to \(e^{\frac{i(n-1)\theta}{2}}\) to obtain the result (AG)✓ 1
Part (c)
M1 \(|S|=\left|\frac{\sin\frac{2\pi}{3}}{\sin\frac{\pi}{6}}\right|\), since \(\left|e^{i\phi}\right|=1\)✓ 1
A1 \(\sqrt3\)✓ 1
Part (d)
M1 requires \(\sin3\theta=0\) with \(\sin\frac{\theta}{2}\neq0\)✓ 1
A1 \(\theta=\frac{\pi}{3}\)✓ 1

Exam tip: Factorising out the 'half-angle' exponential turns a difference of exponentials into a sine. This trick appears whenever a geometric series of e^(ikθ) has to be simplified.

189IBMA2H-045-01-111.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard11 marks

No calculator

In digital signal processing, the fifth roots of unity are used to split a signal into frequency components. Let \(\omega = \text{cis}\,\frac{2\pi}{5}\).

(a)

Show that \(1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0\).

(2)
(b)

Hence show that \(\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}\).

(3)
(c)

Let \(\alpha = \omega + \omega^4\) and \(\beta = \omega^2 + \omega^3\). Show that \(\alpha + \beta = -1\) and \(\alpha\beta = -1\).

(3)
(d)

Hence find the exact value of \(\cos\frac{2\pi}{5}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\omega^5 = 1\), and the sum of the geometric series is \(\frac{\omega^5 - 1}{\omega - 1}\)✓ 1
R1 \(\omega \ne 1\), so the denominator is non-zero and the sum equals 0✓ 1
Part (b)
M1 Take real parts, using de Moivre: \(1 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} + \cos\frac{6\pi}{5} + \cos\frac{8\pi}{5} = 0\)✓ 1
M1 Use \(\cos(2\pi - \theta) = \cos\theta\) to get \(\cos\frac{8\pi}{5} = \cos\frac{2\pi}{5}\) and \(\cos\frac{6\pi}{5} = \cos\frac{4\pi}{5}\)✓ 1
A1 \(1 + 2\left(\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5}\right) = 0\), which gives the result✓ 1
Part (c)
A1 \(\alpha + \beta = \omega + \omega^2 + \omega^3 + \omega^4 = -1\), using \(1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0\)✓ 1
M1 Expand: \(\alpha\beta = \omega^3 + \omega^4 + \omega^6 + \omega^7\)✓ 1
A1 \(\omega^5 = 1\) gives \(\omega^6 = \omega\) and \(\omega^7 = \omega^2\), so \(\alpha\beta = \omega + \omega^2 + \omega^3 + \omega^4 = -1\)✓ 1
Part (d)
M1 \(\alpha\) and \(\beta\) are the roots of \(t^2 + t - 1 = 0\), so \(t = \frac{-1 \pm \sqrt{5}}{2}\)✓ 1
R1 \(\alpha = \omega + \bar{\omega} = 2\cos\frac{2\pi}{5} > 0\), so \(\alpha\) is the positive root (equivalently, \(\beta = 2\cos\frac{4\pi}{5} < 0\))✓ 1
A1 \(\cos\frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4}\)✓ 1

Exam tip: \(\omega^4 = \omega^{-1}\) is the conjugate of \(\omega\), so \(\omega + \omega^4 = 2\cos\frac{2\pi}{5}\) is real. You need this fact to justify choosing the positive root.

190IBMA2H-045-01-121.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard12 marks

No calculator

This question links De Moivre's theorem with the binomial theorem.

(a)

Write \(1 + i\) in the form \(r\,\text{cis}\,\theta\).

(2)
(b)

Hence find \((1 + i)^{10}\) in the form \(a + bi\).

(2)
(c)

By expanding \((1 + i)^{10}\) with the binomial theorem, show that \(\binom{10}{1} - \binom{10}{3} + \binom{10}{5} - \binom{10}{7} + \binom{10}{9} = 32\). Write down the value of \(\binom{10}{0} - \binom{10}{2} + \binom{10}{4} - \binom{10}{6} + \binom{10}{8} - \binom{10}{10}\).

(4)
(d)

Find the least positive integer \(n\) for which \((1 + i)^n\) is a real number greater than 1000, and state the value of \((1 + i)^n\) for this \(n\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = \sqrt{2}\)✓ 1
A1 \(\theta = \frac{\pi}{4}\)✓ 1
Part (b)
M1 \((\sqrt{2})^{10}\,\text{cis}\,\frac{10\pi}{4}\)✓ 1
A1 \(32\,\text{cis}\,\frac{\pi}{2} = 32i\)✓ 1
Part (c)
M1 \((1 + i)^{10} = \sum_{k=0}^{10}\binom{10}{k}i^k\)✓ 1
M1 even \(k\) give real terms with alternating signs (\(i^{2} = -1\), \(i^4 = 1\), ...); odd \(k\) give imaginary terms with signs \(+, -, +, -, +\)✓ 1
R1 equating imaginary parts with \(32i\) gives the required sum \(= 32\)✓ 1
A1 equating real parts gives a sum of 0✓ 1
Part (d)
M1 positive real requires \(\frac{n\pi}{4} = 2k\pi\), so \(n\) is a multiple of 8✓ 1
M1 modulus condition \((\sqrt{2})^n = 2^{\frac{n}{2}} > 1000\)✓ 1
A1 \(n = 8\) gives 16 and \(n = 16\) gives 256, so \(n = 24\)✓ 1
A1 \((1 + i)^{24} = 2^{12} = 4096\)✓ 1

Common error: \(n = 20\) gives \((1 + i)^{20} = -1024\). This is real, but it is not greater than 1000. You need both the argument condition and the modulus condition.

191IBMA2H-045-05-081.10 Counting principles and the extended binomial theorem Paper 1Hard14 marks

No calculator

Four astronauts and four engineers attend a mission briefing and sit at a circular table with eight identical seats. Arrangements that differ only by a rotation are considered the same. One of the astronauts is the commander, A, and one of the engineers is the flight director, E.

(a)

Find the number of different seating arrangements.

(2)
(b)

Find the number of arrangements in which astronauts and engineers sit in alternate seats.

(3)
(c)

Find the number of arrangements in which A and E do not sit next to each other.

(3)
(d)

Find the number of arrangements in which astronauts and engineers alternate and A and E do not sit next to each other.

(3)
(e)

Three of the eight people are chosen to form a crew. Find the number of crews that contain at least one astronaut and at least one engineer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses (n − 1)! for a circular arrangement✓ 1
A1 7! = 5040✓ 1
Part (b)
M1 fixes one astronaut and arranges the other astronauts in 3! ways✓ 1
M1 multiplies by 4! for the engineers in the remaining seats✓ 1
A1 3! × 4! = 144✓ 1
Part (c)
M1 treats A and E as a block: 7 units around a circle, 6! × 2✓ 1
A1 1440 arrangements with A and E together✓ 1
A1 5040 − 1440 = 3600✓ 1
Part (d)
R1 with A fixed, both neighbours of A are engineer seats, so E must take one of the 2 non-adjacent engineer seats✓ 1
M1 2 × 3! × 3!✓ 1
A1 72✓ 1
Part (e)
M1 total \(\binom{8}{3}=56\)✓ 1
M1 subtracts all-astronaut and all-engineer crews: \(\binom{4}{3}+\binom{4}{3}=8\)✓ 1
A1 48✓ 1

Exam tip: In circular problems, fix one person's seat first to remove rotations; every other choice is then made relative to that person.

192IBMA2H-045-05-111.11 Partial fractions Paper 1Hard14 marks

No calculator

Let \(g(x)=(1-4x)^{-\frac{1}{2}}\).

(a)

Find the expansion of \(g(x)\) in ascending powers of x, up to and including the term in \(x^3\).

(3)
(b)

Show that the coefficient of \(x^n\) in the expansion of \(g(x)\) is \(\frac{(2n)!}{(n!)^2}\).

(4)
(c)

A warehouse robot moves along a square grid from its charging point to a shelf located 5 units east and 5 units north. Each move is one unit either east or north. Show that the number of possible routes equals the coefficient of \(x^5\) in the expansion of \(g(x)\), and find this number.

(2)
(d)

The grid point 2 units east and 2 units north of the charging point is blocked. Find the number of routes that avoid this point.

(3)
(e)

Hence find the exact value of \(\displaystyle\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2\,8^n}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 applies the extended binomial theorem with \(n=-\frac{1}{2}\) and \(-4x\)✓ 1
A1 two correct non-constant terms✓ 1
A1 \(1+2x+6x^2+20x^3\)✓ 1
Part (b)
M1 general term: \(\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(-\frac{2n-1}{2})}{n!}(-4)^n\)✓ 1
M1 signs cancel and \(\frac{4^n}{2^n}=2^n\), giving \(\frac{2^n\times1\times3\times5\times\cdots\times(2n-1)}{n!}\)✓ 1
M1 multiplies numerator and denominator by \(2\times4\times\cdots\times2n=2^n n!\)✓ 1
A1 obtains \(\frac{2^n(2n)!}{2^n(n!)^2}\) and cancels \(2^n\) to give \(\frac{(2n)!}{(n!)^2}\) (AG)✓ 1
Part (c)
R1 each route is an arrangement of 5 E moves and 5 N moves, so there are \(\binom{10}{5}=\frac{10!}{(5!)^2}\) routes, which is the coefficient with \(n=5\)✓ 1
A1 252✓ 1
Part (d)
M1 routes through the blocked point: \(\binom{4}{2}\times\binom{6}{3}\)✓ 1
A1 6 × 20 = 120✓ 1
A1 252 − 120 = 132✓ 1
Part (e)
M1 recognises the series as \(g\left(\frac{1}{8}\right)=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}}\), valid since \(\left|\frac{1}{8}\right|<\frac{1}{4}\)✓ 1
A1 \(\sqrt{2}\)✓ 1

Exam tip: To turn a product of odd numbers into factorials, multiply top and bottom by the matching product of even numbers, 2 × 4 × … × 2n = 2ⁿn!. Then cancel any leftover powers of 2.

193IBMA2H-045-07-061.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard14 marks

No calculator

Let \(z=\frac{1}{2}e^{i\theta}\), where \(\theta\in\mathbb{R}\), and consider the infinite series \(S=\sum_{n=0}^{\infty}z^n\).

(a)

Explain why the series \(S\) converges for all values of \(\theta\).

(2)
(b)

Show that \(S=\dfrac{2}{2-e^{i\theta}}\).

(2)
(c)

Hence show that \(\displaystyle\sum_{n=0}^{\infty}\left(\tfrac{1}{2}\right)^n\cos n\theta=\frac{4-2\cos\theta}{5-4\cos\theta}\) and \(\displaystyle\sum_{n=0}^{\infty}\left(\tfrac{1}{2}\right)^n\sin n\theta=\frac{2\sin\theta}{5-4\cos\theta}\).

(5)
(d)

Find the exact value of \(\displaystyle\sum_{n=0}^{\infty}\left(\tfrac{1}{2}\right)^n\sin\frac{n\pi}{3}\).

(3)
(e)

By substituting \(\theta=\pi\) into the cosine result, find \(\displaystyle\sum_{n=0}^{\infty}\left(-\tfrac{1}{2}\right)^n\). Verify your answer by summing this geometric series directly.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 identifies \(S\) as a geometric series with common ratio \(z\), where \(|z|=\frac{1}{2}\)✓ 1
R1 \(|z|<1\) for every \(\theta\), so the series converges✓ 1
Part (b)
M1 \(S=\frac{1}{1-\frac{1}{2}e^{i\theta}}\)✓ 1
A1 multiplies numerator and denominator by 2 to obtain \(\frac{2}{2-e^{i\theta}}\)✓ 1
Part (c)
M1 uses de Moivre: \(z^n=\left(\frac{1}{2}\right)^n(\cos n\theta+i\sin n\theta)\), so the two required sums are \(\mathrm{Re}(S)\) and \(\mathrm{Im}(S)\)✓ 1
M1 multiplies numerator and denominator by the conjugate \(2-e^{-i\theta}\)✓ 1
A1 denominator \((2-\cos\theta)^2+\sin^2\theta=5-4\cos\theta\)✓ 1
A1 numerator \(2(2-\cos\theta)+2i\sin\theta\)✓ 1
A1 equates real and imaginary parts to obtain both results✓ 1
Part (d)
M1 substitutes \(\theta=\frac{\pi}{3}\), using \(\cos\frac{\pi}{3}=\frac{1}{2}\) and \(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)✓ 1
A1 denominator \(5-2=3\)✓ 1
A1 \(\frac{\sqrt3}{3}\)✓ 1
Part (e)
A1 \(\frac{4+2}{5+4}=\frac{2}{3}\)✓ 1
R1 \(\cos n\pi=(-1)^n\), so the series is geometric with ratio \(-\frac{1}{2}\): \(\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\), which agrees✓ 1

Exam tip: To separate the real and imaginary parts of a quotient, multiply the top and bottom by the complex conjugate of the denominator. This makes the denominator real.

194IBMA3H-1-18r5-071.9 The binomial theorem Paper 1Hard7 marks

No calculator


In the expansion of (1 + x)n, where n ≥ 3 is an integer, the coefficients of x, x2 and x3 form an arithmetic sequence.

(a)

Find the value of n.

(5)
(b)

For this value of n, find the coefficient of x3 in the expansion of (2 + x)(1 + x)n.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the arithmetic sequence property: 2 × C(n, 2) = C(n, 1) + C(n, 3)✓ 1
A1 n(n − 1) = n + n(n − 1)(n − 2)/6✓ 1
M1 divide by n and simplify to a quadratic✓ 1
A1 n2 − 9n + 14 = 0✓ 1
R1 n = 7, rejecting n = 2 since n ≥ 3✓ 1
Part (b)
M1 2 × C(7, 3) + C(7, 2) (ECF from (a))✓ 1
A1 91✓ 1

Exam tip: If a quadratic has two roots, check each against the conditions in the question, and say which one you reject and why.

195IBMA3H-1-18r5-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard10 marks

No calculator


Consider the equation z4 = −8 + 8√3 i, where z ∈ ℂ.

(a)

Write −8 + 8√3 i in the form r(cos θ + i sin θ), where r > 0 and −π < θ ≤ π.

(3)
(b)

Find the four solutions of the equation, giving each in the form a + bi with exact values of a and b.

(5)
(c)

The four points representing the solutions on an Argand diagram are the vertices of a square. Find the area of this square.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 modulus √(64 + 192)✓ 1
A1 r = 16✓ 1
A1 θ = 2π/3✓ 1
Part (b)
M1 z4 = 16(cos(2π/3 + 2πk) + i sin(2π/3 + 2πk))✓ 1
A1 modulus of each root is 2✓ 1
A1 arguments π/6 + πk/2, for k = 0, 1, 2, 3✓ 1
M1 evaluate at least one root in Cartesian form✓ 1
A1 √3 + i, −1 + √3 i, −√3 − i, 1 − √3 i✓ 1
Part (c)
M1 diagonal of the square is the diameter of the circle of radius 2, so diagonal = 4✓ 1
A1 area = 42/2 = 8✓ 1

Exam tip: The n roots of a complex number are equally spaced around a circle, so their arguments differ by 2π/n. Use this to check your list.

196IBMA3H-1-18r5-131.15 Proof by induction, contradiction and counterexample Paper 1Hard11 marks

No calculator


Let Sn = ∑r=1n r × 2r, for n ∈ ℤ+.

(a)

Prove by induction that Sn = (n − 1)2n+1 + 2 for all positive integers n.

(7)
(b)

Hence find the value of S10.

(2)
(c)

Hence find the value of ∑r=610 r × 2r.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 n = 1: LHS = 1 × 2 = 2 and RHS = 0 × 4 + 2 = 2, so true for n = 1✓ 1
M1 assume true for n = k: Sk = (k − 1)2k+1 + 2✓ 1
M1 Sk+1 = Sk + (k + 1)2k+1✓ 1
A1 = (k − 1)2k+1 + (k + 1)2k+1 + 2 = 2k × 2k+1 + 2✓ 1
A1 = k × 2k+2 + 2 = ((k + 1) − 1)2(k+1)+1 + 2✓ 1
R1 true for n = 1, and true for n = k implies true for n = k + 1✓ 1
R1 therefore true for all n ∈ ℤ+ by mathematical induction✓ 1
Part (b)
M1 substitute n = 10: 9 × 211 + 2✓ 1
A1 18434✓ 1
Part (c)
M1 S10 − S5 where S5 = 4 × 26 + 2 = 258 (ECF from (b))✓ 1
A1 18176✓ 1

Common error: Writing the conclusion as "true for all n" without stating both the base case and the implication. Both are needed for the final reasoning mark.

197IBMA3H-1-18r6-131.11 Partial fractions Paper 1Hard10 marks

No calculator

Let \(f(x)=\dfrac{5}{(1-2x)(1+3x)}\).

(a)

Express \(f(x)\) in the form \(\dfrac{A}{1-2x}+\dfrac{B}{1+3x}\).

(3)
(b)

Each fraction can be expanded as an infinite geometric series in \(x\). Find the set of values of \(x\) for which both expansions are valid.

(2)
(c)

Show that the coefficient of \(x^n\) in the series for \(f(x)\) is \(2\cdot2^n+3(-3)^n\), and hence find the coefficient of \(x^4\).

(3)
(d)

Hence find the exact value of \(\displaystyle\sum_{n=0}^{\infty}\frac{2\cdot2^n+3(-3)^n}{6^n}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(5=A(1+3x)+B(1-2x)\)✓ 1
A1 \(A=2\)✓ 1
A1 \(B=3\)✓ 1
Part (b)
M1 conditions \(|2x|<1\) and \(|3x|<1\)✓ 1
R1 both hold only when \(|x|<\frac13\)✓ 1
Part (c)
M1 \(\frac{2}{1-2x}=2\sum(2x)^n\) and \(\frac{3}{1+3x}=3\sum(-3x)^n\)✓ 1
A1 coefficient \(2\cdot2^n+3(-3)^n\)✓ 1
A1 \(2(16)+3(81)=275\)✓ 1
Part (d)
M1 recognise the series as \(f(x)\) at \(x=\frac16\) (inside \(|x|<\frac13\)), or sum two geometric series with ratios \(\frac13\) and \(-\frac12\)✓ 1
A1 \(\frac{2}{1-\frac13}+\frac{3}{1+\frac12}=3+2=5\)✓ 1

Exam tip: A geometric series can only be summed if its common ratio has modulus less than 1; check this before substituting a value of x.

198IBMA3H-1-18r6-141.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard10 marks

No calculator

The polynomial \(P(z)=z^4-6z^3+23z^2-34z+26\) has real coefficients and \(z=1+\mathrm{i}\) is a root of \(P(z)=0\).

(a)

Write down another root of \(P(z)=0\).

(1)
(b)

Show that \(z^2-2z+2\) is a factor of \(P(z)\).

(2)
(c)

Find the quadratic \(Q(z)\) such that \(P(z)=(z^2-2z+2)\,Q(z)\).

(3)
(d)

Hence solve \(P(z)=0\) completely.

(2)
(e)

Find the exact value of the product of the moduli of the four roots.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-\mathrm{i}\)✓ 1
Part (b)
M1 \((z-(1+\mathrm{i}))(z-(1-\mathrm{i}))\) expanded✓ 1
A1 \(z^2-2z+2\) (AG)✓ 1
Part (c)
M1 attempt at polynomial division or comparison of coefficients✓ 1
A1 \(z^2\) and \(-4z\) terms found✓ 1
A1 \(Q(z)=z^2-4z+13\)✓ 1
Part (d)
M1 solve \(z^2-4z+13=0\)✓ 1
A1 \(z=2\pm3\mathrm{i}\), together with \(1\pm\mathrm{i}\)✓ 1
Part (e)
M1 moduli \(\sqrt2\) (twice) and \(\sqrt{13}\) (twice)✓ 1
A1 \(\sqrt2\cdot\sqrt2\cdot\sqrt{13}\cdot\sqrt{13}=26\)✓ 1

Common error: A real polynomial of degree 4 needs two conjugate pairs; finding one pair is only half of the roots.

199IBMA3H-1-18r10-131.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard9 marks

No calculator

Let \(\theta\) be a real number.

(a)

Use de Moivre's theorem and a binomial expansion of \((\cos\theta+i\sin\theta)^3\) to show that \(\sin3\theta=3\sin\theta-4\sin^3\theta\).

(4)
(b)

Hence show that \(s=\sin\frac{\pi}{18}\) is a root of \(8s^3-6s+1=0\).

(2)
(c)

The equation \(8s^3-6s+1=0\) has three real roots, \(\sin\frac{\pi}{18}\), \(\sin\frac{5\pi}{18}\) and \(\sin\left(-\frac{7\pi}{18}\right)\). Use the sum of the roots to show that \(\sin\frac{\pi}{18}+\sin\frac{5\pi}{18}=\sin\frac{7\pi}{18}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3\)✓ 1
binomial expansion with terms \(\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\)✓ 1
equate imaginary parts: \(\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta\)✓ 1
use \(\cos^2\theta=1-\sin^2\theta\) to obtain \(3\sin\theta-4\sin^3\theta\)✓ 1
Part (b)
\(\sin\frac{\pi}{6}=\frac12=3s-4s^3\)✓ 1
rearrange to \(8s^3-6s+1=0\)✓ 1
Part (c)
\(3\theta=\frac\pi6+2k\pi\) or \(\frac{5\pi}6+2k\pi\) leads to these three values of \(\theta\) in \(\left(-\frac\pi2,\frac\pi2\right]\)✓ 1
the cubic has no \(s^2\) term, so the sum of the roots is \(0\)✓ 1
\(\sin\left(-\frac{7\pi}{18}\right)=-\sin\frac{7\pi}{18}\) and rearrange✓ 1

Exam tip: The coefficient of \(s^2\) gives the sum of the roots of a cubic: here it is zero.

200IBMA3H-1-18r10-141.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard10 marks

No calculator

A robotic arm's position is modelled by the complex number \(z=1+e^{i\theta}\), where \(-\pi<\theta<\pi\).

(a)

Show that \(z=2\cos\frac\theta2\left(\cos\frac\theta2+i\sin\frac\theta2\right)\).

(3)
(b)

For \(\theta=\frac{2\pi}3\), find the modulus and argument of \(z\).

(2)
(c)

Find the smallest positive integer \(n\) for which \(z^n\) is real and positive when \(\theta=\frac{2\pi}3\).

(2)
(d)

Find all values of \(\theta\) in the interval \(-\pi<\theta<\pi\) for which \(|z|=\sqrt3\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(1+\cos\theta=2\cos^2\frac\theta2\)✓ 1
\(\sin\theta=2\sin\frac\theta2\cos\frac\theta2\)✓ 1
factor out \(2\cos\frac\theta2\)✓ 1
Part (b)
modulus \(=2\cos\frac\pi3=1\)✓ 1
argument \(=\frac\pi3\)✓ 1
Part (c)
\(z^n\) has argument \(\frac{n\pi}3\) which must equal \(2k\pi\)✓ 1
\(n=6\)✓ 1
Part (d)
\(2\cos\frac\theta2=\sqrt3\) (using \(\cos\frac\theta2>0\) on this interval)✓ 1
\(\frac\theta2=\pm\frac\pi6\)✓ 1
\(\theta=\pm\frac\pi3\)✓ 1

Common error: On \(-\pi<\theta<\pi\) the factor \(\cos\frac{\theta}{2}\) is positive, so \(2\cos\frac{\theta}{2}\) is the modulus; check this before stating it.

201IBMA3H-1-19r11-071.11 Partial fractions Paper 1Hard10 marks

No calculator

Let \(f(x)=\frac{4x^2-3x+5}{(x-1)^2(x+2)}\).

(a)

Find the constants \(A\), \(B\) and \(C\) such that \(f(x)=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\).

(5)
(b)

Hence, using the binomial theorem for each partial fraction, find the coefficient of \(x^2\) in the expansion of \(f(x)\) in ascending powers of \(x\), for \(|x|<1\).

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(4x^2-3x+5=A(x-1)(x+2)+B(x+2)+C(x-1)^2\)✓ 1
A1 substitute \(x=1\): \(B=2\)✓ 1
A1 substitute \(x=-2\): \(C=3\)✓ 1
M1 compare \(x^2\) coefficients: \(A+C=4\) (or substitute another value)✓ 1
A1 \(A=1\)✓ 1
Part (b)
M1 rewrites as \(-(1-x)^{-1}+2(1-x)^{-2}+\frac{3}{2}\left(1+\frac{x}{2}\right)^{-1}\)✓ 1
A1 \(x^2\) coefficient from the first term: \(-1\)✓ 1
A1 from the second term: \(2\times3=6\)✓ 1
A1 from the third term: \(\frac{3}{2}\times\frac{1}{4}=\frac{3}{8}\)✓ 1
A1 total \(\frac{43}{8}\)✓ 1

Common error: \(\frac{1}{x-1}\) expands as \(-(1-x)^{-1}\), not \((1-x)^{-1}\); the sign is easy to lose.

202IBMA3H-1-19r11-111.15 Proof by induction, contradiction and counterexample Paper 1Hard11 marks

No calculator

At the start of day 1 a tank holds \(u_1=1\) litre of water. Each day, the volume in the tank is doubled by a process of filtration and then 3 litres are added, so that \(u_{n+1}=2u_n+3\), where \(u_n\) is the volume in litres at the start of day \(n\).

(a)

Find \(u_2\) and \(u_3\).

(2)
(b)

Prove by mathematical induction that \(u_n=2^{n+1}-3\) for all \(n\in\mathbb{Z}^+\).

(6)
(c)

Find the volume of water in the tank at the start of the first day on which it exceeds 1000 litres.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(u_2=2\times1+3=5\)✓ 1
A1 \(u_3=2\times5+3=13\)✓ 1
Part (b)
A1 \(n=1\): \(2^2-3=1=u_1\), so true for \(n=1\)✓ 1
M1 assumes true for \(n=k\): \(u_k=2^{k+1}-3\)✓ 1
M1 \(u_{k+1}=2u_k+3=2\left(2^{k+1}-3\right)+3\)✓ 1
A1 \(=2^{k+2}-6+3\)✓ 1
A1 \(=2^{(k+1)+1}-3\)✓ 1
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{Z}^+\)✓ 1
Part (c)
M1 \(2^{n+1}-3>1000\), e.g. tests \(n=9\) and neighbouring values✓ 1
A1 \(u_8=2^9-3=509\) is too small and \(u_9=2^{10}-3\) is the first to exceed 1000✓ 1
A1 1021 litres✓ 1

Exam tip: For a recurrence proof, substitute the assumed formula for \(u_k\) into the recurrence rule to obtain \(u_{k+1}\).

203IBMA3H-1-18r13-071.11 Partial fractions Paper 1Hard7 marks

No calculator

Consider the function \(g(x) = (2 - x)^3\,(1 + 3x)^{-2}\).

(a)

Find the first four terms of the expansion of \((1+3x)^{-2}\) in ascending powers of \(x\).

(3)
(b)

Hence find the coefficient of \(x^3\) in the expansion of \(g(x)\).

(3)
(c)

State the values of \(x\) for which the expansion of \(g(x)\) is valid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 binomial expansion with \(n=-2\) and \(3x\) in place of \(x\)✓ 1
A1 \(1 - 6x + 27x^2\)✓ 1
A1 \(-108x^3\)✓ 1
Part (b)
M1 \((2-x)^3 = 8 - 12x + 6x^2 - x^3\)✓ 1
M1 identify the four products giving \(x^3\): \(8(-108) + (-12)(27) + 6(-6) + (-1)\times 1\)✓ 1
A1 \(-1225\)✓ 1
Part (c)
A1 \(|x| < \frac13\)✓ 1

Common error: Forgetting one of the four pairings when collecting the \(x^3\) coefficient. List each power of \(x\) from the first bracket against the matching power from the second.

204IBMA3H-1-18r16-111.16 Systems of linear equations Paper 1Hard9 marks

No calculator

Consider the system of equations, where k, m ∈ ℝ:
x + 2y − z = 3
2x − y + z = 1
3x + y + kz = m

(a)

For k = 0 and m = 7, show that the system has no solution.

(3)
(b)

For k = 0 and m = 4, find the general solution of the system in terms of a parameter t, where x = t.

