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ZIMSEC A Level · 9164/2 · N2010

Pure Mathematics Paper 2 November 2010

Questions
57
Total marks
118
Syllabus code
9164/2

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Questions
57
Pass mark
35
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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

[2 marks]Trigonometric identities
Angles AA and BB satisfy A+B=45∘A + B = 45^\circ, and it follows that tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A\tan B = 1. By putting A=B=2212∘A = B = 22\tfrac12^\circ and writing t=tan⁡2212∘t = \tan 22\tfrac12^\circ, write down the resulting quadratic equation in tt.

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[3 marks]Trigonometric identities
For angles with A+B=45∘A + B = 45^\circ it can be shown that tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A\tan B = 1. Use this result to find the exact value of tan⁡2212∘\tan 22\tfrac12^\circ in its simplest surd form.

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[1 marks]Trigonometric identities
Angles AA and BB satisfy A+B=45∘A + B = 45^\circ. Which standard result is the starting point for proving that tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A\tan B = 1?
  1. AThe identity sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A, applied to each of the angles in turn.
  2. BThe compound angle formula for tan⁡(A+B)\tan(A+B), together with tan⁡45∘=1\tan 45^\circ = 1.
  3. CThe sine rule applied to a triangle whose angles are AA, BB and 45∘45^\circ.
  4. DThe double angle formula for tan⁡2A\tan 2A, together with tan⁡45∘=1\tan 45^\circ = 1.

Section A, Question 2

[2 marks]Mathematical induction
The statement '7n−6n−17^n - 6n - 1 is divisible by 36' is to be proved by induction for n∈Nn \in \mathbb{N}, n>1n > 1. Evaluate 7n−6n−17^n - 6n - 1 for the base case n=2n = 2.

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[3 marks]Mathematical induction
In an induction proof that 7n−6n−17^n - 6n - 1 is divisible by 36, assume the result for n=kn = k, so that 7k−6k−1=36p7^k - 6k - 1 = 36p and hence 7k=36p+6k+17^k = 36p + 6k + 1. Simplify 7k+1−6(k+1)−17^{k+1} - 6(k+1) - 1 to a single product showing the factor 36.

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[1 marks]Mathematical induction
In proving by induction that 7n−6n−17^n - 6n - 1 is divisible by 36, the inductive step replaces 7k7^k by 36p+6k+136p + 6k + 1. Why is that substitution the key move?
  1. AIt establishes the base case, which is the part of an induction argument that does the real work.
  2. BIt shows that 7k7^k is itself divisible by 36, so every multiple of it is divisible by 36 too.
  3. CIt brings in the inductive assumption, so the n=k+1n = k+1 expression can be written as a multiple of 36.
  4. DIt removes the variable kk entirely, leaving an expression that is plainly a multiple of 36.

Section A, Question 3

[2 marks]Maclaurin series
Using the Maclaurin series for exe^x, write down the series for e−2xe^{-2x} up to and including the term in x3x^3, with the terms simplified.

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[2 marks]Maclaurin series
Using the Maclaurin series for sin⁡x\sin x, write down the series for sin⁡3x\sin 3x up to and including the term in x3x^3, with the terms simplified.

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[2 marks]Maclaurin series
Given that e−2x=1−2x+2x2−4x33e^{-2x} = 1 - 2x + 2x^2 - \dfrac{4x^3}{3} and sin⁡3x=3x−9x32\sin 3x = 3x - \dfrac{9x^3}{2} up to the x3x^3 terms, find the series expansion of f(x)=e−2xsin⁡3xf(x) = e^{-2x}\sin 3x up to the term in x3x^3.

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[2 marks]Maclaurin series
For f(x)=e−2xsin⁡3xf(x) = e^{-2x}\sin 3x, differentiate once and hence state the value of f′(0)f'(0).

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[1 marks]Maclaurin series
The expansion of f(x)=e−2xsin⁡3xf(x) = e^{-2x}\sin 3x up to x3x^3 is 3x−6x2+32x33x - 6x^2 + \tfrac32 x^3, and differentiating f(x)f(x) directly gives f′(0)=3f'(0) = 3. What does that check confirm?
  1. AThe coefficient of x3x^3, because the highest term of a Maclaurin series is fixed by the first derivative.
  2. BThat the expansion has no constant term, which is the only feature a first derivative can test.
  3. CThat two series may be multiplied term by term without any loss of accuracy in the product.
  4. DThe coefficient of xx in the expansion, because the xx coefficient of a Maclaurin series is f′(0)f'(0).

