WAEC · 2014 · May/June · Further Maths · Paper 2

Question 1

Indices, logarithms & surds

(a)
Solve $(\log_2 m)^2 - \log_2 m^3 = 10$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use $\log_2 m^3 = 3\log_2 m$ : the equation becomes $(\log_2 m)^2 - 3\log_2 m - 10 = 0$ .
Let $y = \log_2 m$ : $y^2 - 3y - 10 = 0$ .
Factorise: $(y - 5)(y + 2) = 0$ , so $y = 5$ or $y = -2$ .
$\log_2 m = 5$ gives $m = 2^5 = 32$ .
$\log_2 m = -2$ gives $m = 2^{-2} = \frac14$ .
So $m = 32$ or $m = \frac14$ .

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Question 2

Binary operations

A binary operation $*$ is defined on the set of rational numbers by $m * n = \dfrac{m^2 - n^2}{2mn}$ , $n \ne 0$ and $m \ne 0$ .

(a)
Find $-3 * 2$ .
(b)
Show whether or not $*$ is associative.
Show the answer
$*$ is not associative

Worked solution (try it first)

(a)

Put $m = -3$ and $n = 2$ : $-3 * 2 = \dfrac{(-3)^2 - 2^2}{2(-3)(2)}$ .
Work it out: $\dfrac{9 - 4}{-12} = -\dfrac{5}{12}$ .

(b)

$*$ is associative if $m * (n * p) = (m * n) * p$ for all $m$ , $n$ and $p$ .
Left side: $n * p = \dfrac{n^2 - p^2}{2np}$ , so $m * (n * p) = \dfrac{m^2 - \left(\frac{n^2 - p^2}{2np}\right)^2}{2m\left(\frac{n^2 - p^2}{2np}\right)}$ .
Right side: $(m * n) * p = \dfrac{\left(\frac{m^2 - n^2}{2mn}\right)^2 - p^2}{2p\left(\frac{m^2 - n^2}{2mn}\right)}$ .
These are not the same expression.
For example, with $m = 1$ , $n = 2$ , $p = 3$ the left side is $-\frac{119}{120}$ and the right side is $\frac{15}{8}$ .
So $*$ is not associative.

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Question 3

Polynomials & quadratic roots

(a)
If $\alpha$ and $\beta$ are the roots of $3x^2 + 5x + 1 = 0$ , evaluate $27(\alpha^3 + \beta^3)$ .

Worked solution (try it first)

(a)

For $3x^2 + 5x + 1 = 0$ : $\alpha + \beta = -\frac{b}{a} = -\frac53$ and $\alpha\beta = \frac{c}{a} = \frac13$ .
Use $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ .
Substitute: $\alpha^3 + \beta^3 = \left(-\frac53\right)^3 - 3\left(\frac13\right)\left(-\frac53\right)$ .
Work it out: $-\frac{125}{27} + \frac53 = -\frac{125}{27} + \frac{45}{27}$
$= -\frac{80}{27}$ .
Multiply by 27: $27(\alpha^3 + \beta^3) = -80$ .

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Question 4

Differentiation

(a)
Find the gradient of $xy^2 + x^2y = 4xy$ at the point $(1, 3)$ .

Worked solution (try it first)

(a)

Differentiate each term with the product rule: $\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$ .
$\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}$ and $\frac{d}{dx}(4xy) = 4y + 4x\frac{dy}{dx}$ .
So $y^2 + 2xy\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 4y + 4x\frac{dy}{dx}$ .
Collect the $\frac{dy}{dx}$ terms: $(2xy + x^2 - 4x)\frac{dy}{dx} = 4y - y^2 - 2xy$ .
Put $x = 1$ , $y = 3$ : $(6 + 1 - 4)\frac{dy}{dx} = 12 - 9 - 6$ , so $3\frac{dy}{dx} = -3$ .
The gradient at $(1, 3)$ is $-1$ .

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Question 5

Vectors

The position vectors of points $P$ , $Q$ and $R$ are $11\mathbf i + \mathbf j$ , $5\mathbf i + \frac{13}{3}\mathbf j$ and $2\mathbf i + 6\mathbf j$ respectively.

(a)
Show that $P$ , $Q$ and $R$ lie on a straight line.
Show the answer
$\overrightarrow{PQ} = 2\overrightarrow{QR}$ , so $P$ , $Q$ , $R$ are collinear
(b)
Find the ratio $|\overrightarrow{PQ}| : |\overrightarrow{QR}|$ .

