WAEC · 2012 · Nov/Dec · Further Maths · Paper 2

1. Write the surds as powers: $7\sqrt7 = 7^1 \times 7^{\frac12}$
   $= 7^{\frac32}$ and $5\sqrt5 = 5^{\frac32}$.
2. Bring the powers down in front: the top is $\frac32\log 7 - \frac32\log 5 = \frac32(\log 7 - \log 5)$.
3. Write the bottom the same way: $1.4 = \frac75$, so $\log 1.4 = \log 7 - \log 5$.
4. The bracket $(\log 7 - \log 5)$ cancels, so the value is $\dfrac32$.

1. Read off the coefficients: $a = k + 3$, $b = 6 - 2k$, $c = k - 1$.
2. Real roots need $b^2 - 4ac \ge 0$: $(6 - 2k)^2 - 4(k + 3)(k - 1) \ge 0$.
3. Expand the square: $(6 - 2k)^2 = 36 - 24k + 4k^2$.
4. Expand the product: $4(k + 3)(k - 1) = 4(k^2 + 2k - 3)$
   $= 4k^2 + 8k - 12$.
5. Take one from the other: $36 - 24k + 4k^2 - 4k^2 - 8k + 12 \ge 0$, so $48 - 32k \ge 0$.
6. Add $32k$ to both sides: $48 \ge 32k$.
7. Divide by 32: $k \le \frac32$.
8. For the equation to be a quadratic, $a \ne 0$, so also $k \ne -3$.
9. The roots are real for $k \le \frac32$ ($k \ne -3$).

(a)

1. Draw $2x - y = -4$, $x + y = 10$, $y = x$ (dashed, since $y - x > 0$), $y = 2$ and $x = 0$, and shade the region that satisfies all five.
2. Its corners: $(0, 2)$.
3. $y = 2$ and $y = x$ give $(2, 2)$.
4. $y = x$ and $x + y = 10$ give $(5, 5)$.
5. $x + y = 10$ and $2x - y = -4$ give $3x = 6$, so $(2, 8)$.
6. $2x - y = -4$ and $x = 0$ give $(0, 4)$.

(b)

1. $10x + 5y$ at the corners: $10$, $30$, $75$, $60$ and $20$.
2. The maximum is $75$, at $x = 5$, $y = 5$.

(i)

1. $\mathbf p \cdot \mathbf q = (3)(2) + (-2)(3)$
   $= 6 - 6$
   $= 0$, so they are perpendicular.

(ii)

1. $|\mathbf p| = \sqrt{9 + 4} = \sqrt{13}$ and $|\mathbf q| = \sqrt{4 + 9} = \sqrt{13}$.
2. The right angle is between them, so they are the base and height: area $= \frac12 \times \sqrt{13} \times \sqrt{13}$
   $= 6.5$ square units.

(iii)

1. One angle is $90^\circ$.
2. The two sides are equal, so the other angles are equal: $\frac{180^\circ - 90^\circ}{2} = 45^\circ$ each.

1. $f(x + h) = 3(x + h) - \dfrac{1}{2(x + h)}$.
2. Take away $f(x)$: $f(x + h) - f(x) = 3h - \left[\dfrac{1}{2(x + h)} - \dfrac{1}{2x}\right]$.
3. Put the fractions over $2x(x + h)$: $\dfrac{x - (x + h)}{2x(x + h)} = \dfrac{-h}{2x(x + h)}$.
4. So $f(x + h) - f(x) = 3h + \dfrac{h}{2x(x + h)}$.
5. Divide by $h$: $3 + \dfrac{1}{2x(x + h)}$.
6. Let $h \to 0$: $f'(x) = 3 + \dfrac{1}{2x^2}$.

