WAEC · 2023 · Private · Further Maths · Paper 2

1. $\,{}^nC_2 = \dfrac{n(n - 1)}{2} = 66$, so $n(n - 1) = 132$.
2. Rearrange: $n^2 - n - 132 = 0$.
3. Factorise: $(n - 12)(n + 11) = 0$.
4. $n$ is a positive whole number, so $n = 12$.

1. Five ordinates means four strips, so $h = \dfrac{2 - 0}{4} = 0.5$.
2. The ordinates of $\dfrac{4}{1 + x^2}$ at $x = 0, 0.5, 1, 1.5, 2$ are $4,\ 3.2,\ 2,\ 1.2308,\ 0.8$.
3. First and last: $4.8$.
4. Twice the rest: $2(3.2 + 2 + 1.2308) = 12.8615$.
5. Trapezium rule: $\dfrac{0.5}{2}(4.8 + 12.8615) = 0.25 \times 17.6615$
   $= 4.4154$, about $4.42$.

(a)

1. Multiply the top and the bottom by the conjugate of the bottom, $9\sqrt5 - 5\sqrt2$.
2. The bottom: $(9\sqrt5)^2 - (5\sqrt2)^2 = 405 - 50$
   $= 355$.
3. The top: $(7 + 2\sqrt{10})(9\sqrt5 - 5\sqrt2) = 63\sqrt5 - 35\sqrt2 + 18\sqrt{50} - 10\sqrt{20}$.
4. Simplify the surds: $18\sqrt{50} = 90\sqrt2$ and $10\sqrt{20} = 20\sqrt5$.
5. So the top is $43\sqrt5 + 55\sqrt2$.
6. Divide each term by 355: $p = \dfrac{43}{355}$ and $q = \dfrac{55}{355} = \dfrac{11}{71}$.

(b)

1. $2p - q = \dfrac{86}{355} - \dfrac{55}{355}$
   $= \dfrac{31}{355}$.

1. Divide by 3: $x^2 + y^2 + 2x + \frac{2m}{3}y - 6 = 0$, so $g = 1$ and $f = \frac{m}{3}$.
2. $r^2 = g^2 + f^2 - c$: $10 = 1 + \dfrac{m^2}{9} + 6$.
3. $\dfrac{m^2}{9} = 3$, so $m^2 = 27$ and $m = 3\sqrt3 \approx 5.196$ (the positive value).

1. $\sum f = 8m - 5$ and $\sum fx = (m + 2) + 2(m - 1) + 3(2m - 3) + 4(m + 1) + 5(3m - 4)$
   $= 28m - 25$.
2. The mean is $\dfrac{28m - 25}{8m - 5} = \dfrac{31}{11}$, so $11(28m - 25) = 31(8m - 5)$.
3. $308m - 275 = 248m - 155$, so $60m = 120$ and $m = 2$.
4. The frequencies are $4, 1, 1, 3, 2$ (total 11).
5. The median is the 6th value: the cumulative frequencies $4, 5, 6$ reach 6 at $x = 3$, so the median is 3.

1. $X \sim B(8, 0.8)$.

(a)

1. $P(7) = \binom87(0.8)^7(0.2)$
   $= 0.33554$
   $\approx 0.336$.

(b)

1. More than three-fourths of 8 is more than 6, so 7 or 8: $0.33554 + 0.8^8 = 0.33554 + 0.16777$
   $= 0.50332$
   $\approx 0.503$.

1. East parts: $8\sin80^\circ + 18\sin240^\circ + 6\sin300^\circ = 7.878 - 15.588 - 5.196$
   $= -12.906$.
2. North parts: $8\cos80^\circ + 18\cos240^\circ + 6\cos300^\circ = 1.389 - 9 + 3$
   $= -4.611$.
3. $|\mathbf R| = \sqrt{12.906^2 + 4.611^2}$
   $= \sqrt{187.83}$
   $\approx 13.71\text{ N}$.

(a)

1. $v = \displaystyle\int (20t - 3t^2)\,dt$
   $= 10t^2 - t^3 + c$.
2. It starts from rest, so $c = 0$.
3. At rest again when $v = 0$: $t^2(10 - t) = 0$, so $t = 10$ s.

(b)

1. $v$ is positive from $t = 0$ to $10$, so the distance in 5 s is $\displaystyle\int_0^5 (10t^2 - t^3)\,dt$.
2. $= \left[\dfrac{10t^3}{3} - \dfrac{t^4}{4}\right]_0^5$
   $= \dfrac{1250}{3} - \dfrac{625}{4}$
   $= \dfrac{3125}{12}$
   $\approx 260.42$ m.

(a)(i)

1. $g \circ h(x) = g(x + 3)$: replace every $x$ in $g$ by $x + 3$.
2. The top: $2 + (x + 3) = x + 5$.
3. The bottom: $(x + 3)^2 + (x + 3) - 20 = x^2 + 6x + 9 + x + 3 - 20$
   $= x^2 + 7x - 8$.
4. So $g \circ h(x) = \dfrac{x + 5}{x^2 + 7x - 8}$.

(ii)

1. Factorise the bottom: $x^2 + 7x - 8 = (x + 8)(x - 1)$.
2. It is zero, so $g \circ h$ is undefined, at $x = 1$ and $x = -8$.