(4)
(c)

Find the values of k for which the system has a unique solution for every value of m.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 add the first two equations✓ 1
A1 3x + y = 4✓ 1
R1 contradicts 3x + y = 7 from the third equation, so no solution✓ 1
Part (b)
R1 the third equation is a consequence of the first two so there are infinitely many solutions✓ 1
M1 use 3x + y = 4 or eliminate✓ 1
A1 y = 4 − 3t✓ 1
A1 z = 5 − 5t✓ 1
Part (c)
M1 the first two equations add to 3x + y = 4, so the third gives kz = m − 4✓ 1
A1 k ≠ 0✓ 1

Common error: A system with no solution and a system with infinitely many solutions can have identical left-hand sides; only the constants decide which case occurs.

205IBMA3H-1-18r16-131.11 Partial fractions Paper 1Hard6 marks

No calculator

Consider the function \(f(x) = (2 - x)(1 + 3x)^{-2}\).

(a)

Find the expansion of \((1+3x)^{-2}\) up to and including the term in \(x^2\).

(3)
(b)

Hence find the coefficient of \(x^2\) in the expansion of f(x).

(2)
(c)

State the values of x for which the expansion of \((1+3x)^{-2}\) is valid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 binomial theorem with \(n=-2\)✓ 1
A1 \(-6x\)✓ 1
A1 \(+27x^2\)✓ 1
Part (b)
M1 \(2\times 27 + (-1)\times(-6)\)✓ 1
A1 60✓ 1
Part (c)
A1 \(|x| < \frac{1}{3}\)✓ 1

Common error: When multiplying two series, collect every pair of terms that gives the required power: here both 2 × x² and (−x) × x contribute.

206IBMA3H-1-20r29-081.15 Proof by induction, contradiction and counterexample Paper 1Hard9 marks

No calculator

This question is about the inequality \(2^n>n^2\).

(a)

Prove by induction that \(2^n>n^2\) for all integers \(n\ge5\).

(7)
(b)

Show by a counterexample that the inequality is not true for all positive integers \(n\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(n=5\): \(32>25\)✓ 1
M1 assume \(2^k>k^2\) for some integer \(k\ge5\)✓ 1
M1 \(2^{k+1}=2\times2^k>2k^2\)✓ 1
M1 attempt to show \(2k^2\ge(k+1)^2\), i.e. \(k^2-2k-1\ge0\)✓ 1
A1 for \(k\ge5\), \(k^2-2k-1=k(k-2)-1\ge14>0\)✓ 1
A1 so \(2^{k+1}>(k+1)^2\)✓ 1
R1 true for \(n=5\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\ge5\)✓ 1
Part (b)
M1 choose \(n=2\), \(3\) or \(4\)✓ 1
A1 e.g. \(n=3\): \(2^3=8\), which is not greater than \(3^2=9\)✓ 1

Exam tip: In the step, bound the new expression by the assumed one, then show the remaining algebra works for every \(k\) in the allowed range.

207IBMA3H-1-20r29-131.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard6 marks

No calculator

The positive numbers \(x\) and \(y\) satisfy \(\log_2x+\log_4y=5\) and \(\log_8(xy)=2\).

(a)

Let \(X=\log_2x\) and \(Y=\log_2y\). Show that \(X+\frac12Y=5\) and \(X+Y=6\).

(3)
(b)

Hence find the value of \(x\) and the value of \(y\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\log_4y=\frac{\log_2y}{\log_24}=\frac{Y}{2}\)✓ 1
M1 \(\log_8(xy)=\frac{\log_2(xy)}{3}=\frac{X+Y}{3}\)✓ 1
A1 both equations obtained: \(X+\frac12Y=5\) and \(X+Y=6\)✓ 1
Part (b)
M1 subtract the equations to eliminate \(X\)✓ 1
A1 \(Y=2\), \(X=4\)✓ 1
A1 \(x=16\), \(y=4\)✓ 1

Common error: Convert every logarithm to the same base before combining. Do not mix \(\log_2\), \(\log_4\) and \(\log_8\) terms directly.

208IBMA3H-1-20r35-091.11 Partial fractions Paper 1Hard6 marks

No calculator

Consider the equation \(\log_x16+\log_4x=3\), where \(x>0\) and \(x\ne1\).

(a)

Let \(u=\log_2x\). Show that the equation can be written as \(u^2-6u+8=0\).

(3)
(b)

Hence solve the equation for \(x\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\log_x16=\dfrac{\log_216}{\log_2x}=\dfrac4u\)✓ 1
M1 \(\log_4x=\dfrac{\log_2x}{\log_24}=\dfrac u2\)✓ 1
A1 \(\dfrac4u+\dfrac u2=3\), so multiplying by \(2u\): \(8+u^2=6u\), giving \(u^2-6u+8=0\) (AG)✓ 1
Part (b)
M1 factorise: \((u-2)(u-4)=0\)✓ 1
A1 \(u=2\) or \(u=4\)✓ 1
A1 \(x=4\) or \(x=16\)✓ 1

Exam tip: When a log equation has different bases, change everything to one common base and substitute a single letter; the equation often becomes a quadratic.

209IBMA3H-1-20r35-131.15 Proof by induction, contradiction and counterexample Paper 1Hard7 marks

No calculator

Consider the number \(\log_23\).

(a)

Prove by contradiction that \(\log_23\) is irrational.

(5)
(b)

Explain why the same method cannot be used to show that \(\log_28\) is irrational.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 assume \(\log_23\) is rational, so \(\log_23=\dfrac pq\) with \(p,q\) positive integers (positive because \(3>1\))✓ 1
M1 rewrite in exponential form: \(2^{p/q}=3\), so \(2^p=3^q\)✓ 1
A1 \(2^p\) is even, because \(p\ge1\)✓ 1
A1 \(3^q\) is odd, for every positive integer \(q\)✓ 1
R1 an even number cannot equal an odd number: contradiction, so \(\log_23\) is irrational✓ 1
Part (b)
A1 \(\log_28=3=\dfrac31\) is rational✓ 1
R1 the equation \(2^p=8^q\) has both sides even (it is satisfied by \(p=3q\)), so no contradiction arises✓ 1

Exam tip: A proof by contradiction has three parts: state the assumption, derive a result that cannot be true, and conclude that the assumption must be false.

210IBMA3H-1-19r39-051.11 Partial fractions Paper 1Hard9 marks

No calculator

Let \(f(x)=\dfrac{x^{2}+7x+1}{(x-1)^{2}(x+2)}\).

(a)

Express \(f(x)\) in the form \(\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}\), where \(A\), \(B\) and \(C\) are constants.

(7)
(b)

Write down the equations of all asymptotes of the graph of \(y=f(x)\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(x^2+7x+1\equiv A(x-1)(x+2)+B(x+2)+C(x-1)^2\)✓ 1
M1 substitutes \(x=1\)✓ 1
A1 \(B=3\)✓ 1
M1 substitutes \(x=-2\)✓ 1
A1 \(C=-1\)✓ 1
M1 equates coefficients of \(x^2\) (\(A+C=1\)) or substitutes another value of \(x\)✓ 1
A1 \(A=2\)✓ 1
Part (b)
A1 vertical asymptotes \(x=1\) and \(x=-2\)✓ 1
A1 horizontal asymptote \(y=0\)✓ 1

Common error: With a repeated factor \((x-1)^2\) you need both \(\frac{A}{x-1}\) and \(\frac{B}{(x-1)^2}\); leaving out the first term makes the identity impossible to satisfy.

211IBMA3H-1-19r39-091.13 Polar and Euler forms Paper 1Hard10 marks

No calculator

A logistics firm places three depots at the points in the complex plane that satisfy \(z^{3}=8i\), with one unit representing 1 km.

(a)

Find the three solutions of \(z^3=8i\), giving each in the form \(a+bi\) with exact values of \(a\) and \(b\).

(5)
(b)

Find the exact distance between the depots at \(\sqrt3+i\) and \(-\sqrt3+i\).

(2)
(c)

The three depots are the vertices of a triangular region. Find the exact area of this region, in km2.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes \(8i=8\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)\)✓ 1
M1 \(z=2\,\mathrm{cis}\left(\dfrac{\pi/2+2k\pi}{3}\right)\) (cube root of modulus, argument divided by 3)✓ 1
A1 \(\sqrt3+i\)✓ 1
A1 \(-\sqrt3+i\)✓ 1
A1 \(-2i\)✓ 1
Part (b)
M1 \(|(\sqrt3+i)-(-\sqrt3+i)|\)✓ 1
A1 \(2\sqrt3\) km✓ 1
Part (c)
M1 recognises the triangle is equilateral (equal sides, or three roots equally spaced on a circle of radius 2)✓ 1
M1 uses area \(=\dfrac{\sqrt3}{4}s^2\) with \(s=2\sqrt3\) (or \(3\times\frac12\times2^2\sin120^\circ\))✓ 1
A1 \(3\sqrt3\) km2✓ 1

Exam tip: Roots of \(z^n=w\) are equally spaced around a circle of radius \(|w|^{1/n}\), separated by \(\frac{2\pi}{n}\); sketching them checks the answers and shows the shape formed.

212IBMA3H-1-19r39-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 1Hard9 marks

No calculator

The quartic equation \(z^{4}+az^{3}+bz^{2}+cz+d=0\), where \(a,b,c,d\in\mathbb{R}\), has roots \(1+i\) and \(2-i\).

(a)

Write down the other two roots of the equation.

(2)
(b)

Find the values of \(a\), \(b\), \(c\) and \(d\).

(5)
(c)

Show that \(d=|1+i|^{2}\times|2-i|^{2}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-i\)✓ 1
A1 \(2+i\)✓ 1
Part (b)
M1 forms quadratic factors from conjugate pairs✓ 1
A1 \(z^2-2z+2\)✓ 1
A1 \(z^2-4z+5\)✓ 1
M1 multiplies the two quadratics✓ 1
A1 \(a=-6,\ b=15,\ c=-18,\ d=10\)✓ 1
Part (c)
M1 \(|1+i|^2=2\) and \(|2-i|^2=5\) (or \((1+i)(1-i)=2\), \((2-i)(2+i)=5\))✓ 1
A1 product is \(10\), which equals \(d\)✓ 1

Exam tip: For a quartic with real coefficients, multiply conjugate pairs to get real quadratic factors; this avoids long expansions with \(i\).

213IBMA3H-1-18r47-041.9 The binomial theorem Paper 1Hard7 marks

No calculator

In the expansion of \((1 + ax)^n\), where \(a\) is a non-zero constant and \(n\) is a positive integer, the coefficient of \(x\) is 12 and the coefficient of \(x^2\) is 60.

(a)

Find the value of \(a\) and the value of \(n\).

(5)
(b)

Find the coefficient of \(x^3\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 coefficient of \(x\): \(na = 12\)✓ 1
M1 coefficient of \(x^2\): \(\binom{n}{2}a^2 = 60\)✓ 1
M1 eliminate one unknown, e.g. \(n = \frac{12}{a}\) giving \(\frac{12}{2}(12 - a) = 60\)✓ 1
A1 \(a = 2\)✓ 1
A1 \(n = 6\)✓ 1
Part (b)
M1 \(\binom{6}{3} \times 2^3\) (ECF from (a))✓ 1
A1 160✓ 1

Exam tip: Write \(\binom{n}{2} = \frac{n(n-1)}{2}\) and use the first equation to replace \(na\); this avoids solving a messy simultaneous equation.

214IBMA3H-1-18r47-081.16 Systems of linear equations Paper 1Hard10 marks

No calculator

Consider the system of equations
\(x + y + z = 6\)
\(2x - y + z = 3\)
\(3x + 2z = k\)
where \(k\) is a constant.

(a)

Show that the system has no solution unless \(k = 9\).

(3)
(b)

For \(k = 9\), find the general solution in terms of a parameter \(\lambda\), where \(z = 3\lambda\).

(4)
(c)

Describe the geometrical arrangement of the three planes when \(k = 9\).

(1)
(d)

Find the value of \(z\) at the point on this line where \(x = 1\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 add the first two equations✓ 1
A1 \(3x + 2z = 9\)✓ 1
R1 if \(k \ne 9\) the third equation contradicts this, so the system is inconsistent✓ 1
Part (b)
M1 substitute \(z = 3\lambda\) into \(3x + 2z = 9\)✓ 1
A1 \(x = 3 - 2\lambda\)✓ 1
M1 substitute into the first (or second) equation to find \(y\)✓ 1
A1 \(y = 3 - \lambda\)✓ 1
Part (c)
R1 the three planes intersect in a common line✓ 1
Part (d)
M1 \(3 - 2\lambda = 1\) so \(\lambda = 1\)✓ 1
A1 \(z = 3\)✓ 1

Common error: When two equations combine to give a multiple or sum of the third, the system is either inconsistent or has infinitely many solutions; check the constant term to decide which.

215AAHL1-q00071.3 Geometric sequences and series Paper 2Easy7 marks

Two colonies of mould, P and Q, are grown in a laboratory. At the start of the first hour, colony P covers 3 mm2 and colony Q covers 200 mm2. At the start of each subsequent hour, the area of P is double its area at the start of the previous hour, and the area of Q is 50% more than its area at the start of the previous hour. Let \(P_{n}\) and \(Q_{n}\) be the areas, in mm2, at the start of the \(n\)th hour.

(a)

Find \(Q_{10}\).

(2)
(b)

Show that \(P_{n}\gt Q_{n}\) when \(\left(\dfrac{4}{3}\right)^{n-1}\gt \dfrac{200}{3}\).

(2)
(c)

Hence find the first hour at the start of which the area of colony P is greater than the area of colony Q.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(200(1.5)^{9}\)(M1)
\(7690\) mm2A1\(7688.67\ldots\)
(b)
\(3(2)^{n-1}\gt 200(1.5)^{n-1}\)M1
\(\left(\dfrac{2}{1.5}\right)^{n-1}\gt \dfrac{200}{3}\)A1
\(\left(\dfrac{4}{3}\right)^{n-1}\gt \dfrac{200}{3}\)AG
(c)
\(n-1\gt \dfrac{\ln(200/3)}{\ln(4/3)}\)(M1)Or a table or graph.
\(n-1\gt 14.598\ldots\), so \(n\gt 15.598\ldots\)(A1)
the 16th hourA1Award A0 for 15 or for 15.6.

Answers: (a) \(7690\) mm2 · (c) the 16th hour

216AAHL1-q00121.5 Exponents and introduction to logarithms Paper 2Easy8 marks

In many large data sets, the proportion of numbers whose first (leading) digit is \(d\) is modelled by \(p(d)=\log_{10}\left(1+\dfrac{1}{d}\right)\), for \(d=1,2,\ldots,9\).

(a)

Find \(p(1)\).

(2)
(b)

Show that \(p(1)+p(2)+\cdots+p(9)=1\).

(2)
(c)

Find the exact proportion of numbers whose leading digit is at least 5, giving your answer as a single logarithm.

(2)
(d)

Find the least value of \(k\) for which the proportion of numbers whose leading digit is at most \(k\) is greater than 0.75.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(\log_{10}2\)(M1)
\(0.301\)A1\(0.30103\ldots\)
(b)
\(\log_{10}\dfrac{2}{1}+\log_{10}\dfrac{3}{2}+\cdots+\log_{10}\dfrac{10}{9}=\log_{10}\left(\dfrac{2}{1}\times\dfrac{3}{2}\times\cdots\times\dfrac{10}{9}\right)\)M1Writing \(1+\frac1d=\frac{d+1}{d}\) and using the product law.
\(=\log_{10}10\)A1
\(=1\)AG
(c)
\(\log_{10}\left(\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times\dfrac{9}{8}\times\dfrac{10}{9}\right)=\log_{10}\dfrac{10}{5}\)(M1)
\(\log_{10}2\)A1Accept \(1-\log_{10}5\).
(d)
\(\log_{10}(k+1)\gt0.75\), so \(k+1\gt10^{0.75}=5.62\ldots\)(M1)Or \(\log_{10}5=0.699\) and \(\log_{10}6=0.778\).
\(k=5\)A1

Answers: (a) \(0.301\) · (c) \(\log_{10}2\) · (d) \(k=5\)

217AAHL1-q00271.10 Counting principles and the extended binomial theorem Paper 2Easy6 marks

In a lottery, a ticket consists of 6 different numbers chosen from the integers 1 to 49. The order of the numbers does not matter. Six winning numbers are drawn.

(a)

Find the number of different tickets possible.

(2)
(b)

Find the number of tickets that match exactly 4 of the 6 winning numbers.

(2)
(c)

The organisers want to choose 6 numbers from the integers 1 to \(n\), so that there are at least 20 000 000 possible tickets. Find the least value of \(n\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\({}^{49}C_{6}\)(M1)
\(13\,983\,816\)A1
(b)
\({}^{6}C_{4}\times{}^{43}C_{2}=15\times903\)(M1)
\(13\,545\)A1
(c)
\({}^{n}C_{6}\ge20\,000\,000\); table: \({}^{51}C_{6}=18\,009\,460\), \({}^{52}C_{6}=20\,358\,520\)(M1)
\(n=52\)A1

Answers: (a) 13 983 816 · (b) 13 545 · (c) \(n=52\)

218AAHL1-q00781.15 Proof by induction, contradiction and counterexample Paper 2Easy6 marks

For each of the following statements, use a counterexample to show that the statement is not always true.

(a)

\(n^{2}+n+17\) is prime for all \(n\in\mathbb{Z}^{+}\).

(2)
(b)

\(2^{p}-1\) is prime for every prime number \(p\).

(2)
(c)

For all real numbers \(a\) and \(b\), if \(a^{2}\gt b^{2}\) then \(a\gt b\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(n=16\): \(16^{2}+16+17=289\)A1Or \(n=17\): \(323=17\times19\).
\(289=17^{2}\), so it is not primeR1Must explain why it is not prime.
(b)
\(p=11\): \(2^{11}-1=2047\)A1
\(2047=23\times89\), so it is not primeR1
(c)
eg \(a=-3,\ b=1\)A1Any \(a\lt0\) with \(|a|\gt|b|\) and \(a\le b\).
\(a^{2}=9\gt1=b^{2}\) but \(-3\lt1\)R1

Answers: (a) \(n=16\): \(289=17^{2}\) · (b) \(p=11\): \(2047=23\times89\) · (c) eg \(a=-3,\ b=1\)

219AAHL1-q00861.16 Systems of linear equations Paper 2Easy6 marks

Mia invests a total of €20 000 in three accounts, A, B and C, which pay simple interest at annual rates of 3%, 4.5% and 6% respectively. She invests €\(x\) in A, €\(y\) in B and €\(z\) in C. The total interest earned in the first year is €1020. The amount invested in C is €2000 less than the total invested in A and B.

(a)

Write down a system of three linear equations in \(x\), \(y\) and \(z\).

(2)
(b)

Find the value of \(x\), of \(y\) and of \(z\).

(2)
(c)

Mia wonders whether she could split €20 000 so that the amount in C is still €2000 less than the total in A and B, but the first-year interest is €1100. Show that this is not possible.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(x+y+z=20\,000\) and \(x+y-z=2000\)A1Both required.
\(0.03x+0.045y+0.06z=1020\)A1
(b)
use of GDC or elimination, eg \(2z=18\,000\)(M1)
\(x=1000,\ y=10\,000,\ z=9000\)A1All three required.
(c)
solving with \(1100\) in place of \(1020\), eg \(z=9000\), \(x+y=11\,000\), \(0.03x+0.045y=560\)M1
\(x=-4333.33\ldots\lt0\) (\(y=15\,333.33\ldots\)), which is impossible as an amount invested cannot be negativeR1

Answers: (b) \(x=1000,\ y=10\,000,\ z=9000\) · (c) gives \(x\approx-4333\lt0\)

220AAHL1-q00881.16 Systems of linear equations Paper 2Easy6 marks

A café supplies three offices with coffees, sandwiches and pastries. Each item has a fixed price. The orders and their total costs are shown in the table.

OfficeCoffeesSandwichesPastriesTotal cost (€)
1421520283.10
2382214297.00
3511825344.80
(a)

Find the price of a coffee, of a sandwich and of a pastry.

(4)
(b)

A fourth office orders 45 coffees, 20 sandwiches and 18 pastries. Find the total cost of this order.

(2)
Show mark scheme
WorkingMarkNotes
(a)
setting up a system, eg \(42c+15s+20p=283.10\), \(38c+22s+14p=297.00\), \(51c+18s+25p=344.80\)M1
coffee €2.80A1
sandwich €6.50A1
pastry €3.40A1
(b)
\(45(2.80)+20(6.50)+18(3.40)\)(M1)FT from (a).
€317.20A1

Answers: (a) coffee €2.80, sandwich €6.50, pastry €3.40 · (b) €317.20

221IBMA2H-045-06-011.16 Systems of linear equations Paper 2Easy5 marks

Calculator

A cable-stayed footbridge is supported by 23 steel cables on one side. The cable lengths form an arithmetic sequence. The shortest cable is 42.5 m long and the longest cable is 128.3 m long.

(a)

Find the common difference of the sequence of cable lengths.

(3)
(b)

Find the total length of steel cable needed for this side of the bridge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses \(u_n=u_1+(n-1)d\) with \(n=23\), or recognises there are 22 common differences between the shortest and longest cable ; A1 correct equation or expression, e.g. \(128.3=42.5+22d\) or \(d=\frac{128.3-42.5}{22}=\frac{85.8}{22}\) ; A1 \(d=3.9\) m✓ 3
Part (b)
M1 substitutes into \(S_n=\frac{n}{2}(u_1+u_n)\), e.g. \(\frac{23}{2}(42.5+128.3)\), or into \(S_n=\frac{n}{2}(2u_1+(n-1)d)\) with their \(d\) ; A1 1964.2 m (accept 1960 m to 3 s.f.)✓ 2

Exam tip: When both the first and last terms are known, \(S_n=\frac{n}{2}(u_1+u_n)\) is quicker than using \(d\). Remember that 23 terms means 22 gaps, not 23.

222IBMA2H-045-06-031.1 Standard form Paper 2Easy5 marks

Calculator

A patient is being weaned off a medication. On day 1 the dose is 60 mg, and the daily dose then decreases by 2.5 mg each day.

(a)

Find the first day on which the dose is less than 20 mg.

(3)
(b)

Find the total amount of medication taken from day 1 up to and including the first day on which the dose is less than 20 mg.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets up \(60-2.5(n-1)<20\), or solves \(60-2.5(n-1)=20\) and checks the next day ; A1 \(n>17\), or \(n=17\) with day 17 dose exactly 20 mg ; A1 day 18✓ 3
Part (b)
M1 \(S_{18}=\frac{18}{2}(60+17.5)\) or \(\frac{18}{2}(120+17(-2.5))\) ; A1 697.5 mg (698 mg to 3 s.f.)✓ 2

Common error: On day 17 the dose is exactly 20 mg, which is not less than 20 mg. Solving the equality and stopping there gives day 17 instead of day 18.

223IBMA2H-045-03-021.7 Rational exponents and laws of logarithms Paper 2Easy6 marks

Calculator

After an injection, the concentration of an anaesthetic in a patient's blood is modelled by \(C=12\times2^{-t/5}\), where \(C\) is in mg L−1 and \(t\) is the time in hours after the injection.

(a)

Write down the concentration at the moment of injection.

(1)
(b)

Find the concentration 8 hours after the injection.

(2)
(c)

The patient may be discharged once the concentration has fallen to 2 mg L−1. Find the earliest time at which the patient may be discharged.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 12 mg L−1✓ 1
Part (b)
M1 substitutes \(t=8\): \(12\times2^{-1.6}\)✓ 1
A1 3.96 mg L−1✓ 1
Part (c)
M1 sets up \(12\times2^{-t/5}=2\), so \(2^{-t/5}=\frac{1}{6}\)✓ 1
M1 uses logarithms: \(\frac{t}{5}=\log_2 6\), or uses a GDC✓ 1
A1 \(t=12.9\) hours✓ 1

Common error: Divide by 12 before taking logarithms. Writing \(\log(12\times2^{-t/5})\) as \(12\log(2^{-t/5})\) is a misuse of the log laws.

224IBMA2H-045-03-031.16 Systems of linear equations Paper 2Easy4 marks

Calculator

In a basketball game a team scored all its points from free throws (1 point each), two-point shots and three-point shots. The team made 40 successful shots in total and scored 80 points. The number of successful two-point shots was 4 more than twice the number of successful three-point shots.
Let \(x\), \(y\) and \(z\) be the numbers of successful free throws, two-point shots and three-point shots respectively.

(a)

Write down a system of three linear equations in \(x\), \(y\) and \(z\).

(2)
(b)

Hence find the number of each type of successful shot.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x+y+z=40\) and \(x+2y+3z=80\)✓ 1
A1 \(y=2z+4\) (or equivalent)✓ 1
Part (b)
M1 attempts to solve the system, by elimination or GDC✓ 1
A1 \(x=9\), \(y=22\), \(z=9\)✓ 1

Exam tip: Turn a phrase such as "4 more than twice" into an equation one step at a time: twice \(z\) is \(2z\), then add 4. Check your solution in all three original statements.

225IBMA2H-045-00-021.4 Financial applications Paper 2Easy7 marks

Calculator

A cycling club deposits $8500 into an account paying a nominal annual interest rate of 3.6%, compounded monthly. No further deposits or withdrawals are made.

(a)

Find the value of the account after 5 years, giving your answer to the nearest dollar.

(2)
(b)

Find the number of complete months after which the value of the account first exceeds $12 000.

(3)
(c)

The annual rate of inflation over the 5 years is 2.5%. Find the real value of the amount in part (a), in terms of today’s money, giving your answer to the nearest dollar.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses 8500(1 + 0.036/12)60 or a TVM solver with N = 60, I% = 3.6, C/Y = 12✓ 1
A1 $10 174✓ 1
Part (b)
M1 sets up 8500(1.003)n > 12 000✓ 1
A1 n > 115.1…✓ 1
A1 116 months✓ 1
Part (c)
M1 divides by 1.0255: 10 173.60 / 1.0255✓ 1
A1 $8992✓ 1

Common error: In part (b) the solver gives 115.1; the account has not yet exceeded $12 000 after 115 months, so the answer must be rounded up to 116.

226IBMA2H-045-00-101.4 Financial applications Paper 2Easy7 marks

Calculator

A hospital research fund invests $2.5 million in an account with interest compounded quarterly. After 6 years, with no further deposits or withdrawals, the investment is worth $3.1 million.

(a)

Find the nominal annual interest rate, giving your answer as a percentage to 3 significant figures.

(3)
(b)

Find the effective annual interest rate.

(2)
(c)

Over the 6 years, inflation is 2.9% per year. Find the value of the $3.1 million in terms of purchasing power at the start of the investment.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets up 2.5(1 + r/400)24 = 3.1, or uses a TVM solver with N = 24, C/Y = 4✓ 1
M1 1 + r/400 = 1.241/24✓ 1
A1 r = 3.60%✓ 1
Part (b)
M1 (1.24)1/6 − 1, or (1 + 0.036013/4)4 − 1✓ 1
A1 3.65%✓ 1
Part (c)
M1 3.1 / 1.0296✓ 1
A1 $2.61 million✓ 1

Common error: Using N = 6 with quarterly compounding gives the rate per quarter over the wrong number of periods; 6 years compounded quarterly means 24 periods.

227IBMA2H-045-02-031.12 Complex numbers in Cartesian form Paper 2Easy6 marks

Calculator

In an alternating-current circuit, the impedance of a speaker coil is \(Z=30+40i\) ohms. A supply voltage, represented by the phasor \(V=230\) volts, is applied, and the current phasor is given by \(I=\frac{V}{Z}\).

(a)

Write \(Z\) in the form \(re^{i\theta}\), giving \(\theta\) in radians.

(2)
(b)

Hence find \(I\) in the form \(re^{i\theta}\).

(2)
(c)

The in-phase (resistive) component of the current is given by the real part of \(I\). Find this value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempts \(|Z|=\sqrt{30^2+40^2}\) and \(\arg Z=\arctan\frac{40}{30}\)✓ 1
A1 \(Z=50e^{0.927i}\)✓ 1
Part (b)
M1 divides the moduli and subtracts the arguments, e.g. \(\frac{230}{50}e^{i(0-0.927)}\)✓ 1
A1 \(I=4.6e^{-0.927i}\)✓ 1
Part (c)
M1 \(\text{Re}(I)=4.6\cos(-0.927)\), or any correct method such as \(\frac{230(30-40i)}{30^2+40^2}\)✓ 1
A1 \(2.76\) A✓ 1

Exam tip: Division is quick in Euler form: divide the moduli and subtract the arguments. Keep the unrounded argument in your calculator memory for the final part.

