WAEC · 2018 · Private, 2nd series · Further Maths · Paper 2

1. Rationalise $\frac1x$: multiply top and bottom by $\sqrt5 - 2$.
2. The bottom is $(2 + \sqrt5)(\sqrt5 - 2) = 5 - 4 = 1$, so $\dfrac1x = \sqrt5 - 2$.
3. Subtract: $x - \dfrac1x = (2 + \sqrt5) - (\sqrt5 - 2)$
   $= 4$.
4. In the form $m + n\sqrt5$, this is $4 + 0\sqrt5$.

1. Integrate: $\displaystyle\int (2x + 5)\,dx = x^2 + 5x$.
2. Apply the limits: $(t^2 + 5t) - (1^2 + 5 \times 1) = t^2 + 5t - 6$.
3. Set it equal to 18: $t^2 + 5t - 6 = 18$, so $t^2 + 5t - 24 = 0$.
4. Factorise: $(t + 8)(t - 3) = 0$, so $t = 3$ or $t = -8$.
5. The upper limit $t$ is meant to be above the lower limit 1, so $t = 3$.

1. $B^2 = \begin{pmatrix} 4 + 3 & 6 + 12 \\ 2 + 4 & 3 + 16 \end{pmatrix}$
   $= \begin{pmatrix} 7 & 18 \\ 6 & 19 \end{pmatrix}$.
2. $3B = \begin{pmatrix} 6 & 9 \\ 3 & 12 \end{pmatrix}$ and $2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$.
3. $3N = B^2 + 3B + 2I$
   $= \begin{pmatrix} 15 & 27 \\ 9 & 33 \end{pmatrix}$.
4. So $N = \begin{pmatrix} 5 & 9 \\ 3 & 11 \end{pmatrix}$.

(a)

1. $2g = -4$ and $2f = -2$, so the centre is $(2, 1)$.
2. $r^2 = g^2 + f^2 - C$: $18 = 4 + 1 - C$, so $C = -13$.

(b)

1. A diameter passes through the centre $(2, 1)$.
2. Through $(2, 1)$ and $(9, 2)$: gradient $\frac{1}{7}$, so $y - 1 = \frac17(x - 2)$.
3. $7y - 7 = x - 2$, so $7y - x - 5 = 0$.

1. There are $4 + 3 = 7$ oranges and 5 bananas.

(a)

1. $\,{}^7C_1 \times {}^5C_1 = 7 \times 5$
   $= 35$ ways.

(b)

1. $\,{}^7C_3 \times {}^5C_4 = 35 \times 5$
   $= 175$ ways.

(c)

1. $\,{}^3C_2 \times {}^4C_1 \times {}^5C_3 = 3 \times 4 \times 10$
   $= 120$ ways.

1. Draw the histogram: the modal class is 35–39, with the tallest bar (38).
2. On the modal bar, join each top corner to the top corner of the opposite neighbour (30 on the left, 36 on the right).
3. Read down from where the lines cross.
4. By calculation: $34.5 + \dfrac{38 - 30}{(38 - 30) + (38 - 36)} \times 5 = 34.5 + \dfrac{8}{10} \times 5$
   $= 38.5$ years.

1. Angle $PQR$ is at $Q$, so use vectors starting at $Q$.
2. $\overrightarrow{QP} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ and $\overrightarrow{QR} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
3. Their scalar product is $(-1)(1) + (3)(1) = 2$.
4. $\cos\theta = \dfrac{2}{\sqrt{10}\sqrt2}$
   $= \dfrac{1}{\sqrt5}$, so $\theta \approx 63.43^\circ$.

1. With uniform acceleration, distance = average velocity × time.
2. $2700 = \dfrac{8 + 13}{2}t = 10.5t$.
3. $t = \dfrac{2700}{10.5} \approx 257.14$, so about $257\text{ s}$.

1. The line joining the midpoints of two sides is parallel to the third side, so $LM \parallel AB$.
2. Gradient of $LM$: $\dfrac{7 - 0}{3 + 1} = \dfrac74$.
3. $AB$ passes through its own midpoint $N(5, -2)$: $y + 2 = \frac74(x - 5)$.
4. $4y + 8 = 7x - 35$, so $4y - 7x + 43 = 0$.

(a)(i)

1. Each value gives an equation: $f(1) = p + q + r = 0$, $f(-1) = p - q + r = 4$ and $f(2) = 4p + 2q + r = 7$.
2. Take the second equation from the first: $2q = -4$, so $q = -2$.
3. Put $q = -2$ in the first: $p + r = 2$.
4. Put it in the third: $4p + r = 11$.
5. Take $p + r = 2$ from $4p + r = 11$: $3p = 9$, so $p = 3$.
6. Then $r = 2 - 3 = -1$.

(ii)

1. $f(x) = 3x^2 - 2x - 1$.
2. Two numbers that multiply to $-3$ and add to $-2$ are $-3$ and 1.
3. So $3x^2 - 3x + x - 1 = 3x(x - 1) + 1(x - 1)$
   $= (x - 1)(3x + 1)$.

