WAEC · 2017 · Private · Further Maths · Paper 2

1. Read off the coefficients: $a = m$, $b = 2m$, $c = -3$.
2. Real roots need $b^2 - 4ac \ge 0$: $(2m)^2 - 4(m)(-3) \ge 0$, so $4m^2 + 12m \ge 0$.
3. Factorise: $4m(m + 3) \ge 0$, with roots $m = 0$ and $m = -3$.
4. "$\ge 0$" is outside the roots: $m \le -3$ or $m \ge 0$.
5. But $m = 0$ removes the $x^2$ term and leaves $-3 = 0$, which has no roots at all.
6. So $m \le -3$ or $m > 0$.

1. Find the first two terms: $U_1 = (3 - 2)\log 5 = \log 5$ and $U_2 = (6 - 2)\log 5 = 4\log 5$.
2. The first term is $a = \log 5$ and the common difference is $d = 4\log 5 - \log 5 = 3\log 5$.
3. Use $S_n = \frac n2[2a + (n - 1)d]$: $S_n = \frac n2[2\log 5 + 3(n - 1)\log 5]$.
4. Simplify the bracket: $2 + 3n - 3 = 3n - 1$, so $S_n = \dfrac{n(3n - 1)}{2}\log 5$.

1. $2x + 3y = 5$ has gradient $-\frac23$, so the bisector has gradient $-\frac23$ too (parallel).
2. The segment is perpendicular to it, with gradient $\frac32$: $\dfrac{2 - y}{5 - 3} = \dfrac32$, so $y = -1$.
3. The midpoint of $(3, -1)$ and $(5, 2)$ is $\left(4, \frac12\right)$.
4. $y - \frac12 = -\frac23(x - 4)$.
5. Multiply by 6: $6y - 3 = -4x + 16$.
6. So $6y + 4x - 19 = 0$.

(a)

1. Write $y = 3x^2 - 3x^{-2}$.
2. Differentiate: $\dfrac{dy}{dx} = 6x + 6x^{-3}$
   $= 6x + \dfrac{6}{x^3}$.

(b)

1. Integrate: $\displaystyle\int (3x^2 - 3x^{-2})\,dx = x^3 + 3x^{-1}$
   $= x^3 + \dfrac3x$.
2. At $x = 3$: $27 + 1 = 28$.
3. At $x = 1$: $1 + 3 = 4$.
4. So $\displaystyle\int_1^3 y\,dx = 28 - 4 = 24$.

1. There are $\,{}^{13}C_5 = 1287$ committees.

(a)

1. All women: $\,{}^7C_5 = 21$, so $P = \dfrac{21}{1287}$
   $= \dfrac{7}{429}$
   $\approx 0.0163$.

(b)

1. 2 men and 3 women: $\,{}^6C_2 \times {}^7C_3 = 15 \times 35$
   $= 525$.
2. So $P = \dfrac{525}{1287}$
   $= \dfrac{175}{429}$
   $\approx 0.4079$.

1. Class marks $16.5, 20.5, 24.5, 28.5, 32.5, 36.5$ and $d = x - 24.5$: $-8, -4, 0, 4, 8, 12$.
2. $\sum fd = -320 - 132 + 0 + 40 + 64 + 48$
   $= -300$ and $\sum f = 120$.
3. Mean $= 24.5 + \dfrac{-300}{120}$
   $= 24.5 - 2.5$
   $= 22$.

1. $u = 15$, $v = 0$ and $t = 5$.

(a)

1. $v = u + at$: $0 = 15 + 5a$, so $a = -3$: a retardation of $3\text{ m s}^{-2}$.

(b)

1. $s = \frac12(u + v)t$
   $= \frac12(15)(5)$
   $= 37.5\text{ m}$.

1. $\mathbf m - 5\mathbf n = \begin{pmatrix} -12 - 5 \\ 5 + 5 \end{pmatrix}$
   $= \begin{pmatrix} -17 \\ 10 \end{pmatrix}$.
2. Its length is $\sqrt{289 + 100} = \sqrt{389}$.
3. $\mathbf p = \dfrac{35}{\sqrt{389}}\begin{pmatrix} -17 \\ 10 \end{pmatrix}$
   $\approx \begin{pmatrix} -30.17 \\ 17.75 \end{pmatrix}$.

(a)

1. Expand along the top row: $|F| = 3\big((-1)(-2) - 2(1)\big) - 4\big(0(-2) - 2(2)\big) + 2\big(0(1) - (-1)(2)\big)$.
2. $= 3(0) - 4(-4) + 2(2) = 20$.

(b)

1. Replace each column in turn by $-1, -8, 6$: $\Delta_x = -20$, $\Delta_y = 40$ and $\Delta_z = -60$.
2. So $x = \dfrac{-20}{20} = -1$, $y = \dfrac{40}{20} = 2$ and $z = \dfrac{-60}{20} = -3$.
3. Check: $3(-1) + 4(2) + 2(-3) = -1$ ✓, $-2 + 2(-3) = -8$ ✓ and $2(-1) + 2 - 2(-3) = 6$ ✓.

(a)

1. Differentiate both sides with respect to $x$, using the product rule on $3xy$: $\dfrac{dy}{dx} = 3y + 3x\dfrac{dy}{dx}$.
2. Collect the $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(1 - 3x) = 3y$, so $\dfrac{dy}{dx} = \dfrac{3y}{1 - 3x}$.
3. At $(-1, 4)$, substitute $x = -1$ and $y = 4$: $\dfrac{dy}{dx} = \dfrac{3 \times 4}{1 - 3(-1)}$.
4. Work out the top and the bottom: $\dfrac{12}{1 + 3} = \dfrac{12}{4}$, so $\dfrac{dy}{dx} = 3$ at $(-1, 4)$.

