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IB Maths Analysis & Approaches · Higher Level · Topic 1

Number and algebra

Original questions in the style of Papers 1, 2 and 3, each with a full worked mark scheme. Tick questions to build a printable test. Topic 1 at Standard Level →

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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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)
Show mark scheme
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)
Show mark scheme
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 \((2,1,1)\)

19AAHL1-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) \(66\,832\)

20AAHL1-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) \(n=8\)

21AAHL1-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) \(192\)

22AAHL1-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) \(4\,102\,649^{2}\)

23AAHL1-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) \(x=3\)

24AAHL1-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) \(k=\dfrac{\ln3}{2}\)

25AAHL1-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) \(3\)

26AAHL1-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 \(0\)

27AAHL1-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

28AAHL1-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

29AAHL1-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) \(2.449488\)

30AAHL1-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) \(k=-\tfrac{16}{3}\)

31AAHL1-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) \(\tfrac43\)

32AAHL1-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) \(1+3i,\ -3-3i\)

33AAHL1-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) \(k=4\)

34AAHL1-q00411.13 Polar and Euler forms Paper 1Medium7 marks

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

(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) \(2\mathrm{e}^{-\pi i/2}=-2i\)

35AAHL1-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) \((2-4\sqrt3)+(4+2\sqrt3)i\)

36AAHL1-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) \(-2\cos\frac{\theta}{2},\ \frac{\theta}{2}-\pi\)

37AAHL1-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) \(\cos\frac{11\pi}{12}=-\frac{\sqrt2+\sqrt6}{4}\), \(\sin\frac{11\pi}{12}=\frac{\sqrt6-\sqrt2}{4}\)

38AAHL1-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) \(a=-6,\ b=23,\ c=-34,\ d=26\)

39AAHL1-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) \(\sqrt3+i,\ -1+\sqrt3 i,\ -\sqrt3-i,\ 1-\sqrt3 i\)

40AAHL1-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

41AAHL1-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) \(\dfrac{8+3\pi}{32}\)

42AAHL1-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) \(19\,447\times2^{11}\) \((=39\,827\,456)\)

43AAHL1-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: \(5^{2(k+1)}+3(k+1)-1=9(25m-8k+3)\).

44AAHL1-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) \(x=-\tfrac{n}{2}\)

45AAHL1-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) \(81\lt96\)

46AAHL1-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) \(n=501\)

47AAHL1-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.

48AAHL1-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) \(\tfrac{n(n+1)(n+2)}{3}\)

49AAHL1-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.

50AAHL1-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.

51AAHL1-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) \(f'(x)=3x^{2}+1\gt0\); \(f(-1)=-1,\ f(0)=1\)

52AAHL1-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

53AAHL1-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

54AAHL1-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) \(k=8\)

55AAHL1-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) \(59+61\); \(27+29+31+33\); \(15+17+\cdots+25\); \(3+5+\cdots+21\)

56AAHL1-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

57AAHL1-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) \(\log_{12}45=\dfrac{p(2+q)}{p+2}\)

58AAHL1-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

59AAHL1-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) \(n=11\), \(r=3\)

60AAHL1-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) \(n=10\)

61AAHL1-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) \(0,\ 1,\ -\tfrac12+\tfrac{\sqrt3}{2}i,\ -\tfrac12-\tfrac{\sqrt3}{2}i\)

62AAHL1-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) \(5\sqrt3\)

63AAHL1-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

64AAHL1-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) \(\dfrac{\sqrt5-1}{4}\)

65AAHL1-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) \(n=12,\ w^{12}=-64\)

66AAHL1-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) \(\dfrac{\sqrt2}{4}\)

67AAHL1-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) \(x=\tfrac{\pi}{4},\ \tfrac{\pi}{2},\ \tfrac{3\pi}{4}\)

68AAHL1-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.

69AAHL1-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) \(103\)

70AAHL1-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) \(N=923=13\times71\)

71AAHL1-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) \(\left(\tfrac{9}{5},-\tfrac{1}{5},-\tfrac{1}{5}\right)\)

72AAHL1-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

73AAHL1-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) \(k=5\)

74AAHL1-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) \(n=52\)

75AAHL1-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 \(a=-3,\ b=1\)

76AAHL1-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 \(x\approx-4333\lt0\)

77AAHL1-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

78AAHL1-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) \(25.8\%\)

79AAHL1-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) \(r=4.82\)

80AAHL1-q00461.13 Polar and Euler forms Paper 2Medium6 marks

Let \(z_{1}=3\mathrm{e}^{0.6i}\) and \(z_{2}=5\mathrm{e}^{2.1i}\).

(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) \(6.01\mathrm{e}^{1.58i}\)

81AAHL1-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

82AAHL1-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.

(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

83AAHL1-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

84AAHL1-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

85AAHL1-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%

86AAHL1-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.

(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

87AAHL1-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

88AAHL1-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

89AAHL1-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) \(\dfrac{1}{m\times m!}\)

90AAHL1-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) \(n=12\)

91AAHL1-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 \(3\pmod 5\)