(b)

1. Write $\dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 2)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 2}$, so $2x^2 - x + 3 = A(x^2 + 2) + (Bx + C)(x + 1)$.
2. Put $x = -1$: $2 + 1 + 3 = 3A$, so $A = 2$.
3. Compare the $x^2$ terms: $2 = A + B$, so $B = 0$.
4. Compare the constants: $3 = 2A + C$, so $C = -1$.
5. So $f(x) = \dfrac{2}{x + 1} - \dfrac{1}{x^2 + 2}$.

(a)

1. $f(x) = 5x - 6x^{-2}$, so $f(x + h) - f(x) = 5h - 6\left[\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2}\right]$.
2. Over one denominator: $\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2} = \dfrac{x^2 - (x + h)^2}{x^2(x + h)^2}$
   $= \dfrac{-h(2x + h)}{x^2(x + h)^2}$.
3. So $f(x + h) - f(x) = 5h + \dfrac{6h(2x + h)}{x^2(x + h)^2}$.
4. Divide by $h$: $5 + \dfrac{6(2x + h)}{x^2(x + h)^2}$.
5. Let $h \to 0$: $f'(x) = 5 + \dfrac{12x}{x^4}$
   $= 5 + \dfrac{12}{x^3}$.

(b)

1. Let $u = 1 - x$.
2. Then $x = 1 - u$ and $dx = -du$.
3. $\displaystyle\int x\sqrt{1 - x}\,dx = -\int (1 - u)u^{\frac12}\,du$
   $= -\int \left(u^{\frac12} - u^{\frac32}\right)du$.
4. Integrate: $-\left(\frac23u^{\frac32} - \frac25u^{\frac52}\right) + c = \frac25u^{\frac52} - \frac23u^{\frac32} + c$.
5. Put $u$ back: $\frac25(1 - x)^{\frac52} - \frac23(1 - x)^{\frac32} + c$.

(a)

1. The scores are already ranks.
2. $d$ = Test 1 rank − Test 2 rank: $0, -3, 1, 1, 1, 0, 3, 3, -1, -5$.
3. $\sum d^2 = 0 + 9 + 1 + 1 + 1 + 0 + 9 + 9 + 1 + 25$
   $= 56$.
4. $\rho = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)}$
   $= 1 - \dfrac{6 \times 56}{10 \times 99}$
   $= 1 - \dfrac{336}{990}$
   $\approx 0.6606$.

(b)(i)

1. $\,{}^4C_3 \times {}^6C_2 = 4 \times 15$
   $= 60$ committees.

(ii)

1. Committees with both the man and the woman: the other 2 men from 3 and the other woman from 5, $\,{}^3C_2 \times {}^5C_1 = 15$.
2. So $60 - 15 = 45$ committees keep them apart.

1. $X \sim B\left(4, \frac13\right)$, with $q = \frac23$.

(a)

1. $P(X = 2)$
   $= \binom42\left(\frac13\right)^2\left(\frac23\right)^2$
   $= \dfrac{24}{81}$
   $= \dfrac{8}{27}$.

(b)

1. At least 1: $1 - \left(\frac23\right)^4 = 1 - \dfrac{16}{81}$
   $= \dfrac{65}{81}$.

(c)

1. At most 2: $P(0) + P(1) + P(2) = \dfrac{16 + 32 + 24}{81}$
   $= \dfrac{72}{81}$
   $= \dfrac89$.

(a)(i)

1. The forces are on bearings $000^\circ$ (6 N), $120^\circ$ (4.5 N) and $210^\circ$ (8 N).
2. East: $0 + 4.5\sin120^\circ + 8\sin210^\circ = 3.897 - 4$
   $= -0.103$.
3. North: $6 + 4.5\cos120^\circ + 8\cos210^\circ = 6 - 2.25 - 6.928$
   $= -3.178$.
4. $|\mathbf R| = \sqrt{0.103^2 + 3.178^2}$
   $\approx 3.18\text{ N}$.

(ii)

1. $a = \dfrac{F}{m}$
   $= \dfrac{3.18}{1.2}$
   $\approx 2.65\text{ m s}^{-2}$.

(b)

1. $v^2 = u^2 + 2gh = 0 + 2(10)(80) = 1600$, so $v = 40\text{ m s}^{-1}$.

(a)

1. Match the parts: $6x + 2y = 8$ and $-5x + 7y = 2$.
2. From the first, $y = 4 - 3x$.
3. Substitute: $-5x + 28 - 21x = 2$, so $x = 1$ and $y = 1$.

(b)(i)

1. $|\mathbf a + \mathbf b|^2 = |\mathbf a|^2 + |\mathbf b|^2 + 2|\mathbf a||\mathbf b|\cos\theta$.
2. $139 = 9 + 100 + 60\cos\theta$, so $\cos\theta = \frac{30}{60} = \frac12$ and $\theta = 60^\circ$.

(ii)

1. $\mathbf a \cdot \mathbf b = |\mathbf a||\mathbf b|\cos\theta$
   $= 3 \times 10 \times \frac12$
   $= 15$.