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ZIMSEC A Level · 9164/2 · J2018

Pure Mathematics Paper 2 June 2018

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

[2 marks]Circular measure
ABCD is a square of side xx cm, drawn with B and C as the upper vertices and A and D as the lower vertices, and O is the midpoint of AD. Find angle AOB in radians, correct to 3 decimal places.

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[2 marks]Circular measure
ABCD is a square of side xx cm with B and C the upper vertices, A and D the lower vertices, and O the midpoint of AD. OBC is a sector of a circle centre O and radius OB. Given that angle AOB is 1,10714871{,}1071487 radians, find angle BOC in radians, correct to 3 decimal places.

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[2 marks]Circular measure
ABCD is a square of side 5 cm with B and C the upper vertices, A and D the lower vertices, and O the midpoint of AD. OBC is a sector of a circle centre O and radius OB, and angle BOC is 0,92729520{,}9272952 radians. Find the area, in cm2^{2}, of the region between the chord BC and the arc BC, correct to 3 decimal places.

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Section a, Question 2

[2 marks]Logarithms
The equation log⁡2(8y)=x+5\log_{2}(8y)=x+5 is to be written without logarithms. Which of these is it equivalent to?
  1. Ay=2x+2y=2^{x+2}
  2. By=2x+5y=2^{x+5}
  3. Cy=2x−2y=2^{x-2}
  4. Dy=8×2x+5y=8\times 2^{x+5}
[3 marks]Logarithms and exponentials
Solve the simultaneous equations log⁡2(8y)=x+5\log_{2}(8y)=x+5 and 4x=3y4^{x}=3y, and state the value of yy.

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[2 marks]Logarithms and exponentials
Solve the simultaneous equations log⁡2(8y)=x+5\log_{2}(8y)=x+5 and 4x=3y4^{x}=3y, and state the value of xx correct to 3 decimal places.

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Section a, Question 3

[3 marks]Implicit differentiation
Find dydx\dfrac{dy}{dx} in terms of xx and yy for the curve y3+3xy2−x3=3y^{3}+3xy^{2}-x^{3}=3.

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[2 marks]Turning points
For the curve y3+3xy2−x3=3y^{3}+3xy^{2}-x^{3}=3 the gradient is dydx=x2−y2y2+2xy\dfrac{dy}{dx}=\dfrac{x^{2}-y^{2}}{y^{2}+2xy}, so at a turning point y=xy=x or y=−xy=-x. Taking y=xy=x, find the value of xx at that turning point.

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[2 marks]Turning points
For the curve y3+3xy2−x3=3y^{3}+3xy^{2}-x^{3}=3 the gradient is dydx=x2−y2y2+2xy\dfrac{dy}{dx}=\dfrac{x^{2}-y^{2}}{y^{2}+2xy}, so at a turning point y=xy=x or y=−xy=-x. Taking y=−xy=-x, find the exact value of xx at that turning point.

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Section a, Question 4

[2 marks]Matrices
Find the determinant of the matrix A=(314231211)A=\begin{pmatrix}3&1&4\\2&3&1\\2&1&1\end{pmatrix}.

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[3 marks]Matrices
For A=(314231211)A=\begin{pmatrix}3&1&4\\2&3&1\\2&1&1\end{pmatrix}, which matrix is A−1A^{-1}?
  1. A(0,20,3−1,10−0,50,5−0,4−0,10,7)\begin{pmatrix}0{,}2&0{,}3&-1{,}1\\0&-0{,}5&0{,}5\\-0{,}4&-0{,}1&0{,}7\end{pmatrix}
  2. B(−0,2−0,31,100,5−0,50,40,1−0,7)\begin{pmatrix}-0{,}2&-0{,}3&1{,}1\\0&0{,}5&-0{,}5\\0{,}4&0{,}1&-0{,}7\end{pmatrix}
  3. C(−0,200,4−0,30,50,11,1−0,5−0,7)\begin{pmatrix}-0{,}2&0&0{,}4\\-0{,}3&0{,}5&0{,}1\\1{,}1&-0{,}5&-0{,}7\end{pmatrix}
  4. D(314231211)\begin{pmatrix}3&1&4\\2&3&1\\2&1&1\end{pmatrix}
[2 marks]Simultaneous equations
Solve the simultaneous equations 3x+y+4z=153x+y+4z=15, 2x+3y+z=122x+3y+z=12 and 2x+y+z=102x+y+z=10, and state the value of xx.

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[2 marks]Simultaneous equations
Solve the simultaneous equations 3x+y+4z=153x+y+4z=15, 2x+3y+z=122x+3y+z=12 and 2x+y+z=102x+y+z=10, and state the value of zz.

