WAEC · 2020 · Private, 1st series · Further Maths · Paper 2

1. Differentiate each term with respect to $x$.
2. $x^2$ gives $2x$.
3. $-xy$ needs the product rule and gives $-\left(y + x\dfrac{dy}{dx}\right)$.
4. $-y^2$ gives $-2y\dfrac{dy}{dx}$.
5. $-31$ gives 0.
6. So $2x - y - x\dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0$.
7. Collect: $2x - y = (x + 2y)\dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{2x - y}{x + 2y}$.
8. At $(2, 5)$: $\dfrac{dy}{dx} = \dfrac{4 - 5}{2 + 10}$
   $= -\dfrac{1}{12}$.

1. $|Q| = (-2)(1) - (5)(-3) = -2 + 15 = 13$.
2. $Q^{-1} = \frac{1}{13}\begin{pmatrix} 1 & -5 \\ 3 & -2 \end{pmatrix}$.
3. $PQ^{-1} = \frac{1}{13}\begin{pmatrix} 3 + 12 & -15 - 8 \\ 5 + 18 & -25 - 12 \end{pmatrix}$
   $= \frac{1}{13}\begin{pmatrix} 15 & -23 \\ 23 & -37 \end{pmatrix}$.

1. The first term is $a = 3$ and the common difference is $d = 8 - 3 = 5$.
2. The sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$
   $= \frac{n}{2}(5n + 1)$.
3. The sum must be greater than 250: $\frac{n}{2}(5n + 1) > 250$, so $5n^2 + n - 500 > 0$.
4. Solve $5n^2 + n - 500 = 0$ with the formula: $n = \frac{-1 \pm \sqrt{1 + 10\,000}}{10}$, so $n \approx 9.90$ (the negative root doesn't apply to a number of terms).
5. $n$ must be a whole number greater than 9.90, so the least number of terms is $10$.
6. Check: $S_9 = \frac92 \times 46 = 207$, which is not enough, and $S_{10} = 5 \times 51 = 255$, which is greater than 250 ✓.

(a)

1. Draw the three boundary lines as solid lines: $x + y = 5$ through $(0, 5)$ and $(5, 0)$.
2. $2y - x = 0$, that is $y = \frac{x}{2}$, through $(0, 0)$ and $(6, 3)$.
3. And $x + 5y = 21$ through $(1, 4)$ and $(6, 3)$.
4. Test a point, such as $(3, 3)$: $3 + 3 = 6 \ge 5$, $6 - 3 = 3 \ge 0$ and $3 + 15 = 18 \le 21$, so it is in $P$.
5. The region $P$ is the triangle containing $(3, 3)$.
6. Shade it.
7. Its corners are where the lines meet: $x + y = 5$ and $x + 5y = 21$ give $4y = 16$, so $(1, 4)$.
8. $y = \frac{x}{2}$ and $x + y = 5$ give $\frac{3x}{2} = 5$, so $\left(\frac{10}{3}, \frac53\right)$.
9. $y = \frac{x}{2}$ and $x + 5y = 21$ give $\frac{7x}{2} = 21$, so $(6, 3)$.

(b)(i)

1. The point of $P$ furthest to the left is the corner $(1, 4)$, so the minimum value of $x$ is $1$.

(ii)

1. The largest value of $3x + 2y$ on the region is at a corner.
2. At $(1, 4)$ it is $11$.
3. At $\left(\frac{10}{3}, \frac53\right)$ it is $13\frac13$.
4. At $(6, 3)$ it is $24$.
5. So the maximum is $24$, at $(6, 3)$.

1. $X \sim B(12, 0.3)$, with $q = 0.7$.

(a)

1. $P(X = 6)$
   $= \binom{12}{6}(0.3)^6(0.7)^6$
   $= 924 \times 0.000729 \times 0.117649$
   $\approx 0.0792$.

(b)

1. More than 2 but fewer than 5 means 3 or 4: $\binom{12}{3}(0.3)^3(0.7)^9 + \binom{12}{4}(0.3)^4(0.7)^8$.
2. $\approx 0.2397 + 0.2311 = 0.4708$.

1. With an assumed mean $A = 14$, work with the deviations $d = x - 14$ instead of the numbers themselves.
2. Set them out in a table:

3. $x$	43	12	19	14	35	22	13	16	26
   $d = x - 14$	29	$-2$	5	0	21	8	$-1$	2	12

4. $\sum d = 29 - 2 + 5 + 0 + 21 + 8 - 1 + 2 + 12$
   $= 74$, and $n = 9$.
5. Mean $= A + \frac{\sum d}{n}$
   $= 14 + \frac{74}{9}$
   $= 14 + 8.2222$
   $= 22.2222$ (to 4 d.p.).

1. $\mathbf p + \mathbf q = (2 + 3)\mathbf i + (4 + 1)\mathbf j$
   $= 5\mathbf i + 5\mathbf j$.

(a)

1. $|\mathbf p + \mathbf q| = \sqrt{25 + 25}$
   $= \sqrt{50}$
   $\approx 7.07$.

(b)

1. $\tan\theta = \frac55 = 1$, so $\theta = 45^\circ$ above the positive $x$-axis (a bearing of $045^\circ$).

1. Take up as positive: $u = 10$, $a = -10$, and the ground is $75\text{ m}$ below, so $s = -75$.
2. $s = ut + \frac12at^2$: $-75 = 10t - 5t^2$, so $t^2 - 2t - 15 = 0$.
3. $(t - 5)(t + 3) = 0$, so $t = 5\text{ s}$ (time cannot be negative).

