WAEC · 2008 · Nov/Dec · Further Maths · Paper 2

(a)

1. Let $y = \dfrac{x + 1}{x - 2}$ and make $x$ the subject.
2. Multiply up: $y(x - 2) = x + 1$.
3. Expand: $xy - 2y = x + 1$.
4. Collect the $x$ terms: $xy - x = 2y + 1$, so $x(y - 1) = 2y + 1$.
5. Divide: $x = \dfrac{2y + 1}{y - 1}$.
6. So $f^{-1}(x) = \dfrac{2x + 1}{x - 1}$, $x \neq 1$.

(b)

1. First find $f^{-1}(4)$: put $x = 4$ into $\dfrac{2x + 1}{x - 1}$ to get $\dfrac93$, which is 3.
2. Then $g(3) = \dfrac{3k + 3}{3} = k + 1$.
3. Set $k + 1 = 6$, so $k = 5$.

1. From $2x^2 + 5x - 6 = 0$: $\alpha + \beta = -\frac52$ and $\alpha\beta = -\frac62 = -3$.
2. New sum: $(\alpha - 2) + (\beta - 2) = (\alpha + \beta) - 4$
   $= -\frac52 - 4$
   $= -\frac{13}{2}$.
3. New product: $(\alpha - 2)(\beta - 2) = \alpha\beta - 2(\alpha + \beta) + 4$.
4. Substitute: $-3 - 2\left(-\frac52\right) + 4$ is $-3 + 5 + 4$, which is 6.
5. The equation is $x^2 - (\text{sum})x + \text{product} = 0$: $x^2 + \frac{13}{2}x + 6 = 0$.
6. Multiply by 2: $2x^2 + 13x + 12 = 0$.

1. Write $px^4 + qx^3 - 8x^2 + 6 = (x^2 - 1)Q(x) + 2x + 1$.
2. At $x = \pm1$, $x^2 - 1 = 0$, so only the remainder is left.
3. Put $x = 1$: $p + q - 8 + 6 = 2 + 1$, so $p + q = 5$.
4. Put $x = -1$: $p - q - 8 + 6 = -2 + 1$, so $p - q = 1$.
5. Add the equations: $2p = 6$, so $p = 3$.
6. Then $q = 5 - 3 = 2$.

1. Let $y = 2^x$.
2. Then $2^{x + 1} = 2y$ and $2^{-x} = \frac1y$.
3. The equation becomes $2y - \frac4y - 7 = 0$.
4. Multiply by $y$: $2y^2 - 7y - 4 = 0$.
5. Factorise: $(2y + 1)(y - 4) = 0$, so $y = -\frac12$ or $y = 4$.
6. $2^x$ is always positive, so reject $y = -\frac12$.
7. $2^x = 4 = 2^2$, so $x = 2$.

(a)

1. There are $n = 8$ deviations.
2. Their sum is $-5 - 3 - 1 + 0 + 1 + 3 + 5 + 7 = 7$.
3. Mean deviation: $\bar d = \frac78 = 0.875$.
4. Mean $= 55 + \bar d = 55.875$.

(b)

1. Square the deviations: $25, 9, 1, 0, 1, 9, 25, 49$.
2. Their sum is $\sum d^2 = 119$.
3. Variance $= \dfrac{\sum d^2}{n} - \bar d^{\,2}$
   $= \dfrac{119}{8} - 0.875^2$.
4. $14.875 - 0.765625 = 14.109375$, so the variance is $14.11$ to two decimal places.

(a)

1. $\overrightarrow{XY} = Y - X$
   $= \begin{pmatrix} 6 - 4 \\ 2 - 0 \end{pmatrix}$
   $= \begin{pmatrix} 2 \\ 2 \end{pmatrix}$.
2. $\overrightarrow{YZ} = Z - Y$
   $= \begin{pmatrix} -2 - 6 \\ 1 - 2 \end{pmatrix}$
   $= \begin{pmatrix} -8 \\ -1 \end{pmatrix}$.
3. $2\overrightarrow{XY} + 3\overrightarrow{YZ} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} -24 \\ -3 \end{pmatrix}$
   $= \begin{pmatrix} -20 \\ 1 \end{pmatrix}$.

(b)

1. $\overrightarrow{XZ} = Z - X$
   $= \begin{pmatrix} -6 \\ 1 \end{pmatrix}$.
2. Its magnitude is $|\overrightarrow{XZ}| = \sqrt{36 + 1}$
   $= \sqrt{37}$.
3. Divide the vector by its magnitude: the unit vector is $\dfrac{1}{\sqrt{37}}(-6\mathbf i + \mathbf j)$.
4. Rationalise: $\dfrac{\sqrt{37}}{37}(-6\mathbf i + \mathbf j)$.

