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 seriesPaper 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
Working
Mark
Notes
(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 logarithmsPaper 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
Working
Mark
Notes
(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 \(\dfrac{8\times3^{n}}{8\times3^{n-1}}\)
8AAHL1-q00351.12 Complex numbers in Cartesian formPaper 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.
(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
Working
Mark
Notes
(a)
\(z_{1}=1+3i,\ z_{2}=4+2i\)
A1
Both 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 isosceles
R1
\(\text{OA}^{2}+\text{AB}^{2}=10+10=20=\text{OB}^{2}\), so (by the converse of Pythagoras) angle OAB is a right angle
20AAHL1-q00041.2 Arithmetic sequences and seriesPaper 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
Working
Mark
Notes
(a)
\(\dfrac{n}{2}\big(2(100)+10(n-1)\big)=180(n-2)\)
M1
Equating the arithmetic series to the angle sum.
\(5n^{2}+95n=180n-360\)
A1
Or \(n(5n+95)=180n-360\).
\(5n^{2}-85n+360=0\), and dividing by 5
A1
\(n^{2}-17n+72=0\)
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 polygon
R1
\(n=8\)
A1
Largest angle \(170^{\circ}\). Award A0 for \(n=8\) without rejecting \(n=9\).
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 seriesPaper 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.
\(\dfrac{Q(t)}{P(t)}=\dfrac{1}{4}\mathrm{e}^{t\left(\frac{\ln3}{2}-0.35\right)}\) 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 afterwards
R1
Answers: (a) \(k=\dfrac{\ln3}{2}\)
25AAHL1-q00181.8 Infinite geometric seriesPaper 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
Working
Mark
Notes
(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 converges
R1
\(\dfrac{u_{1}}{1-r^{2}}\)
AG
(b)
\(\dfrac{u_{1}}{1-r}\div\dfrac{u_{1}}{1-r^{2}}=\dfrac{12}{9}\), so \(1+r=\dfrac{4}{3}\)
M1
Or substituting \(u_{1}=12(1-r)\) into \(\dfrac{u_{1}}{1-r^{2}}=9\).
\(-1\lt r\lt1\Rightarrow0\lt1+r\lt2\), so \(\dfrac{1}{1+r}\gt\dfrac{1}{2}\); as \(S_{\infty}\gt0\), the odd-position sum is more than \(\tfrac12S_{\infty}\)
30AAHL1-q00261.10 Counting principles and the extended binomial theoremPaper 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.
31AAHL1-q00291.11 Partial fractionsPaper 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)
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Working
Mark
Notes
(a)
\(4\equiv A(2k+3)+B(2k-1)\)
M1
\(A=1\), \(B=-1\)
A1
From \(k=\tfrac12\) and \(k=-\tfrac32\).
\(\dfrac{1}{2k-1}-\dfrac{1}{2k+3}\)
A1
(b)
writing out terms: \(\left(1-\tfrac15\right)+\left(\tfrac13-\tfrac17\right)+\left(\tfrac15-\tfrac19\right)+\cdots+\left(\tfrac{1}{2n-1}-\tfrac{1}{2n+3}\right)\)
M1
At least the first three and the last term.
all terms cancel except \(1+\dfrac13\) and \(-\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\)
32AAHL1-q00341.12 Complex numbers in Cartesian formPaper 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
Working
Mark
Notes
(a)
\(a^{2}-b^{2}+2abi=-5+12i\)
M1
Attempt 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
\(z+1=\pm(2+3i)\)
A1
FT 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 formPaper 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
Working
Mark
Notes
(a)
\((a+bi)(c-di)+(a-bi)(c+di)\)
M1
Attempt 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 formsPaper 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
Working
Mark
Notes
(a)
\(|z_{1}|=4,\ |z_{2}|=2\)
A1
\(\arg z_{1}=\dfrac{2\pi}{3}\)
A1
\(\arg z_{2}=-\dfrac{5\pi}{6}\)
A1
Award 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
\(2\mathrm{e}^{-\frac{\pi}{2}i}=-2i\)
A1
Both forms required. Award A0 for \(2\mathrm{e}^{\frac{3\pi}{2}i}\).