Section A, Question 4

[1 marks]Vectors in three dimensions
The points A and B have coordinates (1;2;4)(1; 2; 4) and (3;−2;1)(3; -2; 1) respectively. Find AB→\overrightarrow{AB}.

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[2 marks]Vectors in three dimensions
The points A and B have coordinates (1;2;4)(1; 2; 4) and (3;−2;1)(3; -2; 1) respectively. Find ∣AB→∣\left|\overrightarrow{AB}\right|, leaving your answer in surd form.

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[1 marks]Vectors in three dimensions
The points A and B have coordinates (1;2;4)(1; 2; 4) and (3;−2;1)(3; -2; 1) respectively. Write down the vector equation of the straight line that passes through A and B.

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[3 marks]Vectors in three dimensions
Plane pp has equation r.(2i+j−3k)=28\mathbf{r}.(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) = 28 and plane qq has equation r.(4i−7j+k)=31\mathbf{r}.(4\mathbf{i} - 7\mathbf{j} + \mathbf{k}) = 31. Find a direction vector of their line of intersection, in its simplest integer form.

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[3 marks]Vectors in three dimensions
Plane pp has equation r.(2i+j−3k)=28\mathbf{r}.(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) = 28 and plane qq has equation r.(4i−7j+k)=31\mathbf{r}.(4\mathbf{i} - 7\mathbf{j} + \mathbf{k}) = 31. Find the angle between plane pp and plane qq, giving your answer correct to the nearest 0,1∘0,1^\circ.

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Section A, Question 5

[1 marks]Matrices and transformations
A sequence of matrices is given by (0110)n\begin{pmatrix}0&1\\1&0\end{pmatrix}^{n}, n∈Nn \in \mathbb{N}. Evaluate the second term of the sequence, that is the case n=2n = 2.

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[2 marks]Matrices and transformations
The sequence of matrices (0110)n\begin{pmatrix}0&1\\1&0\end{pmatrix}^{n}, n∈Nn \in \mathbb{N}, has first three terms (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}, (1001)\begin{pmatrix}1&0\\0&1\end{pmatrix}, (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}. State the behaviour of the sequence as nn increases.
  1. AIts entries grow without bound, so the sequence diverges as nn increases without limit.
  2. BIt converges to the zero matrix, because the repeated swapping cancels the entries out.
  3. CIt converges to the identity matrix, because the even powers come to dominate as nn grows.
  4. DIt oscillates with period 2 between (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix} and the identity, so it never settles.
[3 marks]Matrices and transformations
Given that A=(1xy−213−324)A = \begin{pmatrix}1&x&y\\-2&1&3\\-3&2&4\end{pmatrix}, show that ∣A∣=2|A| = 2 reduces to the equation of a straight line, and state that equation.

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[2 marks]Matrices and transformations
Describe completely the single transformation represented by the matrix T=(1021)T = \begin{pmatrix}1&0\\2&1\end{pmatrix}.
  1. AA shear of factor 2 parallel to the yy-axis, with the yy-axis invariant.
  2. BA shear of factor 2 parallel to the xx-axis, with the xx-axis invariant.
  3. CA stretch of factor 2 parallel to the yy-axis, with the xx-axis invariant.
  4. DA reflection in the line y=2xy = 2x, which leaves every point of that line fixed.
[2 marks]Matrices and transformations
Describe completely the single transformation represented by the matrix M=(01−10)M = \begin{pmatrix}0&1\\-1&0\end{pmatrix}.
  1. AA reflection in the line y=xy = x, which leaves every point of that line fixed.
  2. BA rotation of 90∘90^\circ clockwise about the origin, that is 270∘270^\circ anticlockwise.
  3. CA reflection in the xx-axis followed by a stretch of factor 2 parallel to it.
  4. DA rotation of 90∘90^\circ anticlockwise about the origin, that is 270∘270^\circ clockwise.
[3 marks]Matrices and transformations
T=(1021)T = \begin{pmatrix}1&0\\2&1\end{pmatrix} and M=(01−10)M = \begin{pmatrix}0&1\\-1&0\end{pmatrix}. Find the coordinates of the point whose image under (TM)−1(TM)^{-1} is (3;1)(3; 1).

