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ZIMSEC A Level · 9164/2 · J2017

Pure Mathematics Paper 2 June 2017

Questions
52
Total marks
120
Syllabus code
9164/2

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Questions
52
Pass mark
32
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section A

Section A, Question 1

[3 marks]Algebraic manipulation
Express 3x2+6x+7x2+2x+1\dfrac{3x^{2}+6x+7}{x^{2}+2x+1} in the form A(x+B)2+C\dfrac{\mathrm{A}}{(x+\mathrm{B})^{2}}+\mathrm{C}, stating the values of the constants A, B and C.

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[2 marks]Graph transformations
The graph of y=4(x+1)2+3y=\dfrac{4}{(x+1)^{2}}+3 is obtained from the graph of y=1x2y=\dfrac{1}{x^{2}}. Which sequence of transformations does this?
  1. ATranslate 1 unit in the negative xx direction, then stretch parallel to the xx-axis with scale factor 4, then translate 3 units in the negative yy direction.
  2. BTranslate 1 unit in the negative xx direction, then stretch parallel to the yy-axis with scale factor 4, then translate 3 units in the positive yy direction.
  3. CTranslate 1 unit in the positive xx direction, then stretch parallel to the yy-axis with scale factor 4, then translate 3 units in the positive yy direction.
  4. DStretch parallel to the xx-axis with scale factor 4, then translate by (−13)\begin{pmatrix}-1\\3\end{pmatrix}.
[1 marks]Asymptotes
State the equation of the horizontal asymptote of the curve y=4(x+1)2+3y=\dfrac{4}{(x+1)^{2}}+3.

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Section A, Question 2

[3 marks]Coordinate geometry: circles
Find the equation of the circle which passes through the points (−1; 0)(-1;\,0), (1; 2)(1;\,2) and (−5; 4)(-5;\,4), giving your answer in the form x2+y2+2gx+2fy+c=0x^{2}+y^{2}+2gx+2fy+c=0.

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[2 marks]Coordinate geometry: circles
A circle has equation x2+y2+4x−6y+3=0x^{2}+y^{2}+4x-6y+3=0. Find the coordinates of its centre.

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[1 marks]Coordinate geometry: circles
A circle has equation x2+y2+4x−6y+3=0x^{2}+y^{2}+4x-6y+3=0. Find its radius, giving an exact answer.

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Section A, Question 3

[2 marks]Series
Given that ∑r=1nr2=16n(n+1)(2n+1)\displaystyle\sum_{r=1}^{n}r^{2}=\dfrac16 n(n+1)(2n+1), evaluate ∑r=120r2\displaystyle\sum_{r=1}^{20}r^{2}.

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[2 marks]Proof by mathematical induction
In proving by induction that ∑r=1nr2=16n(n+1)(2n+1)\displaystyle\sum_{r=1}^{n}r^{2}=\dfrac16 n(n+1)(2n+1), the inductive step forms 16k(k+1)(2k+1)+(k+1)2\dfrac16 k(k+1)(2k+1)+(k+1)^{2}. Which expression does this simplify to?
  1. A16(k+1)(k+2)(2k+3)\dfrac16(k+1)(k+2)(2k+3)
  2. B16(k+1)(k+2)(2k+1)\dfrac16(k+1)(k+2)(2k+1)
  3. C16k(k+2)(2k+3)\dfrac16 k(k+2)(2k+3)
  4. D16(k+1)2(2k+3)\dfrac16(k+1)^{2}(2k+3)
[2 marks]Proof by mathematical induction
A proof by induction of a formula for ∑r=1nr2\displaystyle\sum_{r=1}^{n}r^{2} begins by checking the case n=1n=1. What does that step establish?
  1. AThat the sum converges to a finite limit as nn becomes arbitrarily large.
  2. BThat the formula holds for the first value of nn.
  3. CThat the formula is true for every positive integer nn, so no further work is needed.
  4. DThat the formula is the only possible closed form for this particular sum of squares.

Section A, Question 4

[3 marks]Turning points
Find the coordinates of the turning point of the curve y=ex+4e−2xy=e^{x}+4e^{-2x}.

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[2 marks]Second derivative test
The curve y=ex+4e−2xy=e^{x}+4e^{-2x} has a turning point at x=ln⁡2x=\ln 2. Determine the nature of that turning point.

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[2 marks]Implicit differentiation
Find the gradient of the curve x2−3xy+y2=5x^{2}-3xy+y^{2}=5 at the point (1; 4)(1;\,4).

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[3 marks]Normals to a curve
Find the equation of the normal to the curve x2−3xy+y2=5x^{2}-3xy+y^{2}=5 at the point (1; 4)(1;\,4), giving your answer in the form ax+by+c=0ax+by+c=0.

