WAEC · 2018 · Private · Further Maths · Paper 2

Question 1

(a)
If $\dfrac{\tan\theta + \tan30^\circ}{1 - \tan\theta\tan30^\circ} + 1 = 0$ , find $\tan\theta$ , leaving the answer in surd form.

Worked solution (try it first)

The left side is the formula for $\tan(\theta + 30^\circ)$ , so $\tan(\theta + 30^\circ) = -1$ .
We can also solve directly: put $\tan 30^\circ = \frac{1}{\sqrt3}$ and write $t = \tan\theta$ .
Multiply the top and bottom of the fraction by $\sqrt3$ : $\dfrac{\sqrt3\,t + 1}{\sqrt3 - t} = -1$ .
Multiply both sides by $\sqrt3 - t$ : $\sqrt3\,t + 1 = t - \sqrt3$ .
Collect the $t$ terms: $\sqrt3\,t - t = -\sqrt3 - 1$ , so $t(\sqrt3 - 1) = -(\sqrt3 + 1)$ and $t = -\dfrac{\sqrt3 + 1}{\sqrt3 - 1}$ .
Rationalise by multiplying top and bottom by $\sqrt3 + 1$ : $t = -\dfrac{(\sqrt3 + 1)^2}{3 - 1}$
$= -\dfrac{4 + 2\sqrt3}{2}$ .
So $\tan\theta = -2 - \sqrt3$ .

Question 2

Integration

Using the trapezium rule with ordinates at $x = 1, 2, 3, 4$ and $5$ , calculate, correct to two decimal places, the value of $\displaystyle\int_1^5 \left(x + \frac{2}{x^2}\right)dx$ .

(a)
Approximate value

Try it on a graph

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

Four strips, $h = 1$ .
The ordinates of $x + \dfrac{2}{x^2}$ at $x = 1, 2, 3, 4, 5$ are $3,\ 2.5,\ 3.2222,\ 4.125,\ 5.08$ .
First and last: $3 + 5.08 = 8.08$ .
Twice the rest: $2(2.5 + 3.2222 + 4.125) = 19.6944$ .
Trapezium rule: $\frac12(8.08 + 19.6944) = 13.8872$ , about $13.89$ .

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Question 3

Sequences, series & binomial expansion

(a)
The sum of the second and third terms of a Geometric Progression (G.P.) is 48. If the sum of the third and fourth terms is 144, find the first term of the progression.

Worked solution (try it first)

The second and third terms: $ar + ar^2 = 48$ , which factorises to $ar(1 + r) = 48$ .
The third and fourth terms: $ar^2 + ar^3 = 144$ , which factorises to $ar^2(1 + r) = 144$ .
Divide the second equation by the first: $r = \dfrac{144}{48} = 3$ .
Substitute: $a(3)(4) = 48$ , so the first term is $a = 4$ .

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Question 4

(a)
If $\sqrt{3x + 1} - \sqrt{2x - 1} = 1$ , find the values of $x$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Get one root on its own: $\sqrt{3x + 1} = 1 + \sqrt{2x - 1}$ .
Square both sides, keeping the middle term on the right: $3x + 1 = 1 + 2\sqrt{2x - 1} + 2x - 1$ .
Simplify and get the root on its own: $x + 1 = 2\sqrt{2x - 1}$ .
Square again: $x^2 + 2x + 1 = 4(2x - 1)$ , so $x^2 - 6x + 5 = 0$ .
Factorise: $(x - 1)(x - 5) = 0$ , so $x = 1$ or $x = 5$ .
Check in the original equation: $x = 1$ gives $2 - 1 = 1$ ✓ and $x = 5$ gives $4 - 3 = 1$ ✓.
So $x = 1$ or $x = 5$ .

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Question 5

Probability & distributions

The probability that a boy wins a race is 0.8. In four different races, what is the probability that he wins:

(a)
all races;
(b)
no race?

Worked solution (try it first)

The races are independent, with $P(\text{win}) = 0.8$ and $P(\text{lose}) = 0.2$ .

(a)

All four: $0.8^4 = 0.4096$ .

(b)

None: $0.2^4 = 0.0016$ .

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Question 6

Statistics & correlation

Geography	6	5	4	3	2	7	1
Economics	7	6	2	4	1	5	3

The table shows the ranks of marks scored by 7 candidates in Geography and Economics tests.