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Section a, Question 5

[2 marks]Algebraic expansion
Expand and simplify (2r+1)3−(2r−1)3(2r+1)^{3}-(2r-1)^{3}.

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[2 marks]Proof by induction
A proof by induction that Un=26n+32n−2U_{n}=2^{6n}+3^{2n-2} is divisible by 5 begins with the base case n=1n=1. Find the value of U1U_{1}.

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[2 marks]Proof by induction
In a proof by induction for Un=26n+32n−2U_{n}=2^{6n}+3^{2n-2}, the term Uk+1=26(k+1)+32(k+1)−2U_{k+1}=2^{6(k+1)}+3^{2(k+1)-2} is rewritten using 26k2^{6k} and 32k−23^{2k-2}. Which expression is it equal to?
  1. A64⋅26k+3⋅32k−264\cdot 2^{6k}+3\cdot 3^{2k-2}
  2. B26k+32k−22^{6k}+3^{2k-2}
  3. C6⋅26k+2⋅32k−26\cdot 2^{6k}+2\cdot 3^{2k-2}
  4. D64⋅26k+9⋅32k−264\cdot 2^{6k}+9\cdot 3^{2k-2}
[3 marks]Proof by induction
Assume 26k+32k−2=5m2^{6k}+3^{2k-2}=5m for an integer mm, so 26k=5m−32k−22^{6k}=5m-3^{2k-2}. Substituting into Uk+1=64⋅26k+9⋅32k−2U_{k+1}=64\cdot 2^{6k}+9\cdot 3^{2k-2}, which form shows that Uk+1U_{k+1} is divisible by 5?
  1. A5(64m+11⋅32k−2)5\left(64m+11\cdot 3^{2k-2}\right)
  2. B5(64m−32k−2)5\left(64m-3^{2k-2}\right)
  3. C5(64m−11⋅32k−2)5\left(64m-11\cdot 3^{2k-2}\right)
  4. D5(9m−64⋅32k−2)5\left(9m-64\cdot 3^{2k-2}\right)

Section a, Question 6

[2 marks]Complex numbers
Find (4+i)2(4+i)^{2}, giving the answer in the form a+bia+bi.

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[3 marks]Complex numbers
Find the real part of (2+12i)8\left(2+\frac{1}{2}i\right)^{8}, giving the answer in exact form.

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[2 marks]Polynomials
Both (x+22)\left(x+2\sqrt{2}\right) and (x−22)\left(x-2\sqrt{2}\right) are factors of a polynomial. Write down, in expanded form, the quadratic factor formed by their product.

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[3 marks]Polynomials
It is given that (x+22)\left(x+2\sqrt{2}\right) and (x−22)\left(x-2\sqrt{2}\right) are factors of f(x)=x4−6x3+ax2+bx−104f(x)=x^{4}-6x^{3}+ax^{2}+bx-104. Find the value of aa.

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[2 marks]Polynomials
It is given that (x+22)\left(x+2\sqrt{2}\right) and (x−22)\left(x-2\sqrt{2}\right) are factors of f(x)=x4−6x3+ax2+bx−104f(x)=x^{4}-6x^{3}+ax^{2}+bx-104. Find the value of bb.

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[3 marks]Polynomials
The polynomial f(x)=x4−6x3+5x2+48x−104f(x)=x^{4}-6x^{3}+5x^{2}+48x-104 factorises as (x2−8)(x2−6x+13)\left(x^{2}-8\right)\left(x^{2}-6x+13\right). Find the two non real roots of f(x)=0f(x)=0.

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Section a, Question 7

[3 marks]Vectors and planes
A plane has vector equation r⋅(2−31)=1\mathbf{r}\cdot\begin{pmatrix}2\\-3\\1\end{pmatrix}=1 and the point P has coordinates (−9;17;−2)(-9;17;-2). Find the coordinates of the foot of the perpendicular from P to the plane.

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[3 marks]Vectors and planes
The foot of the perpendicular from the point P(−9;17;−2)P(-9;17;-2) to a plane is F(97;117;227)F\left(\frac{9}{7};\frac{11}{7};\frac{22}{7}\right). Find the coordinates of the image of P when reflected in that plane.

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[3 marks]Vectors, lines and planes
The plane π\pi has equation r⋅(2−1−1)=6\mathbf{r}\cdot\begin{pmatrix}2\\-1\\-1\end{pmatrix}=6 and the line ll has equation x−21=y−32=z+1−2\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z+1}{-2}. Find the coordinates of the point where ll meets π\pi.

