WAEC · 2022 · Private, 2nd series · Further Maths · Paper 2

(a)(i)

1. The identity satisfies $p\,\Delta\,e = p$: $pe + p + e = p$, so $e(p + 1) = 0$.
2. This must hold for every $p$, so $e = 0$.

(ii)

1. The inverse satisfies $p\,\Delta\,p^{-1} = 0$: $pp^{-1} + p + p^{-1} = 0$.
2. So $p^{-1}(p + 1) = -p$ and $p^{-1} = -\dfrac{p}{p + 1}$, for $p \ne -1$.

(b)

1. $m\,\Delta\,8 = 8m + m + 8 = 35$, so $9m = 27$ and $m = 3$.

1. For $2x^2 - x - 2 = 0$: $\alpha + \beta = -\frac{-1}{2} = \frac12$ and $\alpha\beta = \frac{-2}{2} = -1$.
2. Write the cube in terms of them: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$.
3. First part: $\left(\frac12\right)^3 = \frac18$.
4. Second part: $3\alpha\beta(\alpha + \beta) = 3(-1)\left(\frac12\right)$
   $= -\frac32$.
5. So $\alpha^3 + \beta^3 = \frac18 - \left(-\frac32\right)$
   $= \frac18 + \frac{12}{8}$
   $= \frac{13}{8}$.

1. Five ordinates means four strips, so $h = \dfrac{3 - 1}{4} = 0.5$.
2. The ordinates of $\dfrac{2}{x + 4}$ at $x = 1, 1.5, 2, 2.5, 3$ are $0.4,\ 0.3636,\ 0.3333,\ 0.3077,\ 0.2857$.
3. First and last: $0.6857$.
4. Twice the rest: $2(0.3636 + 0.3333 + 0.3077) = 2.0093$.
5. Trapezium rule: $\dfrac{0.5}{2}(0.6857 + 2.0093) = 0.25 \times 2.6950$
   $= 0.6738$, about $0.67$.

1. Treat it as a quadratic in $\cos\theta$: $(3\cos\theta - 2)(\cos\theta + 4) = 0$.
2. $\cos\theta = -4$ is impossible, so $\cos\theta = \frac23$.
3. $\theta = \cos^{-1}\frac23$
   $\approx 48.19^\circ$ or $360^\circ - 48.19^\circ = 311.81^\circ$.
4. To the nearest degree: $\theta = 48^\circ$ or $312^\circ$.

1. Class marks $43, 50, 57, 64, 71, 78, 85, 92$ and $d = x - 71$: $-28, -21, -14, -7, 0, 7, 14, 21$.
2. $\sum fd = -140 - 42 - 42 - 56 + 0 + 42 + 112 + 21$
   $= -105$ and $\sum f = 40$.
3. Mean $= 71 + \dfrac{-105}{40}$
   $= 71 - 2.625$
   $= 68.375$
   $\approx 68.4$ N.

1. $p = \frac14$, $q = \frac34$ and $n = 10$.
2. At least three correct $= 1 - [P(0) + P(1) + P(2)]$.
3. $P(0) = \left(\frac34\right)^{10}$
   $\approx 0.0563$, $P(1) = 10\left(\frac14\right)\left(\frac34\right)^9$
   $\approx 0.1877$ and $P(2) = 45\left(\frac14\right)^2\left(\frac34\right)^8$
   $\approx 0.2816$.
4. So $P(X \ge 3) \approx 1 - 0.5256 = 0.474$.

1. The weight is $1.5 \times 10 = 15\text{ N}$.
2. The ropes meet at $180^\circ - 30^\circ - 60^\circ = 90^\circ$.
3. The rope at $30^\circ$ makes $120^\circ$ with the weight.
4. The rope at $60^\circ$ makes $150^\circ$.
5. Lami: $\dfrac{T_1}{\sin150^\circ} = \dfrac{T_2}{\sin120^\circ}$
   $= \dfrac{15}{\sin90^\circ}$.
6. $T_1 = 15\sin150^\circ = 7.5\text{ N}$ (the rope at $30^\circ$).
7. $T_2 = 15\sin120^\circ$
   $\approx 12.99\text{ N}$ (the rope at $60^\circ$).