(b)

1. Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 8}$ and multiply through: $x + 5 = A(x + 8) + B(x - 1)$.
2. Put $x = 1$: $6 = 9A$, so $A = \frac23$.
3. Put $x = -8$: $-3 = -9B$, so $B = \frac13$.
4. So $g \circ h(x) = \dfrac{2}{3(x - 1)} + \dfrac{1}{3(x + 8)}$.

(a)(i)

1. Expand along the top row: $3(-3 + 10) - 5(6 - 10) + (-4)(12 - 6) = 21 + 20 - 24$
   $= 17$.

(ii)

1. By Cramer's rule, replace each column by $(1, -15, 5)$ in turn: $\Delta_x = 17$, $\Delta_y = 34$ and $\Delta_z = 51$.
2. So $x = \dfrac{17}{17} = 1$, $y = \dfrac{34}{17} = 2$ and $z = \dfrac{51}{17} = 3$.
3. Check in the third equation: $-2 + 4 + 3 = 5$ ✓.

(b)

1. Differentiate both sides with respect to $x$: $2x + 2y\dfrac{dy}{dx} = 2p\left(y + x\dfrac{dy}{dx}\right)$.
2. Collect the $\dfrac{dy}{dx}$ terms: $2y\dfrac{dy}{dx} - 2px\dfrac{dy}{dx} = 2py - 2x$.
3. Factorise and divide by 2: $(y - px)\dfrac{dy}{dx} = py - x$, so $\dfrac{dy}{dx} = \dfrac{py - x}{y - px}$.

(a)

1. Draw the lines $4y - x = 4$, $3x - 2y = 6$ and $x = -2$.
2. The origin satisfies all three inequalities, so keep the side of each line that contains it.
3. The region is the triangle with vertices $\left(-2, \frac12\right)$, $(-2, -6)$ and $(3.2, 1.8)$.
4. See the workspace.

(b)

1. The 48 payments add up to the debt: $\dfrac{48}{2}(2a + 47d) = 472\,560$, so $2a + 47d = 19\,690$.
2. With $a = 6508$: $13\,016 + 47d = 19\,690$, so $47d = 6674$ and $d = 142$.
3. Paid in the first 20 weeks: $S_{20} = \dfrac{20}{2}(2 \times 6508 + 19 \times 142)$
   $= 10(13\,016 + 2698)$
   $= 157\,140$.
4. Amount left after the 20th week: $472\,560 - 157\,140 =$ ₦315 420.

(a)

1. $p = \frac{15}{100} = 0.15$ and $n = 20$: $P(3) = \binom{20}{3}(0.15)^3(0.85)^{17}$
   $\approx 0.2428$.

(b)

1. $\lambda = 1.8$: $P(X \le 2) = e^{-1.8}\left(1 + 1.8 + \dfrac{1.8^2}{2}\right)$
   $= 4.42e^{-1.8}$
   $\approx 0.7306$.

(a)

1. Four consecutive odd numbers with mean 6 are $3, 5, 7, 9$.
2. Their deviations from 6 are $-3, -1, 1, 3$, so the variance is $\dfrac{9 + 1 + 1 + 9}{4} = 5.0$.

(b)

1. The modal class is 40–59, with 18 of the 50 candidates.
2. There are $\binom{50}{3} = 19\,600$ ways to choose 3.
3. Two from the modal class and one from the other 32: $\binom{18}{2} \times 32 = 153 \times 32$
   $= 4896$.
4. So $P = \dfrac{4896}{19600} \approx 0.250$.

(a)(i)

1. $H = \dfrac{u^2\sin^2\theta}{2g}$
   $= \dfrac{68^2\sin^2 42^\circ}{20}$
   $\approx 103.5\text{ m}$.

(ii)

1. $T = \dfrac{2u\sin\theta}{g}$
   $= \dfrac{136\sin42^\circ}{10}$
   $\approx 9.1\text{ s}$.

(iii)

1. $R = \dfrac{u^2\sin2\theta}{g}$
   $= \dfrac{4624\sin84^\circ}{10}$
   $\approx 459.9\text{ m}$.

(b)

1. The weight acts at the centre of gravity, $15 - 3.5 = 11.5\text{ m}$ from $P$.
2. Moments about $P$: $15R_Q = 120 \times 11.5 = 1380$, so $R_Q = 92\text{ N}$.

(a)

1. In $PQRS$, $\overrightarrow{PS} = \overrightarrow{QR}$
   $= (4 - 2)\mathbf i + (3 + 4)\mathbf j$
   $= 2\mathbf i + 7\mathbf j$.
2. So $\mathbf s = \mathbf p + \overrightarrow{PS}$
   $= (-5 + 2)\mathbf i + (1 + 7)\mathbf j$
   $= -3\mathbf i + 8\mathbf j$.

(b)

1. $\overrightarrow{PQ} = (2 + 5)\mathbf i + (-4 - 1)\mathbf j$
   $= 7\mathbf i - 5\mathbf j$.
2. $\overrightarrow{PS} \cdot \overrightarrow{PQ} = 14 - 35$
   $= -21$, $|\overrightarrow{PS}| = \sqrt{53}$ and $|\overrightarrow{PQ}| = \sqrt{74}$.
3. $\cos\theta = \dfrac{-21}{\sqrt{53}\sqrt{74}}$
   $\approx -0.3353$, so $\theta \approx 109.6^\circ$.