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[2 marks]Differential equations
Find the general solution of the differential equation xdθdx=cos⁡2θx\dfrac{d\theta}{dx}=\cos^{2}\theta.

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Section A, Question 5

[2 marks]Matrices: determinants
Find the determinant of the matrix M=(11111−1−1−23)\mathrm{M}=\begin{pmatrix}1&1&1\\1&1&-1\\-1&-2&3\end{pmatrix}.

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[3 marks]Matrices: simultaneous equations
Solve the simultaneous equations
x+y+z=−2x+y+z=-2
x+y−z=2x+y-z=2
−x−2y+3z=3-x-2y+3z=3

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[2 marks]Matrix equations
Matrices M and N satisfy MN=P\mathrm{MN}=\mathrm{P}, where M is a non-singular 3×33\times3 matrix. Which expression gives N?
  1. APM\mathrm{P}\mathrm{M}
  2. BPM\dfrac{\mathrm{P}}{\mathrm{M}}
  3. CM−1P\mathrm{M}^{-1}\mathrm{P}
  4. DPM−1\mathrm{P}\mathrm{M}^{-1}
[2 marks]Matrix equations
Given that M=(11111−1−1−23)\mathrm{M}=\begin{pmatrix}1&1&1\\1&1&-1\\-1&-2&3\end{pmatrix} and MN=(11120101−2)\mathrm{MN}=\begin{pmatrix}1&1&1\\2&0&1\\0&1&-2\end{pmatrix}, find the element of N in the first row and first column.

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[2 marks]Matrices
Why can the equation MN=P\mathrm{MN}=\mathrm{P} be solved for N when M=(11111−1−1−23)\mathrm{M}=\begin{pmatrix}1&1&1\\1&1&-1\\-1&-2&3\end{pmatrix}?
  1. ABecause det⁡M≠0\det\mathrm{M}\neq0, so M−1\mathrm{M}^{-1} exists.
  2. BBecause M is a square matrix, and every square matrix of order 3 has an inverse.
  3. CBecause every entry of M is a whole number, so its inverse has whole number entries.
  4. DBecause M and P have the same number of rows and the same number of columns.

Section A, Question 6

[2 marks]Vectors: points on a plane
The plane P has equation r⋅(2−4−1)=8\mathbf{r}\cdot\begin{pmatrix}2\\-4\\-1\end{pmatrix}=8 and the point A has position vector 3i−j+2k3\mathbf{i}-\mathbf{j}+2\mathbf{k}. Evaluate OA→⋅(2−4−1)\overrightarrow{\mathrm{OA}}\cdot\begin{pmatrix}2\\-4\\-1\end{pmatrix} and hence state whether A lies in P.

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[2 marks]Vectors
Points A and B have position vectors 3i−j+2k3\mathbf{i}-\mathbf{j}+2\mathbf{k} and 7i−9j7\mathbf{i}-9\mathbf{j} respectively. Find BA→\overrightarrow{\mathrm{BA}}.

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[3 marks]Vectors: angle between two vectors
Points A and B have position vectors 3i−j+2k3\mathbf{i}-\mathbf{j}+2\mathbf{k} and 7i−9j7\mathbf{i}-9\mathbf{j} respectively, and O is the origin. Calculate the angle OB^A\mathrm{O\hat{B}A}, giving your answer to the nearest degree.

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[3 marks]Vectors: angle between a line and a plane
The line ll has equation r=(21−1)+λ(2−12)\mathbf{r}=\begin{pmatrix}2\\1\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix} and the plane P has equation r⋅(2−4−1)=8\mathbf{r}\cdot\begin{pmatrix}2\\-4\\-1\end{pmatrix}=8. Find the acute angle between ll and P, to the nearest degree.

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[2 marks]Vectors: distance from a point to a plane
The plane P has equation r⋅(2−4−1)=8\mathbf{r}\cdot\begin{pmatrix}2\\-4\\-1\end{pmatrix}=8 and the point B has position vector 7i−9j7\mathbf{i}-9\mathbf{j}. Find the perpendicular distance from B to P, correct to 3 significant figures.

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Section A, Question 7

[3 marks]Complex numbers: equating parts
The complex numbers W1=1+ixW_{1}=1+ix and W2=x+iyW_{2}=x+iy, where xx and yy are real, satisfy W1−W2=3iW_{1}-W_{2}=3i. Find the value of xx and the value of yy.

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[2 marks]De Moivre's theorem
Use De Moivre's theorem to find the value of (cos⁡14π+isin⁡14π)12\left(\cos\dfrac14\pi+i\sin\dfrac14\pi\right)^{12}.