228IBMA2H-045-05-021.7 Rational exponents and laws of logarithms Paper 2Easy6 marks

Calculator

A patient is given a 200 mg dose of a medicine at the same time every day. By the time the next dose is due, 35% of the medicine present in the bloodstream just after the previous dose remains. Immediately after the first dose, the amount in the bloodstream is 200 mg.

(a)

Find the amount of medicine in the bloodstream immediately after the third dose.

(2)
(b)

Find the long-term amount of medicine in the bloodstream immediately after a dose, giving your answer to 3 significant figures.

(2)
(c)

A different medicine is also given in 200 mg daily doses. Its long-term amount immediately after a dose is 250 mg. Find the proportion of this medicine that remains in the bloodstream after one day.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 200 + 200(0.35) + 200(0.35)2✓ 1
A1 294.5 mg✓ 1
Part (b)
M1 uses sum to infinity: \(\frac{200}{1-0.35}\)✓ 1
A1 308 mg✓ 1
Part (c)
M1 sets \(\frac{200}{1-r}=250\)✓ 1
A1 r = 0.2 (i.e. 20% remains)✓ 1

Common error: Using 0.65 (the proportion removed) as the common ratio instead of 0.35 (the proportion remaining). The ratio is always what carries over to the next term.

229IBMA2H-045-05-051.10 Counting principles and the extended binomial theorem Paper 2Easy4 marks

Calculator

A basketball squad has 12 players, exactly 2 of whom are centres. The coach selects a starting five.

(a)

Find the number of possible starting fives that contain exactly one centre.

(2)
(b)

After choosing a starting five with exactly one centre, the coach assigns the five players to five distinct positions (point guard, shooting guard, small forward, power forward, centre). The selected centre must play the centre position. Find the total number of ways of selecting and assigning the starting five.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\binom{2}{1}\times\binom{10}{4}\)✓ 1
A1 420✓ 1
Part (b)
M1 multiplies by 4! for the remaining four players in four positions✓ 1
A1 420 × 24 = 10080✓ 1

Exam tip: Separate the problem into 'choose' (combinations) and 'arrange' (permutations) stages, then multiply. Only the unrestricted players are permuted.

230IBMA2H-045-07-021.3 Geometric sequences and series Paper 2Easy5 marks

Calculator

A cyclist preparing for a race rides 25 km in the first week of training. Each week after that, the distance ridden is 8% more than in the previous week.

(a)

Find the distance the cyclist rides in week 10.

(2)
(b)

Find the total distance ridden over the first 10 weeks of training.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_{10}=25\times1.08^{9}\)✓ 1
A1 50.0 km (49.975…)✓ 1
Part (b)
M1 recognises a geometric series with \(u_1=25\) and \(r=1.08\)✓ 1
M1 substitutes into \(S_{10}=\frac{25(1.08^{10}-1)}{1.08-1}\)✓ 1
A1 362 km (362.16…)✓ 1

Exam tip: An increase of 8% means the common ratio is 1.08, not 0.08. Week 10 is the 10th term, so the power is 9.

231IBMA2H-045-07-031.1 Standard form Paper 2Easy5 marks

Calculator

A typical human red blood cell has a diameter of 7.5×10−6 m. An adult body contains about 2.5×1013 red blood cells.

(a)

Suppose all the red blood cells in one adult were placed in a single straight line, each touching the next. Find the length of this line in kilometres. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\), exactly or correct to 3 significant figures.

(3)
(b)

The circumference of the Earth is approximately 4.0×104 km. Find how many times the line of cells would wrap around the Earth. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies 7.5×10−6 × 2.5×1013✓ 1
A1 1.875×108 m✓ 1
A1 1.875×105 km (accept 1.88×105 km)✓ 1
Part (b)
M1 divides their length in km by 4.0×104, e.g. \(\frac{1.875\times10^{5}}{4.0\times10^{4}}\)✓ 1
A1 4.69 (accept 4.70 from using 1.88×105)✓ 1

Exam tip: Change units before you compare quantities. Here the length must be in km to match the Earth's circumference. Keep the unrounded value, 1.875×10⁵, for the next step to avoid rounding errors.

232IBMA3H-1-18r5-011.1 Standard form Paper 2Easy4 marks

Calculator


A Formula 1 circuit has a lap length of 5.8 × 103 m. A race consists of 56 laps.

(a)

Find the total distance of the race. Give your answer in the form a × 10k, where 1 ≤ a < 10 and k ∈ ℤ, with a correct to 3 significant figures.

(2)
(b)

The winning car takes 5.4 × 103 s. Using the exact total distance, find its mean speed in m s−1, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 5.8 × 103 × 56 (= 324 800)✓ 1
A1 3.25 × 105 m✓ 1
Part (b)
M1 324 800 ÷ 5400✓ 1
A1 60.1 m s−1✓ 1

Common error: Rounding the distance to 3.25 × 10⁵ before dividing can change the third significant figure of the speed. Keep the unrounded value in your calculator.

233IBMA3H-1-18r5-021.1 Standard form Paper 2Easy6 marks

Calculator


A hospital laboratory analyses a blood sample. The patient's blood contains 5.2 × 109 red blood cells per millilitre and 6.5 × 106 white blood cells per millilitre.

(a)

A test uses 3.5 × 10−2 ml of blood. Find the number of red blood cells in the test sample, giving your answer in the form a × 10k, where 1 ≤ a < 10 and k ∈ ℤ.

(2)
(b)

The patient has 5.0 × 103 ml of blood in total. Find the total number of red blood cells, in the form a × 10k.

(2)
(c)

Find how many red blood cells there are for each white blood cell.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 5.2 × 109 × 3.5 × 10−2✓ 1
A1 1.82 × 108✓ 1
Part (b)
M1 5.2 × 109 × 5.0 × 103✓ 1
A1 2.6 × 1013✓ 1
Part (c)
M1 (5.2 × 109) ÷ (6.5 × 106)✓ 1
A1 800✓ 1

Exam tip: When multiplying numbers in standard form, multiply the decimal parts and add the powers of ten, then adjust so the first factor lies between 1 and 10.

234IBMA3H-1-18r6-011.8 Infinite geometric series Paper 2Easy4 marks

Calculator

A runner's weekly training distance follows a geometric sequence. In week 1 she runs 14 km and the distance increases by 10% each week.

(a)

Find the distance run in week 8, to 3 significant figures.

(2)
(b)

Find the total distance run in the first 8 weeks, to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(14\times1.1^{7}\)✓ 1
A1 27.3 km✓ 1
Part (b)
M1 \(S_8=\frac{14(1.1^8-1)}{1.1-1}\)✓ 1
A1 160.1 km✓ 1

Common error: A 10% increase gives a common ratio of 1.1, not 0.1.

235IBMA3H-1-18r6-021.1 Standard form Paper 2Easy6 marks

Calculator

The orbital period of a comet shortens by 6% with each successive orbit. Its first orbit lasts 76 years, and the periods form a geometric sequence.

(a)

Find the period of the fifth orbit, to 3 significant figures.

(2)
(b)

Find the number of the first orbit whose period is less than 50 years.

(3)
(c)

Write down the period of that orbit, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(76\times0.94^{4}\)✓ 1
A1 59.3 years✓ 1
Part (b)
M1 \(76\times0.94^{n-1}<50\)✓ 1
A1 \(n-1>6.77\) (or valid trial of terms: \(u_7=52.4\), \(u_8=49.3\))✓ 1
A1 \(n=8\)✓ 1
Part (c)
A1 \(76\times0.94^{7}=49.3\) years✓ 1

Exam tip: Using logarithms to solve \(ar^{n-1}<b\) reverses the inequality when you divide by a negative \(\ln r\).

236IBMA3H-1-18r10-011.4 Financial applications Paper 2Easy4 marks

Calculator

A fishing cooperative invests €8000 for 5 years in an account paying a nominal annual interest rate of 3.6%, compounded monthly.

(a)

Find the value of the investment at the end of the 5 years. Give your answer to the nearest cent.

(3)
(b)

Hence find the interest earned, using the value from part (a) to the nearest cent.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
attempt to use \(8000\left(1+\frac{3.6}{100\times12}\right)^{n}\)✓ 1
\(n=60\) (or \(12\times5\))✓ 1
€9575.16✓ 1
Part (b)
€1575.16✓ 1

Common error: Divide the annual rate by the number of compounding periods per year, and multiply the years by the same number.

237IBMA3H-1-18r10-021.4 Financial applications Paper 2Easy5 marks

Calculator

A fashion workshop buys a sewing machine for €2400. Its value depreciates by 12% each year.

(a)

Find the value of the machine after 4 years, to the nearest cent.

(2)
(b)

Find how many years, to three significant figures, it takes for the value to fall to €1000.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(2400\times0.88^{4}\)✓ 1
€1439.27✓ 1
Part (b)
\(2400\times0.88^{n}=1000\)✓ 1
\(n=\frac{\ln(1000/2400)}{\ln0.88}\)✓ 1
\(n=6.85\) years✓ 1

Exam tip: A 12% fall each year means multiplying by \(0.88\), not by \(0.12\).

238IBMA3H-1-19r11-011.16 Systems of linear equations Paper 2Easy6 marks

Calculator

A patient follows a tapering course of a drug. The mass taken on day 1 is 80 mg, and the mass taken on each later day is 75% of the mass taken on the previous day. The masses taken on successive days form a geometric sequence.

(a)

Find the mass of drug taken on day 4.

(2)
(b)

Show that the total mass taken over the first 6 days is 263 mg, correct to 3 significant figures.

(2)
(c)

The course is extended without end. Find the greatest total mass of the drug that could ever be taken.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(u_n=ar^{n-1}\) with \(a=80\), \(r=0.75\), \(n=4\)✓ 1
A1 33.75 mg✓ 1
Part (b)
M1 use of \(S_n=\frac{a(1-r^n)}{1-r}\) with \(n=6\)✓ 1
A1 \(S_6=\frac{80(1-0.75^6)}{0.25}=263.047\) mg, which is 263 mg to 3 significant figures✓ 1
Part (c)
M1 use of \(S_\infty=\frac{a}{1-r}\) with \(|r|<1\)✓ 1
A1 \(\frac{80}{1-0.75}=320\) mg✓ 1

Exam tip: A sum to infinity exists only when \(|r|<1\); say so before using \(\frac{a}{1-r}\).

239IBMA3H-1-18r13-081.4 Financial applications Paper 2Easy5 marks

Calculator

A fishery estimates that its cod stock is 12 000 fish at the start of a year, and that the stock falls by 8% each year.

(a)

Find the number of complete years after which the stock first falls below 8000 fish.

(3)
(b)

Find the stock at that time, to the nearest whole number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(12000\times 0.92^n < 8000\)✓ 1
M1 attempt to solve, e.g. \(n > \frac{\ln(2/3)}{\ln 0.92} = 4.86\), or compare \(n=4\) (8597) and \(n=5\)✓ 1
A1 \(n = 5\)✓ 1
Part (b)
M1 \(12000\times 0.92^5\)✓ 1
A1 7909✓ 1

Exam tip: A fall of 8% each year multiplies by 0.92, not 0.08.

240IBMA3H-1-18r13-091.4 Financial applications Paper 2Easy4 marks

Calculator

A fashion label deposits €4500 in an account that pays a nominal annual interest rate of 3.2%, compounded quarterly.

(a)

Find the amount in the account after 6 years, to the nearest cent.

(2)
(b)

Find the interest earned over the 6 years, using the unrounded value from part (a) and giving your answer to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(4500\left(1+\frac{3.2}{400}\right)^{24}\)✓ 1
A1 €5448.35✓ 1
Part (b)
M1 \(5448.3536\ldots - 4500\)✓ 1
A1 €948.35✓ 1

Common error: Using 3.2% each quarter. The quarterly rate is \(\frac{3.2}{4}\)% and there are \(4\times 6\) periods.

241IBMA3H-1-18r14-071.16 Systems of linear equations Paper 2Easy5 marks

Calculator

A fashion designer buys three fabrics: silk, cotton and linen, priced at €\(x\), €\(y\) and €\(z\) per metre respectively. Three orders cost: 2 m silk, 3 m cotton and 1 m linen for €69; 1 m silk, 2 m cotton and 3 m linen for €68; 3 m silk, 1 m cotton and 2 m linen for €85.

(a)

Write down a system of three linear equations for \(x\), \(y\) and \(z\), and solve it.

(3)
(b)

Find the cost of a cloak using 3 m of silk, 2.5 m of cotton and 1.5 m of linen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forming the equations \(2x+3y+z=69\), \(x+2y+3z=68\), \(3x+y+2z=85\)✓ 1
A1 \(x=18\)✓ 1
A1 \(y=7\) and \(z=12\)✓ 1
Part (b)
M1 \(3(18)+2.5(7)+1.5(12)\)✓ 1
A1 €89.50 (ECF from (a))✓ 1

Exam tip: Check your solution by substituting it back into all three original equations.

242IBMA3H-1-18r16-081.7 Rational exponents and laws of logarithms Paper 2Easy7 marks

Calculator

A nutritionist plans a meal from three foods, X, Y and Z. Each 100 g serving contains the following (in grams):
X: 20 protein, 50 carbohydrate, 5 fat
Y: 10 protein, 60 carbohydrate, 15 fat
Z: 30 protein, 10 carbohydrate, 20 fat
The meal uses x servings of X, y servings of Y and z servings of Z and must contain exactly 200 g protein, 310 g carbohydrate and 125 g fat.

(a)

Write down a system of three linear equations in x, y and z.

(3)
(b)

Solve the system to find x, y and z.

(2)
(c)

Hence find the total mass of the meal, in grams.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 20x + 10y + 30z = 200✓ 1
A1 50x + 60y + 10z = 310✓ 1
A1 5x + 15y + 20z = 125✓ 1
Part (b)
M1 attempt to solve (e.g. matrix or solver)✓ 1
A1 x = 3, y = 2, z = 4✓ 1
Part (c)
M1 (3 + 2 + 4) × 100✓ 1
A1 900 g (ECF from (b))✓ 1

Exam tip: Write one equation per nutrient and keep the unknowns in the same order in each equation so the system can be entered into a calculator without errors.

243IBMA3H-1-18r24-011.1 Standard form Paper 2Easy4 marks

Calculator

Light from a star travels 2.4 × 1016 m to reach a telescope. The speed of light is 3.0 × 108 m s−1.

(a)

Find the time taken, in seconds, giving your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer.

(2)
(b)

One year is 3.15 × 107 seconds. Convert your answer to part (a) into years, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 time = distance ÷ speed = 2.4 × 1016 ÷ 3.0 × 108✓ 1
A1 8.0 × 107 s✓ 1
Part (b)
M1 8.0 × 107 ÷ 3.15 × 107✓ 1
A1 2.5397 = 2.54 years✓ 1

Common error: When dividing numbers in standard form, divide the coefficients and subtract the powers of ten separately; check that the coefficient ends up between 1 and 10.

244IBMA3H-1-18r24-021.4 Financial applications Paper 2Easy4 marks

Calculator

A rail company buys a track-maintenance machine for €85 000. The machine loses value at 12% of its current value each year.

(a)

Find the value of the machine after 5 years, to the nearest euro.

(2)
(b)

A second machine also costs €85 000 and is worth €40 000 after 6 years. It loses the same percentage of its value each year. Find this annual percentage, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 85 000 × 0.885✓ 1
A1 €44857.21 = €44857✓ 1
Part (b)
M1 85 000(1 − r/100)6 = 40 000, so 1 − r/100 = (40/85)1/6✓ 1
A1 r = 11.806 = 11.8%✓ 1

Exam tip: For a loss of 12% each year the multiplier is 1 − 0.12 = 0.88, not 0.12; check that the multiplier is below 1 for depreciation.

245IBMA3H-1-20r29-031.13 Polar and Euler forms Paper 2Easy5 marks

Calculator

A robotic arm tip is at position \(z=5e^{0.9i}\) metres in the complex plane, with the argument in radians.

OReImOz = 5e^{0.9i}w = 4 − 3i
Not to scale
(a)

Find the real part and the imaginary part of \(z\), each to 3 significant figures.

(2)
(b)

The arm moves to \(w=4-3i\). Write \(w\) in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\), giving \(\theta\) to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 real part \(5\cos0.9=3.11\)✓ 1
A1 imaginary part \(5\sin0.9=3.92\)✓ 1
Part (b)
M1 \(r=\sqrt{4^2+3^2}\)✓ 1
A1 \(r=5\)✓ 1
A1 \(\theta=\arctan\left(-\frac34\right)=-0.644\)✓ 1

Exam tip: Make sure the calculator is in radian mode when the argument is given without degrees.

246IBMA3H-1-20r29-101.1 Standard form Paper 2Easy5 marks

Calculator

The charge level of a delivery robot's battery is modelled by \(C=100e^{-0.15t}\), where \(C\) is the percentage charge and \(t\) is the time in minutes after leaving the dock.

(a)

Find the charge level after 5 minutes, to 3 significant figures.

(2)
(b)

Find the time for the charge level to fall to 25%. Give your answer in minutes to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(100e^{-0.15\times5}\)✓ 1
A1 47.2%✓ 1
Part (b)
M1 \(100e^{-0.15t}=25\), so \(e^{-0.15t}=0.25\)✓ 1
M1 \(t=\frac{\ln4}{0.15}\)✓ 1
A1 9.24 minutes✓ 1

Exam tip: To solve \(e^{kt}=c\), take natural logs: \(kt=\ln c\). Note that \(\ln0.25=-\ln4\), so the negative signs cancel.

247IBMA3H-1-20r35-011.1 Standard form Paper 2Easy5 marks

Calculator

A marine biologist counts phytoplankton cells in a bay. One litre of seawater contains 4.2 × 106 cells, and each cell has a mass of 3.5 × 10−9 g. The bay holds 6.0 × 1010 litres of water, and the cell concentration is the same throughout.

(a)

Find the mass of phytoplankton in one litre of water, in grams. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(b)

Find the total mass of phytoplankton in the bay, in kilograms, giving your answer in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 4.2 × 106 × 3.5 × 10−9✓ 1
A1 1.47 × 10−2 g✓ 1
Part (b)
M1 multiply the mass per litre by 6.0 × 1010✓ 1
A1 8.82 × 108 g✓ 1
A1 8.82 × 105 kg✓ 1

Common error: When converting grams to kilograms, divide by 1000 (subtract 3 from the power of ten) and then check that the number before the power of ten is still between 1 and 10.

248IBMA3H-1-20r35-031.7 Rational exponents and laws of logarithms Paper 2Easy5 marks

Calculator

The pH of seawater is given by \(\mathrm{pH}=-\log_{10}[\mathrm{H}^+]\), where \([\mathrm{H}^+]\) is the concentration of hydrogen ions in mol per litre. Surface seawater has \([\mathrm{H}^+]=7.9\times10^{-9}\).

(a)

Find the pH of the seawater, to 3 significant figures.

(2)
(b)

Ocean acidification lowers the pH by 0.12. Find the percentage increase in \([\mathrm{H}^+]\), to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(-\log_{10}(7.9\times10^{-9})\)✓ 1
A1 pH = 8.10✓ 1
Part (b)
M1 new pH = old pH − 0.12 so \([\mathrm{H}^+]\) is multiplied by \(10^{0.12}\)✓ 1
A1 \(10^{0.12}=1.318\)✓ 1
A1 31.8% increase✓ 1

Common error: A fall in pH means a rise in hydrogen-ion concentration; because the scale is logarithmic, a change of 0.12 in pH is a multiplication of the concentration, not an addition.

249IBMA3H-1-19r39-011.1 Standard form Paper 2Easy5 marks

Calculator

A marine biologist estimates a phytoplankton bloom has a mean density of 3.2×104 cells per m3 over a body of water of volume 5.6×107 m3.

(a)

Calculate the total number of cells, N. Give your answer in the form a×10k, where 1 ≤ a < 10 and k is an integer, correct to 3 significant figures.

(2)
(b)

Each cell has a mass of 4.5×10−7 g. Using the unrounded value of N from part (a), find the total mass of the cells in kg. Give your answer in standard form correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies 3.2×104 by 5.6×107✓ 1
A1 1.79×1012 (1.792×1012 unrounded also accepted)✓ 1
Part (b)
M1 multiplies number of cells by 4.5×10−7 g✓ 1
M1 converts grams to kilograms (divides by 103)✓ 1
A1 8.06×102 kg (806.4 kg unrounded)✓ 1

Common error: When converting units inside standard form, change the power of ten only once; converting grams to kilograms means dividing by 10³, not multiplying.

250IBMA3H-1-19r39-121.1 Standard form Paper 2Easy6 marks

Calculator

A brick tower contains 8.4×106 bricks, each of mass 2.3 kg. Its foundation can safely carry a load of 2.0×107 kg.

(a)

Calculate the total mass of the bricks, giving your answer in standard form correct to 3 significant figures.

(2)
(b)

Using the unrounded total mass from part (a), find the remaining load the foundation can carry. Give your answer in standard form.

(2)
(c)

Find the percentage of the foundation's safe load used by the bricks, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(8.4\times10^6\times2.3\)✓ 1
A1 \(1.93\times10^7\) kg (1.932×107 unrounded)✓ 1
Part (b)
M1 \(2.0\times10^7-1.932\times10^7\)✓ 1
A1 \(6.8\times10^5\) kg✓ 1
Part (c)
M1 \(\dfrac{1.932\times10^7}{2.0\times10^7}\times100\)✓ 1
A1 96.6 %✓ 1

Exam tip: Keep the unrounded value from a previous part in your calculator so that later answers do not pick up rounding errors.

251IBMA3H-1-18r47-051.16 Systems of linear equations Paper 2Easy7 marks

Calculator

A cycle shop sells helmets at €h each and lights at €l each. Three helmets and two lights cost €153 in total. Two helmets and five lights cost €190 in total.

(a)

Write down two equations in \(h\) and \(l\).

(2)
(b)

Solve the equations to find the price of one helmet and the price of one light.

(3)
(c)

Find the cost of four helmets and three lights.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(3h + 2l = 153\)✓ 1
A1 \(2h + 5l = 190\)✓ 1
Part (b)
M1 eliminate a variable or use the equation solver on a calculator✓ 1
A1 \(h = 35\)✓ 1
A1 \(l = 24\)✓ 1
Part (c)
M1 \(4(35) + 3(24)\) (ECF from (b))✓ 1
A1 €212✓ 1

Exam tip: Check your solution by substituting it back into both original equations.

252AAHL1-q00011.1 Standard form Paper 2Medium9 marks

A password is a string of characters, each chosen from 94 possible keyboard symbols. Symbols may be repeated. A computer can test \(1.2\times10^{18}\) passwords each second. Take one year to be \(3.15\times10^{7}\) seconds.

(a)

Find the number of different passwords of length 16. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\), \(k\in\mathbb{Z}\) and \(a\) is given correct to three significant figures.

(2)
(b)

Find the time, in years, that the computer needs to test every password of length 16. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt 10\) and \(k\in\mathbb{Z}\).

(2)
(c)

Find the least password length for which the computer needs more than \(10^{9}\) years to test every password of that length.

(3)
(d)

Find the percentage of passwords of length 16 in which no symbol is used more than once.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(94^{16}\)(M1)
\(3.72\times10^{31}\)A1\(3.7157\ldots\times10^{31}\). Award A0 for 3.72E31.
(b)
\(\dfrac{3.7157\ldots\times10^{31}}{1.2\times10^{18}\times3.15\times10^{7}}\)(M1)
\(9.83\times10^{5}\) yearsA1\(983\,000.7\ldots\) Accept \(9.83\times10^{5}\) or \(9.84\times10^{5}\) from a rounded (a).
(c)
\(\dfrac{94^{n}}{1.2\times10^{18}\times3.15\times10^{7}}\gt10^{9}\), ie \(94^{n}\gt3.78\times10^{34}\)(M1)Or a table of values of the time for successive lengths.
\(n\gt\dfrac{\log_{10}\left(3.78\times10^{34}\right)}{\log_{10}94}=17.52\ldots\), or length 17: \(9.24\times10^{7}\) years; length 18: \(8.69\times10^{9}\) years(A1)
18A1Award A0 for 17.5.
(d)
\(\dfrac{{}^{94}P_{16}}{94^{16}}\) or \(\dfrac{94\times93\times\cdots\times79}{94^{16}}\)(M1)
\(25.8\%\)A1\(25.8416\ldots\%\)

Answers: (a) \(3.72\times10^{31}\) · (b) \(9.83\times10^{5}\) years · (c) 18 · (d) \(25.8\%\)

253AAHL1-q00081.4 Financial applications Paper 2Medium8 marks

Ana invests €10 000 in an account that pays a nominal annual interest rate of 6%, compounded monthly. Ben invests €12 000 in an account that pays a nominal annual interest rate of 3.5%, compounded half-yearly.

(a)

Find the value of Ana’s investment after 5 years. Give your answer to the nearest cent.

(2)
(b)

Find the least number of complete years after which the value of Ana’s investment is greater than the value of Ben’s investment.

(3)
(c)

Carla invests €11 000 in an account that pays a nominal annual interest rate of \(r\%\), compounded quarterly. After 8 years the value of Carla’s investment is equal to the value of Ana’s investment. Find the value of \(r\), correct to two decimal places.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(10\,000\left(1+\dfrac{6}{1200}\right)^{60}\), or TVM with \(N=60\), \(I\%=6\), \(PV=-10\,000\), \(P/Y=C/Y=12\)(M1)
€13 488.50A1
(b)
\(10\,000(1.005)^{12t}\gt12\,000(1.0175)^{2t}\)(M1)Inequality, table or graphs.
after 7 years: Ana €15 203.70 < Ben €15 299.00; after 8 years: Ana €16 141.43 > Ben €15 839.15(A1)Or \(t\gt7.248\ldots\).
8 yearsA1Award A0 for 7.25.
(c)
Ana after 8 years: \(10\,000(1.005)^{96}=16\,141.43\)(M1)
\(11\,000\left(1+\dfrac{r}{400}\right)^{32}=16\,141.43\), or TVM with \(N=32\), \(PV=-11\,000\), \(FV=16\,141.43\), \(P/Y=C/Y=4\)(M1)
\(r=4.82\)A1\(4.8225\ldots\)

Answers: (a) €13 488.50 · (b) 8 years · (c) \(r=4.82\)

254AAHL1-q00461.13 Polar and Euler forms Paper 2Medium6 marks

Let \(z_{1}=3\mathrm{e}^{0.6i}\) and \(z_{2}=5\mathrm{e}^{2.1i}\).

OReImOz₁z₂z₁ + z₂0.62.1
Not to scale
(a)

Find \(z_{1}+z_{2}\) in the form \(a+bi\).

(2)
(b)

Hence write \(z_{1}+z_{2}\) in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(2)
(c)

By considering the parallelogram formed by \(z_{1}\), \(z_{2}\) and \(z_{1}+z_{2}\) on an Argand diagram, explain why \(|z_{1}+z_{2}|^{2}=34+30\cos1.5\).

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(3(\cos0.6+i\sin0.6)+5(\cos2.1+i\sin2.1)\;(=2.476+1.694i-2.524+4.316i)\)(M1)
\(-0.0482+6.01i\)A1\(-0.048224\ldots+6.00997\ldots i\)
(b)
\(r=6.01\)A1\(6.01017\ldots\)
\(\theta=1.58\)A1\(1.57882\ldots\) Award A0 for \(-1.56\) (from \(\arctan\) without quadrant adjustment).
(c)
the angle between \(z_{1}\) and \(z_{2}\) is \(2.1-0.6=1.5\), so the angle of the parallelogram opposite the diagonal \(z_{1}+z_{2}\) is \(\pi-1.5\)R1
cosine rule: \(|z_{1}+z_{2}|^{2}=3^{2}+5^{2}-2(3)(5)\cos(\pi-1.5)=34+30\cos1.5\)R1

Answers: (a) \(-0.0482+6.01i\) · (b) \(6.01\mathrm{e}^{1.58i}\)

255AAHL1-q00471.13 Polar and Euler forms Paper 2Medium7 marks

In an electrical installation the complex power is \(S=P+Qi\), where \(P\) is the real power in kW and \(Q\) is the reactive power in kVAr. The apparent power is \(|S|\) and the power factor is \(\cos(\arg S)\).

A workshop has complex power \(S=4.2+3.1i\).

(a)

Find the apparent power.

(2)
(b)

Find the power factor.