(a)

1. $\dfrac{dy}{dx} = 3x^2 - 2x + 1$, so integrate: $y = x^3 - x^2 + x + c$.
2. The curve passes through $(1, 3)$: $3 = 1 - 1 + 1 + c$, so $c = 2$.
3. The equation is $y = x^3 - x^2 + x + 2$.

(b)(i)

1. Use $S_n = \frac{n}{2}(a + l)$: $255 = \frac{n}{2}(3 + 48)$.
2. So $255 = \frac{51n}{2}$, giving $n = \frac{510}{51} = 10$.

(ii)

1. Find $d$ from the 10th term: $3 + 9d = 48$, so $d = 5$.
2. The $r$th term is $3 + 5(r - 1)$.
3. It exceeds 149 when $5(r - 1) > 146$, so $r - 1 > 29.2$.
4. The smallest whole number is $r = 31$ (the 30th term is 148 and the 31st is 153).

(a)

1. Complete the squares: $(x - 2)^2 - 4 + (y + 1)^2 - 1 + c = 0$.
2. So $(x - 2)^2 + (y + 1)^2 = 5 - c$, and the radius squared is $5 - c$.
3. $r^2 = (2\sqrt3)^2 = 12$, so $5 - c = 12$ and $c = -7$.

(b)

1. For $T$: $\frac{dy}{dx} = 2x + 6$, which is $8$ at $x = 1$.
2. Tangent through $(1, 3)$ with gradient 8: $y - 3 = 8(x - 1)$, so $y = 8x - 5$.
3. For $N$: $\frac{dy}{dx} = 2x - 6$, which is $2$ at $x = 4$.
4. The normal has gradient $-\frac12$.
5. Normal through $(4, 10)$: $y - 10 = -\frac12(x - 4)$, so $x + 2y = 24$.
6. Substitute $y = 8x - 5$: $x + 16x - 10 = 24$, so $17x = 34$ and $x = 2$.
7. Then $y = 8(2) - 5 = 11$.
8. $T$ and $N$ meet at $(2, 11)$.

(a)(i)

1. Differentiate each term with respect to $x$, using the product rule on $4xy$: $2x + 4y + 4x\dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0$.
2. Collect the $\dfrac{dy}{dx}$ terms: $(4x - 2y)\dfrac{dy}{dx} = -2x - 4y$.
3. Divide and simplify by $-2$: $\dfrac{dy}{dx} = \dfrac{x + 2y}{y - 2x}$.

(ii)

1. Put $x = -2$, $y = 4$: $\dfrac{-2 + 8}{4 + 4} = \dfrac68$
   $= \dfrac34$.

(b)

1. Factorise: $x^2 - 9x + 18 = (x - 6)(x - 3)$.
2. Write $\dfrac{x + 3}{(x - 6)(x - 3)} = \dfrac{A}{x - 6} + \dfrac{B}{x - 3}$.
3. Multiply up: $x + 3 = A(x - 3) + B(x - 6)$.
4. Put $x = 6$: $9 = 3A$, so $A = 3$.
5. Put $x = 3$: $6 = -3B$, so $B = -2$.
6. So $\dfrac{x + 3}{x^2 - 9x + 18} = \dfrac{3}{x - 6} - \dfrac{2}{x - 3}$.

(a)

1. $M$ times a $2 \times 2$ matrix gives a $1 \times 2$ matrix, so $M$ is $1 \times 2$.
2. Let $M = \begin{pmatrix} a & b \end{pmatrix}$.
3. Multiply: $\begin{pmatrix} 2a + 8b & 9a + 6b \end{pmatrix} = \begin{pmatrix} 14 & 3 \end{pmatrix}$.
4. So $2a + 8b = 14$, which is $a + 4b = 7$, and $9a + 6b = 3$, which is $3a + 2b = 1$.
5. From the first, $a = 7 - 4b$.
6. Substitute: $21 - 12b + 2b = 1$, so $b = 2$.
7. Then $a = 7 - 8 = -1$, so $M = \begin{pmatrix} -1 & 2 \end{pmatrix}$.