35AAHL1-q00421.13 Polar and Euler formsPaper 1Medium6 marks
The following Argand diagram shows a square OABC, with vertices labelled anticlockwise. The point A represents the complex number \(3+i\).
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)
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Working
Mark
Notes
(a)
multiplying by \(i\) (rotation of \(\frac{\pi}{2}\) anticlockwise): \(i(3+i)\)
37AAHL1-q00451.13 Polar and Euler formsPaper 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)
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Working
Mark
Notes
(a)
\((-1-\sqrt3)+(1-\sqrt3)i\)
A1
(b)
\(u=\sqrt2\,\mathrm{e}^{\frac{3\pi}{4}i}\) and \(v=2\mathrm{e}^{\frac{\pi}{3}i}\)
A1
Both required.
\(uv=2\sqrt2\,\mathrm{e}^{\frac{13\pi}{12}i}\), argument adjusted by \(-2\pi\)
M1
Multiplying moduli and adding arguments.
\(uv=2\sqrt2\,\mathrm{e}^{-\frac{11\pi}{12}i}\)
A1
Award 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 \(2\sqrt2\sin\left(-\dfrac{11\pi}{12}\right)=1-\sqrt3\)
38AAHL1-q00511.14 Conjugate roots, De Moivre’s theorem, powers and rootsPaper 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)
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Working
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Notes
(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 expand
M1
\(a=-6,\ b=23,\ c=-34,\ d=26\)
A1
All four required.
(b)
the four roots are all non-real, so the graph of \(y=p(x)\) does not meet the \(x\)-axis
R1
\(p\) 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 rootsPaper 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)
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Working
Mark
Notes
(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 argument
40AAHL1-q00541.14 Conjugate roots, De Moivre’s theorem, powers and rootsPaper 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)
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Working
Mark
Notes
(a)
\(n=1\): \((\operatorname{cis}\theta)^{1}=\operatorname{cis}\theta\), so true for \(n=1\)
A1
assume true for \(n=k\): \((\operatorname{cis}\theta)^{k}=\cos k\theta+i\sin k\theta\)
\((\operatorname{cis}\theta)^{-m}=\big(\operatorname{cis}(-\theta)\big)^{m}=\operatorname{cis}(-m\theta)\), using part (a) with \(-\theta\) in place of \(\theta\)
R1
Answers: Proof
41AAHL1-q00551.14 Conjugate roots, De Moivre’s theorem, powers and rootsPaper 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
Working
Mark
Notes
(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 cancel
A1
\(z^{n}+z^{-n}=2\cos n\theta\)
AG
(b)
binomial expansion with coefficients \(1,4,6,4,1\)
42AAHL1-q00611.15 Proof by induction, contradiction and counterexamplePaper 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)
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Working
Mark
Notes
(a)
\(n=1\): LHS \(=1\times2=2\), RHS \(=0\times4+2=2\), so true for \(n=1\)
A1
Must see both sides evaluated.
assume true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}r\,2^{r}=(k-1)2^{k+1}+2\)
45AAHL1-q00651.15 Proof by induction, contradiction and counterexamplePaper 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
Working
Mark
Notes
(a)
\(3^{4}=81\) and \(2^{4}+5(16)=96\); \(81\lt96\), so \(P(4)\) is false
A1
Both values required.
(b)
\(n=5\): \(3^{5}=243\gt2^{5}+5(25)=157\), so \(P(5)\) is true
A1
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. \(2^{k}+10k^{2}-10k-5\ge0\)
M1
Award for comparing with the required RHS.
\(10k^{2}-10k-5=10k(k-1)-5\gt0\) for \(k\ge5\) (and \(2^{k}\gt0\)), so \(3^{k+1}\gt2^{k+1}+5(k+1)^{2}\)
A1
Any 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 \(n\ge5\)
R1
Award 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 \(n=5\)
R1
Accept “there is no valid base case below 5”.
Answers: (a) \(81\lt96\)
46AAHL1-q00671.15 Proof by induction, contradiction and counterexamplePaper 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)
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Working
Mark
Notes
(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 \(k+1\) \(=\dfrac{k+1}{2k}\left(1-\dfrac{1}{(k+1)^{2}}\right)\)
Award A0 for \(n=500\) (product exactly \(0.501\)).