1. From $2y = x - 13$, $x = 2y + 13$.
2. Substitute: $3y + 2y + 13 + 12 = 0$, so $y = -5$ and $x = 3$.
3. $L_1$ through $(3, -5)$ and $(-4, -7)$: $m_1 = \dfrac{-7 + 5}{-4 - 3} = \dfrac27$.
4. $2x - 5y = 4$ has gradient $\frac25$, so $L_2$ has $m_2 = -\frac52$.
5. $\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1m_2}\right|$
   $= \left|\dfrac{\frac27 + \frac52}{1 - \frac57}\right|$
   $= \dfrac{39/14}{2/7}$
   $= \dfrac{39}{4}$.
6. $\theta = \tan^{-1} 9.75$
   $\approx 84.14^\circ$.

(a)

1. Split into primes: $15^y = 3^y5^y$ and $9^{y - 1} = 3^{2y - 2}$.
2. Collect the powers of 3: $3^y \times 3^{2y - 2} = 3^{3y - 2}$.
3. Collect the powers of 5: $5^y \times 5^{2y - 2} = 5^{3y - 2}$.
4. So the left side is $3^{3y - 2}5^{3y - 2} = 15^{3y - 2}$, and $1 = 15^0$.
5. Compare the indices: $3y - 2 = 0$, so $y = \frac23$.

(b)

1. Use $\cos 2x = \cos^2 x - \sin^2 x$: the left side is $\sin^2 x + \cos^2 x - \sin^2 x = \cos^2 x$.
2. So $\cos^2 x = \frac{1}{16}$ and $\cos x = \frac14$ or $\cos x = -\frac14$.
3. $\cos x = \frac14$: $x = 75.52^\circ$ or $360^\circ - 75.52^\circ = 284.48^\circ$.
4. $\cos x = -\frac14$: $x = 180^\circ - 75.52^\circ$
   $= 104.48^\circ$ or $180^\circ + 75.52^\circ = 255.52^\circ$.

(a)

1. Expand along the top row: $y(y^2 - 1) - 1(y - 0) + 0 = y^3 - 2y$.
2. Set it to zero: $y(y^2 - 2) = 0$, so $y = 0$ or $y = \pm\sqrt2$.
3. The positive value is $y = \sqrt2 \approx 1.4142$.

(b)

1. Let $u = 3 + 2\sin\theta$.
2. Then $du = 2\cos\theta\,d\theta$, so $4\cos\theta\,d\theta = 2\,du$.
3. Change the limits: $\theta = -\frac\pi2$ gives $u = 1$, and $\theta = \frac\pi2$ gives $u = 5$.
4. $\displaystyle\int_1^5 \frac{2}{u}\,du = [2\ln u]_1^5$
   $= 2\ln5 - 2\ln1$
   $= 2\ln5$
   $\approx 3.2189$.

1. $n = 150$ is large and $p = 0.05$ is small, so use the Poisson approximation with $\lambda = np = 7.5$.
2. $e^{-7.5} \approx 0.000553$.

(a)

1. $P(3) = e^{-7.5} \times \dfrac{7.5^3}{3!}$
   $\approx 0.0389$.

(b)

1. Between 3 and 6 means 4 or 5: $P(4) + P(5) \approx 0.0729 + 0.1094$
   $= 0.1823$.

(c)

1. At most 4: $P(0) + \ldots + P(4) \approx 0.0006 + 0.0041 + 0.0156 + 0.0389 + 0.0729$
   $= 0.1321$.

1. Class marks $5.5, 15.5, \ldots, 65.5$ with $\sum f = 50$ and $\sum fx = 1475$, so the mean is $29.5$.
2. Distances from 29.5: $24, 14, 4, 6, 16, 26, 36$.
3. $\sum f|x - \bar x| = 192 + 168 + 40 + 48 + 48 + 52 + 252$
   $= 800$.
4. Mean deviation $= \dfrac{800}{50} = 16$.

1. $\tan\theta = \frac{5}{12}$ gives a 5–12–13 triangle: $\sin\theta = \frac{5}{13}$ and $\cos\theta = \frac{12}{13}$.
2. $R = 100\cos\theta$
   $= \frac{1200}{13}$
   $\approx 92.31\text{ N}$, so limiting friction is $\frac34 \times 92.31 \approx 69.23\text{ N}$.
3. The weight down the plane is $100\sin\theta \approx 38.46\text{ N}$.

(a)

1. About to move down: friction acts up the plane, so $P + 38.46 = 69.23$ and $P \approx 30.77\text{ N}$.

(b)

1. About to move up: friction acts down the plane, so $P = 38.46 + 69.23 \approx 107.69\text{ N}$.

(a)(i)

1. Times: $3\text{ min} = 180\text{ s}$ and $2\text{ min} = 120\text{ s}$.
2. The graph rises from 0 to $133$ over $0$–$180\text{ s}$, is level to $360\text{ s}$, then falls to 0 at $480\text{ s}$.

(ii)

1. (α)** The area is a trapezium with parallel sides $180$ and $480$: $\frac12(180 + 480) \times 133 = 43\,890\text{ m}$.

(β)

1. Average velocity $= \dfrac{43\,890}{480}$
   $\approx 91.44\text{ m s}^{-1}$.

(b)

1. Horizontally: $x = (60\cos60^\circ)t = 30t$, so $t = 2.5\text{ s}$ when $x = 75$.
2. Vertically: $y = (60\sin60^\circ)(2.5) - 5(2.5)^2$
   $= 75\sqrt3 - 31.25$
   $\approx 98.65\text{ m}$.