1. $\overrightarrow{NQ} = \mathbf q - \mathbf n$
   $= (x - 2)\mathbf i + 2\mathbf j$.
2. $\overrightarrow{MR} = \mathbf r - \mathbf m$
   $= x\mathbf i - 4\mathbf j$.
3. Perpendicular: $x(x - 2) + (2)(-4) = 0$, so $x^2 - 2x - 8 = 0$.
4. Factorise: $(x - 4)(x + 2) = 0$, so $x = 4$ or $x = -2$.

(a)

1. The common denominator is $(2 - \sqrt3)(2 + \sqrt3) = 4 - 3 = 1$.
2. The top is $7(2 + \sqrt3) - 7(2 - \sqrt3) = 14 + 7\sqrt3 - 14 + 7\sqrt3$
   $= 14\sqrt3$.
3. So the value is $14\sqrt3$, that is $0 + 14\sqrt3$.

(b)

1. Let the first term be $a$ and the common ratio $r$.
2. The first and third terms add up to 20: $a + ar^2 = 20$.
3. The product of the first and fourth terms is 18 times the second: $a \times ar^3 = 18ar$.
4. Divide both sides by $ar$: $ar^2 = 18$.
5. Substitute into the first equation: $a + 18 = 20$, so $a = 2$.
6. Then $2r^2 = 18$, so $r^2 = 9$ and $r = 3$ (the ratio is positive).
7. $S_8 = \dfrac{a(r^8 - 1)}{r - 1}$
   $= \dfrac{2(6561 - 1)}{2}$
   $= 6560$.

(a)

1. The top has degree 3 and the bottom degree 2, so divide first.
2. $x(x^2 - 1) = x^3 - x$.
3. Take it away: $-x^2 + x - 4$ is left.
4. $-1(x^2 - 1) = -x^2 + 1$.
5. Take it away: $x - 5$ is left.
6. So the fraction is $x - 1 + \dfrac{x - 5}{x^2 - 1}$.
7. Split: $x - 5 = A(x - 1) + B(x + 1)$ for $\dfrac{A}{x + 1} + \dfrac{B}{x - 1}$.
8. Put $x = 1$: $-4 = 2B$, so $B = -2$.
9. Put $x = -1$: $-6 = -2A$, so $A = 3$.
10. So the answer is $x - 1 + \dfrac{3}{x + 1} - \dfrac{2}{x - 1}$.

(b)

1. Use the quotient rule with $u = \sin x$ and $v = 1 + \cos x$: $u' = \cos x$ and $v' = -\sin x$.
2. $\dfrac{dy}{dx} = \dfrac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2}$
   $= \dfrac{\cos x + \cos^2x + \sin^2x}{(1 + \cos x)^2}$.
3. Use $\cos^2x + \sin^2x = 1$: the top is $1 + \cos x$.
4. Cancel one factor: $\dfrac{dy}{dx} = \dfrac{1}{1 + \cos x}$.

(a)

1. Read the coefficients: $M = \begin{pmatrix} 3 & 1 \\ -2 & 3 \end{pmatrix}$.

(b)(i)

1. $|M| = 9 - (-2) = 11$, so $M^{-1} = \frac{1}{11}\begin{pmatrix} 3 & -1 \\ 2 & 3 \end{pmatrix}$.

(ii)

1. $\begin{pmatrix} x \\ y \end{pmatrix} = M^{-1}\begin{pmatrix} 9 \\ 5 \end{pmatrix}$
   $= \frac{1}{11}\begin{pmatrix} 22 \\ 33 \end{pmatrix}$
   $= \begin{pmatrix} 2 \\ 3 \end{pmatrix}$: the point $(2, 3)$.

(c)

1. The first circle has centre $(5, 4)$.
2. Divide the second by 3: $x^2 + y^2 + 2x - 3y + \frac13 = 0$, with centre $\left(-1, \frac32\right)$.
3. Gradient: $\dfrac{4 - \frac32}{5 + 1} = \dfrac{5}{12}$, so $y - 4 = \frac{5}{12}(x - 5)$.
4. $12y - 48 = 5x - 25$, so $12y - 5x - 23 = 0$.