Answers: (b) \(n=501\)
47AAHL1-q00691.15 Proof by induction, contradiction and counterexamplePaper 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
Working
Mark
Notes
(a)
\(n=1\): \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x)=1=1\times x^{0}\), so true for \(n=1\)
A1
assume true for \(n=k\): \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{k}\right)=kx^{k-1}\)
53AAHL1-q00841.16 Systems of linear equationsPaper 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
Working
Mark
Notes
(a)
\((0,3)\): \(9+3E+F=0\Rightarrow3E+F=-9\)
A1
\(7D+2E+F=-53\) and \(-2D-E+F=-5\)
A1
Both required.
(b)
eliminating \(F\), eg \(7D-E=-44\) and \(-2D-4E=4\)
M1
one correct equation in two variables
A1
\(D=-6,\ E=2\)
A1
\(F=-15\)
A1
(c)
completing the square: \((x-3)^{2}+(y+1)^{2}=15+9+1\)
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 \(k\)
R1
Accept 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 seriesPaper 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
Working
Mark
Notes
(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 even
R1
if both were odd their product would be odd, but 120 is even; so both are even
R1
(c)
factor pairs of 120 with both factors even and \(n\le n+2k\): \(2\times60,\ 4\times30,\ 6\times20,\ 10\times12\)
M1
Award 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 ways
A1
Award the final A1 only if exactly these four sums are given (no others, e.g. not \(n=1\) or \(n=12\) with \(k\lt 0\)).
At 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, 9
R1
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 5
R1
(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 25
R1
this would need \(n^{2}+2\) to be a multiple of 5, which contradicts (b); so the sum is never a perfect square
R1
Answers: (a)–(c) proofs
57AAHL1-q00151.7 Rational exponents and laws of logarithmsPaper 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
Working
Mark
Notes
(a)
\(\log_{3}5=\dfrac{\log_{2}5}{\log_{2}3}\)
M1
Change 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
\(\dfrac{2p+pq}{2+p}\)
A1
Accept \(\dfrac{p(2+q)}{p+2}\).
Answers: (b) \(\log_{12}45=\dfrac{p(2+q)}{p+2}\)
58AAHL1-q00191.8 Infinite geometric seriesPaper 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.
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 \(n\to\infty\).
59AAHL1-q00211.9 The binomial theoremPaper 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 \(r\).
62AAHL1-q00431.13 Polar and Euler formsPaper 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.
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
Working
Mark
Notes
(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 this
63AAHL1-q00491.13 Polar and Euler formsPaper 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.
(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
Working
Mark
Notes
(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 equal
A1
(d)
area \(=\dfrac12|w^{n}|^{2}=\dfrac12\left(\sqrt2\right)^{2n}\)
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 \(0\lt\frac{2\pi}{5}\lt\frac{\pi}{2}\)
A1
Reason for rejecting the negative root required.
Answers: (c) \(\dfrac{\sqrt5-1}{4}\)
65AAHL1-q00571.14 Conjugate roots, De Moivre’s theorem, powers and rootsPaper 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
Working
Mark
Notes
(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 12
\(\cos\) 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 solutions
R1
\(x=\cos\dfrac{k\pi}{8},\ k=1,3,5,7\)
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 \(\cos\frac{\pi}{8}\gt\cos\frac{3\pi}{8}\gt 0\)
A1
Reason 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 \(=\left(\cos\frac{\pi}{8}\cos\frac{3\pi}{8}\right)^{2}\)
70AAHL1-q00791.15 Proof by induction, contradiction and counterexamplePaper 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
Working
Mark
Notes
(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 impossible
R1
(c)(iii)
by (b), \(N\) has a prime factor of the form \(4k+3\), and by (c)(ii) this prime is not in the list
R1
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 prime
R1
Must 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 false
72AAHL1-q00071.3 Geometric sequences and seriesPaper 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.
73AAHL1-q00121.5 Exponents and introduction to logarithmsPaper 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.
74AAHL1-q00271.10 Counting principles and the extended binomial theoremPaper 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 \(n\).