(2)
(c)

A capacitor bank is installed. It reduces \(Q\) but leaves \(P\) unchanged, and the power factor becomes 0.95. Find the new value of \(Q\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(\sqrt{4.2^{2}+3.1^{2}}\)(M1)
\(5.22\) (kVA)A1\(5.22015\ldots\)
(b)
\(\arg S=\arctan\dfrac{3.1}{4.2}=0.6358\ldots\)(M1)Or \(\dfrac{4.2}{|S|}\).
\(0.805\)A1\(0.804574\ldots\)
(c)
\(\arg S_{\text{new}}=\arccos0.95=0.31756\ldots\)(M1)
\(Q=4.2\tan(\arccos0.95)\)(M1)Or \(|S_{\text{new}}|=\dfrac{4.2}{0.95}\) and Pythagoras.
\(Q=1.38\) (kVAr)A1\(1.38047\ldots\)

Answers: (a) 5.22 kVA · (b) 0.805 · (c) 1.38 kVAr

256AAHL1-q00481.13 Polar and Euler forms Paper 2Medium6 marks

The points Z and W on an Argand diagram represent \(z=5+2i\) and \(w=-1+4i\) respectively. O is the origin.

OReImOZ(5 + 2i)W(−1 + 4i)
Not to scale
(a)

Find \(\dfrac{w}{z}\) in the form \(a+bi\).

(2)
(b)

Hence find the size of angle ZOW.

(2)
(c)

Find the area of triangle OZW.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{(-1+4i)(5-2i)}{29}\)(M1)
\(\dfrac{3}{29}+\dfrac{22}{29}i\)A1Accept \(0.103+0.759i\).
(b)
angle ZOW \(=\arg\dfrac{w}{z}=\arctan\dfrac{22}{3}\)(M1)Or \(\arg w-\arg z=1.8158\ldots-0.3805\ldots\)
\(1.44\)A1\(1.43527\ldots\) Accept \(82.2^{\circ}\).
(c)
\(\dfrac12|z||w|\sin(\text{ZOW})=\dfrac12\sqrt{29}\sqrt{17}\sin1.435\ldots\)(M1)
\(11\)A1Accept 11.0.

Answers: (a) \(\tfrac{3}{29}+\tfrac{22}{29}i\) · (b) 1.44 · (c) 11

257AAHL1-q00591.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 2Medium7 marks

Consider the equation \(z^{5}=3+4i\).

(a)

Find the modulus and argument of \(3+4i\).

(2)
(b)

Find the five roots of the equation, giving your answers in the form \(r\mathrm{e}^{i\theta}\), where \(r\gt 0\) and \(-\pi\lt\theta\le\pi\).

(3)
(c)

The roots are represented on an Argand diagram by the vertices of a regular pentagon. Find the perimeter of the pentagon.

(2)
Show mark scheme
WorkingMarkNotes
(a)
modulus \(5\)A1
argument \(0.927\)A1\(0.927295\ldots\)
(b)
\(z=5^{\frac15}\mathrm{e}^{\frac{0.9273+2k\pi}{5}i}\)M1De Moivre’s theorem with a rational exponent.
\(1.38\mathrm{e}^{0.185i}\)A1\(5^{0.2}=1.37972\ldots,\ 0.185459\ldots\)
\(1.38\mathrm{e}^{1.44i},\ 1.38\mathrm{e}^{2.70i},\ 1.38\mathrm{e}^{-2.33i},\ 1.38\mathrm{e}^{-1.07i}\)A1\(1.44210\ldots,\ 2.69873\ldots,\ -2.32782\ldots,\ -1.07118\ldots\) Award A0 for 3.96 or 5.21 (outside the interval).
(c)
side \(=2\times5^{\frac15}\sin\dfrac{\pi}{5}\;(=1.6219\ldots)\)(M1)Or the distance between two adjacent roots.
\(8.11\)A1\(8.10964\ldots\)

Answers: (a) 5, 0.927 · (b) \(1.38\mathrm{e}^{i\theta}\), \(\theta=0.185,\,1.44,\,2.70,\,-2.33,\,-1.07\) · (c) 8.11

258AAHL1-q00871.16 Systems of linear equations Paper 2Medium9 marks

A ball is thrown upwards. Its height above the ground, \(h\) metres, \(t\) seconds after it is thrown is modelled by \(h(t)=at^{2}+bt+c\). The following table shows three measurements.

\(t\) (s)0.41.22.0
\(h\) (m)5.307.984.40
(a)

Write down a system of three linear equations in \(a\), \(b\) and \(c\).

(2)
(b)

Find the value of \(a\), of \(b\) and of \(c\).

(2)
(c)

Find the maximum height of the ball according to the model.

(2)
(d)

Find the time at which the ball reaches the ground.

(2)
(e)

Interpret the value of \(c\) in this context.

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(0.16a+0.4b+c=5.30\)A1At least one correct equation.
\(1.44a+1.2b+c=7.98\) and \(4a+2b+c=4.40\)A1All three correct.
(b)
use of GDC to solve the system(M1)
\(a=-4.89,\ b=11.2,\ c=1.61\)A1Exact values \(a=-4.890625,\ b=11.175,\ c=1.6125\).
(c)
\(t=-\dfrac{b}{2a}=1.14\ldots\), or GDC maximum(M1)
\(8.00\) mA1\(7.99617\ldots\) Accept \(7.99\)–\(8.03\) (8.02 from 3 s.f. coefficients).
(d)
solving \(h(t)=0\)(M1)
\(t=2.42\) sA1\(2.42116\ldots\) Accept \(2.42\)–\(2.43\) (2.43 from 3 s.f. coefficients). Award A0 if the negative root is also given as the answer.
(e)
the height (1.61 m) from which the ball was thrownA1

Answers: (b) \(a=-4.89,\ b=11.2,\ c=1.61\) · (c) 8.00 m · (d) 2.42 s · (e) release height

259IBMA2H-045-06-061.4 Financial applications Paper 2Medium9 marks

Calculator

Two athletes follow progressive training plans. Athlete A runs 12 km in week 1 and increases the weekly distance by 1.5 km each week. Athlete B runs 20 km in week 1 and increases the weekly distance by 0.8 km each week.

(a)

Find the first week in which Athlete A runs a greater distance than Athlete B in that week.

(3)
(b)

Find the first week by the end of which Athlete A's total distance since week 1 is greater than Athlete B's.

(4)
(c)

Find Athlete A's total distance at the end of the week found in the previous part.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(12+1.5(n-1)>20+0.8(n-1)\) ; A1 \(0.7(n-1)>8\), so \(n>12.43\) ; A1 week 13✓ 3
Part (b)
M1 \(S_A=\frac{n}{2}(24+1.5(n-1))\) ; M1 sets \(S_A>S_B\) with \(S_B=\frac{n}{2}(40+0.8(n-1))\) ; A1 \(0.7(n-1)>16\), so \(n>23.86\) ; A1 week 24✓ 4
Part (c)
M1 \(S_{24}=\frac{24}{2}(24+1.5\times23)\) ; A1 702 km✓ 2

Exam tip: When comparing the totals, the factor \(\frac{n}{2}\) is positive, so it can be cancelled from both sides of the inequality. This leaves a linear inequality and avoids a quadratic.

260IBMA2H-045-06-091.3 Geometric sequences and series Paper 2Medium10 marks

Calculator

A streaming service models the number of new subscribers gained in month \(n\) as an arithmetic sequence. In the first 6 months it gained a total of 9300 new subscribers, and in the first 15 months it gained a total of 36 750 new subscribers.

(a)

Find the number of new subscribers gained in month 1 and the common difference.

(4)
(b)

Find the first month in which more than 5000 new subscribers are gained.

(2)
(c)

Find the first month by the end of which the total number of new subscribers exceeds 200 000, and state this total.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(3(2a+5d)=9300\) ; M1 \(7.5(2a+14d)=36750\) ; A1 \(d=200\) ; A1 \(a=1050\)✓ 4
Part (b)
M1 \(1050+200(n-1)>5000\) ; A1 month 21✓ 2
Part (c)
M1 \(S_n=\frac{n}{2}(2100+200(n-1))=100n^2+950n\) ; M1 attempts to solve \(100n^2+950n>200000\) ; A1 \(n>40.2\), so month 41 ; A1 total \(=207\,050\)✓ 4

Exam tip: A quadratic inequality in \(n\) will usually give a non-integer root. State the root, then round up to the next whole month. Checking \(S_{40}\) and \(S_{41}\) confirms the answer.

261IBMA2H-045-03-061.5 Exponents and introduction to logarithms Paper 2Medium8 marks

Calculator

The sound level \(L\) decibels (dB) of a sound with intensity \(I\) W m−2 is given by \(L=10\log_{10}\left(\frac{I}{I_0}\right)\), where \(I_0=10^{-12}\) W m−2.

(a)

A hairdryer produces sound of intensity 3.2×10−5 W m−2. Find its sound level.

(2)
(b)

Noise-cancelling headphones are tested against a sound level of 110 dB. Find the intensity of this sound.

(2)
(c)

Show that doubling the intensity of any sound increases its sound level by approximately 3.01 dB.

(2)
(d)

Find how many times more intense an 85 dB sound is than a 60 dB sound.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes: \(10\log_{10}(3.2\times10^{7})\)✓ 1
A1 75.1 dB✓ 1
Part (b)
M1 \(\frac{I}{I_0}=10^{11}\)✓ 1
A1 \(I=0.1\) W m−2✓ 1
Part (c)
M1 \(10\log_{10}\left(\frac{2I}{I_0}\right)-10\log_{10}\left(\frac{I}{I_0}\right)=10\log_{10}2\) using log laws✓ 1
A1 \(10\log_{10}2=3.0103\ldots\approx3.01\) dB✓ 1
Part (d)
M1 \(\frac{I_2}{I_1}=10^{(85-60)/10}=10^{2.5}\)✓ 1
A1 316✓ 1

Common error: A 25 dB difference does not mean 25 times the intensity. Every 10 dB multiplies the intensity by 10.

262IBMA2H-045-03-071.16 Systems of linear equations Paper 2Medium7 marks

Calculator

A hospital pharmacist mixes three intravenous solutions, A, B and C, to make a single bag. Each litre of each solution contains:
A: 50 g glucose, 30 mmol sodium, 20 mmol potassium
B: 20 g glucose, 150 mmol sodium, 0 mmol potassium
C: 0 g glucose, 77 mmol sodium, 40 mmol potassium
The bag must contain exactly 85 g glucose, 197 mmol sodium and 70 mmol potassium. Let \(a\), \(b\) and \(c\) be the volumes in litres of A, B and C used.

(a)

Write down a system of three equations in \(a\), \(b\) and \(c\).

(2)
(b)

Find the volume of each solution needed.

(3)
(c)

Solutions A, B and C cost $4.20, $3.10 and $5.60 per litre respectively. Find the cost of the solutions in the bag.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 any two of \(50a+20b=85\), \(30a+150b+77c=197\), \(20a+40c=70\) correct✓ 1
A1 all three equations correct✓ 1
Part (b)
M1 attempts to solve the system, e.g. by GDC or matrix method✓ 1
A1 \(a=1.5\) L✓ 1
A1 \(b=0.5\) L, \(c=1\) L✓ 1
Part (c)
M1 \(1.5(4.20)+0.5(3.10)+1(5.60)\)✓ 1
A1 $13.45✓ 1

Exam tip: Write one equation per nutrient, not one per solution. Each equation sums one nutrient's contribution from all three solutions.

263IBMA2H-045-03-101.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

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The atmospheric pressure \(P\) kPa at a height of \(h\) metres above sea level is modelled by \(P=101.3e^{-0.000125h}\).

(a)

Find the pressure predicted by the model at the summit of a mountain 8849 m high.

(2)
(b)

Find the height at which the pressure is half its sea-level value.

(3)
(c)

A delivery drone's rotors need a pressure of at least 60 kPa. Find the maximum height at which the drone can operate.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes \(h=8849\)✓ 1
A1 33.5 kPa✓ 1
Part (b)
M1 sets up \(e^{-0.000125h}=0.5\)✓ 1
M1 takes natural logarithms: \(h=\frac{\ln2}{0.000125}\)✓ 1
A1 \(h=5550\) m✓ 1
Part (c)
M1 solves \(101.3e^{-0.000125h}=60\), giving \(h=\frac{\ln(101.3/60)}{0.000125}\)✓ 1
A1 4190 m✓ 1

Common error: In part (b), 101.3 cancels, so the answer does not depend on the sea-level pressure. Substituting \(P=0.5\) instead of \(P=50.65\) is a common slip.

264IBMA2H-045-04-081.12 Complex numbers in Cartesian form Paper 2Medium7 marks

Calculator

In an AC circuit the voltage \(V\) volts, current \(I\) amps and impedance \(Z\) ohms are complex numbers related by \(V=IZ\). A supply with \(V=230\) is connected to a motor with impedance \(Z=40+30i\).

(a)

Find \(I\) in the form \(a+bi\).

(2)
(b)

Find \(|I|\).

(1)
(c)

A capacitor is added in series. The new impedance is \(40+(30-X)i\), where \(X>0\). The magnitude of the current is now 5 A. Find the possible values of \(X\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(I=\frac{230}{40+30i}\times\frac{40-30i}{40-30i}\)✓ 1
A1 \(I=3.68-2.76i\)✓ 1
Part (b)
A1 \(|I|=4.6\) A✓ 1
Part (c)
M1 uses \(|Z|=\frac{230}{5}=46\)✓ 1
M1 \(40^2+(30-X)^2=46^2\)✓ 1
A1 \((30-X)^2=516\)✓ 1
A1 \(X=7.28\) or \(X=52.7\)✓ 1

Exam tip: Taking moduli of \(V=IZ\) gives \(|V|=|I||Z|\). This turns a complex condition into one real equation.

265IBMA2H-045-00-051.4 Financial applications Paper 2Medium8 marks

Calculator

A medical clinic borrows $240 000 to buy an ultrasound scanner. The loan has a nominal annual interest rate of 5.4%, compounded monthly. It is repaid by equal payments at the end of each month over 8 years.

$240 000startpayment 1month 1payment 2month 2…payment 36month 36…payment 96month 96interest 5.4% per year, compounded monthly; equal payments at the end of each month
Timeline not to scale
(a)

Find the monthly payment, giving your answer to the nearest cent.

(3)
(b)

Find the total interest paid over the 8 years.

(2)
(c)

Immediately after the 36th payment, the clinic pays off the remaining balance in one sum. Find this amount.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses a TVM solver or formula with N = 96, I% = 5.4, PV = 240 000, FV = 0, P/Y = C/Y = 12✓ 1
M1 e.g. P = 240 000 × 0.0045 / (1 − 1.0045−96)✓ 1
A1 $3084.29✓ 1
Part (b)
M1 96 × 3084.29 − 240 000✓ 1
A1 $56 092 (accept $56 100)✓ 1
Part (c)
M1 recognises balance = present value of the remaining 60 payments, or uses solver with N = 36 to find FV✓ 1
M1 e.g. 3084.29 × (1 − 1.0045−60)/0.0045✓ 1
A1 $161 862 (accept $162 000)✓ 1

Exam tip: For an outstanding balance, either discount the remaining payments back to now or run the solver for the payments already made; both give the same value.

266IBMA2H-045-00-061.4 Financial applications Paper 2Medium8 marks

Calculator

A surveying company buys a drone for $18 500. Its value decreases by a fixed percentage, r%, each year. After 4 years the drone is worth 55% of its purchase price.

(a)

Find the value of r.

(3)
(b)

The company replaces the drone at the end of the first complete year in which its value is less than 20% of the purchase price. Find after how many years the drone is replaced.

(3)
(c)

At the time of purchase, the company invests an amount $D at a nominal annual interest rate of 4.2%, compounded quarterly, so that the investment is worth $18 500 when the drone is replaced. Find D.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets up (1 − r/100)4 = 0.55✓ 1
M1 1 − r/100 = 0.550.25 = 0.86117…✓ 1
A1 r = 13.9✓ 1
Part (b)
M1 sets up 0.86117…n < 0.2✓ 1
A1 n > 10.77… (or 10 years gives 22.4%, 11 years gives 19.3%)✓ 1
A1 11 years✓ 1
Part (c)
M1 D(1 + 0.042/4)44 = 18 500✓ 1
A1 D = $11 700 (11 683.5)✓ 1

Common error: Using 0.55/4 as the yearly loss treats depreciation as linear; a fixed percentage decrease needs the fourth root of 0.55.

267IBMA2H-045-02-071.13 Polar and Euler forms Paper 2Medium12 marks

Calculator

A delivery drone hovers at position \(A\), represented by \(a=300+400i\) metres relative to its base at the origin. The real axis points east and the imaginary axis points north. The drone is rotated anticlockwise about the base, at constant distance from it, to position \(B\), represented by \(b=-140+480i\).

OReImOA (300 + 400i)B (−140 + 480i)
Real axis points east, imaginary axis north; distances in metres. Not to scale
(a)

Write \(a\) in the form \(re^{i\theta}\).

(2)
(b)

Find \(\frac{b}{a}\) in the form \(p+qi\), and hence find the angle of rotation.

(4)
(c)

Starting from \(A\), the drone repeats this same rotation \(n\) times. Find the least value of \(n\) for which the drone lies south-west of the base, that is, with negative real part and negative imaginary part.

(3)
(d)

For this value of \(n\), find the position of the drone and its straight-line distance from \(A\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=500\)✓ 1
A1 \(\theta=\arctan\frac{4}{3}=0.927\), so \(a=500e^{0.927i}\)✓ 1
Part (b)
M1 valid method for the division, e.g. multiplies the numerator and denominator by \(300-400i\), or direct use of a GDC✓ 1
A1 \(\frac{b}{a}=0.6+0.8i\)✓ 1
R1 \(\left|\frac{b}{a}\right|=1\), so the movement is a pure rotation✓ 1
A1 angle \(=\arctan\frac{0.8}{0.6}=0.927\) rad \((53.1^\circ)\)✓ 1
Part (c)
M1 the argument after \(n\) rotations is \(0.927(n+1)\)✓ 1
M1 requires \(\pi<0.927(n+1)<\frac{3\pi}{2}\), so \(3.39<n+1<5.08\)✓ 1
A1 \(n+1=4\), so \(n=3\)✓ 1
Part (d)
M1 evaluates \(500e^{4(0.927)i}\) or \(a(0.6+0.8i)^3\)✓ 1
A1 position \(-421.6-268.8i\)✓ 1
A1 distance \(=|-721.6-668.8i|=984\) m✓ 1

Exam tip: Dividing two complex numbers of equal modulus gives a unit complex number whose argument is the rotation angle. This is quicker and more reliable than subtracting two separately rounded arguments.

268IBMA2H-045-02-101.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 2Medium7 marks

Calculator

Noise-cancelling headphones detect two sound waves, \(4\cos(\omega t)\) and \(3\cos\left(\omega t+\frac{\pi}{3}\right)\). Their sum can be written as \(\text{Re}\left[\left(4+3e^{\frac{\pi i}{3}}\right)e^{i\omega t}\right]\).

(a)

Write \(3e^{\frac{\pi i}{3}}\) in the form \(a+bi\), giving exact values.

(2)
(b)

The combined wave is \(R\cos(\omega t+\alpha)\). Find \(R\) and \(\alpha\).

(3)
(c)

The headphones emit a wave \(R\cos(\omega t+\beta)\) that cancels the combined wave exactly. Find \(\beta\), where \(-\pi<\beta\le\pi\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(3\cos\frac{\pi}{3}+3i\sin\frac{\pi}{3}\)✓ 1
A1 \(\frac32+\frac{3\sqrt3}{2}i\)✓ 1
Part (b)
M1 adds to get \(5.5+\frac{3\sqrt3}{2}i\)✓ 1
A1 \(R=\sqrt{37}\approx6.08\)✓ 1
A1 \(\alpha=0.441\) rad✓ 1
Part (c)
M1 the cancelling phasor is the negative of the combined phasor, so \(\beta=\alpha\pm\pi\)✓ 1
A1 \(\beta=-2.70\) rad✓ 1

Common error: Amplitudes of waves with different phases do not simply add, so 4 + 3 = 7 is wrong. Add the phasors as complex numbers and then take the modulus.

269IBMA2H-045-01-061.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator

After an intravenous dose, the concentration of an antibiotic in a patient's blood is modelled by \(C = 40 \times 2^{-\frac{t}{6}}\), where \(C\) is in mg L−1 and \(t\) is the time in hours after the dose.

(a)

Find the concentration 9 hours after the dose.

(2)
(b)

The model can also be written as \(C = 40e^{-kt}\). Find the value of \(k\).

(2)
(c)

The antibiotic stops being effective once its concentration falls below 3 mg L−1. Find how long the dose remains effective.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(40 \times 2^{-1.5}\)✓ 1
A1 \(14.1\) mg L−1✓ 1
Part (b)
M1 \(e^{-k} = 2^{-\frac{1}{6}}\), so \(k = \frac{\ln 2}{6}\)✓ 1
A1 \(k = 0.116\)✓ 1
Part (c)
M1 \(40 \times 2^{-\frac{t}{6}} = 3\)✓ 1
M1 \(t = 6\log_2\left(\frac{40}{3}\right)\) or \(t = \frac{6\ln(40/3)}{\ln 2}\)✓ 1
A1 \(t = 22.4\) hours✓ 1

Exam tip: Isolate the power before you take logarithms. Divide by 40 first, then take logs of both sides.

270IBMA2H-045-01-101.5 Exponents and introduction to logarithms Paper 2Medium7 marks

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Astronomers compare the brightness of stars using apparent magnitude \(m\). The magnitudes of two stars are related to the light fluxes \(F_1\) and \(F_2\) received from them by \(m_1 - m_2 = -2.5\log_{10}\left(\frac{F_1}{F_2}\right)\). A lower magnitude means a brighter star.

(a)

Star A has magnitude 1.2 and star B has magnitude 4.7. Find \(\frac{F_A}{F_B}\).

(3)
(b)

The faintest stars visible to the naked eye have magnitude 6.0. Find the least number of such stars whose combined flux is at least the flux from star A.

(2)
(c)

Star C delivers 60% of the flux of star A. Find the magnitude of star C.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(1.2 - 4.7 = -2.5\log_{10}\left(\frac{F_A}{F_B}\right)\)✓ 1
M1 \(\log_{10}\left(\frac{F_A}{F_B}\right) = 1.4\)✓ 1
A1 \(\frac{F_A}{F_B} = 10^{1.4} = 25.1\)✓ 1
Part (b)
M1 flux ratio \(= 10^{\frac{6.0 - 1.2}{2.5}} = 10^{1.92} = 83.2\)✓ 1
A1 84 stars✓ 1
Part (c)
M1 \(m_C - 1.2 = -2.5\log_{10}(0.6)\)✓ 1
A1 \(m_C = 1.75\)✓ 1

Common error: A fainter star has a larger magnitude. If your star C comes out with a magnitude below 1.2, a sign has been lost.

271IBMA2H-045-05-071.3 Geometric sequences and series Paper 2Medium6 marks

Calculator

Consider the infinite geometric series \(3+3\ln x+3(\ln x)^2+3(\ln x)^3+\dots\), where \(x>0\).

(a)

Find the set of values of x for which the series has a finite sum.

(3)
(b)

The sum of the series is 5. Find the value of x.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 identifies common ratio \(r=\ln x\)✓ 1
M1 uses \(|\ln x|<1\), i.e. \(-1<\ln x<1\)✓ 1
A1 \(e^{-1}<x<e\) (accept \(0.368<x<2.72\))✓ 1
Part (b)
M1 \(\frac{3}{1-\ln x}=5\)✓ 1
A1 \(\ln x=0.4\)✓ 1
A1 \(x=e^{0.4}=1.49\) (3 s.f.) (accept either form)✓ 1

Common error: Writing ln x −1.

272IBMA2H-045-05-091.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 2Medium11 marks

Calculator

In special relativity, an object moving at speed v has Lorentz factor \(\gamma=(1-u)^{-\frac{1}{2}}\), where \(u=\frac{v^2}{c^2}\) and c is the speed of light. Its kinetic energy is \((\gamma-1)mc^2\), where m is its mass.

(a)

Expand \((1-u)^{-\frac{1}{2}}\) in ascending powers of u, up to and including the term in \(u^3\).

(3)
(b)

State the condition on v for the expansion to be valid, and comment on its physical meaning.

(1)
(c)

Show that, for speeds small compared with c, the kinetic energy is approximately \(\frac{1}{2}mv^2+\frac{3mv^4}{8c^2}\).

(2)
(d)

A probe travels at \(v=0.6c\). Use your four-term expansion of \((1-u)^{-\frac{1}{2}}\) to estimate γ, and find the percentage error compared with the exact value of γ.

(3)
(e)

A probe of mass 1200 kg travels at \(3.0\times10^{7}\) m s−1. Taking \(c=3.0\times10^{8}\) m s−1, use the approximation \(\frac{1}{2}mv^2+\frac{3mv^4}{8c^2}\) to estimate its kinetic energy.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 applies extended binomial theorem with \(n=-\frac{1}{2}\) and \(-u\)✓ 1
A1 two correct non-constant terms✓ 1
A1 \(1+\frac{1}{2}u+\frac{3}{8}u^2+\frac{5}{16}u^3\)✓ 1
Part (b)
A1 \(|u|<1\), i.e. \(v<c\), which holds for every object with mass, so the expansion is always valid physically✓ 1
Part (c)
M1 \(\gamma-1\approx\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}\)✓ 1
A1 multiplies by \(mc^2\) to obtain \(\frac{1}{2}mv^2+\frac{3mv^4}{8c^2}\) (AG)✓ 1
Part (d)
M1 substitutes \(u=0.36\)✓ 1
A1 γ ≈ 1 + 0.18 + 0.0486 + 0.01458 = 1.24318✓ 1
A1 exact γ = \(\frac{1}{\sqrt{0.64}}\) = 1.25, percentage error ≈ 0.546% (the expansion is an underestimate)✓ 1
Part (e)
M1 substitutes: \(0.5(1200)(3.0\times10^7)^2+\frac{3(1200)(3.0\times10^7)^4}{8(3.0\times10^8)^2}\)✓ 1
A1 5.44 ×1017 J✓ 1

Common error: Expanding (1 − u)⁻¹/² as if it were (1 + u)⁻¹/². The minus sign inside the bracket must be raised to each power, which here makes every term positive.

273IBMA2H-045-05-121.5 Exponents and introduction to logarithms Paper 2Medium7 marks

Calculator

A genetics laboratory synthesises short DNA strands. Each strand is a sequence of 8 bases, and each base is one of A, C, G or T.

(a)

Find the number of different 8-base sequences.

(1)
(b)

Find the number of sequences that contain exactly three A bases.

(2)
(c)

Find the number of sequences that contain exactly two of each base.

(2)
(d)

Of the sequences that contain exactly two of each base, find the number in which the two G bases are adjacent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 48 = 65536✓ 1
Part (b)
M1 \(\binom{8}{3}\times3^5\)✓ 1
A1 56 × 243 = 13608✓ 1
Part (c)
M1 \(\frac{8!}{(2!)^4}\)✓ 1
A1 2520✓ 1
Part (d)
M1 treats GG as a single unit, giving 7 objects with three repeated pairs: \(\frac{7!}{(2!)^3}\)✓ 1
A1 630✓ 1

Common error: When the two G's are adjacent, some students divide by (2!)⁴ again. Once the two G's form a single block, only three repeated pairs remain, so divide by (2!)³.

274IBMA2H-045-07-051.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator

A hospital patient receives a 200 mg dose of an antibiotic every 6 hours. In the 6 hours between doses, the body removes 65% of the antibiotic that is present. The amount of antibiotic in the body immediately after the \(n\)th dose is \(A_n=200+200(0.35)+200(0.35)^2+\dots+200(0.35)^{n-1}\) mg.

(a)

Find the amount of antibiotic in the body immediately after the 4th dose.

(2)
(b)

The treatment becomes effective once the amount immediately after a dose is more than 300 mg. Find the least number of doses needed.

(3)
(c)

Find the amount of antibiotic that \(A_n\) approaches in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(A_4=\frac{200(1-0.35^4)}{1-0.35}\)✓ 1
A1 303 mg (303.075…)✓ 1
Part (b)
M1 sets up \(\frac{200(1-0.35^n)}{0.65}>300\)✓ 1
M1 simplifies to \(0.35^n<0.025\) and solves, giving \(n>3.51\) (or checks \(A_3=294.5\) and \(A_4=303.1\))✓ 1
A1 4 doses✓ 1
Part (c)
M1 \(S_\infty=\frac{200}{1-0.35}\)✓ 1
A1 308 mg (307.69…)✓ 1

Exam tip: Dividing by \(\ln0.35\) reverses the inequality, because \(\ln0.35<0\). Checking the values of \(A_3\) and \(A_4\) is a quick way to confirm your answer.

