WAEC · 2017 · May/June · Further Maths · Paper 2

(a)

1. Find $g^{-1}$ first: $y = x + 6$ gives $x = y - 6$, so $g^{-1}(x) = x - 6$.
2. $g^{-1}$ acts first: $f \circ g^{-1}(x) = f(x - 6)$
   $= \dfrac{4 - 5(x - 6)}{2}$.
3. Multiply out the top: $4 - 5x + 30 = 34 - 5x$.
4. Divide each term by 2: $f \circ g^{-1}(x) = 17 - \frac52x$.

(b)

1. $P$ divides $A(7, -5)$ to $B(-2, 7)$ in the ratio $5 : 4$, so $P = \left(\dfrac{4x_A + 5x_B}{9}, \dfrac{4y_A + 5y_B}{9}\right)$.
2. The $x$-coordinate: $\dfrac{4(7) + 5(-2)}{9} = \dfrac{18}{9}$, which is 2.
3. The $y$-coordinate: $\dfrac{4(-5) + 5(7)}{9} = \dfrac{15}{9}$
   $= \dfrac53$.
4. So $P\left(2, \frac53\right)$.

1. Split into partial fractions: $\dfrac{x - 1}{(x + 1)^2} = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2}$, so $x - 1 = A(x + 1) + B$.
2. Put $x = -1$: $B = -2$.
3. Compare the $x$ terms: $A = 1$.
4. So the integral is $\displaystyle\int_1^3 \left(\frac{1}{x + 1} - \frac{2}{(x + 1)^2}\right)dx$.
5. Integrate each piece: $\left[\ln(x + 1) + \dfrac{2}{x + 1}\right]_1^3$.
6. At $x = 3$: $\ln4 + \frac12$.
7. At $x = 1$: $\ln2 + 1$.
8. Subtract: $\ln4 - \ln2 - \frac12 = \ln2 - \frac12$
   $\approx 0.6931 - 0.5$
   $= 0.193$.

(a)

1. Split the log of the fraction: $\log_{10}\dfrac{p^{\frac12}q^4}{s^2} = \log_{10} p^{\frac12} + \log_{10} q^4 - \log_{10} s^2$.
2. Bring each power down in front: $\frac12\log_{10} p + 4\log_{10} q - 2\log_{10} s$.
3. Replace the logs with $a$, $b$ and $c$: $\dfrac a2 + 4b - 2c$.

(b)

1. The area is $A = \pi r^2$, so $\dfrac{dA}{dr} = 2\pi r$.
2. By the chain rule, $\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}$: $20 = 2\pi(6) \times \dfrac{dr}{dt}$.
3. So $\dfrac{dr}{dt} = \dfrac{20}{12\pi}$
   $= \dfrac{5}{3\pi}\text{ cm s}^{-1}$.

1. $X \sim B(10, 0.08)$, where $8\% = 0.08$.
2. $P(0) = 0.92^{10} \approx 0.4344$, $P(1) = 10(0.08)(0.92)^9 \approx 0.3777$ and $P(2) = 45(0.08)^2(0.92)^8 \approx 0.1478$.
3. $P(X \ge 3) = 1 - 0.9599$
   $= 0.0401$
   $\approx 0.04$.

(a)

1. Plot the cumulative frequencies $3, 12, 33, 45, 50$ at the upper boundaries $40.5, 45.5, 50.5, 55.5, 60.5$, starting from $(35.5, 0)$, and join with a smooth curve.

(b)

1. $N = 50$.
2. Read across from 12.5 and 37.5: $Q_1 \approx 45.6$ and $Q_3 \approx 52.4$.
3. Semi-interquartile range $= \frac12(52.4 - 45.6) \approx 3.4$.

1. In $MNQR$, $\overrightarrow{MN} = \overrightarrow{RQ}$.
2. $\overrightarrow{MN} = \begin{pmatrix} 10 - 4 \\ 2 + 6 \end{pmatrix}$
   $= \begin{pmatrix} 6 \\ 8 \end{pmatrix}$ and $\overrightarrow{RQ} = \begin{pmatrix} 8 - x \\ 16 - y \end{pmatrix}$.
3. So $8 - x = 6$ and $16 - y = 8$: $R(2, 8)$.

1. $F_1$: $18\sin330^\circ = -9$ east and $18\cos330^\circ = 15.588$ north.
2. $F_2$: $10$ east and $0$ north.
3. $F_3$: $0$ east and $-25$ north.
4. Add: east $-9 + 10 = 1$.
5. North $15.588 - 25 = -9.412$.
6. $|\mathbf R| = \sqrt{1 + 88.59}$
   $\approx 9.5\text{ N}$.
7. It points slightly east of south: $\tan^{-1}\frac{1}{9.412} = 6.07^\circ$ east of south, a bearing of $180^\circ - 6.07^\circ \approx 173.9^\circ$.

(a)(i)

1. Factorise: $(x^2 - 1)(x + 2) = (x - 1)(x + 1)(x + 2)$.
2. $f(x)$ is undefined where this is zero: $x = 1$, $-1$ or $-2$.