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Section A, Question 6

[1 marks]Complex numbers
The complex number z=a+biz = a + bi, where aa and bb are positive real numbers, and w=izw = iz. Write down ww in terms of aa and bb.

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[2 marks]Complex numbers
The complex number z=a+biz = a + bi, where aa and bb are positive real numbers, and w=iz=−b+aiw = iz = -b + ai. Explain the geometrical relationship between zz and ww on an Argand diagram.
  1. Aww is zz enlarged about the origin by scale factor ii, so that ∣w∣=∣z∣2|w| = |z|^2.
  2. Bww is zz rotated 90∘90^\circ anticlockwise about the origin, and ∣w∣=∣z∣|w| = |z|.
  3. Cww is zz rotated 90∘90^\circ clockwise about the origin, and ∣w∣=∣z∣|w| = |z|.
  4. Dww is the reflection of zz in the real axis, so ww is the conjugate of zz.
[2 marks]Complex numbers
Given that w=izw = iz and z=3+2iz = 3 + 2i, find ww in the form x+yix + yi.

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[3 marks]Complex numbers
The complex numbers ww and vv are defined by w=izw = iz and v=12z+wv = \tfrac12 z + w. Find vv in the form x+yix + yi when z=3+2iz = 3 + 2i.

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[2 marks]Complex numbers
The complex number z=a+biz = a + bi has w=iz=−b+aiw = iz = -b + ai, and v=12z+wv = \tfrac12 z + w. Write down the coordinates of the point representing vv on an Argand diagram, in terms of aa and bb.

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[3 marks]Complex numbers
Use De Moivre's theorem to find the 4 roots of unity, giving your answers in exponential form.

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Section A, Question 7

[2 marks]Trigonometric identities and three dimensional trigonometry
Simplify cos⁡4θ−sin⁡4θsin⁡θcos⁡3θ+sin⁡3θcos⁡θ\dfrac{\cos^4\theta - \sin^4\theta}{\sin\theta\cos^3\theta + \sin^3\theta\cos\theta} by factorising the numerator and the denominator.

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[2 marks]Trigonometric identities and three dimensional trigonometry
Express cos⁡2θ−sin⁡2θsin⁡θcos⁡θ\dfrac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta} as a single trigonometric function of 2θ2\theta.

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[2 marks]Trigonometric identities and three dimensional trigonometry
A rectangular piece of cardboard ABCD is fixed against a vertical wall. The edge AB is inclined at θ∘\theta^\circ and the edge BC at α∘\alpha^\circ to the horizontal. Why must θ+α=90\theta + \alpha = 90?
  1. AThe three heights 41, 62 and 104 are in arithmetic progression, which forces a right angle at B.
  2. BThe wall is vertical, so every edge of the cardboard makes an angle of 45∘45^\circ with the ground.
  3. CABCD is a rectangle, so AB and BC are perpendicular and their inclinations must add to 90∘90^\circ.
  4. DAB is three times BC, and the ratio of two sides always fixes their inclinations at 90∘90^\circ.
[2 marks]Trigonometric identities and three dimensional trigonometry
A rectangular cardboard ABCD leans against a vertical wall with A, B and C at heights 41 cm, 62 cm and 104 cm above the ground. AB and BC are inclined at θ∘\theta^\circ and α∘\alpha^\circ to the horizontal, with θ+α=90\theta + \alpha = 90 and 6sin⁡θ=sin⁡α6\sin\theta = \sin\alpha. Find tan⁡θ\tan\theta.

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[2 marks]Trigonometric identities and three dimensional trigonometry
For a rectangular cardboard leaning on a wall, the inclinations satisfy θ+α=90\theta + \alpha = 90 and 6sin⁡θ=sin⁡α6\sin\theta = \sin\alpha. Solve for θ\theta, giving your answer in degrees correct to 2 decimal places.

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[3 marks]Trigonometric identities and three dimensional trigonometry
A rectangular cardboard ABCD leans against a vertical wall, with A at height 41 cm and B at height 62 cm above the ground. AB is inclined at θ∘\theta^\circ to the horizontal where tan⁡θ=16\tan\theta = \tfrac16. Calculate the length of AB, in cm, correct to 1 decimal place.