275IBMA2H-045-07-081.8 Infinite geometric series Paper 2Medium13 marks

Calculator

An engineer tests a shock absorber by attaching a mass to a damped spring. The first oscillation has an amplitude of 5.0 cm, and each oscillation has an amplitude 4% smaller than the one before. During each oscillation, the mass travels a distance equal to four times that oscillation's amplitude.

(a)

Find the amplitude of the 10th oscillation.

(2)
(b)

Find the number of the first oscillation whose amplitude is less than 1 cm.

(3)
(c)

Find the total distance travelled by the mass during the first 20 oscillations.

(3)
(d)

Show that the total distance travelled by the mass can never reach 5 m.

(2)
(e)

The damping is changed so that the first amplitude is still 5.0 cm, but the total distance ever travelled is exactly 3.0 m. Find the new percentage decrease in amplitude per oscillation.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(A_{10}=5\times0.96^{9}\)✓ 1
A1 3.46 cm✓ 1
Part (b)
M1 sets up \(5\times0.96^{n-1}<1\)✓ 1
M1 solves using logarithms or a table: \(n-1>39.4\)✓ 1
A1 41st oscillation✓ 1
Part (c)
M1 distance in the \(n\)th oscillation is \(20\times0.96^{n-1}\) cm, a geometric sequence with first term 20✓ 1
M1 \(S_{20}=\frac{20(1-0.96^{20})}{1-0.96}\)✓ 1
A1 279 cm✓ 1
Part (d)
M1 \(S_\infty=\frac{20}{1-0.96}=500\) cm✓ 1
R1 all terms are positive, so the partial sums increase towards 500 cm = 5 m but never reach it✓ 1
Part (e)
M1 \(\frac{20}{1-r}=300\)✓ 1
A1 \(r=\frac{14}{15}\)✓ 1
A1 6.67%✓ 1

Common error: Check which sequence each part is about. The amplitudes start at 5 cm, but the distances travelled start at 4 × 5 = 20 cm. Using 5 as the first term in the distance sums gives answers that are four times too small.

276IBMA2H-045-07-091.1 Standard form Paper 2Medium7 marks

Calculator

A computer processor performs 3.2×109 operations per second. A brute-force search algorithm with input size \(n\) needs \(2^n\) operations.

(a)

Find the largest value of \(n\) for which the algorithm finishes within one hour.

(3)
(b)

For \(n=50\), find the time the algorithm takes in seconds. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(c)

Convert your answer to the previous part into days.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 operations in one hour: 3.2×109 × 3600 = 1.152×1013✓ 1
M1 \(2^n\le1.152\times10^{13}\), so \(n\le\frac{\log(1.152\times10^{13})}{\log2}\approx43.4\), or a check that \(2^{43}\approx8.80\times10^{12}\) and \(2^{44}\approx1.76\times10^{13}\)✓ 1
A1 \(n=43\)✓ 1
Part (b)
M1 \(\frac{2^{50}}{3.2\times10^9}\)✓ 1
A1 3.52×105 s✓ 1
Part (c)
M1 divides their time in seconds by 86 400 (= 24 × 3600)✓ 1
A1 4.07 days (follow through from their previous answer)✓ 1

Exam tip: Since \(n\) must be a whole number, round down here: \(n=44\) would need more than one hour.

277IBMA3H-1-18r5-031.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator


A beekeeper estimates that each forager bee collects 3.6 × 10−5 kg of nectar per trip and makes 12 trips per day. The hive has 2.8 × 104 foragers.

(a)

Find the mass of nectar the hive collects per day. Give the exact value in kg, and then write it in the form a × 10k with a correct to 3 significant figures.

(3)
(b)

Every 3.0 kg of nectar makes 1 kg of honey. Using the exact daily mass from (a), find the number of days needed to collect enough nectar for 100 kg of honey. Give your answer to 3 significant figures.

(3)
(c)

The beekeeper plans for 25 days rather than 24.8 days. Explain why 24 days would not be enough.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 3.6 × 10−5 × 12 × 2.8 × 104✓ 1
A1 12.096 kg✓ 1
A1 1.21 × 101 kg✓ 1
Part (b)
M1 300 kg of nectar needed✓ 1
M1 300 ÷ 12.096✓ 1
A1 24.8 days✓ 1
Part (c)
R1 in 24 days only 24 × 12.096 ≈ 290 kg of nectar is collected, which is less than the 300 kg needed, so the 24.8 days must be rounded up✓ 1

Common error: Rounding a value up or down is a modelling decision. Say which way and why, rather than just rounding to the nearest whole number.

278IBMA3H-1-18r5-041.4 Financial applications Paper 2Medium6 marks

Calculator


A Formula 1 car starts a race with 1.1 × 102 kg of fuel. The race lasts 3.1 × 103 s.

(a)

Find the greatest average rate, in kg s−1, at which the car can burn fuel without running out. Give your answer in the form a × 10k, where a is correct to 3 significant figures.

(3)
(b)

At full throttle the engine burns fuel at 4.4 × 10−2 kg s−1. Find, in the form a × 10k, the time for which the car could run at full throttle before the fuel is used up.

(2)
(c)

Find by how many seconds this time is shorter than the race.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 (1.1 × 102) ÷ (3.1 × 103)✓ 1
A1 0.0355✓ 1
A1 3.55 × 10−2 kg s−1✓ 1
Part (b)
M1 (1.1 × 102) ÷ (4.4 × 10−2)✓ 1
A1 2.5 × 103 s✓ 1
Part (c)
A1 3100 − 2500 = 600 s✓ 1

Exam tip: Dividing by a number with a negative power of ten adds to the exponent: 10² ÷ 10⁻² = 10⁴.

279IBMA3H-1-18r6-031.3 Geometric sequences and series Paper 2Medium6 marks

Calculator

A train brakes so that the distance it travels in each successive 10-second interval is 0.8 times the distance in the previous interval. In the first interval it travels 250 m. Treat the distances as a geometric sequence.

(a)

Find the total distance the train would travel if the braking continued indefinitely.

(2)
(b)

Find the smallest number of intervals \(n\) for which the distance travelled is greater than 99% of this total.

(3)
(c)

Find the distance travelled in these \(n\) intervals, to 1 decimal place.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(S_\infty=\frac{250}{1-0.8}\)✓ 1
A1 1250 m✓ 1
Part (b)
M1 \(\frac{250(1-0.8^n)}{0.2}>0.99\times1250\)✓ 1
A1 \(0.8^n<0.01\) so \(n>20.6\)✓ 1
A1 \(n=21\)✓ 1
Part (c)
A1 1238.5 m (ECF from (b))✓ 1

Exam tip: Check the integer from an inequality by testing n and n − 1 in the original sum.

280IBMA3H-1-18r6-041.8 Infinite geometric series Paper 2Medium6 marks

Calculator

The number of meteors counted on successive nights of a shower forms a geometric sequence. On night 2 the count is 540 and on night 5 the count is 160.

(a)

Find the common ratio and the count on night 1.

(4)
(b)

Find the total number of meteors counted over the first 8 nights. Give your answer to the nearest whole number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_5=u_2r^3\) so \(r^3=\frac{160}{540}\)✓ 1
A1 \(r^3=\frac8{27}\)✓ 1
A1 \(r=\frac23\)✓ 1
A1 \(u_1=810\)✓ 1
Part (b)
M1 \(S_8=\frac{810(1-(2/3)^8)}{1/3}\)✓ 1
A1 2335 (2335.19…)✓ 1

Exam tip: To find r from two non-adjacent terms, divide the terms and take the root of the gap between their positions.

281IBMA3H-1-18r6-051.5 Exponents and introduction to logarithms Paper 2Medium6 marks

Calculator

A basketball is dropped from a height of 2.4 m onto a floor. After each bounce it rises to 75% of the height it fell from. Model the motion as vertical only.

(a)

Find the height reached after the 4th bounce, to 3 significant figures.

(2)
(b)

Find the total vertical distance travelled by the ball from release until it comes to rest.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(2.4\times0.75^{4}\)✓ 1
A1 0.759 m✓ 1
Part (b)
M1 recognise the rebound heights form a geometric series with \(a=1.8\) and \(r=0.75\)✓ 1
M1 sum of rebound heights \(\frac{1.8}{1-0.75}=7.2\)✓ 1
M1 each rebound is travelled twice (up and down) and add the initial 2.4 m drop✓ 1
A1 \(2.4+2(7.2)=16.8\) m✓ 1

Common error: After the first drop the ball travels each height twice, so doubling the series is needed.

282IBMA3H-1-18r10-031.4 Financial applications Paper 2Medium6 marks

Calculator

A fisheries company invests £15000 in an account with a nominal annual interest rate of \(r\%\), compounded quarterly. After 6 years the account holds £19500.

(a)

Write down an equation for \(r\).

(2)
(b)

Find \(r\), to three significant figures.

(2)
(c)

Find the effective annual interest rate, as a percentage to three significant figures. Use the exact quarterly growth factor, not the rounded value of \(r\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(15000\left(1+\frac{r}{400}\right)^{24}\)✓ 1
\(=19500\)✓ 1
Part (b)
\(1+\frac r{400}=1.3^{1/24}\)✓ 1
\(r=4.40\)✓ 1
Part (c)
annual factor \(=1.3^{1/6}\) (or \((1.3^{1/24})^4\))✓ 1
effective rate \(=4.47\%\)✓ 1

Common error: A nominal rate of \(r\%\) compounded quarterly gives \(\frac{r}{4}\%\) per quarter; the effective annual rate is a different, larger figure.

283IBMA3H-1-18r10-041.4 Financial applications Paper 2Medium5 marks

Calculator

A robotics start-up borrows €20000 to be repaid in equal monthly payments, made at the end of each month for 5 years. The nominal annual interest rate is 6%, compounded monthly.

(a)

Find the monthly payment, to the nearest cent. You may use a financial solver, with N = 60, I% = 6, PV = 20000, FV = 0, P/Y = C/Y = 12.

(3)
(b)

Using your answer to (a) rounded to the nearest cent, find the total interest paid over the loan.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
correct set-up of values in a solver (or the equation \(20000=P\times\frac{1-1.005^{-60}}{0.005}\))✓ 1
\(I\% = 6\) and \(P/Y = 12\) (monthly rate \(0.5\%\))✓ 1
€386.66✓ 1
Part (b)
total repaid \(=60\times386.66=23199.60\)✓ 1
interest \(=\) €3199.60✓ 1

Exam tip: Check that the payments-per-year and compounding-per-year settings in your calculator both match the monthly pattern.

284IBMA3H-1-18r10-051.4 Financial applications Paper 2Medium5 marks

Calculator

A fashion designer deposits £200 at the start of every month for 3 years into an account paying a nominal annual rate of 4.8%, compounded monthly (so the monthly rate is \(0.4\%\)). No withdrawals are made.

(a)

Write down, in terms of powers of \(1.004\), the value at the end of the 3 years of the first deposit and of the last deposit.

(2)
(b)

Hence find the total value of the account at the end of the 3 years, to the nearest penny.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
first deposit: \(200\times1.004^{36}\)✓ 1
last deposit: \(200\times1.004^{1}\)✓ 1
Part (b)
geometric series with first term \(200\times1.004\) and common ratio \(1.004\), with 36 terms✓ 1
\(S=\frac{200(1.004)(1.004^{36}-1)}{0.004}\)✓ 1
£7758.53✓ 1

Common error: Deposits at the start of each month earn interest for the full month, so the last deposit still gains one month of interest.

285IBMA3H-1-18r10-091.16 Systems of linear equations Paper 2Medium5 marks

Calculator

A fish farm mixes three feeds, A, B and C. Per kilogram, the feeds supply nutrient 1: 2, 1 and 3 units; nutrient 2: 1, 2 and 1 units; nutrient 3: 3, 1 and 2 units respectively. A daily mix of \(x\) kg of A, \(y\) kg of B and \(z\) kg of C must supply exactly 17 units of nutrient 1, 12 units of nutrient 2 and 19 units of nutrient 3.

(a)

Write the situation as a matrix equation.

(2)
(b)

Solve the system to find the mass of each feed in the mix.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
correct coefficient matrix \(\begin{pmatrix}2&1&3\\1&2&1\\3&1&2\end{pmatrix}\)✓ 1
correct unknowns and right-hand side \(\begin{pmatrix}17\\12\\19\end{pmatrix}\)✓ 1
Part (b)
\(x=4\) kg✓ 1
\(y=3\) kg✓ 1
\(z=2\) kg✓ 1

Exam tip: Keep the rows in the same order as the equations and check your solution in each original equation.

286IBMA3H-1-19r11-031.2 Arithmetic sequences and series Paper 2Medium7 marks

Calculator

In a video game, a healing spell restores \(H\) health points on the first cast, and each further cast restores 60% of the health restored by the previous cast. A player notes that casting the spell without end would restore a total of 250 health points.

(a)

Find the value of \(H\).

(2)
(b)

Find the least number of casts for which the total health restored exceeds 98% of the total from endless casting.

(4)
(c)

Find the total health restored by this number of casts, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{H}{1-0.6}=250\)✓ 1
A1 \(H=100\)✓ 1
Part (b)
M1 \(S_n=\frac{100(1-0.6^n)}{0.4}=250(1-0.6^n)\)✓ 1
M1 \(250(1-0.6^n)>245\), so \(0.6^n<0.02\)✓ 1
M1 take logarithms and reverse the inequality: \(n>\frac{\ln 0.02}{\ln 0.6}=7.66\ldots\)✓ 1
A1 \(n=8\)✓ 1
Part (c)
A1 \(250(1-0.6^8)=245.8\ldots=246\) health points✓ 1

Common error: When dividing both sides of an inequality by \(\ln 0.6\) (a negative number), the inequality sign reverses.

287IBMA3H-1-19r11-041.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator

A pump moves 40 litres of water in the first minute. In each later minute it moves \(\frac{3-2x}{4}\) times the volume moved in the previous minute, where \(x\) is a control setting. The volumes moved in successive minutes form a geometric sequence.

(a)

Find the range of values of \(x\) for which the total volume moved over an unlimited time is finite.

(3)
(b)

The operator wants the total volume moved over an unlimited time to be exactly 100 litres. Find the value of \(x\) and justify that it is valid.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 convergence requires \(|r|<1\), that is \(-1<\frac{3-2x}{4}<1\)✓ 1
A1 \(-4<3-2x<4\)✓ 1
A1 \(-\frac{1}{2}<x<\frac{7}{2}\)✓ 1
Part (b)
M1 \(\frac{40}{1-r}=100\)✓ 1
A1 \(r=0.6\)✓ 1
A1 \(\frac{3-2x}{4}=0.6\), so \(x=0.3\)✓ 1
R1 \(0.3\) lies in \(-\frac{1}{2}<x<\frac{7}{2}\), so the series converges✓ 1

Exam tip: Whenever the common ratio contains an unknown, solve for it first and then check the answer against \(|r|<1\).

288IBMA3H-1-19r11-121.16 Systems of linear equations Paper 2Medium10 marks

Calculator

A pharmacist combines three types of tablet, A, B and C, to supply a patient's daily requirement of three vitamins, X, Y and Z. Each tablet of type A contains 5 mg of X, 3 mg of Y and 2 mg of Z. Each tablet of type B contains 2 mg of X, 6 mg of Y and 1 mg of Z. Each tablet of type C contains 4 mg of X, 2 mg of Y and 5 mg of Z. The patient needs exactly 33 mg of X, 43 mg of Y and 21 mg of Z each day, taking \(a\) tablets of type A, \(b\) of type B and \(c\) of type C.

(a)

Write down a system of three linear equations in \(a\), \(b\) and \(c\).

(3)
(b)

Solve the system.

(2)
(c)

Type C tablets are withdrawn, so only types A and B can be used. Show that the patient's requirements cannot then be met.

(3)
(d)

Tablets of types A, B and C cost £0.40, £0.25 and £0.60 each. Find the daily cost for the solution to part (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(5a+2b+4c=33\)✓ 1
A1 \(3a+6b+2c=43\)✓ 1
A1 \(2a+b+5c=21\)✓ 1
Part (b)
M1 solves the system (by elimination or technology)✓ 1
A1 \(a=3\), \(b=5\), \(c=2\)✓ 1
Part (c)
M1 solves two equations with \(c=0\), e.g. \(5a+2b=33\) and \(3a+6b=43\)✓ 1
A1 \(a=\frac{14}{3}\), \(b=\frac{29}{6}\)✓ 1
R1 the third equation gives \(2a+b=\frac{85}{6}\neq21\), so the system is inconsistent and the requirements cannot be met✓ 1
Part (d)
M1 \(3(0.40)+5(0.25)+2(0.60)\)✓ 1
A1 £3.65✓ 1

Exam tip: Substitute your solution back into all three original equations to catch a slip in the setting-up.

289IBMA3H-1-19r11-131.16 Systems of linear equations Paper 2Medium12 marks

Calculator

An engineer models three calibration tests on valve settings \(x\), \(y\) and \(z\) in a water-supply network using the system
\(x+2y-z=4\)
\(2x-y+3z=3\)
\(3x+y+kz=m\)
where \(k\) and \(m\) are constants.

(a)

When \(k=2\) and \(m=7\), show that the system has infinitely many solutions and find the general solution in vector form.

(5)
(b)

When \(k=2\) and \(m\neq7\), show that there is no solution and describe the geometrical arrangement of the three planes.

(4)
(c)

When \(k=5\) and \(m=7\), find the solution of the system.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 adds the first two equations: \(3x+y+2z=7\)✓ 1
R1 this is the third equation, so it is redundant and there are infinitely many solutions✓ 1
M1 sets \(z=t\) and solves the first two equations for \(x\) and \(y\)✓ 1
A1 \(x=2-t\), \(y=1+t\)✓ 1
A1 \(\mathbf{r}=\begin{pmatrix}2\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\1\\1\end{pmatrix}\)✓ 1
Part (b)
M1 first two equations give \(3x+y+2z=7\), which contradicts the third equation when \(m\neq7\)✓ 1
A1 no solution✓ 1
R1 the normals \((1,2,-1)\), \((2,-1,3)\) and \((3,1,2)\) are not parallel, so no two planes are parallel✓ 1
A1 the planes form a triangular prism (no common point)✓ 1
Part (c)
M1 subtracts \(3x+y+2z=7\) from the third equation: \(3z=0\)✓ 1
A1 \(z=0\)✓ 1
A1 \(x=2\), \(y=1\)✓ 1

Common error: If two equations add to give the third on the left side, check the right side too: it decides between infinitely many solutions and none.

290IBMA3H-1-18r13-021.10 Counting principles and the extended binomial theorem Paper 2Medium7 marks

Calculator

A fashion designer has 7 different tops, 5 different pairs of trousers and 4 different jackets. She is choosing garments for a capsule collection.

(a)

She chooses 3 tops, 2 pairs of trousers and 1 jacket. Find the number of different selections.

(2)
(b)

One particular top, the signature top, must be included among the 3 tops. Find the number of selections.

(2)
(c)

The designer instead chooses any 6 of the 16 garments. Find the number of selections that contain at least one jacket.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\binom{7}{3}\times\binom{5}{2}\times\binom{4}{1}\)✓ 1
A1 \(35\times 10\times 4 = 1400\)✓ 1
Part (b)
M1 \(\binom{6}{2}\times 10\times 4\)✓ 1
A1 600✓ 1
Part (c)
M1 \(\binom{16}{6} - \binom{12}{6}\)✓ 1
A1 \(8008\) and \(924\)✓ 1
A1 7084✓ 1

Exam tip: "At least one" is often quickest as total minus none.

291IBMA3H-1-18r13-031.11 Partial fractions Paper 2Medium7 marks

Calculator

The function \(f(x) = (8 + x)^{1/3}\) can be expanded as a series in ascending powers of \(x\).

(a)

Find the first three terms of the expansion of \(f(x)\).

(4)
(b)

State the values of \(x\) for which the expansion is valid.

(1)
(c)

Use your expansion with \(x = 1\) to estimate \(\sqrt[3]{9}\) as a single fraction, and give its value to 5 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((8+x)^{1/3} = 8^{1/3}\left(1+\frac{x}{8}\right)^{1/3} = 2\left(1+\frac{x}{8}\right)^{1/3}\)✓ 1
M1 binomial expansion with \(n=\frac13\): \(1 + \frac13\cdot\frac{x}{8} + \frac{\frac13\left(-\frac23\right)}{2}\left(\frac{x}{8}\right)^2\)✓ 1
A1 \(2 + \frac{x}{12}\)✓ 1
A1 \(-\frac{x^2}{288}\)✓ 1
Part (b)
A1 \(|x| < 8\)✓ 1
Part (c)
M1 \(2 + \frac{1}{12} - \frac{1}{288}\)✓ 1
A1 \(\frac{599}{288} = 2.0799\)✓ 1

Common error: Expanding \((8+x)^{1/3}\) directly with \(x\) as the second term. The form \((1+u)^n\) is needed, so factor out \(8^{1/3}\) first.

292IBMA3H-1-18r13-101.4 Financial applications Paper 2Medium7 marks

Calculator

A warehouse robot costs €48 000 when new. Its value depreciates by 14% each year.

(a)

Find the value of the robot after 4 years, to the nearest euro.

(2)
(b)

Find the least whole number of years after which the value is below €15 000.

(2)
(c)

A second robot of the same price has a value of €25 000 after 5 years, with the same percentage depreciation each year. Find this annual rate of depreciation as a percentage, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(48000\times 0.86^4\)✓ 1
A1 €26 256✓ 1
Part (b)
M1 solve \(48000\times 0.86^n < 15000\) (logarithms or table), \(n > 7.71\)✓ 1
A1 \(n = 8\)✓ 1
Part (c)
M1 \(48000(1-r)^5 = 25000\)✓ 1
M1 \(1 - r = \left(\frac{25000}{48000}\right)^{1/5} = 0.8777\)✓ 1
A1 12.2%✓ 1

Exam tip: Depreciation of \(r\)% a year multiplies the value by \(1-\frac{r}{100}\) each year.

293IBMA3H-1-18r13-111.4 Financial applications Paper 2Medium7 marks

Calculator

A fashion company invests €25 000 at a nominal annual interest rate of 4.5%, compounded monthly.

(a)

Find the value of the investment after 3 years, to the nearest euro.

(2)
(b)

Find the least whole number of months after which the value of the investment exceeds €30 000.

(3)
(c)

Find the effective annual interest rate, as a percentage to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(25000\left(1+\frac{4.5}{1200}\right)^{36}\)✓ 1
A1 €28 606✓ 1
Part (b)
M1 \(25000(1.00375)^m > 30000\)✓ 1
M1 solve, e.g. \(m > \frac{\ln 1.2}{\ln 1.00375} = 48.7\)✓ 1
A1 \(m = 49\)✓ 1
Part (c)
M1 \((1.00375)^{12} - 1\)✓ 1
A1 4.59%✓ 1

Exam tip: The effective annual rate compares one full year of growth with the starting amount, so it is always at least the nominal rate.

294IBMA3H-1-18r13-141.4 Financial applications Paper 2Medium9 marks

Calculator

The price per metre of a fabric rose by the same percentage each year. It cost €15.00 in 2020 and €21.00 in 2025.

(a)

Find the annual percentage increase, to 3 significant figures.

(3)
(b)

Assuming the same annual percentage increase continues, find the price in 2030. Use the unrounded value of \(r\).

(2)
(c)

Find the first year in which the price per metre exceeds €45, and the price in that year, to the nearest cent. Use the unrounded value of \(r\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(15r^5 = 21\)✓ 1
M1 \(r = 1.4^{1/5} = 1.0696\)✓ 1
A1 6.96%✓ 1
Part (b)
M1 \(15\times 1.4^2\) or \(21\times r^5\)✓ 1
A1 €29.40✓ 1
Part (c)
M1 \(15\times 1.4^{n/5} > 45\), where \(n\) is the number of years after 2020✓ 1
M1 solve, e.g. \(n > \frac{5\ln 3}{\ln 1.4} = 16.3\)✓ 1
A1 2037 (\(n=17\))✓ 1
A1 €47.09✓ 1

Common error: Using \(r^6\) for a gap of 2020 to 2025. A gap of 5 years means 5 growth factors, not 6.

295IBMA3H-1-18r14-031.1 Standard form Paper 2Medium6 marks

Calculator

A fashion house studies a yarn made of identical circular fibres, each of diameter \(1.2\times10^{-5}\) m. The fibres in a cross-section of the yarn have a total area of \(2.0\times10^{-7}\) m2.

(a)

Find the cross-sectional area of one fibre, in m2, in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\), with \(a\) to 3 significant figures.

(2)
(b)

Find the number of fibres in the yarn, to 3 significant figures. Use the unrounded value from (a).

(2)
(c)

Because of twisting, each fibre is 1.12 m long for every metre of yarn. Find the total length of fibre in 50 m of yarn, in the form \(a\times10^k\) with \(a\) to 3 significant figures. Use the unrounded value from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\pi\times(6.0\times10^{-6})^2\)✓ 1
A1 \(1.13\times10^{-10}\) m2✓ 1
Part (b)
M1 \(\frac{2.0\times10^{-7}}{\pi(6.0\times10^{-6})^2}\) (\(=1768.4\ldots\))✓ 1
A1 \(1.77\times10^3\)✓ 1
Part (c)
M1 \(1768.4\ldots\times1.12\times50\) (\(=99029.7\ldots\))✓ 1
A1 \(9.90\times10^4\) m✓ 1

Exam tip: Carry unrounded values between parts and round only the final answer to the accuracy requested.

296IBMA3H-1-18r14-041.1 Standard form Paper 2Medium6 marks

Calculator

A very large swarm of Antarctic krill occupies a volume of \(2.5\times10^8\) m3. The swarm contains \(1.6\times10^4\) krill per m3, and one krill has an average mass of \(4.5\times10^{-4}\) kg.

(a)

Find the number of krill in the swarm, in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(b)

Find the total mass of the swarm, in kg, in standard form.

(2)
(c)

A fishery has an annual catch limit of \(6.3\times10^5\) tonnes, where 1 tonne = \(10^3\) kg. Express this limit as a percentage of the mass of the swarm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(2.5\times10^8\times1.6\times10^4\)✓ 1
A1 \(4.0\times10^{12}\)✓ 1
Part (b)
M1 \(4.0\times10^{12}\times4.5\times10^{-4}\) (ECF from (a))✓ 1
A1 \(1.8\times10^9\) kg✓ 1
Part (c)
M1 converting: \(6.3\times10^5\times10^3=6.3\times10^8\) kg✓ 1
A1 \(\frac{6.3\times10^8}{1.8\times10^9}\times100=35\%\)✓ 1

Common error: Convert tonnes to kilograms before comparing masses; mixing units changes the answer by a factor of 1000.

297IBMA3H-1-18r14-131.13 Polar and Euler forms Paper 2Medium6 marks

Calculator

A robot arm makes two displacements from the origin, in metres, represented in the complex plane by \(w_1=4e^{i\pi/6}\) and \(w_2=5e^{2i\pi/3}\).

OReImOw₁w₂
Distances in metres. Not to scale
(a)

Find the total displacement \(w_1+w_2\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(3)
(b)

Find the distance of the arm's end point from the origin, and the argument of \(w_1+w_2\), each to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 using \(re^{i\theta}=r(\cos\theta+i\sin\theta)\)✓ 1
A1 \(w_1=3.46+2i\) and \(w_2=-2.50+4.33i\)✓ 1
A1 \(w_1+w_2=0.964+6.33i\)✓ 1
Part (b)
M1 using \(|w|=\sqrt{a^2+b^2}\) or \(\arg w=\tan^{-1}\frac ba\) (ECF from (a))✓ 1
A1 6.40 m✓ 1
A1 1.42 radians✓ 1

Exam tip: Convert from Euler form to Cartesian form before adding complex numbers, since addition works on real and imaginary parts.

298IBMA3H-1-18r16-031.1 Standard form Paper 2Medium6 marks

Calculator

A space probe is powered by a radioactive source. When launched the source provides 150 W, and its power output falls by 5% each year, so after n years the output is 150 × 0.95n W.

(a)

Find the power output after 10 years, correct to 3 significant figures.

(2)
(b)

The probe needs at least 75 W to operate. Find the number of complete years after which the output is first below 75 W.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 150 × 0.9510✓ 1
A1 89.8 W✓ 1
Part (b)
M1 150 × 0.95n < 75✓ 1
M1 attempt to solve with logarithms or graph✓ 1
A1 n > 13.5✓ 1
A1 14 years✓ 1

Exam tip: Dividing both sides of an inequality by log 0.95, which is negative, reverses the inequality sign.