1. Add the frequencies: $3 + 8 + 24 + 37 + 10 + 60 + 79 + 11 + 9 = 241$ tosses.

(a)

1. Exactly 8 heads: $\dfrac{9}{241} \approx 0.0373$.

(b)

1. Between 2 and 5 means 3 or 4 heads: $\dfrac{37 + 10}{241} = \dfrac{47}{241}$
   $\approx 0.1950$.

(c)

1. At most 1 head means 0 or 1: $\dfrac{3 + 8}{241} = \dfrac{11}{241}$
   $\approx 0.0456$.

1. The weight is $2.4 \times 10 = 24\text{ N}$.
2. Across: $T_1\cos25^\circ = T_2\cos54^\circ$, so $T_2 = 1.5419\,T_1$.
3. Up: $T_1\sin25^\circ + T_2\sin54^\circ = 24$, so $T_1(0.4226 + 1.2474) = 24$.
4. $T_1 = \dfrac{24}{1.6700}$
   $\approx 14.37\text{ N}$ (the string at $25^\circ$).
5. $T_2 = 1.5419 \times 14.37$
   $\approx 22.16\text{ N}$ (the string at $54^\circ$).

(a)

1. $(1 + mx)^n = 1 + n(mx) + \dfrac{n(n - 1)}{2}(mx)^2 + \ldots$
2. Match the $x$ terms: $nm = 14$, so $m = \dfrac{14}{n}$.
3. Match the $x^2$ terms: $\dfrac{n(n - 1)}{2}m^2 = 84$.
4. Substitute $m$: $\dfrac{n(n - 1)}{2} \times \dfrac{196}{n^2} = 84$.
5. Simplify: $\dfrac{98(n - 1)}{n} = 84$, so $98n - 98 = 84n$, and $14n = 98$.
6. So $n = 7$ and $m = 2$.

(b)

1. $(1.06)^7 = (1 + 2x)^7$ with $2x = 0.06$, so $x = 0.03$.
2. $(1 + 2x)^7 = 1 + 14x + 84x^2 + 280x^3 + \ldots$
   $= 1 + 0.42 + 0.0756 + 0.00756 + \ldots$
3. That is $1.503\ldots$, so $(1.06)^7 = 1.50$ to three significant figures.

(a)

1. Multiply the top and the bottom by the conjugate of the bottom, $10 - 2\sqrt{14}$.
2. The bottom: $10^2 - (2\sqrt{14})^2 = 100 - 56 = 44$.
3. The top: $(4\sqrt7 + 3\sqrt2)(10 - 2\sqrt{14}) = 40\sqrt7 - 8\sqrt{98} + 30\sqrt2 - 6\sqrt{28}$.
4. Simplify the surds: $\sqrt{98} = 7\sqrt2$ and $\sqrt{28} = 2\sqrt7$, so the top is $40\sqrt7 - 56\sqrt2 + 30\sqrt2 - 12\sqrt7 = 28\sqrt7 - 26\sqrt2$.
5. Divide by 44 and simplify: $\dfrac{28\sqrt7 - 26\sqrt2}{44} = \dfrac{14\sqrt7 - 13\sqrt2}{22}$.

(b)

1. Write $y$ as powers of $x$: $y = px^{-2} + qx^{-4}$.
2. Differentiate: $\dfrac{dy}{dx} = -2px^{-3} - 4qx^{-5}$, and again: $\dfrac{d^2y}{dx^2} = 6px^{-4} + 20qx^{-6}$.
3. Multiply: $x^2\dfrac{d^2y}{dx^2} = 6px^{-2} + 20qx^{-4}$, $7x\dfrac{dy}{dx} = -14px^{-2} - 28qx^{-4}$ and $8y = 8px^{-2} + 8qx^{-4}$.
4. Add the three: the $px^{-2}$ terms give $6 - 14 + 8 = 0$ and the $qx^{-4}$ terms give $20 - 28 + 8 = 0$.
5. So $x^2\dfrac{d^2y}{dx^2} + 7x\dfrac{dy}{dx} + 8y = 0$, as required.