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[3 marks]Vectors, lines and planes
The plane π\pi has equation r⋅(2−1−1)=6\mathbf{r}\cdot\begin{pmatrix}2\\-1\\-1\end{pmatrix}=6 and the line ll has equation x−21=y−32=z+1−2\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z+1}{-2}. Find the acute angle between π\pi and ll, correct to the nearest degree.

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[3 marks]Vectors and lines
Find the shortest distance from the origin to the line x−21=y−32=z+1−2\dfrac{x-2}{1}=\dfrac{y-3}{2}=\dfrac{z+1}{-2}, giving the exact value.

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[2 marks]Vectors, lines and planes
When the angle between a line and a plane is found from the dot product of the line's direction vector and the plane's normal vector, why does that dot product give the sine of the required angle rather than its cosine?
  1. ABecause a plane's normal vector is always a unit vector, and that converts every cosine into a sine.
  2. BBecause a line and a plane can never be parallel, so the cosine of the angle would be undefined.
  3. CBecause it gives the angle to the normal, and the angle with the plane is its complement.
  4. DBecause a dot product of two direction vectors always returns a sine value and never a cosine value.

Section b

Section b, Question 8

[3 marks]Statics
A particle P is in equilibrium under four forces. One force of magnitude TT N acts along a ray at 45∘45^{\circ} above the horizontal to the left of P. A second force of magnitude TT N acts on the lower left, making 60∘60^{\circ} with the downward vertical through P. A force of magnitude QQ N acts horizontally to the right, and a force of 6 N acts vertically downwards. Find the exact value of TT.

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[2 marks]Statics
A particle P is in equilibrium under four forces: TT N at 45∘45^{\circ} above the horizontal to the left of P, TT N on the lower left making 60∘60^{\circ} with the downward vertical, QQ N horizontally to the right, and 6 N vertically downwards. Given that T=12(2+1)T=12\left(\sqrt{2}+1\right) N, find QQ correct to 3 significant figures.

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Section b, Question 9

[2 marks]Kinematics
The velocity-time graph of a particle moving in a straight line is a single straight line from the point (0;2)(0;2) to the point (8;−6)(8;-6), where vv is in ms−1^{-1} and tt is in seconds. Find the acceleration of the particle, in ms−2^{-2}.

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[2 marks]Kinematics
A particle moves in a straight line with velocity falling steadily from 2 ms−1^{-1} at t=0t=0 to −6-6 ms−1^{-1} at t=8t=8 s. Find the average velocity of the particle over the 8 seconds, in ms−1^{-1}.

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[1 marks]Kinematics
A particle moves in a straight line with velocity falling steadily from 2 ms−1^{-1} at t=0t=0 to −6-6 ms−1^{-1} at t=8t=8 s, starting from the origin. Which describes its displacement-time graph?
  1. AAn upward parabola through the origin, dipping to −2-2 m at t=2t=2 s and then rising to reach 16 m at t=8t=8 s.
  2. BA downward parabola through the origin, peaking at 2 m at t=2t=2 s and reaching −16-16 m at t=8t=8 s.
  3. CA straight line falling steadily from the origin to reach −16-16 m at t=8t=8 s, with one gradient throughout.
  4. DA horizontal line resting at zero displacement for the whole of the 8 seconds of the motion.

Section b, Question 10

[2 marks]Friction
A particle of mass 8 kg rests on a rough plane of inclination tan⁡−1(34)\tan^{-1}\left(\frac{3}{4}\right) to the horizontal, and the coefficient of friction between the particle and the plane is 15\frac{1}{5}. Taking g=9,81g=9{,}81 ms−2^{-2}, find the limiting friction force on the particle, in N, correct to 3 significant figures.

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[2 marks]Connected particles
A particle P of mass 8 kg rests on a rough plane of inclination tan⁡−1(34)\tan^{-1}\left(\frac{3}{4}\right), with coefficient of friction 15\frac{1}{5}. It is joined by a light inextensible string over a smooth pulley at the top of the plane to a particle Q of mass 2 kg hanging freely, and P slides down the plane. Taking g=9,81g=9{,}81 ms−2^{-2}, find the acceleration of the particles, in ms−2^{-2}, correct to 3 significant figures.

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[2 marks]Connected particles
A particle P of mass 8 kg on a rough plane is joined by a light inextensible string over a smooth pulley to a particle Q of mass 2 kg hanging freely. P slides down the plane and the system accelerates at 1,491121{,}49112 ms−2^{-2}. Taking g=9,81g=9{,}81 ms−2^{-2}, find the tension in the string, in N, correct to 3 significant figures.

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Section b, Question 11

[2 marks]Projectiles
A particle is projected with an initial speed of 60 ms−1^{-1} at an angle of elevation sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right). Find the time, in seconds, it takes to reach the point whose horizontal displacement from the point of projection is 144 m.