Worked solution (try it first)

(a)

$\overrightarrow{PQ} = \mathbf q - \mathbf p$ , which is $(5 - 11)\mathbf i + \left(\frac{13}{3} - 1\right)\mathbf j = -6\mathbf i + \frac{10}{3}\mathbf j$ .
$\overrightarrow{QR} = \mathbf r - \mathbf q$ , which is $(2 - 5)\mathbf i + \left(6 - \frac{13}{3}\right)\mathbf j = -3\mathbf i + \frac53\mathbf j$ .
So $\overrightarrow{PQ} = 2\overrightarrow{QR}$ : the two vectors are parallel.
They share the point $Q$ , so $P$ , $Q$ and $R$ lie on a straight line.

(b)

$|\overrightarrow{PQ}| = \sqrt{36 + \frac{100}{9}}$
$= \frac{2\sqrt{106}}{3}$ .
$|\overrightarrow{QR}| = \sqrt{9 + \frac{25}{9}}$
$= \frac{\sqrt{106}}{3}$ .
So $|\overrightarrow{PQ}| : |\overrightarrow{QR}| = 2 : 1$ .

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Question 6

A committee of 3 is formed from a panel of 5 men and 3 women. Find the:

(a)
number of ways of forming the committee;
(b)
probability that at least one woman is on the committee.

Worked solution (try it first)

(a)

The panel has $5 + 3 = 8$ people.
Choose 3 of them: ${}^8C_3 = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}$
$= 56$ ways.

(b)

A committee with no woman is chosen from the 5 men: ${}^5C_3 = 10$ ways.
So P(no woman) $= \dfrac{10}{56}$ .
P(at least one woman) $= 1 - \dfrac{10}{56} = \dfrac{46}{56}$ .
Simplify: the probability is $\dfrac{23}{28}$ .

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Question 7

Statistics & correlation

The table shows the distribution of the lengths of 20 iron rods measured in metres.

Length (m)	1.0 – 1.1	1.2 – 1.3	1.4 – 1.5	1.6 – 1.7	1.8 – 1.9
Frequency	2	3	8	5	2

(a)
Using an assumed mean of 1.45, calculate the mean of the distribution.

Worked solution (try it first)

(a)

The class mid-points are 1.05, 1.25, 1.45, 1.65 and 1.85.
With $A = 1.45$ , the deviations $d = x - A$ are $-0.4$ , $-0.2$ , $0$ , $0.2$ and $0.4$ .
Multiply by the frequencies: $fd = -0.8$ , $-0.6$ , $0$ , $1.0$ and $0.8$ .
Add: $\sum fd = 0.4$ and $\sum f = 20$ .
Mean $= A + \dfrac{\sum fd}{\sum f}$ , so the mean is $1.45 + \dfrac{0.4}{20}$ .
That is $1.45 + 0.02 = 1.47$ .
The mean length is $1.47\text{ m}$ .

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Question 8 ✱

(a)
Find the direction of the resultant of the forces in the diagram. Give the direction as a bearing.

Worked solution (try it first)

(a)

Take $x$ to the right and $y$ upwards.
The 10 N force is $30^\circ$ from the $y$ -axis, the $8\sqrt2$ N force is $45^\circ$ from the $y$ -axis, and the 20 N force is $60^\circ$ below the negative $x$ -axis.
Vertical components: $Y = 10\cos30^\circ + 8\sqrt2\cos45^\circ - 20\sin60^\circ$ .
That is $Y = 5\sqrt3 + 8 - 10\sqrt3$
$= 8 - 5\sqrt3$
$\approx -0.660\text{ N}$ .
Horizontal components: $X = 8\sqrt2\sin45^\circ - 10\sin30^\circ - 20\cos60^\circ$ .
That is $X = 8 - 5 - 10 = -7\text{ N}$ .
Both components are negative, so the resultant points down and to the left.
$\tan\theta = \dfrac{0.660}{7}$
$\approx 0.0943$ , so $\theta \approx 5.39^\circ$ below the negative $x$ -axis.
As a bearing, this is $270^\circ - 5.39^\circ \approx 264.6^\circ$ , about $265^\circ$ (S $84.6^\circ$ W).

