WAEC · 2013 · May/June · Further Maths · Paper 2

(a)

1. The $x^2$ term of $(p + qx)^7$ is $\binom72p^5(qx)^2 = 21p^5q^2x^2$.
2. The $x^3$ term is $\binom73p^4(qx)^3 = 35p^4q^3x^3$.
3. Set the coefficients equal: $21p^5q^2 = 35p^4q^3$.
4. Divide both sides by $7p^4q^2$: $3p = 5q$, so $q = \frac35p$.

(b)

1. The payments form an A.P. with $a = 180$ and $d = 80$.
2. The total in 16 weeks is $S_{16}$.
3. $S_{16} = \dfrac{16}{2}[2(180) + 15(80)]$
   $= 8(360 + 1200)$
   $= 8 \times 1560$.
4. So he contributes ₦12 480.

1. Differentiate each term with respect to $x$: $3x^2 + 3y^2\dfrac{dy}{dx} - 2\left(y + x\dfrac{dy}{dx}\right) = 0$.
2. Collect the $\dfrac{dy}{dx}$ terms: $(3y^2 - 2x)\dfrac{dy}{dx} = 2y - 3x^2$.
3. So $\dfrac{dy}{dx} = \dfrac{2y - 3x^2}{3y^2 - 2x}$.
4. At $(2, -1)$: $\dfrac{dy}{dx} = \dfrac{-2 - 12}{3 - 4}$
   $= \dfrac{-14}{-1}$
   $= 14$.

1. Let the sides be $x$ and $3x$.
2. The perimeter is $P = 8x$ and the area is $A = 3x^2$.
3. $P$ is a fixed multiple of $x$, so a 2% increase in $P$ means a 2% increase in $x$.
4. For a small change, $\dfrac{\delta A}{A} \approx \dfrac{6x\,\delta x}{3x^2}$
   $= 2\dfrac{\delta x}{x}$.
5. So the area increases by about $2 \times 2\% = 4\%$.
6. (Exactly, $1.02^2 - 1 = 4.04\%$.)

1. From the line, $x = 2y - 3$.
2. Substitute into the circle.
3. $(2y - 3)^2 + y^2 - 2(2y - 3) + 6y - 15 = 0$.
4. $4y^2 - 12y + 9 + y^2 - 4y + 6 + 6y - 15 = 0$, so $5y^2 - 10y = 0$.
5. $5y(y - 2) = 0$: $y = 0$ gives $x = -3$, and $y = 2$ gives $x = 1$.
6. $N$ is in the first quadrant, so $N(1, 2)$ and $M(-3, 0)$.

1. There are $\,{}^9C_3 = 84$ ways to choose 3 prefects from 9.

(a)

1. Only boys: $\,{}^5C_3 = 10$, so $P = \dfrac{10}{84}$
   $= \dfrac{5}{42}$
   $\approx 0.119$.

(b)

1. More girls than boys means 3 girls, or 2 girls and 1 boy.
2. 3 girls: $\,{}^4C_3 = 4$. 2 girls and 1 boy: $\,{}^4C_2 \times {}^5C_1 = 6 \times 5$
   $= 30$.
3. So $P = \dfrac{4 + 30}{84}$
   $= \dfrac{34}{84}$
   $= \dfrac{17}{42}$
   $\approx 0.405$.

1. Class marks $13, 16, 19, 22, 25$.
2. $d = x - 19$: $-6, -3, 0, 3, 6$.
3. Frequencies $6, 10, 3, 2, 1$.
4. $\sum fd = -36 - 30 + 0 + 6 + 6 = -54$ and $\sum fd^2 = 216 + 90 + 0 + 18 + 36 = 360$.

(a)

1. $\bar x = 19 + \dfrac{-54}{22}$
   $= 19 - 2.4545$
   $\approx 16.5$ years.

(b)

1. $\sigma = \sqrt{\dfrac{360}{22} - 2.4545^2}$
   $= \sqrt{16.3636 - 6.0248}$
   $\approx 3.22$ years.

1. Change in momentum $= m\mathbf v - m\mathbf u$.
2. $= 0.1(30\mathbf i) - 0.1(40\mathbf j)$
   $= 3\mathbf i - 4\mathbf j$.
3. Its magnitude is $\sqrt{9 + 16} = 5\text{ kg m s}^{-1}$.