1. The mean is 7, so the eight marks add to $8 \times 7 = 56$.
2. The six known marks add to 38, so the missing ones satisfy $a + b = 18$.
3. $\sigma^2 = 3 = \dfrac{\sum x^2}{8} - 7^2$, so $\sum x^2 = 8 \times 52 = 416$.
4. The six known squares add to 252, so $a^2 + b^2 = 164$.
5. Substitute $b = 18 - a$: $a^2 + (18 - a)^2 = 164$, so $2a^2 - 36a + 160 = 0$, that is $a^2 - 18a + 80 = 0$.
6. Factorise: $(a - 8)(a - 10) = 0$.
7. The two marks are 8 and 10.

1. $X \sim B\left(6, \frac25\right)$.
2. Over 15 625: $P(0) = 729$, $P(1) = 2916$, $P(2) = 4860$, $P(3) = 4320$.

(a)

1. $P(6) = \left(\frac25\right)^6$
   $= \dfrac{64}{15625}$
   $\approx 0.004$.

(b)

1. At most 3: $\dfrac{729 + 2916 + 4860 + 4320}{15625} = \dfrac{12825}{15625}$
   $\approx 0.821$.

(c)

1. At least 2: $1 - \dfrac{729 + 2916}{15625} = \dfrac{11980}{15625}$
   $\approx 0.767$.

(a)(i)

1. Draw the beam $ST$ with the support $P$, 40 kg at $T$, 50 kg at $y$ m from $S$, and the beam's weight at its midpoint.

(ii)

1. $|SP| = \frac35 \times 7 = 4.2\text{ m}$ and $|PT| = 2.8\text{ m}$.
2. The midpoint is 3.5 m from $S$, so the beam's weight ($170\text{ N}$) acts $0.7\text{ m}$ from $P$ on the $S$ side.
3. Moments about $P$: $500(4.2 - y) + 170(0.7) = 400(2.8)$.
4. $2100 - 500y + 119 = 1120$, so $500y = 1099$ and $y \approx 2.2\text{ m}$.

(b)

1. $(\mathbf i + k\mathbf j) \cdot (3\mathbf i - 4\mathbf j) = 3 - 4k$, and the lengths are $\sqrt{1 + k^2}$ and 5.
2. So $\dfrac{3 - 4k}{5\sqrt{1 + k^2}} = \dfrac{11\sqrt5}{25}$.
3. Square both sides: $\dfrac{(3 - 4k)^2}{25(1 + k^2)} = \dfrac{121}{125}$, so $5(3 - 4k)^2 = 121(1 + k^2)$.
4. Expand: $45 - 120k + 80k^2 = 121 + 121k^2$, so $41k^2 + 120k + 76 = 0$.
5. Factorise: $(41k + 38)(k + 2) = 0$, so $k = -2$ or $k = -\frac{38}{41}$.
6. Both make $3 - 4k$ positive, as the positive cosine needs, so both are valid.

(a)

1. Take $P$'s direction as positive.
2. Momentum before $= 3(5) + 1.5(-7) = 15 - 10.5 = 4.5$.
3. Momentum after $= 3V_P + 1.5V_Q$.
4. Momentum is conserved: $3V_P + 1.5V_Q = 4.5$, so $2V_P + V_Q = 3$.
5. The velocities differ by $\frac34$: $V_P - V_Q = \frac34$.
6. Add the equations: $3V_P = \frac{15}{4}$, so $V_P = \frac54\text{ m s}^{-1}$ and $V_Q = \frac12\text{ m s}^{-1}$.

(b)(i)

1. $v = \dfrac{dh}{dt} = 30 - 10t$.
2. At $t = 2$: $v = 10\text{ m s}^{-1}$.

(ii)

1. The top is where $v = 0$: $t = 3$.
2. Then $h = 5 + 90 - 45 = 50$ m.