(a)
Calculate the Spearman's rank correlation coefficient (4 d.p.).

Worked solution (try it first)

The table already gives ranks.
$d$ = Geography − Economics: $-1, -1, 2, -1, 1, 2, -2$ .
$\sum d^2 = 1 + 1 + 4 + 1 + 1 + 4 + 4 = 16$ .
$\rho = 1 - \dfrac{6 \times 16}{7 \times 48}$
$= 1 - \dfrac{96}{336}$
$= \dfrac57$
$\approx 0.7143$ .

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Question 7

(a)
Two points $X$ and $Y$ have position vectors $\mathbf x = 2\mathbf i - 3\mathbf j$ and $\mathbf y = -\mathbf i + 2\mathbf j$ . Find the position vector of the point $M$ on $\overline{XY}$ such that $|\overrightarrow{XM}| : |\overrightarrow{MY}| = 3 : 2$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$XM : MY = 3 : 2$ , so $M$ is $\frac35$ of the way from $X$ to $Y$ .
$\overrightarrow{OM} = \dfrac{2\mathbf x + 3\mathbf y}{5}$ .
$= \dfrac{2(2\mathbf i - 3\mathbf j) + 3(-\mathbf i + 2\mathbf j)}{5}$
$= \dfrac{\mathbf i + 0\mathbf j}{5}$ .
So $\overrightarrow{OM} = \frac15\mathbf i$ .

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Question 8

A body is thrown vertically upwards. Its height $h$ metres at time $t$ seconds is given by $h = \left(12t - \frac25t^2\right)\text{ m}$ . Find:

(a)
the time at which it is momentarily at rest;
(b)
the maximum height reached by the body.

Worked solution (try it first)

(a)

The body is momentarily at rest when its velocity is zero: $\dfrac{dh}{dt} = 12 - \dfrac45t = 0$ .
So $\dfrac45t = 12$ and $t = 15$ s.

(b)

Put $t = 15$ into the height: $h = 12(15) - \frac25(15)^2$
$= 180 - 90$
$= 90$ m.

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Question 9

(a)
The radius of a sphere increased by $2\frac12\%$ . Find the percentage increase in the volume.
(b)
If the coefficients of $x^2$ and $x^3$ in the binomial expansion of $(p + qx)^7$ are equal, express $q$ in terms of $p$ .

Worked solution (try it first)

(a)

$V = \frac43\pi r^3$ , so for a small change $\dfrac{\delta V}{V} \approx 3\dfrac{\delta r}{r}$ .
With $\dfrac{\delta r}{r} = 2\frac12\%$ : $\dfrac{\delta V}{V} \approx 3 \times 2\frac12\%$
$= 7\frac12\%$ .
(Exactly, $1.025^3 = 1.0769$ , an increase of about $7.69\%$ .)

(b)

The $x^2$ term of $(p + qx)^7$ is $21p^5q^2x^2$ and the $x^3$ term is $35p^4q^3x^3$ .
Set the coefficients equal: $21p^5q^2 = 35p^4q^3$ .
Divide by $7p^4q^2$ : $3p = 5q$ , so $q = \frac35p$ .

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Question 10

(a)
Find the equation of the normal to the curve $x^2 + xy + 2y^2 = 8$ at the point $(-3, 1)$ .
Show the answer
$5y + x - 2 = 0$
(b)
Find the equation of a line passing through the midpoint of the line joining $A(5, 1)$ and $B(-1, -5)$ and perpendicular to line $AB$ .
Show the answer
$y + x = 0$

Try it on a graph

The curve x² + xy + 2y² = 8 (upper and lower halves) and the normal at (−3, 1).

Worked solution (try it first)

(a)

Differentiate implicitly: $2x + \left(y + x\dfrac{dy}{dx}\right) + 4y\dfrac{dy}{dx} = 0$ .
Collect: $(x + 4y)\dfrac{dy}{dx} = -(2x + y)$ , so $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 4y}$ .
At $(-3, 1)$ : $\dfrac{dy}{dx} = -\dfrac{-6 + 1}{-3 + 4}$
$= 5$ , so the normal's gradient is $-\frac15$ .
The normal: $y - 1 = -\frac15(x + 3)$ .
Multiply by 5: $5y - 5 = -x - 3$ , so $5y + x - 2 = 0$ .