1. Take down as positive: $u = 20$, $a = 10$ and $s = 45$.

(a)

1. $s = ut + \frac12at^2$: $45 = 20t + 5t^2$, so $t^2 + 4t - 9 = 0$.
2. $t = \dfrac{-4 + \sqrt{16 + 36}}{2}$
   $= -2 + \sqrt{13}$
   $\approx 1.61\text{ s}$ (the other root is negative).

(b)

1. $v^2 = u^2 + 2as = 400 + 900 = 1300$, so $v = \sqrt{1300}$
   $\approx 36.06\text{ m s}^{-1}$.

(a)

1. Quotient rule with $u = x^2 + 1$ and $v = (x + 1)^2$: $u' = 2x$ and $v' = 2(x + 1)$.
2. $\dfrac{dy}{dx} = \dfrac{(x + 1)^2 \cdot 2x - (x^2 + 1) \cdot 2(x + 1)}{(x + 1)^4}$.
3. Take out $2(x + 1)$ from the top: $2(x + 1)[x(x + 1) - (x^2 + 1)] = 2(x + 1)(x - 1)$.
4. Cancel one $(x + 1)$: $\dfrac{dy}{dx} = \dfrac{2(x - 1)}{(x + 1)^3}$.

(b)(i)

1. Expand along the top row: $1(9 + 1) - 2(6 - 1) + (-1)(2 + 3) = 10 - 10 - 5$
   $= -5$.

(ii)

1. By Cramer's rule, replace each column by $(4, 2, -1)$ in turn: $\Delta_x = 25$, $\Delta_y = -15$ and $\Delta_z = 15$.
2. So $x = \dfrac{25}{-5} = -5$, $y = \dfrac{-15}{-5} = 3$ and $z = \dfrac{15}{-5} = -3$.
3. Check in the first equation: $-5 + 6 + 3 = 4$ ✓.

(a)(i)

1. Write the three terms as $\dfrac ar$, $a$ and $ar$.
2. Their product is $a^3 = 8$, so $a = 2$.
3. Their sum: $\dfrac2r + 2 + 2r = 7$.
4. Multiply by $r$: $2 + 2r + 2r^2 = 7r$, so $2r^2 - 5r + 2 = 0$.
5. Factorise: $(2r - 1)(r - 2) = 0$, so $r = \frac12$ or $r = 2$.
6. The sequence is decreasing, so $r = \frac12$.

(ii)

1. The terms are $\dfrac{2}{\frac12} = 4$, then $2$, then $1$.

(b)

1. $h = 1$.
2. $y = x + \dfrac{2}{x^2}$ gives $3,\ 2.5,\ 3.2222,\ 4.125,\ 5.08$ at $x = 1, \ldots, 5$.
3. $\frac12[(3 + 5.08) + 2(2.5 + 3.2222 + 4.125)] = \frac12[8.08 + 19.6944]$
   $= 13.8872$, about $13.89$.

(a)

1. Differentiate: $\dfrac{dy}{dx} = 6x^2 - 6x - 12$
   $= 6(x - 2)(x + 1)$.
2. Stationary points where $\dfrac{dy}{dx} = 0$: $x = 2$ or $x = -1$.
3. The second derivative is $\dfrac{d^2y}{dx^2} = 12x - 6$.
4. At $x = -1$: $\dfrac{d^2y}{dx^2} = -18 < 0$, a maximum, with $y = -2 - 3 + 12 + 4 = 11$.
5. So the maximum point is $(-1, 11)$.
6. At $x = 2$: $\dfrac{d^2y}{dx^2} = 18 > 0$, a minimum, with $y = 16 - 12 - 24 + 4 = -16$.
7. So the minimum point is $(2, -16)$.

(b)

1. The curve is a cubic with a positive $x^3$ term: it rises to the maximum $(-1, 11)$, falls through $(0, 4)$ to the minimum $(2, -16)$, then rises again.

(a)

1. Plot the eight points with the given scales.

(b)

1. $\bar x = \dfrac{360}{8} = 45$ and $\bar y = \dfrac{12.4}{8} = 1.55$.
2. Plot $(45, 1.55)$.