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Question 9

(a)
Differentiate $(x - 3)(x^2 + 5)$ with respect to $x$ .
(b)(i)
If $(x + 1)^2$ is a factor of $f(x) = x^3 + ax^2 + bx + 3$ , where $a$ and $b$ are constants, find the values of $a$ and $b$ .
Separate values with commas, e.g. 3, −2
(b)(ii)
Find the zeros of $f(x)$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Expand first: $(x - 3)(x^2 + 5) = x^3 - 3x^2 + 5x - 15$ .
Differentiate term by term: $\dfrac{dy}{dx} = 3x^2 - 6x + 5$ .

(b)(i)

Divide $x^3 + ax^2 + bx + 3$ by $(x + 1)^2 = x^2 + 2x + 1$ : the quotient is $x + (a - 2)$ .
The remainder is $(b - 2a + 3)x + (5 - a)$ .
$(x + 1)^2$ is a factor, so the remainder is zero: $5 - a = 0$ and $b - 2a + 3 = 0$ .
So $a = 5$ , and then $b = 2(5) - 3 = 7$ .

(ii)

$f(x) = x^3 + 5x^2 + 7x + 3$
$= (x + 1)^2(x + 3)$ , since the quotient is $x + (5 - 2) = x + 3$ .
So the zeros of $f(x)$ are $-1$ (twice) and $-3$ .

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Question 10

(a)
Express $\dfrac{5 + \sqrt2}{3 - \sqrt2} - \dfrac{5 - \sqrt2}{3 + \sqrt2}$ in the form $a + b\sqrt2$ .
(b)
Solve the following equations simultaneously using the determinant method:

$3x - y - z = -2$

$x + 5y + 2z = 5$

$2x + 3y + z = 0$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use the common denominator $(3 - \sqrt2)(3 + \sqrt2) = 9 - 2 = 7$ .
First numerator: $(5 + \sqrt2)(3 + \sqrt2) = 15 + 5\sqrt2 + 3\sqrt2 + 2$
$= 17 + 8\sqrt2$ .
Second numerator: $(5 - \sqrt2)(3 - \sqrt2) = 15 - 5\sqrt2 - 3\sqrt2 + 2$
$= 17 - 8\sqrt2$ .
Subtract: $\dfrac{(17 + 8\sqrt2) - (17 - 8\sqrt2)}{7} = \dfrac{16\sqrt2}{7}$ .
In the form $a + b\sqrt2$ : $0 + \frac{16}{7}\sqrt2$ , so $a = 0$ and $b = \frac{16}{7}$ .

(b)

$\Delta = \begin{vmatrix} 3 & -1 & -1 \\ 1 & 5 & 2 \\ 2 & 3 & 1 \end{vmatrix}$
$= 3(5 - 6) + 1(1 - 4) - 1(3 - 10)$ .
So $\Delta = -3 - 3 + 7 = 1$ .
Replace the first column by $(-2, 5, 0)$ : $\Delta_x = -2(5 - 6) + 1(5 - 0) - 1(15 - 0)$
$= -8$ .
Replace the second column: $\Delta_y = \begin{vmatrix} 3 & -2 & -1 \\ 1 & 5 & 2 \\ 2 & 0 & 1 \end{vmatrix}$
$= 15 - 6 + 10$
$= 19$ .
Replace the third column: $\Delta_z = \begin{vmatrix} 3 & -1 & -2 \\ 1 & 5 & 5 \\ 2 & 3 & 0 \end{vmatrix}$
$= -45 - 10 + 14$
$= -41$ .
Divide each by $\Delta = 1$ : $x = -8$ , $y = 19$ and $z = -41$ .
Check in the third equation: $2(-8) + 3(19) + (-41) = 0$ ✓.

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Question 11

(a)
If $f(x) = \dfrac{x - 3}{2x - 1}$ , $x \ne \frac12$ and $g(x) = \dfrac{x - 1}{x + 1}$ , $x \ne -1$ , find $g \circ f$ .
(b)(i)
Sketch the curve $y = 9x - x^3$ .
Model answer
Mark the crossings at $x = -3$ , $0$ and $3$ , the minimum $(-\sqrt3, -6\sqrt3)$ and the maximum $(\sqrt3, 6\sqrt3)$ , and join them with a smooth curve that rises from the bottom right to the top left overall.
(b)(ii)
Calculate the total area bounded by the $x$ -axis and the curve $y = 9x - x^3$ .