(b)

The midpoint of $AB$ is $\left(\dfrac{5 - 1}{2}, \dfrac{1 - 5}{2}\right) = (2, -2)$ .
The gradient of $AB$ is $\dfrac{-5 - 1}{-1 - 5} = 1$ , so a perpendicular line has gradient $-1$ .
The line: $y + 2 = -(x - 2)$ , so $y = -x$ , that is $y + x = 0$ .

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Question 11

(a)
Given the matrix $A = \begin{pmatrix} 3 & 8 \\ 5 & -2 \end{pmatrix}$ , find its inverse.
Show the answer
$A^{-1} = \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$
(b)
Solve the simultaneous equations $\log(x - 2) + \log 2 = 2\log y$ and $\log(x - 3y + 3) = 0$ .
Show the answer
$x = 4, y = 2$ or $x = 10, y = 4$

Worked solution (try it first)

(a)

The determinant is $|A| = 3(-2) - 8(5) = -6 - 40 = -46$ .
Swap the leading diagonal and change the signs of the other two entries: $\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$ .
Divide by the determinant: $A^{-1} = -\dfrac{1}{46}\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$
$= \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$ .

(b)

First equation: $\log[2(x - 2)] = \log y^2$ , so $2(x - 2) = y^2$ .
Second equation: $\log(x - 3y + 3) = 0$ means $x - 3y + 3 = 10^0 = 1$ , so $x = 3y - 2$ .
Substitute: $2(3y - 4) = y^2$ , so $y^2 - 6y + 8 = 0$ and $(y - 2)(y - 4) = 0$ .
$y = 2$ gives $x = 4$ , and $y = 4$ gives $x = 10$ .
Both keep every log positive, so $x = 4, y = 2$ or $x = 10, y = 4$ .

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Question 12

Statistics & correlation

The data show the production of mobile phones per day over a thirty-day period by a company.

53 26 21 28 38 46 35 31 51 35 23 38 46 24 44 36 39 40 45 33 32 54 51 49 20 42 46 38 39 40

(a)

Construct a frequency distribution table for the data using 7 class intervals of equal widths of 5.

Model answer

Class interval	Frequency
$20\text{–}24$	$4$
$25\text{–}29$	$2$
$30\text{–}34$	$3$
$35\text{–}39$	$8$
$40\text{–}44$	$4$
$45\text{–}49$	$5$
$50\text{–}54$	$4$
Total	$30$

Start at the smallest value, 20, and take classes of width 5 up to 50–54, which holds the largest value, 54.

(b)
Draw a histogram for the distribution.
Model answer
Draw bars on the class boundaries, not the class limits (19.5, 24.5, …, 54.5), with no gaps between them. The height of each bar is the frequency, and each axis is labelled.

To estimate the mode, take the tallest bar. Join its top-left corner to the top-left corner of the bar on its right, and its top-right corner to the top-right corner of the bar on its left. Read down from where the two lines cross: the mode is about 37.3.
(c)
Use the histogram to estimate the mode.
(d)
Find, correct to two decimal places, the percentage of production above the modal class.

Try it on a graph

Histogram with the crossed lines that locate the mode.

Worked solution (try it first)

(a)

Tally the 30 values into 20–24, 25–29, …, 50–54: frequencies $4, 2, 3, 8, 4, 5, 4$ .

(b)

Draw bars of these heights over the boundaries $19.5, 24.5, \ldots, 54.5$ .

(c)

The modal class is 35–39.
Mode $= 34.5 + \dfrac{8 - 3}{(8 - 3) + (8 - 4)} \times 5$
$= 34.5 + \dfrac59 \times 5$
$\approx 37.3$ .

(d)

Above the modal class: $4 + 5 + 4 = 13$ of 30 days, which is $\dfrac{13}{30} \times 100 \approx 43.33\%$ .