(a)

1. $(x + 2)$ is a factor, so $f(-2) = 0$: $96 - 8m - 52 - 2n + 14 = 0$, i.e. $4m + n = 29$.
2. $(x - 1)$ is a factor, so $f(1) = 0$: $6 + m - 13 + n + 14 = 0$, i.e. $m + n = -7$.
3. Subtract: $3m = 36$, so $m = 12$ and $n = -19$.

(b)

1. Put $x = 1$ in the circle: $1 + y^2 - 2 + 2y + 1 = 0$, so $y^2 + 2y = 0$ and $y = 0$ or $y = -2$.
2. Differentiate: $2x + 2y\dfrac{dy}{dx} - 2 + 2\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = \dfrac{1 - x}{y + 1}$.
3. At $(1, 0)$ and at $(1, -2)$ the top is $1 - 1 = 0$, so the gradient is $0$ at both points (the tangents are parallel to the $x$-axis).

(a)

1. Read the coefficients of $x$ and $y$ in each new coordinate: $P = \begin{pmatrix} 3 & -4 \\ -1 & 0 \end{pmatrix}$ and $Q = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix}$.

(b)

1. Multiply rows of $P$ by columns of $Q$: $PQ = \begin{pmatrix} 0 + 8 & 3 - 4 \\ 0 + 0 & -1 + 0 \end{pmatrix}$
   $= \begin{pmatrix} 8 & -1 \\ 0 & -1 \end{pmatrix}$.
2. Double $Q$: $2Q = \begin{pmatrix} 0 & 2 \\ -4 & 2 \end{pmatrix}$.
3. Subtract: $PQ - 2Q = \begin{pmatrix} 8 & -3 \\ 4 & -3 \end{pmatrix}$.

(c)

1. Multiply: $\begin{pmatrix} 8 & -3 \\ 4 & -3 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 8 + 6 \\ 4 + 6 \end{pmatrix}$
   $= \begin{pmatrix} 14 \\ 10 \end{pmatrix}$.
2. The image is the point $(14, 10)$.

1. The number of hunters who hit is binomial with $n = 4$, $p = \frac35 = 0.6$, $q = 0.4$.
2. $P(0) = 0.4^4 = 0.0256$, $P(1) = 4(0.6)(0.4)^3 = 0.1536$, $P(4) = 0.6^4 = 0.1296$.

(a)(i)

1. $P(\text{none}) = 0.02560$.

(ii)

1. $P(1 \le X \le 3) = 1 - P(0) - P(4)$
   $= 1 - 0.0256 - 0.1296$
   $= 0.8448$.

(iii)

1. $P(X \ge 2) = 1 - P(0) - P(1)$
   $= 1 - 0.0256 - 0.1536$
   $= 0.8208$.

(b)

1. The target is hit means $X \ge 1$: $P(X \ge 1) = 1 - 0.0256 = 0.9744$.
2. At most 3 and at least 1 is $1 \le X \le 3$, with probability $0.8448$ from (a)(ii).
3. Conditional probability: $\dfrac{0.8448}{0.9744} \approx 0.8670$.

(a)

1. The frequencies add to 40: $34 + m + n = 40$, so $m + n = 6$.
2. Add up $fx$: $2 + 6 + 3m + 32 + 50 + 30 + 21 + 24 + 9n = 165 + 3m + 9n$.
3. Mean: $\dfrac{165 + 3m + 9n}{40} = 4.725$, so $165 + 3m + 9n = 189$ and $m + 3n = 8$.
4. Subtract $m + n = 6$: $2n = 2$, so $n = 1$ and $m = 5$.

(b)

1. Scored more than 5 (marks 6 to 9): $5 + 3 + 3 + 1 = 12$ of the 40.
2. Probability $= \dfrac{12}{40} = \dfrac{3}{10}$.

1. Choosing 3 from 16: ${}^{16}C_3 = \dfrac{16 \times 15 \times 14}{6}$
   $= 560$ ways.

(a)

1. All boys: ${}^{10}C_3 = 120$ ways, so the probability is $\dfrac{120}{560} = \dfrac{3}{14}$.

(b)

1. 2 girls and 1 boy: ${}^6C_2 \times {}^{10}C_1 = 15 \times 10$
   $= 150$ ways, so $\dfrac{150}{560} = \dfrac{15}{56}$.