(ii)

1. Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2}$ and multiply through: $2x - 3 = A(x + 1)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x + 1)$.
2. Put $x = 1$: $-1 = 6A$, so $A = -\frac16$.
3. Put $x = -1$: $-5 = -2B$, so $B = \frac52$.
4. Put $x = -2$: $-7 = 3C$, so $C = -\frac73$.
5. So $f(x) = -\dfrac{1}{6(x - 1)} + \dfrac{5}{2(x + 1)} - \dfrac{7}{3(x + 2)}$.

(b)

1. The radius is the distance from $(-3, 1)$ to $(3, 1)$: $r = 6$.
2. The equation: $(x + 3)^2 + (y - 1)^2 = 36$.
3. Expand: $x^2 + 6x + 9 + y^2 - 2y + 1 = 36$, so $x^2 + y^2 + 6x - 2y - 26 = 0$.

(a)

1. Integrate: $f(x) = 2x^2 - \dfrac{x^3}{3} + c$.
2. Use $f(3) = 21$: $18 - 9 + c = 21$, so $c = 12$.
3. Then $f(x) = 2x^2 - \dfrac{x^3}{3} + 12$.

(b)(i)

1. The 2nd, 4th and 8th terms of the A.P. are $a + d$, $a + 3d$ and $a + 7d$.
2. They form a G.P., so $(a + 3d)^2 = (a + d)(a + 7d)$.
3. Expand: $a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2$, so $2d^2 = 2ad$ and $d(d - a) = 0$.
4. An A.P. with $d = 0$ would be constant, so $d = a$.
5. The 3rd and 5th terms add to 20: $(a + 2d) + (a + 4d) = 20$, so $2a + 6d = 20$.
6. With $d = a$: $8a = 20$, so $a = d = \frac52$.
7. The first four terms are $\frac52, 5, \frac{15}{2}, 10$.

(ii)

1. $S_{10} = \dfrac{10}{2}\left[2\left(\frac52\right) + 9\left(\frac52\right)\right]$
   $= 5\left(5 + \dfrac{45}{2}\right)$
   $= 137.5$.

(a)

1. $P(\text{pass}) = \frac45 = 0.8$, with $n = 7$.

(i)

1. At least 3 passed $= 1 - [P(0) + P(1) + P(2)]$
   $= 1 - 0.0047$
   $= 0.9953$
   $\approx 1.00$.

(ii)

1. Between 3 and 6 failed means 4 or 5 failed: $\binom74(0.2)^4(0.8)^3 + \binom75(0.2)^5(0.8)^2 = 0.0287 + 0.0043$
   $\approx 0.03$.

(b)

1. $P(\text{six}) = \frac16$: $\binom53\left(\frac16\right)^3\left(\frac56\right)^2 = \dfrac{250}{7776}$
   $\approx 0.0322$.

(a)

1. Class widths (from the boundaries): $3, 3, 6, 6, 9$.
2. Frequency densities: $\frac63 = 2$, $\frac93 = 3$, $\frac{12}{6} = 2$, $\frac{18}{6} = 3$, $\frac{18}{9} = 2$.
3. Draw each bar with this height over its class boundaries.

(b)

1. Class marks $18, 21, 25.5, 31.5, 39$.
2. $\sum fx = 108 + 189 + 306 + 567 + 702$
   $= 1872$ and $\sum f = 63$.
3. Mean $= \dfrac{1872}{63} \approx 29.71$ years.

(a)(i)

1. Draw the body hanging from $B$, with $AB$ at $30^\circ$ and $DB$ at $60^\circ$ to the ceiling, and its weight $10q$ downwards.

(ii)

1. Across: $48\cos30^\circ = T\cos60^\circ$.
2. $T = \dfrac{48\cos30^\circ}{\cos60^\circ}$
   $= 48\sqrt3$
   $\approx 83.14\text{ N}$.

(iii)

1. Up: $48\sin30^\circ + T\sin60^\circ = 10q$, so $24 + 72 = 10q$ and $q = 9.6$.

(b)(i)

1. At the top $v = 0$: $0 = 20^2 - 2(10)h$, so $h = 20.00\text{ m}$.

(ii)

1. $0 = 20 - 10t$, so $t = 2.00\text{ s}$.

(a)(i)

1. $|\mathbf m| = 10$.
2. $\mathbf m$ points 6 east and 8 north, a bearing of $\tan^{-1}\frac68 \approx 037^\circ$.
3. $|\mathbf n| = \sqrt{64 + \frac{49}{9}}$
   $= \frac{25}{3}$.
4. $\mathbf n$ points 8 west and $\frac73$ north.
5. So its bearing is $360^\circ - \tan^{-1}\frac{24}{7} = 360^\circ - 73.7^\circ$
   $\approx 286^\circ$.

(ii)

1. $\mathbf m \cdot \mathbf n = -48 + \frac{56}{3}$
   $= -\frac{88}{3}$.
2. $\cos\theta = \dfrac{-88/3}{10 \times 25/3}$
   $= -\dfrac{88}{250}$
   $= -0.352$, so $\theta \approx 110.6^\circ$.

(b)

1. $\overrightarrow{PQ} = \begin{pmatrix} 12 \\ 1 \end{pmatrix}$ and $\overrightarrow{RS} = \begin{pmatrix} 1 \\ -12 \end{pmatrix}$.
2. $\overrightarrow{PQ} \cdot \overrightarrow{RS} = 12 - 12$
   $= 0$, so they are perpendicular.