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Question 13

(a)
At the end of full time, a football match between teams $A$ and $B$ was goalless. The winning team had to be decided through penalty kicks. Each team had to take five kicks. The probability that a kick by team $A$ will result in a goal was 0.8 and for team $B$ the probability was 0.3. Find, correct to three decimal places, the probability that the final scores would be: (i) 4 – 2 in favour of team $B$ ; (ii) 5 – 0 in favour of team $A$ .
Separate values with commas, e.g. 3, −2
(b)
Three students are to be selected from 4 boys and 5 girls to represent their school in a Mathematics quiz. (i) In how many ways can the three students be selected? (ii) What is the probability that more boys will be selected than girls?
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

4–2 to $B$ : $B$ scores 4 of 5 and $A$ scores 2 of 5.
$P(B \text{ scores } 4) = {}^5C_4(0.3)^4(0.7)$
$= 0.02835$ and $P(A \text{ scores } 2) = {}^5C_2(0.8)^2(0.2)^3$
$= 0.0512$ .
Multiply (the teams are independent): $0.02835 \times 0.0512 \approx 0.00145$ , which is $0.001$ to three decimal places.

(ii)

5–0 to $A$ : $A$ scores all 5 and $B$ misses all 5: $0.8^5 \times 0.7^5 = 0.32768 \times 0.16807$
$\approx 0.055$ .

(b)(i)

$\,{}^9C_3 = 84$ ways.

(ii)

More boys: 3 boys, or 2 boys and 1 girl: $\dfrac{{}^4C_3 + {}^4C_2 \times {}^5C_1}{84} = \dfrac{4 + 30}{84}$
$= \dfrac{17}{42}$
$\approx 0.405$ .

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Question 14

(a)
A body is under the action of forces $M = (4\text{ N}, 030^\circ)$ , $N = (10\text{ N}, 300^\circ)$ and $P = (a\text{ N}, x^\circ)$ which keep it in equilibrium. Find the force $P$ .
Separate values with commas, e.g. 3, −2
(b)
The initial velocity of a particle of mass $0.2\text{ kg}$ is $40\text{ m s}^{-1}$ in the direction of the unit vector $\mathbf j$ . The velocity of the particle changed to $30\text{ m s}^{-1}$ in the direction of the unit vector $\mathbf i$ . Find the change in momentum.

Worked solution (try it first)

(a)

$M$ : $4\sin30^\circ = 2$ east and $4\cos30^\circ = 3.464$ north.
$N$ : $10\sin300^\circ = -8.660$ east and $10\cos300^\circ = 5$ north.
$M + N = -6.660\mathbf i + 8.464\mathbf j$ .
In equilibrium $P = -(M + N)$
$= 6.660\mathbf i - 8.464\mathbf j$ .
$|P| = \sqrt{44.36 + 71.64}$
$= \sqrt{116.0}$
$\approx 10.77\text{ N}$ .
$P$ points south-east: $\tan^{-1}\frac{6.660}{8.464} = 38.2^\circ$ east of south, a bearing of $180^\circ - 38.2^\circ = 141.8^\circ$ .
So $P \approx (10.77\text{ N}, 141.8^\circ)$ .

(b)

Change in momentum $= m\mathbf v - m\mathbf u$
$= 0.2(30\mathbf i) - 0.2(40\mathbf j)$
$= 6\mathbf i - 8\mathbf j$ .
Its magnitude is $\sqrt{36 + 64} = 10\text{ kg m s}^{-1}$ .

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Question 15

Vectors

(a)
Given that $\mathbf m = 6\mathbf i + 4\mathbf j$ and $\mathbf n = 3\mathbf i - 4\mathbf j$ , find, correct to the nearest degree, the angle between $\mathbf m$ and $\mathbf n$ .
(b)
$A(-1, 1)$ , $B(1, 3)$ , $C(x, y)$ and $D(3, -3)$ are the vertices of a parallelogram $ABCD$ . Calculate the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$\mathbf m \cdot \mathbf n = 18 - 16 = 2$ , $|\mathbf m| = \sqrt{52} = 2\sqrt{13}$ and $|\mathbf n| = 5$ .
$\cos\theta = \dfrac{2}{10\sqrt{13}}$
$\approx 0.0555$ , so $\theta \approx 87^\circ$ .

(b)

In $ABCD$ , $\overrightarrow{AB} = \overrightarrow{DC}$ : $\begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} x - 3 \\ y + 3 \end{pmatrix}$ .
So $x = 5$ and $y = -1$ .

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