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[3 marks]Projectiles
A particle is projected from the top of a vertical dam wall, 10 m above water level, with an initial speed of 60 ms−1^{-1} at an angle of elevation sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right). Taking g=9,81g=9{,}81 ms−2^{-2}, find the time, in seconds, the particle takes to reach the water level, correct to 3 significant figures.

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[3 marks]Projectiles
A particle is projected from the top of a vertical dam wall, 10 m above water level, with an initial speed of 60 ms−1^{-1} at an angle of elevation sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right). Taking g=9,81g=9{,}81 ms−2^{-2}, find the maximum height, in metres, attained by the particle above the water level, correct to 3 significant figures.

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Section c

Section c, Question 12

[2 marks]Discrete random variables
A dice is weighted so that the probability of each face coming up is proportional to the face value xx, where x=1,2,3,4,5,6x=1,2,3,4,5,6. Write down P(X=x)P(X=x) in terms of xx.

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[2 marks]Discrete random variables
A random variable XX takes the values 1,2,3,4,5,61,2,3,4,5,6 with P(X=x)=x21P(X=x)=\dfrac{x}{21}. Calculate E(X)E(X), giving the answer as an exact fraction.

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Section c, Question 13

[2 marks]Binomial probability
A and B played 12 games of chess of which 6 were won by A, 4 were won by B and 2 ended in a draw. Using these results as the probabilities for a single game, they play a tournament of 3 games. Find the probability that exactly two of the games end in a draw, giving the answer as an exact fraction.

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[3 marks]Conditional probability
A and B played 12 games of chess of which 6 were won by A, 4 were won by B and 2 ended in a draw. Using these results as the probabilities for a single game, they play a tournament of 3 games. Find the probability that B wins the tournament given that two of the games ended in a draw.

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Section c, Question 14

[2 marks]Binomial distribution
The probability that an adult chosen at random from a particular group is short sighted is 15\frac{1}{5}. A committee of 7 of the adults in that group is chosen at random. Find the probability that at least one adult on the committee is short sighted, correct to 3 significant figures.

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[3 marks]Normal approximation to the binomial
The probability that an adult from a particular group is short sighted is 15\frac{1}{5}. A bus carried 60 adult passengers from that group. Using a normal approximation with a continuity correction, find the probability that at least 15 of them were short sighted, correct to 3 significant figures.

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Section c, Question 15

[2 marks]Continuous random variables
A continuous random variable XX has probability density function f(x)=ce−2xf(x)=ce^{-2x} for x>0x>0 and f(x)=0f(x)=0 otherwise, where cc is a constant. Find the value of cc.

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[2 marks]Continuous random variables
A continuous random variable XX has probability density function f(x)=2e−2xf(x)=2e^{-2x} for x>0x>0 and f(x)=0f(x)=0 otherwise. Calculate the median of XX, correct to 3 significant figures.

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[1 marks]Continuous random variables
For a function ff to be a probability density function on x>0x>0, which condition fixes the value of the constant multiplying it?
  1. AThe integral of f(x)f(x) over its whole range equals 1.
  2. BThe largest value that ff reaches anywhere on its range must come out exactly equal to 1.
  3. CThe derivative of ff must come out as exactly zero at the median of the distribution.
  4. DThe value that ff takes at the left hand end of its range must come out exactly equal to 1.

Section c, Question 16

[2 marks]Measures of location

The sales, in US$, made by a vendor on 30 days were:

11, 37, 20, 15, 16, 23, 20, 26, 10, 21, 28, 13, 36, 18, 15, 32, 31, 16, 28, 26, 45, 46, 23, 58, 15, 54, 43, 37, 10, 32.

Find the median.

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[2 marks]Measures of spread

The sales, in US$, made by a vendor on 30 days were:

11, 37, 20, 15, 16, 23, 20, 26, 10, 21, 28, 13, 36, 18, 15, 32, 31, 16, 28, 26, 45, 46, 23, 58, 15, 54, 43, 37, 10, 32.

Find the interquartile range.

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[1 marks]Skewness
A set of 30 sales figures has minimum 10, lower quartile 16, median 24,5, upper quartile 36 and maximum 58. What does its box and whisker diagram show about the shape of the distribution?
  1. AIt is uniform, because the whiskers and the two halves of the box all come out equal in length to each other.
  2. BIt is negatively skewed: the median sits nearer the upper quartile and the left whisker is much the longer of the two.
  3. CIt is symmetrical, because the median lies exactly halfway between the two quartiles and the whiskers match.
  4. DIt is positively skewed: the median sits nearer the lower quartile and the right whisker is longer.

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