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[2 marks]Complex numbers: loci
On an Argand diagram, which region satisfies both π4≤arg⁡z≤π2\dfrac{\pi}{4}\le\arg z\le\dfrac{\pi}{2} and ∣z−3i∣≤3|z-3i|\le3?
  1. AThe whole closed disc of radius 3 centred at the point (0; 3)(0;\,3), including its boundary circle.
  2. BThe whole sector between the half line y=xy=x, x≥0x\ge0, and the positive imaginary axis.
  3. CThe part of the disc ∣z−3i∣≤3|z-3i|\le3 that lies in the first quadrant, boundary included.
  4. DThe part of the disc ∣z−3i∣≤3|z-3i|\le3 inside the sector π4≤arg⁡z≤π2\tfrac{\pi}{4}\le\arg z\le\tfrac{\pi}{2}.
[2 marks]De Moivre's theorem
Expanding (cos⁡θ+isin⁡θ)4(\cos\theta+i\sin\theta)^{4} by the binomial theorem and writing c=cos⁡θc=\cos\theta, s=sin⁡θs=\sin\theta, which expression is sin⁡4θ\sin4\theta?
  1. Ac4−6c2s2+s4c^{4}-6c^{2}s^{2}+s^{4}
  2. B4c3s+4cs34c^{3}s+4cs^{3}
  3. C4c2s2−6cs4c^{2}s^{2}-6cs
  4. D4c3s−4cs34c^{3}s-4cs^{3}
[2 marks]De Moivre's theorem
De Moivre's theorem gives cos⁡4θ=cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos4\theta=\cos^{4}\theta-6\cos^{2}\theta\sin^{2}\theta+\sin^{4}\theta. Dividing this by cos⁡4θ\cos^{4}\theta gives the denominator of the standard expression for tan⁡4θ\tan4\theta in terms of tan⁡θ\tan\theta. Write down that denominator.

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Section B

Section B, Question 8

[2 marks]Projectile motion
A particle is projected horizontally from a point O which is at a height of 4545 m vertically above a point P on level ground, and it hits the ground at Q (see diagram). Taking g=9.81 ms−2g=9.81\ \mathrm{ms^{-2}}, calculate the time taken by the particle to reach Q, correct to 3 significant figures.

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[2 marks]Projectile motion
A particle is projected horizontally from a point O which is 4545 m vertically above a point P on level ground, and it lands at Q with PQ=15\mathrm{PQ}=15 m (see diagram). Taking g=9.81 ms−2g=9.81\ \mathrm{ms^{-2}}, find the horizontal speed of projection, correct to 3 significant figures.

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[2 marks]Projectile motion
A particle is projected horizontally from a point O which is 4545 m vertically above a point P on level ground, and it lands at Q with PQ=15\mathrm{PQ}=15 m (see diagram). Taking g=9.81 ms−2g=9.81\ \mathrm{ms^{-2}}, find its speed at Q, correct to 3 significant figures.

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Section B, Question 9

[1 marks]Newton's second law
A particle B of mass 2.52.5 kg is accelerated from rest along a smooth horizontal surface at 4 ms−24\ \mathrm{ms^{-2}}. Find the magnitude of the force acting on B.

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[2 marks]Motion under gravity
A particle A is dropped from rest from a point (9g2)\left(\dfrac{9g}{2}\right) metres vertically above a point R on a horizontal surface, where gg is the acceleration due to gravity. Find the time it takes to reach R.

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[2 marks]Uniform acceleration
A particle B starts from rest and accelerates uniformly at 4 ms−24\ \mathrm{ms^{-2}} along a smooth horizontal surface for 3 seconds. Find the distance it covers.

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[1 marks]Motion under gravity
A particle is dropped from a point (9g2)\left(\dfrac{9g}{2}\right) metres above a horizontal surface. Taking g=9.81 ms−2g=9.81\ \mathrm{ms^{-2}}, find that height in metres, correct to 3 significant figures.

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[2 marks]Displacement-time graphs
Two particles each start from rest under constant acceleration: A falls freely with acceleration 9.81 ms−29.81\ \mathrm{ms^{-2}} and B accelerates at 4 ms−24\ \mathrm{ms^{-2}}. Their displacements are plotted against time on the same axes. What do the two graphs look like?
  1. ABoth are parabolas through the origin, with A's the steeper.
  2. BBoth are straight lines through the origin, with A's line the steeper of the two.
  3. CA is a parabola and B is a straight line, and both of them pass through the origin.
  4. DBoth are parabolas through the origin, with B's the steeper of the two throughout.

Section B, Question 10

[2 marks]Friction: limiting equilibrium
A particle rests in limiting equilibrium on a rough plane inclined at 35∘35^{\circ} to the horizontal. Calculate the coefficient of friction between the particle and the plane, correct to three decimal places.