(c)

1. The only committee with no boy is 3 girls: ${}^6C_3 = 20$ ways.
2. $P(\text{at least one boy}) = 1 - \dfrac{20}{560}$
   $= \dfrac{540}{560}$
   $= \dfrac{27}{28}$.

(a)

1. Write $\mathbf r = \alpha\mathbf p + \beta\mathbf q$: $\begin{pmatrix} 5 \\ 17 \end{pmatrix} = \begin{pmatrix} 3\alpha + 2\beta \\ 5\alpha - \beta \end{pmatrix}$.
2. Compare components: $3\alpha + 2\beta = 5$ and $5\alpha - \beta = 17$.
3. Double the second and add: $13\alpha = 39$, so $\alpha = 3$ and $\beta = 5(3) - 17 = -2$.
4. So $\mathbf r = 3\mathbf p - 2\mathbf q$.

(b)

1. $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}$
   $= \begin{pmatrix} -6 + 5 \\ -9 + 1 \end{pmatrix}$
   $= \begin{pmatrix} -1 \\ -8 \end{pmatrix}$.
2. $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}$
   $= \begin{pmatrix} -5 + 4 \\ -1 - 7 \end{pmatrix}$
   $= \begin{pmatrix} -1 \\ -8 \end{pmatrix}$
   $= \overrightarrow{BC}$, so $AD$ is equal and parallel to $BC$.
3. $\overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD}$
   $= \begin{pmatrix} -6 + 1 \\ -9 + 8 \end{pmatrix}$
   $= \begin{pmatrix} -5 \\ -1 \end{pmatrix}$
   $= \overrightarrow{AB}$, so $AB$ is equal and parallel to $DC$.
4. Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

1. Bearings are measured clockwise from north, so a force $(F, \theta)$ has east part $F\sin\theta$ and north part $F\cos\theta$.
2. East: $30\sin30^\circ + 40\sin60^\circ = 15 + 20\sqrt3$
   $= 49.641$ N.
3. North: $30\cos30^\circ + 40\cos60^\circ = 15\sqrt3 + 20$
   $= 45.981$ N.

(a)

1. Magnitude: $\sqrt{49.641^2 + 45.981^2} = 67.66$ N.

(b)

1. Angle east of north: $\tan^{-1}\dfrac{49.641}{45.981} = 47.2^\circ$, so the resultant acts on a bearing of $047^\circ$.

(c)

1. $F = ma$: $a = \dfrac{67.66}{20} = 3.383$ m s$^{-2}$.

(a)

1. Distance of the first body after $t$ s: $s_1 = 8t + \frac12(20)t^2 = 8t + 10t^2$.
2. Distance of the second body: $s_2 = 22t + \frac12(18)t^2 = 22t + 9t^2$.
3. The first body is 15 m ahead: $s_1 - s_2 = 15$, so $t^2 - 14t = 15$.
4. Rearrange and factorise: $t^2 - 14t - 15 = 0$, i.e. $(t - 15)(t + 1) = 0$.
5. Time is positive, so $t = 15$ s.

(b)

1. Take the 25 g ball's direction as positive.
2. Momentum before: $25 \times 8 + 15 \times (-3) = 155$ g m s$^{-1}$.
3. Momentum after: $25 \times 5 + 15v = 125 + 15v$.
4. Momentum is conserved: $125 + 15v = 155$, so $v = 2$ m s$^{-1}$.
5. Change in momentum of the 15 g ball: $15(2 - (-3)) = 75$ g m s$^{-1}$ ($0.075$ kg m s$^{-1}$).