299IBMA3H-1-18r16-041.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator

An athlete's training plan starts with a run of 4 km in week 1. The distance run in each following week is 6% more than in the previous week, so the weekly distances form a geometric sequence.

(a)

Find the distance run in week 10, correct to 3 significant figures.

(2)
(b)

Find the total distance run in the first 12 weeks, correct to 3 significant figures.

(2)
(c)

Find the smallest number of weeks after which the total distance run first exceeds 100 km.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 4 × 1.069✓ 1
A1 6.76 km✓ 1
Part (b)
M1 \(\frac{4(1.06^{12}-1)}{0.06}\)✓ 1
A1 67.5 km✓ 1
Part (c)
M1 \(\frac{4(1.06^{n}-1)}{0.06} > 100\)✓ 1
A1 \(n > 15.7\)✓ 1
A1 16 weeks✓ 1

Common error: For the total of the first n terms use \(S_n = \frac{u_1(r^n-1)}{r-1}\); the 10th term uses the power n − 1 = 9, not 10.

300IBMA3H-1-18r16-051.4 Financial applications Paper 2Medium7 marks

Calculator

A freight train brakes so that, in each minute, it travels 90% of the distance it covered in the previous minute. In the first minute of braking it travels 600 m.

(a)

Find the total distance travelled in the first 5 minutes of braking, to the nearest metre.

(2)
(b)

Find the greatest distance the train could travel while braking, according to this model.

(2)
(c)

Find the number of whole minutes after which the train has covered more than 95% of the distance in part (b).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{600(1-0.9^5)}{1-0.9}\)✓ 1
A1 2457 m✓ 1
Part (b)
M1 \(S_\infty = \frac{600}{1-0.9}\)✓ 1
A1 6000 m✓ 1
Part (c)
M1 \(0.05 > 0.9^n\), or \(600\frac{(1-0.9^n)}{0.1} > 5700\)✓ 1
A1 n > 28.4✓ 1
A1 29 minutes✓ 1

Exam tip: A sum to infinity exists only when |r| < 1; state this before using \(S_\infty = \frac{u_1}{1-r}\).

301IBMA3H-1-18r16-071.4 Financial applications Paper 2Medium6 marks

Calculator

An astronomy club buys a telescope for £4200. Its value falls by 12% each year.

(a)

Find its value after 4 years, to the nearest pound.

(2)
(b)

Find the number of complete years after which the value first falls below £1600.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 4200 × 0.884✓ 1
A1 £2519✓ 1
Part (b)
M1 4200 × 0.88n < 1600✓ 1
M1 attempt to solve with logarithms or graph✓ 1
A1 n > 7.55✓ 1
A1 8 years✓ 1

Exam tip: State the inequality, solve for n, then round up to a whole number of years because the value must already be below the target.

302IBMA3H-1-18r16-101.16 Systems of linear equations Paper 2Medium6 marks

Calculator

The cross-section of a rail tunnel arch is modelled by the curve \(h = ax^2 + bx + c\), where h metres is the height above the track at a horizontal distance of x metres from the left wall. The arch has height 3 m when x = 1, height 5 m when x = 3 and height 3 m when x = 5.

(a)

Write down three linear equations for a, b and c.

(2)
(b)

Solve the equations to find a, b and c.

(2)
(c)

Hence find the maximum height of the arch above the track.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a + b + c = 3✓ 1
A1 9a + 3b + c = 5 and 25a + 5b + c = 3✓ 1
Part (b)
M1 attempt to solve✓ 1
A1 a = −0.5, b = 3, c = 0.5✓ 1
Part (c)
M1 vertex at \(x = -\frac{b}{2a} = 3\)✓ 1
A1 5 m (ECF from (b))✓ 1

Exam tip: When a model passes through given points, substitute each point into the general form to obtain one equation per point.

303IBMA3H-1-18r16-121.11 Partial fractions Paper 2Medium6 marks

Calculator

Consider the expansion of \((1+2x)^{-3}\) in ascending powers of x.

(a)

Find the first four terms of the expansion.

(3)
(b)

State the values of x for which the expansion is valid.

(1)
(c)

By substituting a suitable value of x, estimate \(1.02^{-3}\). Give your answer to 5 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the binomial theorem with \(n=-3\)✓ 1
A1 coefficients 1, −3, 6, −10 with \((2x)^k\)✓ 1
A1 \(1 - 6x + 24x^2 - 80x^3\)✓ 1
Part (b)
A1 \(|x| < \frac{1}{2}\)✓ 1
Part (c)
M1 substitute x = 0.01✓ 1
A1 0.94232✓ 1

Exam tip: Use only the terms asked for and make sure the substituted value lies inside the validity range.

304IBMA3H-1-18r24-051.4 Financial applications Paper 2Medium8 marks

Calculator

A sports club invests €12 000 in an account that pays a nominal annual interest rate of 3.6%, compounded monthly.

(a)

Find the value of the investment after 4 years, giving your answer to the nearest cent.

(3)
(b)

Find the number of years it takes for the investment to double in value, to 3 significant figures.

(3)
(c)

Find the effective annual interest rate, as a percentage to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses 12 000(1 + 0.036/12)n with monthly rate 0.3%✓ 1
A1 n = 48✓ 1
A1 €13855.62✓ 1
Part (b)
M1 12 000(1.003)12t = 24 000, so (1.003)12t = 2 (accept GDC solver or graph)✓ 1
M1 takes logarithms or solves numerically: 12t = ln 2 / ln 1.003✓ 1
A1 t = 19.283 = 19.3 years✓ 1
Part (c)
M1 (1.003)12 − 1✓ 1
A1 3.6600% = 3.66%✓ 1

Common error: For monthly compounding divide the annual rate by 12 and multiply the number of years by 12; changing only one of them is a frequent slip.

305IBMA3H-1-18r24-081.16 Systems of linear equations Paper 2Medium7 marks

Calculator

A sports scientist designs a daily snack plan from three portions. Portion X contains 12 g protein, 30 g carbohydrate and 4 g fat. Portion Y contains 8 g protein, 45 g carbohydrate and 2 g fat. Portion Z contains 20 g protein, 10 g carbohydrate and 6 g fat. The plan uses x portions of X, y of Y and z of Z, and contains exactly 108 g protein, 290 g carbohydrate and 32 g fat.

(a)

Write down three linear equations for x, y and z.

(3)
(b)

Solve the system.

(2)
(c)

Portions X, Y and Z cost €1.20, €0.85 and €2.10 respectively. Find the cost of the plan.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 12x + 8y + 20z = 108✓ 1
A1 30x + 45y + 10z = 290✓ 1
A1 4x + 2y + 6z = 32✓ 1
Part (b)
M1 solves using a GDC or by elimination✓ 1
A1 x = 3, y = 4, z = 2✓ 1
Part (c)
M1 1.20(3) + 0.85(4) + 2.10(2) (ECF from (b))✓ 1
A1 €11.20✓ 1

Exam tip: Keep the same order of unknowns in every equation (here protein, carbohydrate, fat in the same row order) so that the system can be entered directly into a GDC.

306IBMA3H-1-18r24-091.16 Systems of linear equations Paper 2Medium7 marks

Calculator

The distance d (in millions of km) of a spacecraft from a relay station is modelled by d = at2 + bt + c, where t is the time in days after launch. The distance is 7.5 on day 1, 6 on day 2 and 7.5 on day 5.

(a)

Write down three equations for a, b and c.

(3)
(b)

Solve the system to find a, b and c.

(2)
(c)

Find the minimum distance of the spacecraft from the relay station, according to the model.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a + b + c = 7.5✓ 1
A1 4a + 2b + c = 6✓ 1
A1 25a + 5b + c = 7.5✓ 1
Part (b)
M1 solves using a GDC or by elimination✓ 1
A1 a = 0.5, b = −3, c = 10✓ 1
Part (c)
M1 uses t = −b/(2a) = 3 (or the GDC minimum; ECF from (b))✓ 1
A1 d = 5.5 million km✓ 1

Common error: Each data point gives one equation: substitute both the time and the distance, and square the time correctly in the t² term.

307IBMA3H-1-18r24-101.16 Systems of linear equations Paper 2Medium8 marks

Calculator

A freight train carries three types of cargo. The masses of the cargo, in tonnes, are x, y and z. They satisfy
x + y + z = 6
x + 2y + 3z = 10
2x + 3y + kz = m
where k and m are constants.

(a)

Show that when k = 4 the system does not have a unique solution.

(2)
(b)

Find the value of m for which the system has infinitely many solutions when k = 4.

(2)
(c)

For k = 4 and m = 16, let z = λ. Find x and y in terms of λ.

(2)
(d)

All three masses must be non-negative. Find the greatest possible value of z.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 adds the first two equations to get 2x + 3y + 4z = 16✓ 1
R1 for k = 4 the third equation has the same left-hand side as this sum, so the equations are dependent (no unique solution)✓ 1
Part (b)
M1 compares the third equation with the sum 2x + 3y + 4z = 16✓ 1
A1 m = 16✓ 1
Part (c)
M1 subtracts the first equation from the second: y + 2z = 4✓ 1
A1 y = 4 − 2λ and x = 2 + λ✓ 1
Part (d)
M1 y ≥ 0, so 4 − 2λ ≥ 0 (ECF from (c))✓ 1
A1 z = 2 tonnes✓ 1

Exam tip: If a system has a parameter, look for a combination of two equations that reproduces the third; that tells you both when it fails to be unique and what value makes it consistent.

308IBMA3H-1-18r24-111.16 Systems of linear equations Paper 2Medium8 marks

Calculator

An observatory shares its telescope between three targets for x, y and z hours. The schedule must satisfy
x + 2y + pz = 5
2x + y + z = 5
x + y + z = 4
where p is a constant that depends on the third target.

(a)

Find the value of p for which the system does not have a unique solution.

(2)
(b)

Show that when p takes this value the system has no solution.

(3)
(c)

Solve the system when p = 1.

(1)
(d)

For other values of p the solution gives z = −2/(p − 2). Find the value of p for which z = 2.5.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use the determinant of the coefficient matrix or elimination✓ 1
A1 p = 2✓ 1
Part (b)
M1 subtracts the third equation from the second to get x = 1✓ 1
M1 substitutes into the first and third equations: y + z = 2 and y + z = 3✓ 1
R1 contradiction, so no solution✓ 1
Part (c)
A1 x = 1, y = 1, z = 2✓ 1
Part (d)
M1 −2/(p − 2) = 2.5✓ 1
A1 p = 1.2✓ 1

Common error: Showing that a system has no solution needs a stated contradiction (such as y + z = 2 and y + z = 3); a determinant of zero alone does not distinguish no solution from infinitely many.

309IBMA3H-1-20r29-111.4 Financial applications Paper 2Medium6 marks

Calculator

A fashion retailer buys a fabric from two suppliers. Supplier A charges €120 per roll now and its price rises by 3% each year. Supplier B charges €90 per roll now and its price rises by 7% each year.

(a)

Find the price of a roll from each supplier after 5 years, to the nearest cent.

(2)
(b)

Find the time, in years to 3 significant figures, when the two prices are equal.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A: \(120\times1.03^5=\text{€}139.11\)✓ 1
A1 B: \(90\times1.07^5=\text{€}126.23\)✓ 1
Part (b)
M1 \(120\times1.03^t=90\times1.07^t\)✓ 1
M1 \(\left(\frac{1.07}{1.03}\right)^t=\frac{120}{90}\)✓ 1
M1 \(t=\frac{\ln(4/3)}{\ln(1.07/1.03)}\)✓ 1
A1 \(t=7.55\) years✓ 1

Exam tip: Collect the two exponentials on one side so the equation becomes \(k^t=c\), and then take logarithms.

310IBMA3H-1-20r35-051.4 Financial applications Paper 2Medium7 marks

Calculator

A logistics company records the fuel it saves each week after introducing a new routing system. In week 1 it saves 480 litres, and in each later week the saving is 0.85 times the saving in the previous week.

(a)

Find the saving in week 5, to the nearest litre.

(2)
(b)

Find the smallest number of weeks after which the total saving is more than 3000 litres.

(3)
(c)

If the routing system continued indefinitely, find the total saving, and state what this value represents for the company.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(480\times0.85^{4}\)✓ 1
A1 251 litres✓ 1
Part (b)
M1 \(\dfrac{480(1-0.85^n)}{1-0.85}>3000\)✓ 1
M1 \(0.85^n<0.0625\) and use of logarithms: \(n>\dfrac{\ln0.0625}{\ln0.85}=17.06\)✓ 1
A1 \(n=18\) weeks✓ 1
Part (c)
M1 \(S_\infty=\dfrac{480}{1-0.85}\)✓ 1
A1 3200 litres, the upper limit that the total fuel saved can never exceed✓ 1

Common error: When you divide both sides of an inequality by ln 0.85, remember that ln 0.85 is negative; so reverse the sign, or take logs in a way that avoids the problem.

311IBMA3H-1-20r35-061.1 Standard form Paper 2Medium6 marks

Calculator

A bridge design uses steel cable of diameter 2.8 × 10−2 m. The cable is solid with a circular cross-section, and steel has density 7.85 × 103 kg m−3. The design needs 8.5 × 104 m of cable in total. Give each answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\), to 3 significant figures, and use unrounded values in later parts.

(a)

Find the cross-sectional area of the cable, in m2.

(2)
(b)

Find the mass of one metre of cable, in kg.

(2)
(c)

Find the total mass of cable needed, in kg.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\pi\times(1.4\times10^{-2})^2\)✓ 1
A1 6.16 × 10−4 m2✓ 1
Part (b)
M1 area × 1 m × density = \(6.1575\times10^{-4}\times7.85\times10^{3}\)✓ 1
A1 4.83 kg✓ 1
Part (c)
M1 mass per metre × \(8.5\times10^4\)✓ 1
A1 4.11 × 105 kg✓ 1

Common error: The radius is half the diameter; using 2.8 × 10⁻² m as the radius makes the area, and so the mass, four times too large.

312IBMA3H-1-20r35-071.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

Calculator

The area covered by a coral colony is modelled by \(A=0.80\times1.35^{t}\), where \(A\) is in m2 and \(t\) is the time in years since the colony was first measured.

(a)

Find the area covered when \(t=3\), to 3 significant figures.

(2)
(b)

Find the time taken for the area to reach 6.0 m2, to 3 significant figures.

(3)
(c)

Find the time taken for the area to double, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(0.80\times1.35^3\)✓ 1
A1 1.97 m2✓ 1
Part (b)
M1 \(1.35^t=7.5\)✓ 1
M1 take logarithms: \(t=\dfrac{\ln7.5}{\ln1.35}\)✓ 1
A1 \(t=6.71\) years✓ 1
Part (c)
M1 \(1.35^t=2\)✓ 1
A1 \(t=\dfrac{\ln2}{\ln1.35}=2.31\) years✓ 1

Exam tip: The doubling time of an exponential model does not depend on the starting value: the 0.80 cancels, so only the growth factor matters.

313IBMA3H-1-19r39-021.9 The binomial theorem Paper 2Medium7 marks

Calculator

A freight company finds that its weekly parcel volume grows by 2% each week, so after 6 weeks the volume is \((1.02)^6\) times the starting volume. This can be written as \((1+2x)^6\) with \(x=0.01\).

(a)

Find the first three terms of the expansion of \((1+2x)^6\) in ascending powers of \(x\).

(3)
(b)

Use your answer to part (a) to estimate the value of \((1.02)^6\).

(2)
(c)

Using the value of \((1.02)^6\) given by your calculator as exact, find the percentage error in your estimate. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the binomial theorem with terms \(\binom{6}{r}(2x)^r\)✓ 1
A1 \(1+12x\)✓ 1
A1 \(+\,60x^2\)✓ 1
Part (b)
M1 substitutes \(x=0.01\)✓ 1
A1 1.126✓ 1
Part (c)
M1 \(\dfrac{|1.126-1.02^6|}{1.02^6}\times 100\) with \(1.02^6=1.126162\ldots\)✓ 1
A1 0.0144 %✓ 1

Exam tip: A percentage error is taken relative to the exact value, not the estimate; write the formula before substituting.

314IBMA3H-1-18r47-021.9 The binomial theorem Paper 2Medium8 marks

Calculator

Consider the expansion of \(\left(x^2 + \frac{2}{x}\right)^9\).

(a)

Show that the general term of the expansion can be written as \(\binom{9}{r}2^r x^{18-3r}\).

(3)
(b)

Find the constant term.

(3)
(c)

Find the coefficient of \(x^6\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 general term \(\binom{9}{r}(x^2)^{9-r}\left(\frac{2}{x}\right)^r\)✓ 1
M1 powers of \(x\): \(x^{18-2r} \times x^{-r}\)✓ 1
A1 \(\binom{9}{r}2^r x^{18-3r}\) (AG)✓ 1
Part (b)
M1 set \(18 - 3r = 0\)✓ 1
A1 \(r = 6\)✓ 1
A1 \(\binom{9}{6} \times 2^6 = 5376\)✓ 1
Part (c)
M1 \(18 - 3r = 6\) so \(r = 4\)✓ 1
A1 \(\binom{9}{4} \times 2^4 = 2016\)✓ 1

Exam tip: For a specific term, set the exponent of \(x\) in the general term equal to the target power and solve for \(r\) before evaluating the coefficient.

315IBMA3H-1-18r47-031.9 The binomial theorem Paper 2Medium7 marks

Calculator

The monthly electricity cost of a data centre increases by 2% each month, so after \(n\) months the cost is multiplied by \(1.02^n\).

(a)

Write down the first three terms in the expansion of \((1 + x)^8\) in ascending powers of \(x\).

(2)
(b)

Use your answer to part (a) with \(x = 0.02\) to estimate the factor by which the monthly cost increases after 8 months.

(2)
(c)

Find the percentage error in this estimate, taking the value of \(1.02^8\) from your calculator as exact. Give your answer correct to three significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(\binom{8}{r}x^r\)✓ 1
A1 \(1 + 8x + 28x^2\)✓ 1
Part (b)
M1 \(1 + 8(0.02) + 28(0.02)^2\)✓ 1
A1 1.1712✓ 1
Part (c)
M1 \(1.02^8 = 1.171659\ldots\)✓ 1
M1 \(\frac{|1.171659\ldots - 1.1712|}{1.171659\ldots} \times 100\)✓ 1
A1 0.0392%✓ 1

Common error: Percentage error is measured relative to the exact value, not to the estimate.

316IBMA3H-1-18r47-061.16 Systems of linear equations Paper 2Medium9 marks

Calculator

A data centre charges a monthly cost of €C for storing T terabytes, modelled by \(C = f + kT\), where \(f\) and \(k\) are constants. Storing 40 terabytes costs €650 per month and storing 100 terabytes costs €1400 per month.

(a)

Write down two equations in \(f\) and \(k\).

(2)
(b)

Solve the equations to find \(f\) and \(k\).

(3)
(c)

Interpret the meaning of \(f\) and of \(k\) in this context.

(2)
(d)

A customer has a monthly budget of €2000. Find the greatest amount of data, in terabytes, that can be stored.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(f + 40k = 650\)✓ 1
A1 \(f + 100k = 1400\)✓ 1
Part (b)
M1 subtract the equations (or use the equation solver), e.g. \(60k = 750\)✓ 1
A1 \(k = 12.5\)✓ 1
A1 \(f = 150\)✓ 1
Part (c)
A1 $f = $ €150 is the fixed monthly charge, the cost when no data is stored✓ 1
A1 $k = $ €12.50 is the extra cost per additional terabyte stored✓ 1
Part (d)
M1 \(150 + 12.5T = 2000\) (ECF from (b))✓ 1
A1 148 terabytes✓ 1

Common error: Interpret each constant with its units and its meaning in the situation, rather than just saying it is the gradient or the intercept.

317IBMA3H-1-18r47-071.16 Systems of linear equations Paper 2Medium7 marks

Calculator

An archaeological team recovers 400 pottery sherds from a site. Each sherd is a rim, a base or a body sherd. The mean masses are 12 g for a rim, 30 g for a base and 8 g for a body sherd, and the 400 sherds have a total mass of 5420 g. There are 190 more body sherds than rim sherds. Let \(a\), \(b\) and \(c\) be the numbers of rim, base and body sherds.

(a)

Write down a system of three linear equations for \(a\), \(b\) and \(c\).

(3)
(b)

Solve the system to find \(a\), \(b\) and \(c\).

(3)
(c)

Find the total mass of the body sherds.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(a + b + c = 400\)✓ 1
A1 \(12a + 30b + 8c = 5420\)✓ 1
A1 \(c - a = 190\)✓ 1
Part (b)
M1 eliminate variables or use the equation solver or matrix methods on a calculator✓ 1
A1 two correct values✓ 1
A1 \(a = 60\), \(b = 90\), \(c = 250\)✓ 1
Part (c)
A1 \(8 \times 250 = 2000\) g (ECF from (b))✓ 1

Exam tip: Define your variables clearly at the start, then check the final answers in all three equations.

318IBMA3H-1-18r47-111.10 Counting principles and the extended binomial theorem Paper 2Medium9 marks

Calculator

A cycling club has 7 men and 5 women. A team of 5 riders is to be chosen for a relay.

(a)

Find the number of different teams that can be chosen.

(2)
(b)

Find the number of different teams that include at least 2 women.

(4)
(c)

A particular team has 3 men and 2 women. The 5 riders line up in single file at the start. Find the number of line-ups in which the two women stand next to each other.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\binom{12}{5}\)✓ 1
A1 792✓ 1
Part (b)
M1 consider the complement: teams with 0 or 1 woman✓ 1
A1 0 women: \(\binom{7}{5} = 21\); 1 woman: \(5 \times \binom{7}{4} = 175\)✓ 1
M1 \(792 - (21 + 175)\) (ECF from (a))✓ 1
A1 596✓ 1
Part (c)
M1 treat the two women as one block, giving 4 objects to arrange in \(4!\) ways✓ 1
M1 multiply by 2! for the order of the women within the block✓ 1
A1 48✓ 1

Exam tip: For "at least" conditions, subtracting the unwanted cases from the total is usually quicker than adding up every wanted case.

319AAHL1-q00091.4 Financial applications Paper 2Hard15 marks

Marta buys a new car for €34 000. Four years later the car is worth €16 400. Assume that the car depreciates by the same percentage at the end of every year.

(a)

Show that the annual rate of depreciation is 16.7%, correct to three significant figures.

(2)
(b)

Find the value of the car after 6 years.

(2)
(c)

Marta intends to replace the car with a new model that costs €31 200 today. The price of the new model increases by 2.3% at the end of every year. Find the price of the new model in 6 years’ time.

(2)
(d)

Find the least number of complete years after which the value of Marta’s car is less than half of the price of the new model at that time.

(3)
(e)

Marta will replace the car after 6 years. She will sell her car for its value after 6 years and pay the rest of the price of the new model from savings. Find the amount she must invest today, at a nominal annual interest rate of 4.5% compounded monthly, so that she has exactly the amount she needs after 6 years.

(3)
(f)

Marta instead has €20 000 to invest today, compounded monthly. Find the least nominal annual interest rate, correct to two decimal places, that would give her at least the amount she needs after 6 years.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(34\,000(1-d)^{4}=16\,400\), or TVM solving for \(I\%\) with \(N=4\), \(PV=-34\,000\), \(FV=16\,400\)M1
\(1-d=\left(\dfrac{16\,400}{34\,000}\right)^{1/4}=0.83337\ldots\), so \(d=0.166623\ldots\) (16.66…%)A1
16.7%AG
(b)
\(34\,000(0.83337\ldots)^{6}\) or \(16\,400(0.83337\ldots)^{2}\)(M1)
€11 390.07A1Accept €11 400. Using 16.7% gives €11 359.23: accept this, and follow through in later parts.
(c)
\(31\,200(1.023)^{6}\)(M1)
€35 760.90A1
(d)
\(34\,000(0.83337\ldots)^{n}\lt\dfrac{1}{2}\times31\,200(1.023)^{n}\)(M1)Inequality, table or graphs.
\(n\gt3.80\ldots\), or after 3 years €19 678.98 > €16 701.35; after 4 years €16 400 < €17 085.48(A1)Using 16.7%: €19 652.32 and €16 370.39.
4 yearsA1Award A0 for 3.8.
(e)
amount needed \(=35\,760.90-11\,390.07=24\,370.83\)(M1)FT from (b) and (c).
\(PV\left(1+\dfrac{4.5}{1200}\right)^{72}=24\,370.83\), or TVM with \(N=72\), \(I\%=4.5\), \(FV=24\,370.83\), \(P/Y=C/Y=12\)(M1)
€18 613.59A1Accept €18 600. FT from (b): €18 637.14 using €11 359.23.
(f)
\(20\,000\left(1+\dfrac{r}{1200}\right)^{72}\ge 24\,370.83\), or TVM solving for \(I\%\)(M1)
\(r\ge 3.2987\ldots\)(A1)3.29% gives €24 358.04 (too little); 3.30% gives €24 372.62
\(3.30\%\)A1Award A0 for 3.29. FT from (b): 3.32% using €11 359.23.

Answers: (b) €11 390.07 · (c) €35 760.90 · (d) 4 years · (e) €18 613.59 · (f) 3.30%

320AAHL1-q00101.4 Financial applications Paper 2Hard11 marks

Lina invests €50 000 in an account that pays interest at a rate of 5% per year, compounded annually. At the end of each year, immediately after the interest has been added, she withdraws €3000. Let \(A_{n}\) be the amount in the account immediately after the \(n\)th withdrawal.

€50 000start−€3000end yr 1: A₁−€3000end yr 2: A₂−€3000end yr 3: A₃…−€3000end yr n: Aₙ5% interest added at the end of each year, then withdrawal
Timeline not to scale
(a)

Find \(A_{1}\) and \(A_{2}\).

(2)
(b)

Show that \(A_{n}=60\,000-10\,000(1.05)^{n}\).

(3)
(c)

Find the number of full withdrawals of €3000 that Lina can make.

(3)
(d)

Lina’s brother also invests €50 000 in this account, and withdraws €\(W\) at the end of each year, immediately after the interest has been added. Find the greatest value of \(W\) for which the amount in his account never decreases.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(A_{1}=50\,000(1.05)-3000=\)€49 500A1
\(A_{2}=49\,500(1.05)-3000=\)€48 975A1
(b)
\(A_{n}=50\,000(1.05)^{n}-3000\left(1+1.05+1.05^{2}+\cdots+1.05^{n-1}\right)\)M1Recognising a geometric series of withdrawals with their accumulated interest.
\(1+1.05+\cdots+1.05^{n-1}=\dfrac{1.05^{n}-1}{0.05}=20\left(1.05^{n}-1\right)\)A1
\(A_{n}=50\,000(1.05)^{n}-60\,000(1.05)^{n}+60\,000\)A1
\(A_{n}=60\,000-10\,000(1.05)^{n}\)AG
(c)
\(60\,000-10\,000(1.05)^{n}\ge0\), ie \(1.05^{n}\le6\)(M1)
\(n\le36.72\ldots\), or \(A_{36}=\)€2081.84 and \(A_{36}\times1.05=\)€2185.93 \(\lt3000\)(A1)
36 withdrawalsA1Award A0 for 37 or 36.7.
(d)
\(A_{n}=50\,000(1.05)^{n}-20W\left(1.05^{n}-1\right)=(50\,000-20W)(1.05)^{n}+20W\)M1Or: interest in the first year is \(0.05\times50\,000=2500\).
the amount never decreases if and only if \(50\,000-20W\ge0\) (equivalently, the withdrawal is no more than the interest added each year)R1
€2500A1

Answers: (a) €49 500, €48 975 · (c) 36 · (d) €2500

321AAHL1-q00391.12 Complex numbers in Cartesian form Paper 2Hard14 marks

In an alternating-current circuit, the impedance of a component is a complex number \(Z=R+Xi\) measured in ohms (Ω), where \(R\) is the resistance and \(X\) is the reactance. When two components with impedances \(Z_{1}\) and \(Z_{2}\) are connected in parallel, the total impedance \(Z\) satisfies

\(\dfrac{1}{Z}=\dfrac{1}{Z_{1}}+\dfrac{1}{Z_{2}}.\)

A coil has impedance \(Z_{1}=30+40i\) and a capacitor branch has impedance \(Z_{2}=20-Xi\), where \(X\gt 0\). They are connected in parallel.

(a)

Find \(|Z_{1}|\) and \(\arg Z_{1}\).

(2)
(b)

Show that \(\dfrac{1}{Z_{1}}=0.012-0.016i\).