Worked solution (try it first)

(a)

$g \circ f(x) = g(f(x))$
$= \dfrac{f(x) - 1}{f(x) + 1}$ .
Top: $\dfrac{x - 3}{2x - 1} - 1$ has numerator $x - 3 - (2x - 1) = -x - 2$ over $2x - 1$ .
Bottom: $\dfrac{x - 3}{2x - 1} + 1$ has numerator $x - 3 + (2x - 1) = 3x - 4$ over $2x - 1$ .
Divide.
The $2x - 1$ cancels: $g \circ f(x) = \dfrac{-(x + 2)}{3x - 4}$ , $x \ne \frac43$ .

(b)(i)

On the $x$ -axis $y = 0$ : $x(3 - x)(3 + x) = 0$ , so the curve crosses at $x = -3$ , $0$ and $3$ .
Turning points: $\dfrac{dy}{dx} = 9 - 3x^2 = 0$ gives $x = \pm\sqrt3$ .
At $x = \sqrt3$ : $y = 9\sqrt3 - 3\sqrt3$
$= 6\sqrt3$
$\approx 10.4$ , a maximum.
At $x = -\sqrt3$ : $y = -6\sqrt3 \approx -10.4$ , a minimum.
Sketch a smooth curve from the top left, down through $(-3, 0)$ to $(-\sqrt3, -6\sqrt3)$ , up through the origin to $(\sqrt3, 6\sqrt3)$ , and down through $(3, 0)$ .

(ii)

The curve is below the axis for $-3 < x < 0$ and above it for $0 < x < 3$ , so find the two areas separately.
$\int_0^3 (9x - x^3)\,dx = \left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_0^3$
$= \frac{81}{2} - \frac{81}{4}$
$= \frac{81}{4}$ .
$\int_{-3}^0 (9x - x^3)\,dx = -\frac{81}{4}$ , so that area is also $\frac{81}{4}$ .
Total area $= \frac{81}{4} + \frac{81}{4} = 40.5$ square units.

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Question 12

Statistics & correlation

The histogram represents the scores of some candidates in an examination.

(a)

Using the histogram, construct a frequency distribution table, indicating clearly the class intervals.

Model answer

Class interval	Class boundaries	Frequency
9 – 18	8.5 – 18.5	4
19 – 28	18.5 – 28.5	6
29 – 38	28.5 – 38.5	8
39 – 48	38.5 – 48.5	13
49 – 58	48.5 – 58.5	15
59 – 68	58.5 – 68.5	10
69 – 78	68.5 – 78.5	3
79 – 88	78.5 – 88.5	1
	Total	60

(b)
Draw a cumulative frequency curve of the distribution and use it to estimate the: (i) median; (ii) quartile deviation.
Separate values with commas, e.g. 3, −2
Model answer

Worked solution (try it first)

(a)

The bars stand on the class boundaries 8.5, 18.5, …, 88.5 (width 10), so the class intervals are 9–18, 19–28, 29–38, 39–48, 49–58, 59–68, 69–78 and 79–88.
Read the bar heights for the frequencies: 4, 6, 8, 13, 15, 10, 3 and 1, a total of 60.

(b)

Add up the frequencies for the cumulative frequencies: 4, 10, 18, 31, 46, 56, 59 and 60.
Plot each cumulative frequency at its upper class boundary (18.5, 28.5, …, 88.5), start at $(8.5, 0)$ , and join the points with a smooth curve.

(i)

The median is at the $\frac{60}{2} = 30$ th score: reading across at 30 gives about 47.7.

(ii)

$Q_1$ is at the $\frac{60}{4} = 15$ th score: about 34.8.
$Q_3$ is at the 45th score: about 57.8.
Quartile deviation $= \frac12(Q_3 - Q_1)$ , which is $\frac12(57.8 - 34.8) \approx 11.5$ .
The median is about 47.7 and the quartile deviation is about 11.5.

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The probabilities that Kofi, Kwasi and Ama will pass a certain examination are $\frac{9}{10}$ , $\frac45$ and $x$ respectively. If the probability that only one of them will pass the examination is $\frac{9}{50}$ , find the:

(a)
value of $x$ ;
(b)
probability that at least one of them will pass the examination.

Worked solution (try it first)

(a)

Only one passes means Kofi only, Kwasi only, or Ama only.
In each case the other two fail.
So $\frac{9}{10} \times \frac15(1 - x) + \frac{1}{10} \times \frac45(1 - x) + \frac{1}{10} \times \frac15x = \frac{9}{50}$ .
Simplify: $\frac{9}{50}(1 - x) + \frac{4}{50}(1 - x) + \frac{1}{50}x = \frac{9}{50}$ .
Multiply by 50: $13(1 - x) + x = 9$ , so $13 - 12x = 9$ .
So $12x = 4$ and $x = \frac13$ .