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[2 marks]Trigonometric identities and three dimensional trigonometry
A rectangular piece of cardboard ABCD is fixed against a vertical wall. The heights of A, B and C above the ground are 41 cm, 62 cm and 104 cm. Find the height of D above the horizontal, in cm.

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Section B

Section B, Question 8

[2 marks]Projectiles
A stone is projected from a point A with a speed of 20 ms−120\,\mathrm{ms}^{-1} at 30∘30^\circ to the horizontal, towards a bird 30 m away horizontally. Find the time taken for the stone to travel the 30 m horizontally.

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[3 marks]Projectiles
A stone is projected from A at 20 ms−120\,\mathrm{ms}^{-1} at 30∘30^\circ to the horizontal towards a bird standing 2,5 m above level ground on a pole 30 m away. Taking g=9,81 ms−2g = 9,81\,\mathrm{ms}^{-2} and reaching the pole after 1,73 s, find how far vertically above or below the bird the stone passes.

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Section B, Question 9

[2 marks]Friction on an inclined plane
A block of weight 20 N rests on a rough plane inclined at β\beta where sin⁡β=45\sin\beta = \tfrac45, acted on by a horizontal force of PP N pressing towards the plane. Resolving perpendicular to the plane, express the normal reaction RR in terms of PP.

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[2 marks]Friction on an inclined plane
A block of weight 20 N is on the point of slipping down a rough plane inclined at β\beta where sin⁡β=45\sin\beta = \tfrac45, held by a horizontal force of PP N and by limiting friction with μ=14\mu = \tfrac14. Resolving along the plane, express the normal reaction RR in terms of PP.

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[2 marks]Friction on an inclined plane
For a block on an inclined plane the two resolved equations give R=64−12P5R = 64 - \dfrac{12P}{5} and R=4P5+12R = \dfrac{4P}{5} + 12. Solve for the horizontal force PP, in newtons.

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Section B, Question 10

[2 marks]Connected particles
Masses of 2 kg and 5 kg hang on either side of a smooth fixed pulley, joined by a light inextensible string, and are released from rest. Find the acceleration of the particles in terms of gg.

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[2 marks]Connected particles
Masses of 2 kg and 5 kg hang either side of a smooth fixed pulley and are released from rest, each 2,5 m above the ground, with acceleration 3g7\dfrac{3g}{7}. Taking g=9,81 ms−2g = 9,81\,\mathrm{ms}^{-2}, find the speed of the 5 kg mass as it hits the ground.

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[3 marks]Connected particles
Masses of 2 kg and 5 kg hang either side of a smooth fixed pulley, released from rest 2,5 m above the ground. The 5 kg mass lands at 4,58 ms−14,58\,\mathrm{ms}^{-1} and does not rebound. Taking g=9,81 ms−2g = 9,81\,\mathrm{ms}^{-2}, find the greatest height above the ground reached by the 2 kg mass, assuming it does not reach the pulley.

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Section B, Question 11

[2 marks]Kinematics of a particle
A particle starts from rest and moves in a straight line with acceleration a=3 ms−2a = 3\,\mathrm{ms}^{-2} for 0≤t≤20 \le t \le 2 and a=−3 ms−2a = -3\,\mathrm{ms}^{-2} for 2<t≤82 < t \le 8. Find its velocity when t=2t = 2 seconds.

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[2 marks]Kinematics of a particle
A particle starts from rest and moves in a straight line with acceleration a=3 ms−2a = 3\,\mathrm{ms}^{-2} for 0≤t≤20 \le t \le 2 and a=−3 ms−2a = -3\,\mathrm{ms}^{-2} for 2<t≤82 < t \le 8. Find the total distance it travels in the first 8 seconds.