(b)

1. For $3x^2 - 5x + 1 = 0$: $\alpha + \beta = \frac53$ and $\alpha\beta = \frac13$.
2. New sum: $\left(\alpha - \frac1\beta\right) + \left(\beta - \frac1\alpha\right) = (\alpha + \beta) - \dfrac{\alpha + \beta}{\alpha\beta}$.
3. Substitute: $\frac53 - \dfrac{\frac53}{\frac13} = \frac53 - 5$
   $= -\frac{10}{3}$.
4. New product: $\left(\alpha - \frac1\beta\right)\left(\beta - \frac1\alpha\right) = \alpha\beta - 1 - 1 + \dfrac{1}{\alpha\beta}$.
5. Substitute: $\frac13 - 2 + 3 = \frac43$.
6. The equation: $x^2 + \frac{10}{3}x + \frac43 = 0$.
7. Multiply through by 3: $3x^2 + 10x + 4 = 0$.

(a)

1. The curves meet where $(2x - 5)^3 = \dfrac{1}{(2x - 5)^3}$.
2. Multiply both sides by $(2x - 5)^3$: $(2x - 5)^6 = 1$.
3. So $2x - 5 = 1$ or $2x - 5 = -1$, which gives $x = 3$ or $x = 2$.

(b)

1. Four strips from 1 to 5, so $h = 1$.
2. Work out $y = x + \dfrac{8}{x^2}$ at each ordinate.
3. $x = 1$: $9$.
4. $x = 2$: $4$.
5. $x = 3$: $3.8889$.
6. $x = 4$: $4.5$.
7. $x = 5$: $5.32$.
8. $\frac12[(9 + 5.32) + 2(4 + 3.8889 + 4.5)] = \frac12[14.32 + 24.7778]$
   $= 19.5489$, about $19.55$.

(a)

1. The frequencies add to 30: $5 + 6 + \frac p3 + 5 + 8 + \frac23p + 2 + 1 = 30$, so $27 + p = 30$ and $p = 3$.

(b)

1. The frequencies are $5, 6, 6, 8, 4, 1$ and the class marks $1, 4, 7, 10, 13, 16$.
2. With $d = x - 7$: $\sum fd = 9$ and $\sum fd^2 = 531$.
3. Variance $= \dfrac{531}{30} - \left(\dfrac{9}{30}\right)^2$
   $= 17.7 - 0.09$
   $= 17.61$.

(c)

1. At most 11 covers the first four classes: $\dfrac{5 + 6 + 6 + 8}{30} = \dfrac{25}{30}$
   $= \dfrac56$.

(a)

1. Both black: $\dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90}$.
2. Both brown: $\dfrac{7}{10} \times \dfrac{6}{9} = \dfrac{42}{90}$.
3. Same colour: $\dfrac{6 + 42}{90} = \dfrac{8}{15}$
   $\approx 0.533$.

(b)

1. Each guess is correct with $p = \frac14$, and there are $n = 12$ questions.

(i)

1. $P(3 \text{ correct}) = {}^{12}C_3\left(\frac14\right)^3\left(\frac34\right)^9$
   $\approx 0.258$.

(ii)

1. At most 8 wrong means at least 4 correct: $1 - [P(0) + P(1) + P(2) + P(3)]$.
2. $= 1 - (0.0317 + 0.1267 + 0.2323 + 0.2581)$
   $\approx 0.351$.

(a)(i)

1. $v = u - gt$
   $= 75 - 10(5)$
   $= 25\text{ m s}^{-1}$.

(ii)

1. $v^2 = u^2 - 2gs$: $15^2 = 75^2 - 20s$, so $s = \dfrac{5625 - 225}{20} = 270\text{ m}$.

(iii)

1. $270 = 75t - 5t^2$, so $t^2 - 15t + 54 = 0$.
2. $(t - 6)(t - 9) = 0$: at $t = 6\text{ s}$ (going up) and $t = 9\text{ s}$ (coming down).

(b)

1. $a = \dfrac{42 - 6}{6} = 6\text{ m s}^{-2}$.
2. $q = \dfrac{F}{a}$
   $= \dfrac{24}{6}$
   $= 4\text{ kg}$.

(a)

1. $3\mathbf m + 2\mathbf n - 5\mathbf p = (6 + 12 - 10)\mathbf i + (9 + 46 + 35)\mathbf j$
   $= 8\mathbf i + 90\mathbf j$.
2. Its magnitude is $\sqrt{64 + 8100} = \sqrt{8164}$
   $\approx 90.355$.

(b)

1. Each force is its size times the unit vector of its direction.
2. $F_1 = -18\mathbf i$ and $F_3 = 20\mathbf j$.
3. $|3\mathbf i + 4\mathbf j| = 5$, so $F_2 = 15 \times \frac15(3\mathbf i + 4\mathbf j)$
   $= 9\mathbf i + 12\mathbf j$.
4. Resultant: $(-18 + 9)\mathbf i + (12 + 20)\mathbf j = -9\mathbf i + 32\mathbf j$.
5. $|\mathbf R| = \sqrt{81 + 1024}$
   $= \sqrt{1105}$
   $\approx 33.242\text{ N}$.
6. It points north-west: $\tan^{-1}\frac{9}{32} = 15.71^\circ$ west of north, a bearing of $344.29^\circ$.