(c)

1. Draw the line through $(45, 1.55)$ and $(10, 1)$.

(d)(i)

1. Gradient $= \dfrac{1.55 - 1}{45 - 10}$
   $= \dfrac{11}{700}$, so $y = 1 + \dfrac{11}{700}(x - 10)$, about $y = 0.016x + 0.84$.

(ii)

1. At $x = 75$: $y = 1 + \dfrac{11}{700} \times 65$
   $\approx 2.02$ m.

(a)

1. There are $\binom{12}{3} = 220$ ways to draw 3 of the 12 balls.

(i)

1. All the same colour: $\binom53 + \binom43 + \binom33 = 10 + 4 + 1$
   $= 15$, so $P = \frac{15}{220} = \frac{3}{44}$.

(ii)

1. Exactly two the same: two of one colour and one of another: $\binom52 \times 7 + \binom42 \times 8 + \binom32 \times 9 = 70 + 48 + 27$
   $= 145$.
2. So $P = \dfrac{145}{220} = \dfrac{29}{44}$.

(b)

1. The table already gives ranks.
2. $d$: $-1, -1, 2, -1, 1, 2, -2$, so $\sum d^2 = 16$.
3. $\rho = 1 - \dfrac{6 \times 16}{7 \times 48}$
   $= 1 - \dfrac{96}{336}$
   $= \dfrac57$
   $\approx 0.7143$.

(a)(i)

1. Winning each race is independent: $P(\text{all 4}) = 0.8^4 = 0.4096$.

(ii)

1. $P(\text{none}) = 0.2^4 = 0.0016$.

(iii)

1. At most 3 is everything except winning all 4: $1 - 0.4096 = 0.5904$.

(b)

1. There are $\,{}^{15}C_5 = 3003$ committees.

(i)

1. 3 boys and 2 girls: $\,{}^{10}C_3 \times {}^5C_2 = 120 \times 10$
   $= 1200$, so $P = \dfrac{1200}{3003} \approx 0.3996$.

(ii)

1. At least one girl is everything except all boys: $1 - \dfrac{{}^{10}C_5}{3003} = 1 - \dfrac{252}{3003}$
   $\approx 0.9161$.

(a)

1. $\mathbf P$: $4\sin30^\circ = 2$ east and $4\cos30^\circ = 3.464$ north.
2. $\mathbf R$: $10\sin300^\circ = -8.660$ east and $10\cos300^\circ = 5$ north.
3. The balancing force is $-(\mathbf P + \mathbf R) = 6.660\mathbf i - 8.464\mathbf j$.
4. Its size is $\sqrt{6.660^2 + 8.464^2} \approx 10.77\text{ N}$.
5. It points south-east: $\tan^{-1}\frac{6.660}{8.464} = 38.2^\circ$ east of south, a bearing of $141.8^\circ$.

(b)(i)

1. $s = \frac12(u + v)t$: $1500 = \frac12(45)t$, so $t = 66\frac23\text{ s}$.

(ii)

1. $v^2 = u^2 + 2as$: $0 = 2025 + 3000a$, so $a = -0.675$: a retardation of $0.675\text{ m s}^{-2}$.

(a)

1. Work out $S = t^2 - 6t + 5$ at each whole second: $S = 5, 0, -3, -4, -3, 0, 5$ at $t = 0, 1, 2, 3, 4, 5, 6$.
2. Plot them and join with a smooth U-shaped curve.

(b)(i)

1. The velocity is the gradient of the displacement–time graph.
2. It is zero at the lowest point, $t = 3$ s (also $\dfrac{dS}{dt} = 2t - 6 = 0$).

(ii)

1. Average velocity $= \dfrac{\text{change in displacement}}{\text{time}}$
   $= \dfrac{S(4) - S(0)}{4}$
   $= \dfrac{-3 - 5}{4}$
   $= -2\text{ m s}^{-1}$, that is $2\text{ m s}^{-1}$ back towards $O$.

(iii)

1. From $t = 0$ to $3$ the particle moves from $S = 5$ to $S = -4$: 9 m.
2. From $t = 3$ to $5$ it moves back from $S = -4$ to $S = 0$: 4 m.
3. The total distance is $9 + 4 = 13$ m.