WAEC · 2022 · May/June · Further Maths · Paper 2

(a)

1. Put the row heading in for $p$ and the column heading in for $q$ in $p^2 + 2pq - q^2$.
2. Row $-2$: $8$, $7$, $-1$, $-8$.
3. For example $(-2) * (-2) = 4 + 8 - 4 = 8$.
4. Row $-1$: $1$, $2$, $-2$, $-7$.
5. For example $(-1) * 2 = 1 - 4 - 4 = -7$.
6. Row $1$: $-7$, $-2$, $2$, $1$.
7. Row $2$: $-8$, $-1$, $7$, $8$.

(b)

1. Let $x = -2 * p$.
2. Then $x * 2 = -7$: in the column under 2, $-7$ is in the row of $-1$, so $x = -1$.
3. Now $-2 * p = -1$: in the row of $-2$, $-1$ is under $1$, so $p = 1$.

1. Split the index: $2^{2y + 1} = 2 \times (2^y)^2$.
2. Let $x = 2^y$: $2x^2 - 5x + 2 = 0$.
3. Factorise: $(2x - 1)(x - 2) = 0$, so $x = \frac12$ or $x = 2$.
4. Back to $y$: $2^y = \frac12 = 2^{-1}$ gives $y = -1$, and $2^y = 2$ gives $y = 1$.

1. $f$ acts first: $g \circ f(x) = g(x^2 + 2)$
   $= \dfrac{1}{x^2 + 4}$.
2. Write $y = \dfrac{1}{x^2 + 4}$.
3. Turn both sides over: $x^2 + 4 = \dfrac1y$.
4. Take 4 from both sides: $x^2 = \dfrac{1 - 4y}{y}$, so $x = \pm\sqrt{\dfrac{1 - 4y}{y}}$ and $(g \circ f)^{-1}(x) = \pm\sqrt{\dfrac{1 - 4x}{x}}$.
5. The inverse needs $x \ne 0$ and $\dfrac{1 - 4x}{x} \ge 0$.
6. The top and bottom can't both be negative here, so both are positive: $0 < x \le \frac14$.
7. So the domain is $\left\{x : 0 < x \le \frac14\right\}$.
8. It is also the range of $g \circ f$: $x^2 + 4 \ge 4$, so $0 < \dfrac{1}{x^2 + 4} \le \frac14$.

1. Use $\cos 2x = 1 - 2\sin^2 x$: $3(1 - 2\sin^2 x) - \sin x = 0$.
2. So $6\sin^2 x + \sin x - 3 = 0$.
3. $\sin x = \dfrac{-1 \pm \sqrt{1 + 72}}{12}$
   $= \dfrac{-1 \pm \sqrt{73}}{12}$, so $\sin x \approx 0.6287$ or $-0.7953$.
4. $\sin x = 0.6287$: $x \approx 38.95^\circ$ or $180^\circ - 38.95^\circ = 141.05^\circ$.
5. $\sin x = -0.7953$: the reference angle is $52.69^\circ$, so $x \approx 232.69^\circ$ or $307.31^\circ$.

1. $X \sim B\left(6, \frac25\right)$, with $q = \frac35$.

(a)

1. $P(3) = \binom63\left(\frac25\right)^3\left(\frac35\right)^3$
   $= \dfrac{4320}{15625}$
   $\approx 0.2765$.

(b)

1. $P(0) = \left(\frac35\right)^6$
   $= \dfrac{729}{15625}$
   $\approx 0.04666$.

(c)

1. $P(6) = \left(\frac25\right)^6$
   $= \dfrac{64}{15625}$
   $= 0.004096$.

1. Rank the Vocal scores (1 for the highest): $7, 5, 4, 8, 3, 2, 1, 6$.
2. Rank the Instrumental scores: $5, 4, 3, 8, 7, 2, 1, 6$.
3. $d$: $2, 1, 1, 0, -4, 0, 0, 0$, so $\sum d^2 = 4 + 1 + 1 + 16 = 22$.
4. $\rho = 1 - \dfrac{6 \times 22}{8 \times 63}$
   $= 1 - \dfrac{132}{504}$
   $= \dfrac{31}{42}$
   $\approx 0.738$.

1. The weight is $18 \times 10 = 180\text{ N}$.
2. The string is at $35^\circ$ to the vertical.
3. Up: $T\cos35^\circ = 180$.
4. Across: $T\sin35^\circ = F$.

(a)

1. Divide: $F = 180\tan35^\circ$
   $\approx 126.04\text{ N}$.

(b)

1. $T = \dfrac{180}{\cos35^\circ}$
   $\approx 219.74\text{ N}$.

1. $\mathbf p$: $8\sin30^\circ = 4$ east and $8\cos30^\circ = 6.928$ north.
2. $\mathbf q$: $9\sin150^\circ = 4.5$ east and $9\cos150^\circ = -7.794$ north.
3. $\mathbf p - \mathbf q = -0.5\mathbf i + 14.722\mathbf j$.
4. $|\mathbf p - \mathbf q| = \sqrt{0.25 + 216.74}$
   $\approx 14.731$.
5. Unit vector: $\dfrac{-0.5\mathbf i + 14.722\mathbf j}{14.731} \approx -0.0339\mathbf i + 0.9994\mathbf j$.

(a)

1. For an A.P., twice the middle term equals the sum of the other two: $2\binom n5 = \binom n4 + \binom n6$.
2. Write each in terms of $\binom n4$: $\binom n5 = \binom n4 \times \dfrac{n - 4}{5}$ and $\binom n6 = \binom n4 \times \dfrac{(n - 4)(n - 5)}{30}$.
3. Divide by $\binom n4$: $\dfrac{2(n - 4)}{5} = 1 + \dfrac{(n - 4)(n - 5)}{30}$.
4. Multiply by 30: $12(n - 4) = 30 + (n - 4)(n - 5)$, so $12n - 48 = n^2 - 9n + 50$.
5. Rearrange: $n^2 - 21n + 98 = 0$, so $(n - 7)(n - 14) = 0$, and $n = 7$ or $n = 14$.