(b)

P(none pass) $= \frac{1}{10} \times \frac15 \times \frac23$
$= \frac{1}{75}$ .
P(at least one passes) $= 1 - \frac{1}{75}$
$= \frac{74}{75}$
$\approx 0.987$ .

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Question 14

Vectors

(a)
Given that $\mathbf x = 3\mathbf i - \mathbf j$ , $\mathbf y = 2\mathbf i + k\mathbf j$ and the cosine of the angle between $\mathbf x$ and $\mathbf y$ is $\frac{\sqrt5}{5}$ , find the values of the constant $k$ .
Separate values with commas, e.g. 3, −2
(b)
In the quadrilateral $ABCD$ , $\overrightarrow{AB} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} -6 \\ -9 \end{pmatrix}$ and $\overrightarrow{BD} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}$ . Show whether or not $ABCD$ is a parallelogram.
Show the answer
$ABCD$ is a parallelogram

Worked solution (try it first)

(a)

$\mathbf x \cdot \mathbf y = 3(2) + (-1)(k)$
$= 6 - k$ , $|\mathbf x| = \sqrt{10}$ and $|\mathbf y| = \sqrt{4 + k^2}$ .
So $\dfrac{6 - k}{\sqrt{10}\sqrt{4 + k^2}} = \dfrac{\sqrt5}{5}$ , which gives $5(6 - k) = \sqrt{50}\sqrt{4 + k^2}$ .
Square both sides: $900 - 300k + 25k^2 = 200 + 50k^2$ .
Rearrange: $25k^2 + 300k - 700 = 0$ , so $k^2 + 12k - 28 = 0$ .
Factorise: $(k - 2)(k + 14) = 0$ , so $k = 2$ or $k = -14$ .
Both make $6 - k$ positive (4 and 20), so both satisfy the original equation.

(b)

$\overrightarrow{BC}$ is $\overrightarrow{AC} - \overrightarrow{AB}$ , which is $\begin{pmatrix} -1 \\ -8 \end{pmatrix}$ .
$\overrightarrow{AD}$ is $\overrightarrow{AB} + \overrightarrow{BD}$ , which is $\begin{pmatrix} -1 \\ -8 \end{pmatrix}$ .
So $\overrightarrow{AD} = \overrightarrow{BC}$ : $AD$ and $BC$ are equal and parallel.
Also $\overrightarrow{DC}$ is $\overrightarrow{AC} - \overrightarrow{AD}$ , which is $\begin{pmatrix} -5 \\ -1 \end{pmatrix}$ , the same as $\overrightarrow{AB}$ .
Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

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Question 15 ✱

(a)
$A(-1, 2)$ , $B(3, 5)$ and $C(4, 8)$ are the vertices of triangle $ABC$ . Forces whose magnitudes are $5\text{ N}$ and $3\sqrt{10}\text{ N}$ act along $\overrightarrow{AB}$ and $\overrightarrow{CB}$ respectively. Find the direction of the resultant of the forces. Give the direction as a bearing.
(b)(i)
A particle starts from rest and moves in a straight line. It attains a velocity of $20\text{ m s}^{-1}$ after covering a distance of 8 metres. Calculate its acceleration.
(b)(ii)
Calculate the time it will take to cover a distance of 40 metres.

Worked solution (try it first)

(a)

$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ , of length 5, so the 5 N force is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ N.
$\overrightarrow{CB} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$ , of length $\sqrt{10}$ , so the $3\sqrt{10}$ N force is $3\begin{pmatrix} -1 \\ -3 \end{pmatrix} = \begin{pmatrix} -3 \\ -9 \end{pmatrix}$ N.
Add: the resultant is $\begin{pmatrix} 1 \\ -6 \end{pmatrix}$ N.
It points right and down, at $\tan^{-1}6 \approx 80.5^\circ$ below the positive $x$ -axis.
As a bearing, this is $90^\circ + 80.5^\circ \approx 170.5^\circ$ , about $171^\circ$ (S $9.5^\circ$ E).

(b)(i)

Use $v^2 = u^2 + 2as$ : $20^2 = 0 + 2a(8)$ , so $400 = 16a$ .
So $a = 25\text{ m s}^{-2}$ .

(ii)

Use $s = ut + \frac12at^2$ : $40 = 0 + \frac12(25)t^2$ , so $t^2 = 3.2$ .
So $t = \sqrt{3.2} \approx 1.8\text{ s}$ .

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