WAEC · 2013 · Nov/Dec · Further Maths · Paper 2

(a)

1. The odd numbers $51, 53, \ldots, 99$ form an AP with first term $a = 51$, common difference $d = 2$ and last term $l = 99$.
2. Find the number of terms from $l = a + (n - 1)d$: $99 = 51 + 2(n - 1)$.
3. So $2(n - 1) = 48$, which gives $n - 1 = 24$ and $n = 25$.
4. Use $S_n = \frac{n}{2}(a + l)$: $S_{25} = \frac{25}{2}(51 + 99)$.
5. Work it out: $\frac{25}{2} \times 150 = 1875$.
6. The sum of the odd numbers from 51 to 99 is 1875.

(a)

1. Use $\cos\theta = \dfrac{\mathbf a \cdot \mathbf b}{|\mathbf a||\mathbf b|}$ with $\mathbf a = 2\mathbf i - \mathbf j$ and $\mathbf b = \mathbf i - \mathbf j$.
2. Dot product: $(2)(1) + (-1)(-1) = 3$.
3. Magnitudes: $|\mathbf a| = \sqrt{4 + 1} = \sqrt5$ and $|\mathbf b| = \sqrt{1 + 1} = \sqrt2$.
4. So $\cos\theta = \dfrac{3}{\sqrt5\sqrt2}$
   $= \dfrac{3}{\sqrt{10}}$
   $\approx 0.9487$.
5. Take the inverse cosine: $\theta \approx 18.43^\circ$.
6. The acute angle between the vectors is $18.43^\circ$.

(a)

1. Use ${}^nC_4 = \dfrac{n(n - 1)(n - 2)(n - 3)}{4!}$, and $4! = 24$.
2. So ${}^{p+1}C_4 = \dfrac{(p + 1)p(p - 1)(p - 2)}{24}$.
3. And ${}^{p-1}C_4 = \dfrac{(p - 1)(p - 2)(p - 3)(p - 4)}{24}$.
4. Take out the common factor $\dfrac{(p - 1)(p - 2)}{24}$: the difference is $\dfrac{(p - 1)(p - 2)}{24}\left[(p + 1)p - (p - 3)(p - 4)\right]$.
5. Expand the bracket: $p^2 + p - (p^2 - 7p + 12) = 8p - 12$.
6. Write $8p - 12$ as $4(2p - 3)$ and cancel 4 into 24.
7. So ${}^{p+1}C_4 - {}^{p-1}C_4 = \dfrac{(p - 1)(p - 2)(2p - 3)}{6}$.

1. Use the assumed mean $A = 85$ and the deviations $d$.
2. There are $n = 9$ numbers.
3. Add the deviations: $\sum d = -6 - 4 - 2 + 1 + 2 + 3 + 4 + 5 + 8$
   $= 11$.

(a)

1. Mean $= A + \dfrac{\sum d}{n}$, so the mean is $85 + \dfrac{11}{9}$.
2. That is $85 + 1.222 \approx 86.22$.

(b)

1. Square and add the deviations: $\sum d^2 = 36 + 16 + 4 + 1 + 4 + 9 + 16 + 25 + 64$
   $= 175$.
2. Variance $= \dfrac{\sum d^2}{n} - \left(\dfrac{\sum d}{n}\right)^2$, so it is $\dfrac{175}{9} - \left(\dfrac{11}{9}\right)^2$.
3. Work it out: $19.444 - 1.494 \approx 17.95$.
4. The mean is 86.22 and the variance is 17.95.

1. The weights are $6 \times 10 = 60\text{ N}$ and $5 \times 10 = 50\text{ N}$, so the 6 kg mass moves down and the 5 kg mass moves up.

(a)

1. Let $T$ be the tension.
2. For the 6 kg mass (moving down): $60 - T = 6a$.
3. For the 5 kg mass (moving up): $T - 50 = 5a$.
4. Add the two equations to remove $T$: $10 = 11a$.
5. So $a = \frac{10}{11}$
   $\approx 0.91\text{ m s}^{-2}$ for each mass.

(b)

1. Use $s = ut + \frac12at^2$ with $u = 0$ and $t = 2$.
2. So $s = \frac12 \times \frac{10}{11} \times 4$
   $= \frac{20}{11}$.
3. Each mass moves about $1.82\text{ m}$ in 2 seconds.

(a)

1. Put $x = 0$ in the equation: $y^2 - 6y + 5 = 0$.
2. Factorise: $(y - 1)(y - 5) = 0$, so the points are $(0, 1)$ and $(0, 5)$.
3. Differentiate implicitly: $2x + 2y\dfrac{dy}{dx} - 2 - 6\dfrac{dy}{dx} = 0$.
4. Collect the $\dfrac{dy}{dx}$ terms: $(2y - 6)\dfrac{dy}{dx} = 2 - 2x$, so $\dfrac{dy}{dx} = \dfrac{1 - x}{y - 3}$.
5. At $(0, 1)$ the gradient is $\dfrac{1}{-2} = -\dfrac12$.
6. At $(0, 5)$ the gradient is $\dfrac{1}{2}$.