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[3 marks]Motion on a rough inclined plane
A particle of mass 0.50.5 kg is released from rest on a rough plane inclined at 65∘65^{\circ} to the horizontal, where the coefficient of friction is 0.7000.700. Taking g=9.81 ms−2g=9.81\ \mathrm{ms^{-2}}, find its acceleration down the line of greatest slope, correct to 3 significant figures.

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[3 marks]Motion on a rough inclined plane
A particle is released from rest 0.80.8 m up the line of greatest slope of a rough plane inclined at 65∘65^{\circ} to the horizontal, and slides down with acceleration 5.99 ms−25.99\ \mathrm{ms^{-2}}. Find the time it takes to reach the horizontal surface, correct to 2 significant figures.

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[2 marks]Motion on a rough inclined plane
A particle is released from rest 0.80.8 m up the line of greatest slope of a rough plane and slides down with acceleration 5.99 ms−25.99\ \mathrm{ms^{-2}}. Find its velocity when it reaches the horizontal surface, correct to 2 significant figures.

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Section C

Section C, Question 11

[2 marks]Geometric distribution
The discrete random variable XX has a geometric distribution, X∼Geo(0.4)X\sim\mathrm{Geo}(0.4). Find P(X≤7)\mathrm{P}(X\le7), correct to 3 significant figures.

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[3 marks]Geometric distribution
The discrete random variable XX has a geometric distribution, X∼Geo(0.4)X\sim\mathrm{Geo}(0.4). Find P(X>8∣X>3)\mathrm{P}(X>8\mid X>3).

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Section C, Question 12

[2 marks]Continuous random variables
The continuous random variable XX has probability density function f(x)=2e−kxf(x)=2e^{-kx} for x≥0x\ge0 and f(x)=0f(x)=0 for x<0x<0, where kk is a positive integer. Find the value of kk.

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[2 marks]Cumulative distribution functions
The continuous random variable XX has probability density function f(x)=2e−2xf(x)=2e^{-2x} for x≥0x\ge0 and f(x)=0f(x)=0 for x<0x<0. Find the cumulative distribution function F(x)F(x) for x≥0x\ge0.

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[2 marks]Median of a continuous distribution
The continuous random variable XX has cumulative distribution function F(x)=1−e−2xF(x)=1-e^{-2x} for x≥0x\ge0. Find the exact value of the median of XX.

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Section C, Question 13

[2 marks]Median
The marks obtained by 36 students in a Mathematics test were
59 53 74 55 90 57
88 68 59 67 82 62
61 77 74 86 60 83
92 58 60 72 57 96
56 67 73 78 66 79
51 60 54 67 80 63
Find the median mark.

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[3 marks]Interquartile range
The marks obtained by 36 students in a Mathematics test were
59 53 74 55 90 57
88 68 59 67 82 62
61 77 74 86 60 83
92 58 60 72 57 96
56 67 73 78 66 79
51 60 54 67 80 63
Find the interquartile range of the marks.

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[1 marks]Frequency
The marks obtained by 36 students in a Mathematics test were
59 53 74 55 90 57
88 68 59 67 82 62
61 77 74 86 60 83
92 58 60 72 57 96
56 67 73 78 66 79
51 60 54 67 80 63
How many students scored a mark in the class 65 to 69 inclusive?

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Section C, Question 14

[3 marks]Normal distribution
The random variable XX is normally distributed with mean μ\mu and variance σ2\sigma^{2}. Given that P(X>65)=0.01\mathrm{P}(X>65)=0.01 and P(X<20)=0.02\mathrm{P}(X<20)=0.02, find σ\sigma, correct to 3 significant figures.

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[2 marks]Normal distribution
The random variable XX is normally distributed with mean μ\mu and standard deviation σ=10.27\sigma=10.27. Given that P(X>65)=0.01\mathrm{P}(X>65)=0.01, find μ\mu, correct to 3 significant figures.

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[2 marks]Normal distribution
For a normally distributed XX it is given that P(X>65)=0.01\mathrm{P}(X>65)=0.01 and P(X<20)=0.02\mathrm{P}(X<20)=0.02. Which pair of standardised values does this give?
  1. A65−μσ=0.01\dfrac{65-\mu}{\sigma}=0.01 and 20−μσ=0.02\dfrac{20-\mu}{\sigma}=0.02
  2. B65−μσ=2.054\dfrac{65-\mu}{\sigma}=2.054 and 20−μσ=−2.326\dfrac{20-\mu}{\sigma}=-2.326
  3. C65−μσ=2.326\dfrac{65-\mu}{\sigma}=2.326 and 20−μσ=−2.054\dfrac{20-\mu}{\sigma}=-2.054
  4. D65−μσ=−2.326\dfrac{65-\mu}{\sigma}=-2.326 and 20−μσ=2.054\dfrac{20-\mu}{\sigma}=2.054

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