(b)

1. $n = 14$: the terms are $1001, 2002, 3003$, so the common difference is $1001$.
2. $n = 7$: the terms are $35, 21, 7$, so the common difference is $-14$.

(a)

1. The volume is $3x \times 2x \times y = 6x^2y = 72$, so $y = \dfrac{12}{x^2}$.

(b)(i)

1. Top and bottom: $2 \times 6x^2 = 12x^2$.
2. The four sides: $2(3xy) + 2(2xy) = 10xy$.
3. So $A = 12x^2 + 10xy$
   $= 12x^2 + 10x \times \dfrac{12}{x^2}$
   $= 12x^2 + \dfrac{120}{x}$.

(ii)

1. $\dfrac{dA}{dx} = 24x - \dfrac{120}{x^2}$
   $= 0$, so $24x^3 = 120$ and $x^3 = 5$.
2. So $x = \sqrt[3]5 \approx 1.71$.

(a)

1. Each term of $(1 + 2x)^7$ is $\binom7r(2x)^r$: $1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7$.
2. $(1 - 2x)^7$ has the same terms with the odd powers of $x$ negative: $1 - 14x + 84x^2 - 280x^3 + 560x^4 - 672x^5 + 448x^6 - 128x^7$.

(b)

1. $(1.2)^7 - (0.8)^7$ is $(1 + 2x)^7 - (1 - 2x)^7$ with $x = 0.1$.
2. Subtracting, the even powers cancel and the odd ones double: $28x + 560x^3 + 1344x^5 + 256x^7$.
3. With $x = 0.1$: $2.8 + 0.56 + 0.01344 + 0.0000256 = 3.3734656$, so the value is $3.373$ to three decimal places.

1. The numbers are $x$, $x + 1$ and $x + 2$, with $3x + 3 = 12$, so $x = 3$: 3 oranges, 4 apples and 5 avocado pears.

(a)

1. $P(\text{orange, then avocado}) = \dfrac{3}{12} \times \dfrac{5}{11}$
   $= \dfrac{15}{132}$
   $= \dfrac{5}{44}$
   $\approx 0.1136$.

(b)

1. Same fruit: $\dfrac{3 \times 2 + 4 \times 3 + 5 \times 4}{12 \times 11} = \dfrac{38}{132}$
   $= \dfrac{19}{66}$
   $\approx 0.2879$.

(c)

1. At least one apple is everything except no apple: $1 - \dfrac{8}{12} \times \dfrac{7}{11} = 1 - \dfrac{56}{132}$
   $= \dfrac{19}{33}$
   $\approx 0.5758$.

(a)

1. Plot the ten points $(14, 22), (16, 19), \ldots, (33, 2)$.

(b)(i)

1. $\bar x = \dfrac{230}{10} = 23$ and $\bar y = \dfrac{104}{10} = 10.4$.

(ii)

1. The $X$ values below 23 are $14, 16, 17, 18, 22$: $\bar x_1 = 17.4$, with $\bar y_1 = \dfrac{22 + 19 + 15 + 13 + 10}{5}$
   $= 15.8$.

(c)

1. Draw the line through $(23, 10.4)$ and $(17.4, 15.8)$.

(d)(i)

1. Gradient $= \dfrac{15.8 - 10.4}{17.4 - 23}$
   $= -\dfrac{27}{28}$, so $y - 10.4 = -\dfrac{27}{28}(x - 23)$, which gives $140y + 135x - 4561 = 0$.

(ii)

1. At $x = 20$: $140y = 4561 - 2700 = 1861$, so $y \approx 13.3$.

(a)(i)

1. $v = \displaystyle\int (10t - 4t^2)\,dt$
   $= 5t^2 - \frac43t^3 + k$.
2. It starts at rest, so $k = 0$ and $v = 5t^2 - \frac43t^3$.

(ii)

1. The 4th second runs from $t = 3$ to $t = 4$.
2. $v(3) = 45 - 36 = 9$ and $v(4) = 80 - \frac{256}{3} = -\frac{16}{3}$.
3. Average acceleration $= \dfrac{v(4) - v(3)}{1}$
   $= -\dfrac{16}{3} - 9$
   $= -\dfrac{43}{3}$
   $= -14\frac13\text{ m s}^{-2}$.

(b)(i)

1. At constant velocity there is no acceleration: $R - 120 \times 10 = 0$, so $R = 1200$ N.

(ii)

1. Accelerating upwards: $R - 1200 = 120 \times 3$, so $R = 1560$ N.

(a)

1. $|6\mathbf i + 8\mathbf j| = 10$, so $\overrightarrow{OP} = \frac{80}{10}(6\mathbf i + 8\mathbf j)$
   $= 48\mathbf i + 64\mathbf j$.
2. $|8\mathbf i - 6\mathbf j| = 10$, so $\overrightarrow{OQ} = \frac{120}{10}(8\mathbf i - 6\mathbf j)$
   $= 96\mathbf i - 72\mathbf j$.

(b)

1. $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$
   $= 48\mathbf i - 136\mathbf j$.
2. $|\overrightarrow{PQ}| = \sqrt{2304 + 18\,496}$
   $= \sqrt{20\,800}$.
3. $20\,800 = 1600 \times 13$, so $|\overrightarrow{PQ}| = 40\sqrt{13}$.