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Section C

Section C, Question 12

[2 marks]Binomial and normal distributions
In a large batch of bolts the probability that a randomly chosen bolt is defective is 16\tfrac16, and a random sample of 42 bolts is taken. State the distribution of the number of defective bolts and the normal distribution that approximates it.
  1. AX∼B(42;16)X\sim B\left(42;\tfrac16\right), approximated by Y∼N(7;356)Y\sim N\left(7;\tfrac{35}{6}\right).
  2. BX∼B(42;16)X\sim B\left(42;\tfrac16\right), approximated by Y∼N(7; 7)Y\sim N\left(7;\,7\right).
  3. CX∼B(42;56)X\sim B\left(42;\tfrac56\right), approximated by Y∼N(35;356)Y\sim N\left(35;\tfrac{35}{6}\right).
  4. DX∼Po(7)X\sim Po\left(7\right), approximated by Y∼N(16;356)Y\sim N\left(\tfrac16;\tfrac{35}{6}\right).
[3 marks]Binomial and normal distributions
In a large batch of bolts the probability that a randomly chosen bolt is defective is 16\tfrac16. A random sample of 42 bolts is taken. Find the probability that more than two bolts are defective, correct to 3 decimal places.

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Section C, Question 13

[1 marks]Discrete random variables
A discrete random variable X has P(X=x)=x−116P(X = x) = \dfrac{x-1}{16} for x=2,3,4,5x = 2, 3, 4, 5 and P(X=x)=9−x16P(X = x) = \dfrac{9-x}{16} for x=6,7,8x = 6, 7, 8. Find P(X=4)P(X = 4).

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[2 marks]Discrete random variables
A discrete random variable X takes the values 2, 3, 4, 5, 6, 7, 8 with probabilities 116,216,316,416,316,216,116\tfrac1{16}, \tfrac2{16}, \tfrac3{16}, \tfrac4{16}, \tfrac3{16}, \tfrac2{16}, \tfrac1{16} respectively. Calculate E(X)E(X).

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[1 marks]Discrete random variables
A discrete random variable X takes the values 2, 3, 4, 5, 6, 7, 8 with probabilities 116,216,316,416,316,216,116\tfrac1{16}, \tfrac2{16}, \tfrac3{16}, \tfrac4{16}, \tfrac3{16}, \tfrac2{16}, \tfrac1{16} respectively. State the nature of the distribution.
  1. ASymmetrical about x=5x = 5, which is why the mean is exactly 5.
  2. BUniform, since every one of the seven probabilities is a number of sixteenths.
  3. CPositively skewed, with a long tail running towards the larger values of xx.
  4. DNegatively skewed, with a long tail running towards the smaller values of xx.

Section C, Question 14

[3 marks]Continuous random variables
A continuous random variable X has probability density function f(x)=kxf(x) = kx for 0≤x≤30 \le x \le 3, f(x)=3k(4−x)f(x) = 3k(4-x) for 3≤x≤43 \le x \le 4 and f(x)=0f(x) = 0 otherwise. Find the value of kk.

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[1 marks]Continuous random variables
A continuous random variable X has probability density function f(x)=16xf(x) = \tfrac16 x for 0≤x≤30 \le x \le 3, f(x)=12(4−x)f(x) = \tfrac12(4-x) for 3≤x≤43 \le x \le 4 and f(x)=0f(x) = 0 otherwise. State the greatest value taken by f(x)f(x), and the value of xx at which it occurs.

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[3 marks]Continuous random variables
A continuous random variable X has probability density function f(x)=16xf(x) = \tfrac16 x for 0≤x≤30 \le x \le 3, f(x)=12(4−x)f(x) = \tfrac12(4-x) for 3≤x≤43 \le x \le 4 and f(x)=0f(x) = 0 otherwise. Find the probability that x>2x > 2.

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Section C, Question 15

[1 marks]Probability
Three players A, B and C throw a fair cubical die in that order, and the first to throw a 6 wins. Find the probability that A wins on his first throw.

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[2 marks]Probability
Three players A, B and C throw a fair cubical die in that order, and the first to throw a 6 wins, the game continuing until someone does. Find the probability that A wins on his second throw.

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[3 marks]Probability
Three players A, B and C throw a fair cubical die in that order, and the first to throw a 6 wins, the game continuing indefinitely until one of them does. Find the probability that A wins the game.

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[2 marks]Probability
Three players A, B and C throw a fair cubical die in that order, and the first to throw a 6 wins. The probability that A wins is 3691\tfrac{36}{91} and the probability that B wins is 3091\tfrac{30}{91}. Find the probability that C wins.

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