(b)

1. At $(0, 1)$: $y - 1 = -\frac12(x - 0)$.
2. Multiply by 2 and rearrange to get $x + 2y - 2 = 0$.
3. At $(0, 5)$: $y - 5 = \frac12(x - 0)$.
4. Multiply by 2 and rearrange to get $x - 2y + 10 = 0$.
5. The tangents are $x + 2y - 2 = 0$ and $x - 2y + 10 = 0$.

(a)(i)

1. With no restrictions, choose 6 from 11 people: ${}^{11}C_6 = 462$ ways.

(ii)

1. At least two girls means 2, 3 or 4 girls, since there are only 4 girls.
2. 2 girls and 4 boys: ${}^4C_2 \times {}^7C_4 = 6 \times 35$
   $= 210$.
3. 3 girls and 3 boys: ${}^4C_3 \times {}^7C_3 = 4 \times 35$
   $= 140$.
4. 4 girls and 2 boys: ${}^4C_4 \times {}^7C_2 = 1 \times 21 = 21$.
5. Add the cases and divide by 462: $\dfrac{371}{462}$, which simplifies to $\dfrac{53}{66} \approx 0.80$.

(b)

1. Rank each test from the highest score (rank 1).
2. Theory, A to H: 4, 3, 5, 7, 2, 1, 6, 8.
3. Practical, A to H: 1, 3, 7, 8, 5, 2, 4, 6.
4. Differences $d$: 3, 0, $-2$, $-1$, $-3$, $-1$, 2, 2, so $\sum d^2 = 9 + 0 + 4 + 1 + 9 + 1 + 4 + 4$
   $= 32$.
5. Use $r = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)}$ with $n = 8$: $r = 1 - \dfrac{6 \times 32}{8 \times 63}$.
6. Work it out: $1 - \dfrac{192}{504} \approx 0.62$.
7. Spearman's rank correlation coefficient is about 0.62.

(a)

1. Mean: $\dfrac{2 + 5 + 6 + 8 + x + y}{6} = 6$, so $21 + x + y = 36$.
2. So $x + y = 15$ … (1).
3. The variance is $(\sqrt5)^2 = 5$, so the squared deviations from 6 add up to $6 \times 5 = 30$.
4. The known numbers give $(2 - 6)^2 + (5 - 6)^2 + (6 - 6)^2 + (8 - 6)^2 = 16 + 1 + 0 + 4$
   $= 21$.
5. So $(x - 6)^2 + (y - 6)^2 = 9$ … (2).
6. Put $y = 15 - x$ from (1) into (2): $(x - 6)^2 + (9 - x)^2 = 9$.
7. Expand: $2x^2 - 30x + 117 = 9$, so $x^2 - 15x + 54 = 0$.
8. Factorise: $(x - 6)(x - 9) = 0$, so $x = 6$ or $x = 9$.
9. From (1), $y = 9$ or $y = 6$: the two numbers are 6 and 9.

(a)

1. From $F = ma$: $a = \dfrac{120}{1200}$
   $= 0.1\text{ m s}^{-2}$.
2. Use $s = ut + \frac12at^2$ with $u = 0$ and $t = 6$: $s = \frac12 \times 0.1 \times 36$.
3. So the lorry covers $1.8\text{ m}$.

(b)

1. Let $T_1$ be the tension in the string at $60^\circ$ and $T_2$ the tension in the string at $30^\circ$.
2. Resolve horizontally: $T_1\cos60^\circ = T_2\cos30^\circ$, so $\frac12T_1 = \frac{\sqrt3}{2}T_2$ and $T_1 = \sqrt3\,T_2$.
3. Resolve vertically: $T_1\sin60^\circ + T_2\sin30^\circ = 30$.
4. Substitute $T_1 = \sqrt3\,T_2$: $\sqrt3\,T_2 \times \frac{\sqrt3}{2} + \frac12T_2 = 30$, so $2T_2 = 30$.
5. So $T_2 = 15\text{ N}$ and $T_1 = 15\sqrt3 \approx 25.98\text{ N}$.
6. The tensions are $25.98\text{ N}$ (string at $60^\circ$) and $15\text{ N}$ (string at $30^\circ$).

(a)

1. Momentum is conserved: $0.084u + 20(0) = (0.084 + 20)(0.24)$.
2. The right-hand side is $20.084 \times 0.24 = 4.82016$.
3. So $u = \dfrac{4.82016}{0.084}$
   $\approx 57.38\text{ m s}^{-1}$.

(b)

1. Differentiate to get the velocity: $v = \dfrac{ds}{dt} = 2t^2 - 7t + 5$.
2. The body is momentarily at rest when $v = 0$: $(2t - 5)(t - 1) = 0$, so $t = 1$ or $t = 2.5$.
3. At $t = 1$: $s = \frac23 - \frac72 + 5$
   $= \frac{13}{6}\text{ m}$.
4. At $t = 2.5$: $s = \frac{125}{12} - \frac{175}{8} + \frac{25}{2}$
   $= \frac{25}{24}\text{ m}$.
5. Subtract: $\frac{13}{6} - \frac{25}{24} = \frac{27}{24}$
   $= 1.125$.
6. The distance between the two points is $1.13\text{ m}$ to 3 significant figures.