(2)
(c)

Show that \(\dfrac{1}{Z_{2}}=\dfrac{20+Xi}{400+X^{2}}\).

(1)
(d)

The circuit is tuned so that the total impedance \(Z\) is a real number. Find the two possible values of \(X\).

(4)
(e)

For the larger value of \(X\), find \(Z\).

(2)
(f)

For the smaller value of \(X\), a real voltage of 230 V is applied across the circuit. The current \(I\), in amperes, is given by \(I=\dfrac{230}{Z}\). Find \(I\).

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(|Z_{1}|=50\)A1
\(\arg Z_{1}=0.927\)A1\(0.927295\ldots\) Accept \(53.1^{\circ}\).
(b)
\(\dfrac{1}{30+40i}\times\dfrac{30-40i}{30-40i}\)M1
\(\dfrac{30-40i}{2500}\)A1
\(=0.012-0.016i\)AG
(c)
\(\dfrac{1}{20-Xi}\times\dfrac{20+Xi}{20+Xi}=\dfrac{20+Xi}{400+X^{2}}\)A1
(d)
setting the imaginary part of \(\dfrac{1}{Z}\) equal to zero (since \(Z\) is real if and only if \(\frac{1}{Z}\) is real)M1
\(\dfrac{X}{400+X^{2}}=0.016\)A1Or \(0.016X^{2}-X+6.4=0\) or \(2X^{2}-125X+800=0\).
\(X=7.24\)A1\(7.23828\ldots\)
\(X=55.3\)A1\(55.2617\ldots\)
(e)
\(\dfrac{1}{Z}=0.012+\dfrac{20}{400+55.26^{2}}\;(=0.017790\ldots)\)M1
\(Z=56.2\ \Omega\)A1\(56.2093\ldots\) Accept 56.2 + 0i.
(f)
\(\dfrac{1}{Z}=0.012+\dfrac{20}{400+7.238^{2}}\;(=0.056210\ldots)\)M1
\(Z=17.8\ \Omega\)A1\(17.7906\ldots\)
\(I=12.9\) AA1\(12.9281\ldots\) FT from their \(Z\). Accept \(12.9+0i\).

Answers: (a) 50, 0.927 · (d) \(X=7.24\) or \(55.3\) · (e) \(Z=56.2\ \Omega\) · (f) \(I=12.9\) A

322AAHL1-q00901.16 Systems of linear equations Paper 2Hard14 marks

A gardener mixes three fertilisers, A, B and C. The table shows the mass, in kg, of nitrogen (N), phosphate (P) and potassium (K) in each kilogram of fertiliser.

ABC
N0.100.300.20
P0.200.100.15
K0.100.200.15

The gardener needs a mixture containing exactly 4 kg of nitrogen, 3 kg of phosphate and 3 kg of potassium, using \(x\) kg of A, \(y\) kg of B and \(z\) kg of C.

(a)

Write down a system of three linear equations in \(x\), \(y\) and \(z\).

(2)
(b)

Show that the system does not have a unique solution.

(2)
(c)

Find the general solution of the system, using \(z=\lambda\).

(3)
(d)

Find the range of possible values of \(\lambda\) in this context.

(2)
(e)

Fertilisers A, B and C cost €1.20, €1.50 and €2.00 per kg respectively. Find the minimum cost of a suitable mixture, and the masses of A, B and C used.

(3)
(f)

Show that a mixture containing exactly 4 kg of nitrogen, 3 kg of phosphate and 3.5 kg of potassium cannot be made.

(2)
Show mark scheme
WorkingMarkNotes
(a)
\(0.1x+0.3y+0.2z=4\)A1
\(0.2x+0.1y+0.15z=3\) and \(0.1x+0.2y+0.15z=3\)A1
(b)
row reduction / GDC rref, or showing \(R_{3}=0.6R_{1}+0.2R_{2}\)M1\(0.6(0.1,0.3,0.2\,|\,4)+0.2(0.2,0.1,0.15\,|\,3)=(0.1,0.2,0.15\,|\,3)\)
a row of zeros (third equation is a combination of the first two), so infinitely many solutions, not a unique oneR1
(c)
reduced form \(x+\tfrac{1}{2}z=10\), \(y+\tfrac{1}{2}z=10\)(M1)
\(x=10-\dfrac{\lambda}{2}\)A1
\(y=10-\dfrac{\lambda}{2}\), \(z=\lambda\)A1
(d)
\(x,y,z\ge0\)M1Masses cannot be negative.
\(0\le\lambda\le20\)A1
(e)
cost \(=1.2\left(10-\tfrac{\lambda}{2}\right)+1.5\left(10-\tfrac{\lambda}{2}\right)+2\lambda=27+0.65\lambda\)M1
minimum when \(\lambda=0\): €27A1
10 kg of A, 10 kg of B, no CA1
(f)
\(0.6(4)+0.2(3)=3\neq3.5\), or GDC rref showing a row \(0=1\)M1
the system is inconsistent (no solution), so the mixture cannot be madeR1

Answers: (c) \(x=y=10-\tfrac{\lambda}{2},\ z=\lambda\) · (d) \(0\le\lambda\le20\) · (e) €27: 10 kg A, 10 kg B, 0 kg C · (f) inconsistent

323IBMA2H-045-06-121.8 Infinite geometric series Paper 2Hard13 marks

Calculator

Magnetic tape of thickness 0.010 cm is wound onto an empty reel. The radius of the first turn is 6.0 cm, and each complete turn increases the radius by the thickness of the tape. The length of the \(k\)th turn is modelled as \(L_k=2\pi\left(6.0+0.010(k-1)\right)\) cm, where the radius of the \(k\)th turn is \(6.0+0.010(k-1)\) cm.

(a)

Show that \(L_1, L_2, L_3,\dots\) form an arithmetic sequence and state its common difference in terms of \(\pi\).

(2)
(b)

Find the length of the 500th turn.

(2)
(c)

Find the total length of tape on a reel holding exactly 500 turns, giving your answer in metres.

(3)
(d)

A 150 m length of tape is wound onto an empty reel of the same type. Find the number of the turn on which the tape finishes.

(4)
(e)

Hence find the radius of the turn on which the tape finishes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(L_{k+1}-L_k=2\pi(0.010)\), which is independent of \(k\), so the sequence is arithmetic ; A1 \(d=0.02\pi\) cm✓ 2
Part (b)
M1 \(L_{500}=2\pi(6.0+499\times0.010)\) ; A1 \(21.98\pi\approx69.1\) cm✓ 2
Part (c)
M1 \(S_{500}=\frac{500}{2}(12\pi+21.98\pi)\) ; A1 \(8495\pi\) cm ; A1 267 m✓ 3
Part (d)
M1 \(S_n=\frac{n}{2}(12\pi+0.02\pi(n-1))=n\pi(12+0.01(n-1))\) ; M1 sets \(S_n\ge15000\) (cm), or solves \(S_n=15000\) ; A1 \(n\approx315.3\) ; A1 the tape finishes on the 316th turn✓ 4
Part (e)
M1 radius of the 316th turn \(=6.0+315\times0.010\) ; A1 9.15 cm✓ 2

Common error: Mixing units when finding the number of turns is costly. The model gives lengths in cm, so 150 m must be written as 15 000 cm before you solve for \(n\).

324IBMA2H-045-03-121.7 Rational exponents and laws of logarithms Paper 2Hard13 marks

Calculator

After a laptop is switched off, the temperature of its processor, \(T\) °C, \(t\) minutes later is modelled by \(T=A+Be^{-kt}\), where \(A\), \(B\) and \(k\) are positive constants. Readings of 85 °C, 55 °C and 40 °C are taken at \(t=0\), \(t=5\) and \(t=10\) respectively.

(a)

Write down three equations connecting \(A\), \(B\) and \(k\).

(2)
(b)

Show that \((55-A)^{2}=(85-A)(40-A)\).

(3)
(c)

Hence find the values of \(A\) and \(B\), and the exact value of \(k\).

(4)
(d)

Find the time taken for the processor to cool to 30 °C.

(3)
(e)

Explain why, according to the model, the processor never cools to 25 °C.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(A+B=85\)✓ 1
A1 \(A+Be^{-5k}=55\) and \(A+Be^{-10k}=40\)✓ 1
Part (b)
M1 writes \(e^{-5k}=\frac{55-A}{B}\) and \(e^{-10k}=\frac{40-A}{B}\), with \(B=85-A\)✓ 1
M1 uses \(e^{-10k}=\left(e^{-5k}\right)^{2}\)✓ 1
A1 \(\frac{40-A}{85-A}=\frac{(55-A)^2}{(85-A)^2}\), leading to the given result✓ 1
Part (c)
M1 expands: \(3025-110A+A^2=3400-125A+A^2\)✓ 1
A1 \(A=25\)✓ 1
A1 \(B=60\)✓ 1
A1 \(e^{-5k}=\frac{1}{2}\), so \(k=\frac{\ln2}{5}\)✓ 1
Part (d)
M1 sets up \(25+60e^{-kt}=30\), so \(e^{-kt}=\frac{1}{12}\)✓ 1
M1 takes logarithms: \(t=\frac{\ln12}{k}=\frac{5\ln12}{\ln2}\)✓ 1
A1 \(t=17.9\) minutes✓ 1
Part (e)
R1 \(60e^{-kt}>0\) for all \(t\), so \(T>25\) always (25 °C is the room-temperature asymptote)✓ 1

Exam tip: When readings are taken at equally spaced times, the excess temperatures \(T-A\) form a geometric sequence. That is what removes \(k\) and \(B\) and gives a linear equation in \(A\).

325IBMA2H-045-02-121.11 Partial fractions Paper 2Hard13 marks

Calculator

In a chemical reactor, substance A (initial concentration 4 mol dm−3) reacts with substance B (initial concentration 1 mol dm−3). The concentration of product, \(x\) mol dm−3, after \(t\) minutes satisfies
\(\frac{dx}{dt}=0.2(4-x)(1-x)\), with \(x=0\) when \(t=0\) and \(0\le x<1\).

(a)

Express \(\frac{1}{(4-x)(1-x)}\) in partial fractions.

(3)
(b)

Hence show that \(x=\frac{4\left(e^{0.6t}-1\right)}{4e^{0.6t}-1}\).

(6)
(c)

Find the limiting value of \(x\) as \(t\to\infty\), and interpret this value in context.

(2)
(d)

Find the time taken for the product concentration to reach \(0.9\) mol dm−3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms the identity \(1\equiv A(1-x)+B(4-x)\)✓ 1
A1 \(A=-\frac13\)✓ 1
A1 \(B=\frac13\), so the expression is \(\frac{1}{3(1-x)}-\frac{1}{3(4-x)}\)✓ 1
Part (b)
M1 separates the variables: \(\int\frac{dx}{(4-x)(1-x)}=\int0.2\,dt\)✓ 1
A1 \(\frac13\ln\left(\frac{4-x}{1-x}\right)=0.2t+c\)✓ 1
M1 substitutes \(x=0\), \(t=0\)✓ 1
A1 \(c=\frac13\ln4\)✓ 1
M1 obtains \(\frac{4-x}{4(1-x)}=e^{0.6t}\) and rearranges to make \(x\) the subject✓ 1
A1 \(x(4e^{0.6t}-1)=4e^{0.6t}-4\), giving the result (AG)✓ 1
Part (c)
M1 divides the numerator and denominator by \(e^{0.6t}\), or argues that \(e^{-0.6t}\to0\)✓ 1
A1 \(x\to1\): the reaction stops when substance B is used up✓ 1
Part (d)
M1 substitutes \(x=0.9\): \(e^{0.6t}=\frac{3.1}{0.4}=7.75\)✓ 1
A1 \(t=3.41\) minutes✓ 1

Common error: Integrating 1/(1−x) gives −ln(1−x), not ln(1−x). Losing this minus sign reverses the log quotient and produces a solution that does not satisfy x(0) = 0.

326IBMA2H-045-07-111.1 Standard form Paper 2Hard15 marks

Calculator

A simple model of global data production assumes that 2.0×1022 bytes of new data are created in year 1, and that the amount created each year is 23% more than in the previous year. Let \(u_n\) be the amount of new data created in year \(n\), and let \(S_n\) be the total created in the first \(n\) years.

(a)

Find \(u_8\), giving your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(b)

Find \(S_8\).

(3)
(c)

Find the first year in which the total amount of data created since year 1 exceeds 1.0×1024 bytes.

(4)
(d)

Find the value of \(\dfrac{u_8}{S_7}\).

(2)
(e)

Show that \(\dfrac{u_n}{S_{n-1}}=\dfrac{0.23}{1-1.23^{\,1-n}}\) for \(n\ge2\). Hence find the limit of \(\dfrac{u_n}{S_{n-1}}\) as \(n\to\infty\), and interpret this limit in context.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_8=2.0\times10^{22}\times1.23^{7}\)✓ 1
A1 8.52×1022 bytes✓ 1
Part (b)
M1 uses the geometric sum formula with \(r=1.23\)✓ 1
A1 \(S_8=\frac{2.0\times10^{22}(1.23^8-1)}{0.23}\)✓ 1
A1 3.69×1023 bytes✓ 1
Part (c)
M1 \(\frac{2.0\times10^{22}(1.23^n-1)}{0.23}>1.0\times10^{24}\)✓ 1
A1 \(1.23^n>12.5\)✓ 1
M1 \(n>\frac{\ln12.5}{\ln1.23}=12.2\)✓ 1
A1 year 13✓ 1
Part (d)
M1 \(S_7=\frac{2.0\times10^{22}(1.23^7-1)}{0.23}=2.834\times10^{23}\)✓ 1
A1 0.301✓ 1
Part (e)
M1 \(\frac{u_n}{S_{n-1}}=\frac{2\times10^{22}\times1.23^{n-1}}{2\times10^{22}(1.23^{n-1}-1)/0.23}=\frac{0.23\times1.23^{n-1}}{1.23^{n-1}-1}\)✓ 1
A1 divides the numerator and denominator by \(1.23^{n-1}\) to obtain the given result✓ 1
R1 \(1.23^{1-n}\to0\) as \(n\to\infty\), so the limit is 0.23✓ 1
R1 in the long run, each year's new data is about 23% of the total created in all previous years combined✓ 1

Exam tip: To find a limit, rearrange the expression so that the term which tends to zero is clear. Here that term is \(1.23^{1-n}\). Simply substituting large values of \(n\) does not show that the ratio tends to 0.23.

327IBMA3H-1-18r6-071.4 Financial applications Paper 2Hard10 marks

Calculator

A sports club pays €200 into a savings account at the start of every month. The account pays interest of 0.4% per month, added at the end of each month. Let \(V_n\) be the value, in euros, of the account at the end of month \(n\), just after interest is added.

(a)

Show that \(V_2=402.40\) to the nearest cent.

(2)
(b)

Show that \(V_n=50200\,(1.004^n-1)\).

(3)
(c)

Find the smallest value of \(n\) for which \(V_n\) exceeds €9500.

(3)
(d)

Find \(V_n\) for this value of \(n\), to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(200(1.004)+200(1.004)^2\)✓ 1
A1 402.4032, so €402.40 (AG)✓ 1
Part (b)
M1 \(V_n\) is a geometric series with first term \(200(1.004)\) and ratio 1.004✓ 1
M1 \(V_n=\frac{200(1.004)(1.004^n-1)}{1.004-1}\)✓ 1
A1 \(\frac{200(1.004)}{0.004}=50200\) (AG)✓ 1
Part (c)
M1 \(50200(1.004^n-1)>9500\)✓ 1
A1 \(1.004^n>1.1892\) so \(n>\frac{\ln1.1892}{\ln1.004}=43.4\)✓ 1
A1 \(n=44\)✓ 1
Part (d)
M1 \(50200(1.004^{44}-1)\) (ECF from (c))✓ 1
A1 €9639.38✓ 1

Exam tip: Deposits at the start of each month earn interest for the full month, which is why the first term is \(200(1.004)\).

328IBMA3H-1-18r10-061.4 Financial applications Paper 2Hard10 marks

Calculator

A fishing cooperative sets up a fund that earns 5% per year, compounded annually. At the end of year 1 it withdraws €2000. Each later annual withdrawal is 3% larger than the one before. The fund must pay exactly 10 withdrawals and then be empty. Let \(r=\frac{1.03}{1.05}\).

initial depositstart€2000end yr 1×1.03end yr 2…end yr 10fund earns 5% per year, compounded annually; withdrawals grow by 3% each year
Timeline not to scale
(a)

Show that the present value of the \(k\)th withdrawal is \(\frac{2000}{1.05}\,r^{k-1}\).

(2)
(b)

Find the amount that must be deposited at the start, to the nearest euro.

(3)
(c)

Find the balance in the fund immediately after the fifth withdrawal, to the nearest euro. Use unrounded values throughout.

(3)
(d)

The cooperative instead wants to fund 15 withdrawals. Find the deposit needed, to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(k\)th withdrawal \(=2000\times1.03^{k-1}\)✓ 1
discounted \(k\) years: divide by \(1.05^{k}\), giving the result✓ 1
Part (b)
geometric series with first term \(\frac{2000}{1.05}\) and ratio \(r\)✓ 1
\(S_{10}=\frac{2000}{1.05}\times\frac{1-r^{10}}{1-r}\)✓ 1
€17495✓ 1
Part (c)
balance equals the value at that time of withdrawals 6 to 10✓ 1
first of these, discounted one year, is \(\frac{2000\times1.03^{5}}{1.05}\), with 5 terms and ratio \(r\)✓ 1
€10628✓ 1
Part (d)
use \(n=15\) in the sum✓ 1
€25059✓ 1

Exam tip: Convert each cash flow to a present value, and the total becomes a single geometric series.

329IBMA3H-1-19r11-051.3 Geometric sequences and series Paper 2Hard13 marks

Calculator

A patient takes a 150 mg dose of a drug every 8 hours, the first dose being taken at time 0. During each 8-hour interval, 38% of the drug present in the body is eliminated, so 62% remains. Let \(A_n\) be the mass of drug in the body, in mg, immediately after the \(n\)th dose.

(a)

Show that \(A_2=243\).

(2)
(b)

Show that \(A_n=\frac{150(1-0.62^n)}{0.38}\).

(3)
(c)

Find the value approached by \(A_n\) as \(n\to\infty\), to 3 significant figures.

(2)
(d)

The mass of drug in the body immediately after a dose must not exceed 385 mg. Find the number of the first dose after which this limit is exceeded.

(4)
(e)

The regular dose is changed to \(D\) mg so that \(A_n\) approaches exactly 385 mg in the long term. Find \(D\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(A_2=150\times0.62+150\)✓ 1
A1 \(93+150=243\) mg (AG)✓ 1
Part (b)
M1 \(A_n=150+150(0.62)+\ldots+150(0.62)^{n-1}\)✓ 1
M1 recognises a geometric series with \(a=150\), \(r=0.62\) and uses \(S_n=\frac{a(1-r^n)}{1-r}\)✓ 1
A1 \(A_n=\frac{150(1-0.62^n)}{0.38}\) (AG)✓ 1
Part (c)
M1 \(0.62^n\to0\) (or use \(\frac{a}{1-r}\))✓ 1
A1 \(\frac{150}{0.38}=395\) mg✓ 1
Part (d)
M1 \(\frac{150(1-0.62^n)}{0.38}>385\)✓ 1
M1 \(0.62^n<1-\frac{385\times0.38}{150}=0.0246\ldots\)✓ 1
M1 \(n>\frac{\ln(0.0246\ldots)}{\ln 0.62}=7.74\ldots\)✓ 1
A1 the 8th dose✓ 1
Part (e)
M1 \(\frac{D}{0.38}=385\)✓ 1
A1 \(D=146.3\) mg✓ 1

Common error: \(A_n\) is the mass just after the dose, so the \(n\)th dose is included in full and the first dose has been reduced \(n-1\) times.

330IBMA3H-1-18r13-061.13 Polar and Euler forms Paper 2Hard8 marks

Calculator

Four charging docks in a warehouse are placed at the points in the complex plane, in metres from the origin, that represent the solutions of \(z^4 = -8 + 8\sqrt{3}\,i\).

(a)

Write \(-8 + 8\sqrt{3}\,i\) in the form \(r\,\mathrm{cis}\,\theta\), where \(r>0\) and \(0\leq\theta<2\pi\).

(2)
(b)

Find the four solutions of \(z^4 = -8+8\sqrt3\,i\). Give each in the form \(r\,\mathrm{cis}\,\theta\) with \(0\leq\theta<2\pi\).

(4)
(c)

The four docks are the vertices of a quadrilateral. Find its area in square metres.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = \sqrt{64+192} = 16\)✓ 1
A1 \(\theta = \frac{2\pi}{3}\) (second quadrant)✓ 1
Part (b)
M1 \(z = 16^{1/4}\,\mathrm{cis}\left(\frac{2\pi/3 + 2\pi k}{4}\right)\)✓ 1
A1 modulus 2✓ 1
A1 \(2\,\mathrm{cis}\frac{\pi}{6}\) and \(2\,\mathrm{cis}\frac{2\pi}{3}\)✓ 1
A1 \(2\,\mathrm{cis}\frac{7\pi}{6}\) and \(2\,\mathrm{cis}\frac{5\pi}{3}\)✓ 1
Part (c)
M1 the points lie on a circle of radius 2, equally spaced by \(\frac{\pi}{2}\), so form a square with diagonal 4✓ 1
A1 area \(= \frac12\times 4\times 4 = 8\) m2✓ 1

Exam tip: The four roots have the same modulus and arguments that differ by \(\frac{2\pi}{4}\), so they sit evenly around a circle.

331IBMA3H-1-18r13-121.3 Geometric sequences and series Paper 2Hard11 marks

Calculator

A fish farm has 10 000 salmon at the start of year 0. Each year, 15% of the fish present are harvested, and then 2000 young fish are added. Let \(P_n\) be the number of salmon at the start of year \(n\) after the young fish are added, with \(P_0 = 10\,000\).

(a)

Show that \(P_2 = 10\,925\).

(2)
(b)

Show that \(P_n = \dfrac{40000}{3} - \dfrac{10000}{3}\,(0.85)^n\).

(4)
(c)

Write down the value that \(P_n\) approaches as \(n\) becomes very large.

(1)
(d)

The farm's licence allows at most 13 000 salmon. Find the number of salmon at the first start of a year, after the young fish are added, when \(P_n\) exceeds 13 000. Give your answer to the nearest whole number.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(P_1 = 0.85\times 10000 + 2000 = 10500\)✓ 1
A1 \(P_2 = 0.85\times 10500 + 2000 = 10925\)✓ 1
Part (b)
M1 \(P_n = 10000(0.85)^n + 2000\left(1 + 0.85 + \cdots + 0.85^{n-1}\right)\)✓ 1
M1 geometric sum \(2000\times\frac{1-0.85^n}{1-0.85}\)✓ 1
A1 \(\frac{40000}{3}(1-0.85^n) + 10000(0.85)^n\)✓ 1
A1 \(= \frac{40000}{3} - \frac{10000}{3}(0.85)^n\) (AG)✓ 1
Part (c)
A1 \(\frac{40000}{3} \approx 13\,333\)✓ 1
Part (d)
M1 \(\frac{40000}{3} - \frac{10000}{3}(0.85)^n > 13000\), i.e. \(0.85^n < 0.1\)✓ 1
M1 solve, e.g. \(n > \frac{\ln 0.1}{\ln 0.85} = 14.2\), or compare \(P_{14}\) and \(P_{15}\)✓ 1
A1 \(n = 15\) (\(P_{14} = 12\,991 < 13\,000\))✓ 1
A1 \(P_{15} = 13\,042\)✓ 1

Common error: Treating the 2000 added fish as if they also lose 15% in the year they arrive. Here they are added after the harvest, so the recurrence is \(P_{n+1}=0.85P_n+2000\).

332IBMA3H-1-18r14-091.16 Systems of linear equations Paper 2Hard10 marks

Calculator

A trawler's weekly catch consists of \(x\) tonnes of cod, \(y\) tonnes of haddock and \(z\) tonnes of plaice. The total catch is 40 tonnes. Prices are €3.2 thousand per tonne for cod, €2.5 thousand per tonne for haddock and €1.8 thousand per tonne for plaice, and the total revenue is €107 thousand. The cod catch exceeds the haddock catch by 4 tonnes.

(a)

Write down three linear equations for \(x\), \(y\) and \(z\).

(3)
(b)

Solve the system to find the mass of each species caught.

(3)
(c)

In another week the plaice price is €\(p\) thousand per tonne, and the cod and haddock prices, total catch (40 tonnes), revenue (€107 thousand) and the cod-haddock difference are unchanged. Show that the system has no unique solution for exactly one value of \(p\), find it, and state how many solutions it then has.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x+y+z=40\)✓ 1
A1 \(3.2x+2.5y+1.8z=107\)✓ 1
A1 \(x-y=4\)✓ 1
Part (b)
M1 solving by technology or elimination✓ 1
A1 \(x=18\) and \(y=14\)✓ 1
A1 \(z=8\)✓ 1
Part (c)
M1 eliminating \(x\) and \(z\) using \(x=y+4\) and \(z=36-2y\)✓ 1
A1 \(y(5.7-2p)=94.2-36p\)✓ 1
A1 no unique solution when \(5.7-2p=0\), i.e. \(p=2.85\)✓ 1
R1 then the right-hand side is \(-8.4\ne0\), so there are no solutions✓ 1

Exam tip: With a parameter, reduce the system to a single equation of the form \(ay=b\) and check separately the cases \(a=0\) with \(b=0\) and \(b\ne0\).

333IBMA3H-1-18r14-121.14 Conjugate roots, De Moivre’s theorem, powers and roots Paper 2Hard10 marks

Calculator

(a)

Use de Moivre's theorem to show that \(\cos3\theta=4\cos^3\theta-3\cos\theta\).

(4)
(b)

Show that, if \(x=\cos\theta\), the equation \(8x^3-6x-1=0\) can be written as \(\cos3\theta=\frac12\).

(2)
(c)

Hence find the three roots of \(8x^3-6x-1=0\), giving each to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta\)✓ 1
M1 expanding the cube binomially✓ 1
A1 real part \(\cos^3\theta-3\cos\theta\sin^2\theta\)✓ 1
A1 using \(\sin^2\theta=1-\cos^2\theta\) to obtain \(4\cos^3\theta-3\cos\theta\) (AG)✓ 1
Part (b)
M1 \(8\cos^3\theta-6\cos\theta=2\cos3\theta\)✓ 1
A1 \(2\cos3\theta-1=0\), so \(\cos3\theta=\frac12\)✓ 1
Part (c)
M1 solving \(\cos3\theta=\frac12\) with \(3\theta=\frac\pi3+2k\pi\) or \(3\theta=\frac{5\pi}{3}+2k\pi\)✓ 1
A1 three distinct angles \(\theta=\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}\)✓ 1
A1 \(x=\cos\frac{\pi}{9}=0.940\) and \(x=\cos\frac{5\pi}{9}=-0.174\)✓ 1
A1 \(x=\cos\frac{7\pi}{9}=-0.766\)✓ 1

Common error: Use the angle values that give three different cosines; angles differing by \(2\pi\) give the same root.

334IBMA3H-1-18r16-141.15 Proof by induction, contradiction and counterexample Paper 2Hard7 marks

Calculator

A rail operator models a growing fleet with the geometric series \(5 + 15 + 45 + \dots\) The sum of the first n terms is \(S_n = \sum_{k=0}^{n-1} 5\times 3^k\).

(a)

Prove by mathematical induction that \(S_n = \frac{5(3^n - 1)}{2}\) for all \(n \in \mathbb{Z}^+\).

(5)
(b)

Hence find the sum of the first 10 terms.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 n = 1: LHS = 5, RHS = \(\frac{5(3-1)}{2} = 5\) so true for n = 1✓ 1
M1 assume true for n = m: \(S_m = \frac{5(3^m-1)}{2}\)✓ 1
M1 \(S_{m+1} = \frac{5(3^m-1)}{2} + 5\times 3^m\)✓ 1
A1 \(= \frac{5(3^m - 1 + 2\times 3^m)}{2} = \frac{5(3^{m+1}-1)}{2}\)✓ 1
R1 true for n = 1 and true for n = m implies true for n = m + 1, so true for all \(n\in\mathbb{Z}^+\)✓ 1
Part (b)
M1 \(\frac{5(3^{10}-1)}{2}\)✓ 1
A1 147 620✓ 1

Exam tip: An induction proof needs all three parts: the base case, the assumption for n = m, and a closing statement using the words 'for all'.