76AAHL1-q00861.16 Systems of linear equationsPaper 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
Working
Mark
Notes
(a)
\(x+y+z=20\,000\) and \(x+y-z=2000\)
A1
Both required.
\(0.03x+0.045y+0.06z=1020\)
A1
(b)
use of GDC or elimination, eg \(2z=18\,000\)
(M1)
\(x=1000,\ y=10\,000,\ z=9000\)
A1
All 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 negative
77AAHL1-q00881.16 Systems of linear equationsPaper 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.
Office
Coffees
Sandwiches
Pastries
Total cost (€)
1
42
15
20
283.10
2
38
22
14
297.00
3
51
18
25
344.80
(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
Working
Mark
Notes
(a)
setting up a system, eg \(42c+15s+20p=283.10\), \(38c+22s+14p=297.00\), \(51c+18s+25p=344.80\)
78AAHL1-q00011.1 Standard formPaper 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
Working
Mark
Notes
(a)
\(94^{16}\)
(M1)
\(3.72\times10^{31}\)
A1
\(3.7157\ldots\times10^{31}\). Award A0 for 3.72E31.
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)
18
A1
Award A0 for 17.5.
(d)
\(\dfrac{{}^{94}P_{16}}{94^{16}}\) or \(\dfrac{94\times93\times\cdots\times79}{94^{16}}\)
79AAHL1-q00081.4 Financial applicationsPaper 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
Working
Mark
Notes
(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)
€13 488.50
A1
(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 years
A1
Award 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 \(N=32\), \(PV=-11\,000\), \(FV=16\,141.43\), \(P/Y=C/Y=4\)
80AAHL1-q00461.13 Polar and Euler formsPaper 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 \(|z_{1}+z_{2}|^{2}=34+30\cos1.5\).
81AAHL1-q00471.13 Polar and Euler formsPaper 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
Working
Mark
Notes
(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 formsPaper 2Medium6 marks
The points Z and W on an Argand diagram represent \(z=5+2i\) and \(w=-1+4i\) respectively. O is the origin.
84AAHL1-q00871.16 Systems of linear equationsPaper 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)
0.4
1.2
2.0
\(h\) (m)
5.30
7.98
4.40
(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.
\(7.99617\ldots\) Accept \(7.99\)–\(8.03\) (8.02 from 3 s.f. coefficients).
(d)
solving \(h(t)=0\)
(M1)
\(t=2.42\) s
A1
\(2.42116\ldots\) 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 thrown
A1
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 applicationsPaper 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
Working
Mark
Notes
(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)
€11 390.07
A1
Accept €11 400. Using 16.7% gives €11 359.23: accept this, and follow through in later parts.
86AAHL1-q00101.4 Financial applicationsPaper 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.
87AAHL1-q00391.12 Complex numbers in Cartesian formPaper 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
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 \(I\).
\(12.9281\ldots\) 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 equationsPaper 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.
A
B
C
N
0.10
0.30
0.20
P
0.20
0.10
0.15
K
0.10
0.20
0.15
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
Working
Mark
Notes
(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 \(R_{3}=0.6R_{1}+0.2R_{2}\)
\(0.6(4)+0.2(3)=3\neq3.5\), or GDC rref showing a row \(0=1\)
M1
the system is inconsistent (no solution), so the mixture cannot be made
R1
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 counterexamplePaper 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).
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: \(S_{m}(n)=\dfrac{1}{m}\left(\dfrac{1}{1\times2\times\cdots\times m}-\dfrac{1}{(n+1)(n+2)\cdots(n+m)}\right)\)
90AAHL1-q00811.15 Proof by induction, contradiction and counterexamplePaper 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
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.
\(n=1\): \(1+\sqrt{2}=\left(1+\sqrt{2}\right)^{1}\), so true for \(n=1\)
A1
assume \(a_{k}+b_{k}\sqrt{2}=\left(1+\sqrt{2}\right)^{k}\); then \(\left(1+\sqrt{2}\right)^{k+1}=\left(a_{k}+b_{k}\sqrt{2}\right)\left(1+\sqrt{2}\right)\)
\(2q-p\gt0\) 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 irrational
91AAHL1-q00911.16 Systems of linear equationsPaper 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.
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