335IBMA3H-1-18r24-121.16 Systems of linear equations Paper 2Hard10 marks

Calculator

Three exoplanets have masses m1, m2 and m3, measured in units of 1024 kg. Astronomers know the combined mass of each pair: m1 + m2 = 10.8, m2 + m3 = 5.44 and m1 + m3 = 6.64.

(a)

Write down the three equations as a matrix equation or as a system, ready to be solved.

(2)
(b)

Solve the system.

(3)
(c)

Write the mass of the first planet in kilograms, in the form a × 10k.

(1)
(d)

The first planet is a sphere of radius 6.4 × 106 m. Find its average density, in kg m−3, to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 a correct system of three equations in m1, m2, m3 (or the matrix form with the coefficient matrix [1 1 0; 0 1 1; 1 0 1])✓ 1
A1 right-hand side 10.8, 5.44, 6.64 correctly matched to the equations✓ 1
Part (b)
M1 solves using a GDC or elimination (for example adding all three: m1 + m2 + m3 = 11.44)✓ 1
A1 m1 = 6.0 and m2 = 4.8✓ 1
A1 m3 = 0.64✓ 1
Part (c)
A1 6.0 × 1024 kg (ECF from (b))✓ 1
Part (d)
M1 volume = (4/3)π(6.4 × 106)3✓ 1
A1 1.098 × 1021 m3✓ 1
M1 density = mass ÷ volume = 6.0 × 1024 ÷ volume✓ 1
A1 5464 ≈ 5.46 × 103 = 5460 kg m−3✓ 1

Exam tip: Convert scaled units (such as 10²⁴ kg) to kilograms before using a physical formula, and write the answer with a power of ten.

336IBMA3H-1-18r24-141.1 Standard form Paper 2Hard11 marks

Calculator

The distances between consecutive stations on a new rail line form an arithmetic sequence. The distance from station 3 to station 4 is 1.8 km (the third gap) and the sum of the first 8 gaps is 18.0 km. The first gap is a km and the common difference is d km.

(a)

Find a and d.

(4)
(b)

The line has 25 stations, so 24 gaps. Find the length of the last gap.

(2)
(c)

Find the total length of the line.

(2)
(d)

A train travels at an average speed of 54 km h−1 between stops and also waits 2 minutes at each of the 23 intermediate stations. Find the total journey time from the first to the last station, in minutes.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 a + 2d = 1.8✓ 1
M1 (8/2)(2a + 7d) = 18.0, so 2a + 7d = 4.5✓ 1
A1 d = 0.3✓ 1
A1 a = 1.2✓ 1
Part (b)
M1 u24 = 1.2 + 23(0.3)✓ 1
A1 8.1 km✓ 1
Part (c)
M1 S24 = (24/2)(1.2 + 8.1)✓ 1
A1 111.6 km✓ 1
Part (d)
M1 travel time = 111.6 ÷ 54 h = 124 minutes (ECF from (c))✓ 1
M1 waiting time = 23 × 2 = 46 minutes✓ 1
A1 170 minutes✓ 1

Exam tip: Two unknowns in an arithmetic sequence need two pieces of information: write one equation for a term and one for a sum, then solve them together.

337IBMA3H-1-20r29-141.7 Rational exponents and laws of logarithms Paper 2Hard9 marks

Calculator

The stock \(S\) of a fish species, in tonnes, is modelled by \(S=ab^t\), where \(t\) is the time in years and \(a\) and \(b\) are constants. The stock is 5200 tonnes at \(t=2\) and 3380 tonnes at \(t=5\).

(a)

Show that \(\ln S=\ln a+t\ln b\).

(2)
(b)

Find the value of \(b\) and the value of \(a\), each to 3 significant figures.

(4)
(c)

The fishery closes when the stock falls below 1000 tonnes. Find the time at which this happens, in years to 3 significant figures. Use the unrounded values of \(a\) and \(b\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\ln(ab^t)=\ln a+\ln b^t\)✓ 1
A1 \(=\ln a+t\ln b\)✓ 1
Part (b)
M1 \(5200=ab^2\) and \(3380=ab^5\)✓ 1
M1 divide: \(b^3=\frac{3380}{5200}=0.65\)✓ 1
A1 \(b=0.866\)✓ 1
A1 \(a=6930\)✓ 1
Part (c)
M1 \(ab^t=1000\)✓ 1
M1 \(t=\frac{\ln(1000/a)}{\ln b}\)✓ 1
A1 \(t=13.5\) years✓ 1

Exam tip: Store unrounded \(a\) and \(b\) in your calculator: rounded values give 13.4 here instead of 13.5.

338IBMA3H-1-20r35-121.15 Proof by induction, contradiction and counterexample Paper 2Hard8 marks

Calculator

A logistics planner models the number of possible routing options after \(n\) stages by \(u_n\), where \(u_1=3\), \(u_2=21\) and \(u_{n+2}=7u_{n+1}-10u_n\) for \(n\in\mathbb{Z}^+\).

(a)

Prove by induction that \(u_n=5^n-2^n\) for all \(n\in\mathbb{Z}^+\).

(6)
(b)

Using the result of (a), find the value of \(u_{10}-5u_9\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 for \(n=1\) and \(n=2\): \(5-2=3\) and \(25-4=21\), so true for both✓ 1
M1 assume true for \(n=k\) and \(n=k+1\): \(u_k=5^k-2^k\) and \(u_{k+1}=5^{k+1}-2^{k+1}\)✓ 1
M1 \(u_{k+2}=7(5^{k+1}-2^{k+1})-10(5^k-2^k)\)✓ 1
A1 \(=5^k(35-10)-2^k(14-10)=25\times5^k-4\times2^k\)✓ 1
A1 \(=5^{k+2}-2^{k+2}\)✓ 1
R1 true for \(n=1,2\), and true for \(n=k,k+1\) implies true for \(n=k+2\), so true for all \(n\in\mathbb{Z}^+\) by induction✓ 1
Part (b)
M1 \((5^{10}-2^{10})-5(5^9-2^9)\)✓ 1
A1 1536✓ 1

Common error: With a two-term recurrence, one base case is not enough: you need two starting values, and the inductive step must assume the statement for two consecutive values.

339IBMA3H-1-18r47-121.10 Counting principles and the extended binomial theorem Paper 2Hard8 marks

Calculator

The efficiency factor of a data centre cooling unit is modelled by \(E = \frac{1}{\sqrt{4 - x}}\), where \(x\) is a small adjustment parameter.

(a)

Find the first three terms in the expansion of \((4 - x)^{-\frac{1}{2}}\) in ascending powers of \(x\).

(4)
(b)

State the values of \(x\) for which the expansion is valid.

(1)
(c)

Use the first three terms of your expansion with \(x = 0.1\) to estimate \(\frac{1}{\sqrt{3.9}}\). Give your answer correct to four decimal places.

(2)
(d)

Explain why the expansion cannot be used to estimate \(\frac{1}{\sqrt{4 - 5}}\) by setting \(x = 5\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 rewrite as \(4^{-\frac{1}{2}}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\)✓ 1
M1 use the binomial expansion with \(n = -\frac{1}{2}\) and \(-\frac{x}{4}\) in place of \(x\)✓ 1
A1 \(\frac{1}{2} + \frac{x}{16}\)✓ 1
A1 \(+\,\frac{3x^2}{256}\)✓ 1
Part (b)
A1 \(|x| < 4\)✓ 1
Part (c)
M1 \(\frac{1}{2} + \frac{0.1}{16} + \frac{3(0.1)^2}{256}\) (ECF from (a))✓ 1
A1 0.5064✓ 1
Part (d)
R1 \(x = 5\) lies outside the interval of validity \(|x| < 4\) (accept: \(4 - 5 < 0\) so the square root is not real)✓ 1

Common error: To use the binomial expansion with a non-integer power, the bracket must start with 1: factorise out the 4 first, remembering to raise it to the same power.

340AAHL1-q00801.15 Proof by induction, contradiction and counterexample Paper 3Hard27 marks

This question investigates sums of the form \(S_{m}(n)=\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)\cdots(r+m)}\), where \(m,n\in\mathbb{Z}^{+}\). The denominator of each term is the product of \(m+1\) consecutive integers.

(a)(i)

Consider \(m=1\).

Express \(\dfrac{1}{r(r+1)}\) in partial fractions.

(2)
(a)(ii)

Hence show that \(S_{1}(n)=\dfrac{n}{n+1}\).

(2)
(a)(iii)

Write down \(\displaystyle\lim_{n\to\infty}S_{1}(n)\).

(1)
(b)(i)

Consider \(m=2\).

Show that \(\dfrac{1}{r(r+1)(r+2)}=\dfrac{1}{2}\left(\dfrac{1}{r(r+1)}-\dfrac{1}{(r+1)(r+2)}\right)\).

(2)
(b)(ii)

Hence show that \(S_{2}(n)=\dfrac{1}{4}-\dfrac{1}{2(n+1)(n+2)}\).

(3)
(c)

It is given that \(S_{3}(n)=A-\dfrac{1}{3(n+1)(n+2)(n+3)}\), where \(A\) is a constant. Use the value of \(S_{3}(1)\) to show that \(A=\dfrac{1}{18}\).

(2)
(d)

Prove by mathematical induction that \(S_{3}(n)=\dfrac{1}{18}-\dfrac{1}{3(n+1)(n+2)(n+3)}\), for \(n\in\mathbb{Z}^{+}\).

(6)
(e)

Find the least value of \(n\) for which \(S_{3}(n)\) differs from \(\dfrac{1}{18}\) by less than \(10^{-6}\).

(3)
(f)

Using your answers to parts (a)(iii), (b)(ii) and (c), suggest an expression, in terms of \(m\), for \(\displaystyle\lim_{n\to\infty}S_{m}(n)\).

(2)
(g)

By first showing that \(\dfrac{1}{r(r+1)\cdots(r+m-1)}-\dfrac{1}{(r+1)(r+2)\cdots(r+m)}=\dfrac{m}{r(r+1)\cdots(r+m)}\), prove your suggestion from part (f).

(4)
Show mark scheme
WorkingMarkNotes
(a)(i)
\(1\equiv A(r+1)+Br\)M1Or cover-up rule.
\(\dfrac{1}{r}-\dfrac{1}{r+1}\)A1
(a)(ii)
\(\left(1-\tfrac{1}{2}\right)+\left(\tfrac{1}{2}-\tfrac{1}{3}\right)+\cdots+\left(\tfrac{1}{n}-\tfrac{1}{n+1}\right)\)M1Writing out at least the first two and the last term.
\(=1-\dfrac{1}{n+1}\)A1
\(=\dfrac{n}{n+1}\)AG
(a)(iii)
\(1\)A1
(b)(i)
\(\dfrac{1}{r(r+1)}-\dfrac{1}{(r+1)(r+2)}=\dfrac{(r+2)-r}{r(r+1)(r+2)}\)M1
\(=\dfrac{2}{r(r+1)(r+2)}\), so half of this is \(\dfrac{1}{r(r+1)(r+2)}\)A1
identity as givenAG
(b)(ii)
\(S_{2}(n)=\dfrac{1}{2}\left[\left(\tfrac{1}{1\times2}-\tfrac{1}{2\times3}\right)+\left(\tfrac{1}{2\times3}-\tfrac{1}{3\times4}\right)+\cdots+\left(\tfrac{1}{n(n+1)}-\tfrac{1}{(n+1)(n+2)}\right)\right]\)M1
intermediate terms cancelA1
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{(n+1)(n+2)}\right)\)A1
\(=\dfrac{1}{4}-\dfrac{1}{2(n+1)(n+2)}\)AG
(c)
\(S_{3}(1)=\dfrac{1}{1\times2\times3\times4}=\dfrac{1}{24}\), so \(\dfrac{1}{24}=A-\dfrac{1}{3\times2\times3\times4}\)M1
\(A=\dfrac{1}{24}+\dfrac{1}{72}=\dfrac{4}{72}\)A1
\(A=\dfrac{1}{18}\)AG
(d)
\(n=1\): LHS \(=\dfrac{1}{24}\), RHS \(=\dfrac{1}{18}-\dfrac{1}{72}=\dfrac{1}{24}\), so true for \(n=1\)A1
assume true for \(n=k\)M1
\(S_{3}(k+1)=\dfrac{1}{18}-\dfrac{1}{3(k+1)(k+2)(k+3)}+\dfrac{1}{(k+1)(k+2)(k+3)(k+4)}\)M1
\(=\dfrac{1}{18}-\dfrac{(k+4)-3}{3(k+1)(k+2)(k+3)(k+4)}\)A1Correct common denominator.
\(=\dfrac{1}{18}-\dfrac{1}{3(k+2)(k+3)(k+4)}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(e)
\(\dfrac{1}{3(n+1)(n+2)(n+3)}\lt10^{-6}\), ie \(3(n+1)(n+2)(n+3)\gt10^{6}\)M1
\(n=67\): \(3\times68\times69\times70=985\,320\); \(n=68\): \(3\times69\times70\times71=1\,028\,790\)(A1)Or GDC solution \(n\gt67.3\).
\(n=68\)A1
(f)
limits \(1,\ \dfrac{1}{4},\ \dfrac{1}{18}\) written as \(\dfrac{1}{1\times1!},\ \dfrac{1}{2\times2!},\ \dfrac{1}{3\times3!}\)M1
\(\displaystyle\lim_{n\to\infty}S_{m}(n)=\dfrac{1}{m\times m!}\)A1
(g)
common denominator: \(\dfrac{(r+m)-r}{r(r+1)\cdots(r+m)}=\dfrac{m}{r(r+1)\cdots(r+m)}\)M1
summing from \(r=1\) to \(n\) and telescoping: \(S_{m}(n)=\dfrac{1}{m}\left(\dfrac{1}{1\times2\times\cdots\times m}-\dfrac{1}{(n+1)(n+2)\cdots(n+m)}\right)\)M1
\(=\dfrac{1}{m}\left(\dfrac{1}{m!}-\dfrac{1}{(n+1)\cdots(n+m)}\right)\)A1
as \(n\to\infty\), \(\dfrac{1}{(n+1)\cdots(n+m)}\to0\), so the limit is \(\dfrac{1}{m\times m!}\)R1

Answers: (a)(i) \(\tfrac{1}{r}-\tfrac{1}{r+1}\) · (a)(iii) 1 · (e) \(n=68\) · (f) \(\dfrac{1}{m\times m!}\)

341AAHL1-q00811.15 Proof by induction, contradiction and counterexample Paper 3Hard28 marks

This question investigates a sequence of fractions that approximate \(\sqrt{2}\). Two sequences of positive integers are defined by \(a_{1}=b_{1}=1\) and

\(a_{n+1}=a_{n}+2b_{n},\qquad b_{n+1}=a_{n}+b_{n},\qquad n\in\mathbb{Z}^{+}\).

(a)

Find \(a_{n}\) and \(b_{n}\) for \(n=2,3,4\).

(2)
(b)

Find the value of \(\left|\dfrac{a_{5}}{b_{5}}-\sqrt{2}\right|\).

(2)
(c)

Prove by mathematical induction that \(a_{n}^{2}-2b_{n}^{2}=(-1)^{n}\), for \(n\in\mathbb{Z}^{+}\).

(5)
(d)

Hence show that \(a_{n}\) and \(b_{n}\) have no common factor greater than 1.

(2)
(e)

Use part (c) to show that \(\left|\dfrac{a_{n}}{b_{n}}-\sqrt{2}\right|=\dfrac{1}{b_{n}\left(a_{n}+b_{n}\sqrt{2}\right)}\).

(3)
(f)

Prove by mathematical induction that \(a_{n}+b_{n}\sqrt{2}=\left(1+\sqrt{2}\right)^{n}\), for \(n\in\mathbb{Z}^{+}\).

(4)
(g)(i)

It can be shown in the same way that \(a_{n}-b_{n}\sqrt{2}=\left(1-\sqrt{2}\right)^{n}\).

Hence find an expression for \(a_{n}\) in terms of \(n\).

(2)
(g)(ii)

Find the value of \(a_{20}\).

(1)
(h)

Find the least value of \(n\) for which \(\left|\dfrac{a_{n}}{b_{n}}-\sqrt{2}\right|\lt10^{-8}\).

(2)
(i)(i)

Suppose, for a contradiction, that \(\sqrt{2}=\dfrac{p}{q}\), where \(p,q\in\mathbb{Z}^{+}\) and \(q\) is the smallest positive integer for which this is possible.

Show that \(\sqrt{2}=\dfrac{2q-p}{p-q}\).

(2)
(i)(ii)

Show that \(0\lt p-q\lt q\).

(2)
(i)(iii)

Hence complete the proof that \(\sqrt{2}\) is irrational.

(1)
Show mark scheme
WorkingMarkNotes
(a)
\(a_{2}=3,\ b_{2}=2\)(M1)
\(a_{3}=7,\ b_{3}=5;\ a_{4}=17,\ b_{4}=12\)A1
(b)
\(a_{5}=41,\ b_{5}=29\)(M1)
\(4.20\times10^{-4}\)A1\(0.000420458\ldots\)
(c)
\(n=1\): \(1-2=-1=(-1)^{1}\), so true for \(n=1\)A1
assume \(a_{k}^{2}-2b_{k}^{2}=(-1)^{k}\)M1
\(a_{k+1}^{2}-2b_{k+1}^{2}=(a_{k}+2b_{k})^{2}-2(a_{k}+b_{k})^{2}\)M1
\(=a_{k}^{2}+4a_{k}b_{k}+4b_{k}^{2}-2a_{k}^{2}-4a_{k}b_{k}-2b_{k}^{2}=-\left(a_{k}^{2}-2b_{k}^{2}\right)=-(-1)^{k}=(-1)^{k+1}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least three of the previous marks have been awarded.
(d)
if \(d\) divides both \(a_{n}\) and \(b_{n}\), then \(d\) divides \(a_{n}^{2}-2b_{n}^{2}=\pm1\)M1
so \(d=1\): there is no common factor greater than 1R1
(e)
\(\dfrac{a_{n}}{b_{n}}-\sqrt{2}=\dfrac{a_{n}-b_{n}\sqrt{2}}{b_{n}}\times\dfrac{a_{n}+b_{n}\sqrt{2}}{a_{n}+b_{n}\sqrt{2}}\)M1
\(=\dfrac{a_{n}^{2}-2b_{n}^{2}}{b_{n}\left(a_{n}+b_{n}\sqrt{2}\right)}\)A1
\(=\dfrac{(-1)^{n}}{b_{n}\left(a_{n}+b_{n}\sqrt{2}\right)}\); taking the modulus, with \(a_{n},b_{n}\gt0\)A1
\(\left|\dfrac{a_{n}}{b_{n}}-\sqrt{2}\right|=\dfrac{1}{b_{n}\left(a_{n}+b_{n}\sqrt{2}\right)}\)AG
(f)
\(n=1\): \(1+\sqrt{2}=\left(1+\sqrt{2}\right)^{1}\), so true for \(n=1\)A1
assume \(a_{k}+b_{k}\sqrt{2}=\left(1+\sqrt{2}\right)^{k}\); then \(\left(1+\sqrt{2}\right)^{k+1}=\left(a_{k}+b_{k}\sqrt{2}\right)\left(1+\sqrt{2}\right)\)M1
\(=a_{k}+a_{k}\sqrt{2}+b_{k}\sqrt{2}+2b_{k}=(a_{k}+2b_{k})+(a_{k}+b_{k})\sqrt{2}=a_{k+1}+b_{k+1}\sqrt{2}\)A1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least two of the previous marks have been awarded.
(g)(i)
adding the two results: \(2a_{n}=\left(1+\sqrt{2}\right)^{n}+\left(1-\sqrt{2}\right)^{n}\)M1
\(a_{n}=\dfrac{\left(1+\sqrt{2}\right)^{n}+\left(1-\sqrt{2}\right)^{n}}{2}\)A1
(g)(ii)
\(a_{20}=22\,619\,537\)A1Must be an exact integer.
(h)
\(n=11\): \(1.07\times10^{-8}\); \(n=12\): \(1.84\times10^{-9}\)(M1)Values \(a_{11}=8119,\ b_{11}=5741;\ a_{12}=19\,601,\ b_{12}=13\,860\).
\(n=12\)A1
(i)(i)
\(\dfrac{2q-p}{p-q}=\dfrac{2-\frac{p}{q}}{\frac{p}{q}-1}=\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)M1Or using \(p^{2}=2q^{2}\) to show \((2q-p)q\sqrt{2}\ldots\); any valid method.
\(=\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}\)A1
\(\sqrt{2}=\dfrac{2q-p}{p-q}\)AG
(i)(ii)
\(1\lt\sqrt{2}\lt2\Rightarrow q\lt p\lt2q\)M1
so \(p-q\gt0\) and \(p-q\lt2q-q=q\)A1
(i)(iii)
\(2q-p\gt0\) and \(p-q\) are positive integers with \(p-q\lt q\), so \(\sqrt{2}\) has been written with a smaller positive denominator than \(q\): a contradiction; hence \(\sqrt{2}\) is irrationalR1

Answers: (a) \(3/2,\ 7/5,\ 17/12\) · (b) \(4.20\times10^{-4}\) · (g)(i) \(a_{n}=\tfrac{1}{2}\left[\left(1+\sqrt2\right)^{n}+\left(1-\sqrt2\right)^{n}\right]\) · (g)(ii) \(22\,619\,537\) · (h) \(n=12\)

342AAHL1-q00911.16 Systems of linear equations Paper 3Hard30 marks

This question investigates formulae for sums of powers of the first \(n\) positive integers. Let \(S_{k}(n)=\displaystyle\sum_{r=1}^{n}r^{k}\), where \(k,n\in\mathbb{Z}^{+}\).

(a)

Write down an expression for \(S_{1}(n)\) in terms of \(n\).

(1)
(b)(i)

It is given that \(S_{2}(n)=an^{3}+bn^{2}+cn\) for all \(n\in\mathbb{Z}^{+}\).

By considering \(n=1,2,3\), write down a system of three linear equations in \(a\), \(b\) and \(c\).

(2)
(b)(ii)

Solve your system to find \(a\), \(b\) and \(c\).

(2)
(c)(i)

It is given that \(S_{3}(n)=n^{2}\left(pn^{2}+qn+s\right)\) for all \(n\in\mathbb{Z}^{+}\).

Find the value of \(p\), of \(q\) and of \(s\).

(3)
(c)(ii)

Hence show that \(S_{3}(n)=\big(S_{1}(n)\big)^{2}\).

(1)
(d)(i)

It is given that \(S_{4}(n)=\dfrac{n(n+1)(2n+1)}{30}\left(\alpha n^{2}+\beta n+\gamma\right)\) for all \(n\in\mathbb{Z}^{+}\).

Show that \(\alpha+\beta+\gamma=5\), \(4\alpha+2\beta+\gamma=17\) and \(9\alpha+3\beta+\gamma=35\).

(2)
(d)(ii)

Hence find \(\alpha\), \(\beta\) and \(\gamma\).

(2)
(e)

Verify that your formula for \(S_{4}(n)\) gives the correct value when \(n=4\).

(1)
(f)

Prove by mathematical induction that \(S_{4}(n)=\dfrac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}\), for \(n\in\mathbb{Z}^{+}\).

(7)
(g)

Find the least value of \(n\) for which \(S_{4}(n)\gt10^{9}\).

(2)
(h)(i)

Consider the ratio \(R(n)=\dfrac{S_{4}(n)}{S_{2}(n)}\).

Show that \(R(n)=\dfrac{3n^{2}+3n-1}{5}\).

(2)
(h)(ii)

A student conjectures that \(R(n)\) is an integer for all \(n\in\mathbb{Z}^{+}\). Use a counterexample to show that the conjecture is false.

(2)
(h)(iii)

By considering the remainder when \(n\) is divided by 5, determine the set of values of \(n\) for which \(R(n)\) is an integer.

(3)
Show mark scheme
WorkingMarkNotes
(a)
\(\dfrac{n(n+1)}{2}\)A1
(b)(i)
\(a+b+c=1\)A1
\(8a+4b+2c=5\) and \(27a+9b+3c=14\)A1
(b)(ii)
use of GDC or elimination(M1)
\(a=\dfrac{1}{3},\ b=\dfrac{1}{2},\ c=\dfrac{1}{6}\)A1
(c)(i)
\(p+q+s=1\); \(4(4p+2q+s)=9\); \(9(9p+3q+s)=36\)M1
solving the system(M1)
\(p=\dfrac{1}{4},\ q=\dfrac{1}{2},\ s=\dfrac{1}{4}\)A1
(c)(ii)
\(S_{3}(n)=\dfrac{n^{2}}{4}\left(n^{2}+2n+1\right)=\dfrac{n^{2}(n+1)^{2}}{4}=\left(\dfrac{n(n+1)}{2}\right)^{2}\)A1
\(=\big(S_{1}(n)\big)^{2}\)AG
(d)(i)
\(n=1\): \(\dfrac{6}{30}(\alpha+\beta+\gamma)=1\Rightarrow\alpha+\beta+\gamma=5\)M1
\(n=2\): \(\dfrac{30}{30}(4\alpha+2\beta+\gamma)=17\); \(n=3\): \(\dfrac{84}{30}(9\alpha+3\beta+\gamma)=98\Rightarrow9\alpha+3\beta+\gamma=35\)A1Must see \(S_{4}(2)=17\) and \(S_{4}(3)=98\).
three equations as givenAG
(d)(ii)
use of GDC or elimination, eg \(3\alpha+\beta=12\), \(5\alpha+\beta=18\)(M1)
\(\alpha=3,\ \beta=3,\ \gamma=-1\)A1
(e)
\(\dfrac{4\times5\times9}{30}(48+12-1)=6\times59=354\) and \(1+16+81+256=354\)A1
(f)
\(n=1\): RHS \(=\dfrac{1\times2\times3\times5}{30}=1=\) LHS, so true for \(n=1\)A1
assume true for \(n=k\)M1
\(S_{4}(k+1)=\dfrac{k(k+1)(2k+1)(3k^{2}+3k-1)}{30}+(k+1)^{4}\)M1
\(=\dfrac{(k+1)\left[k(2k+1)(3k^{2}+3k-1)+30(k+1)^{3}\right]}{30}\)A1
\(=\dfrac{(k+1)\left(6k^{4}+39k^{3}+91k^{2}+89k+30\right)}{30}\)A1
\(=\dfrac{(k+1)(k+2)(2k+3)(3k^{2}+9k+5)}{30}\), where \(3(k+1)^{2}+3(k+1)-1=3k^{2}+9k+5\)A1Accept showing that the expanded target equals the expanded expression.
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); therefore true for all \(n\in\mathbb{Z}^{+}\)R1Award the final R1 only if at least four of the previous marks have been awarded.
(g)
\(S_{4}(86)=968\,416\,459\lt10^{9}\) and \(S_{4}(87)=1\,025\,706\,220\gt10^{9}\)(M1)Or GDC solution of \(S_{4}(n)=10^{9}\): \(n=86.4\ldots\)
\(n=87\)A1
(h)(i)
\(S_{2}(n)=\dfrac{n(n+1)(2n+1)}{6}\) from part (b)M1
\(R(n)=\dfrac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}\times\dfrac{6}{n(n+1)(2n+1)}\)A1
\(=\dfrac{3n^{2}+3n-1}{5}\)AG
(h)(ii)
\(n=2\): \(S_{4}(2)=17\), \(S_{2}(2)=5\)A1Or any \(n\) with \(n\equiv0,2,4\pmod5\).
\(R(2)=\dfrac{17}{5}\), which is not an integerR1
(h)(iii)
considering \(n=5j,\,5j+1,\,5j+2,\,5j+3,\,5j+4\) in \(3n^{2}+3n-1\)M1Or the values of \(3n^{2}+3n-1\) for \(n=1,\ldots,5\): \(5,17,35,59,89\), noting the remainder depends only on \(n\) mod 5.
remainders of \(3n^{2}+3n-1\) on division by 5 are \(4,0,2,0,4\) respectivelyA1
\(R(n)\) is an integer exactly when \(n\) leaves remainder 1 or 3 on division by 5, ie \(n=5j+1\) or \(n=5j+3\), \(j\in\mathbb{N}\)A1

Answers: (a) \(\tfrac{n(n+1)}{2}\) · (b)(ii) \(a=\tfrac13,\ b=\tfrac12,\ c=\tfrac16\) · (c)(i) \(p=\tfrac14,\ q=\tfrac12,\ s=\tfrac14\) · (d)(ii) \(\alpha=3,\ \beta=3,\ \gamma=-1\) · (g) \(n=87\) · (h)(ii) \(n=2\) · (h)(iii) \(n\equiv1\) or \(3\pmod 5\)