(b)(i)

1. At $x = 0$: $\frac{1}{\sqrt2} = 0.7071$.
2. At $0.5$: $\frac{1}{\sqrt{2.25}} = 0.6667$.
3. At $0.75$: $\frac{1}{\sqrt{2.5625}} = 0.6247$.
4. At $1$: $\frac{1}{\sqrt3} = 0.5774$.

(ii)

1. Trapezium rule with $h = 0.25$: $\frac{h}{2}\left[y_0 + y_4 + 2(y_1 + y_2 + y_3)\right]$.
2. Substitute: $0.125\left[0.7071 + 0.5774 + 2(0.6963 + 0.6667 + 0.6247)\right] = 0.125 \times 5.2599$.
3. $0.6575$, which is $0.657$ to three significant figures.

(a)

1. Plot the 8 points with $x$ (Algebra) across and $y$ (Statistics) up.

(b)

1. $\sum x = 10 + 24 + 30 + 35 + 48 + 59 + 68 + 70$
   $= 344$, so $\bar x = \frac{344}{8} = 43$.
2. $\sum y = 30 + 47 + 44 + 71 + 60 + 89 + 97 + 74$
   $= 512$, so $\bar y = \frac{512}{8} = 64$.
3. Plot $(43, 64)$.

(c)

1. Draw a straight line through $(43, 64)$ that follows the trend, with the points spread evenly on both sides.

(d)(i)

1. Read a second point on the line, for example $(78, 97)$.
2. Gradient $= \frac{97 - 64}{78 - 43}$
   $= \frac{33}{35}$
   $\approx 0.94$.
3. Use $y - 64 = 0.94(x - 43)$: the line is about $y = 0.94x + 23.5$.

(ii)

1. Read up from $x = 50$ to the line and across: $y \approx 0.94(50) + 23.5 \approx 71$.
2. The Statistics mark is about 71.

(a)(i)

1. Eight boys in a row: $8! = 40\,320$ ways.

(ii)

1. First count the ways with the two boys together: treat them as one unit, giving $7!$ arrangements, and they can swap places in $2$ ways.
2. Together: $2 \times 7! = 10\,080$ ways.
3. Not together: $8! - 2 \times 7! = 40\,320 - 10\,080$
   $= 30\,240$ ways.

(b)

1. Number of ways to choose any 4 of the 10: $^{10}C_4 = 210$.

(i)

1. 3 men and 1 woman: $^7C_3 \times {^3C_1} = 35 \times 3 = 105$.
2. Probability $= \frac{105}{210} = \frac12$.

(ii)

1. At least 2 women means 2 or 3 women. 2 women: $^7C_2 \times {^3C_2} = 21 \times 3 = 63$. 3 women: $^7C_1 \times {^3C_3} = 7$.
2. Probability $= \frac{63 + 7}{210}$
   $= \frac{70}{210}$
   $= \frac13$.

(a)

1. $^nC_r = \dfrac{n!}{r!(n - r)!}$ and $^nC_{r-1} = \dfrac{n!}{(r - 1)!(n - r + 1)!}$.
2. Common denominator $r!(n - r + 1)!$: multiply the first fraction by $\frac{n - r + 1}{n - r + 1}$ and the second by $\frac{r}{r}$.
3. Add the tops: $n!(n - r + 1) + n!\,r = n!(n + 1)$
   $= (n + 1)!$.
4. So $^nC_r + {^nC_{r-1}} = \dfrac{(n + 1)!}{r!(n + 1 - r)!}$
   $= {^{n+1}C_r}$.

(b)(i)

1. Binomial with $n = 6$, $p = 0.04$, $q = 0.96$: $P(2) = {^6C_2}(0.04)^2(0.96)^4$.
2. $15 \times 0.0016 \times 0.84935 \approx 0.0204$.

(ii)

1. "At most 3" is easier as $1 - P(4, 5 \text{ or } 6)$.
2. $P(4) = 15(0.04)^4(0.96)^2$
   $\approx 0.0000354$.
3. $P(5) = 6(0.04)^5(0.96) \approx 0.0000006$.
4. $P(6) = (0.04)^6 \approx 0$.
5. So $P(\text{at most } 3) \approx 1 - 0.0000360$
   $= 0.99996$.

(a)(i)

1. Use $v = u + at$ with $u = 32$, $a = -0.7$, $t = 8$: $v = 32 - 5.6 = 26.4\text{ m s}^{-1}$.

(ii)

1. Use $s = ut + \frac12at^2$: $s = 32(8) - \frac12(0.7)(64)$
   $= 256 - 22.4$
   $= 233.6\text{ m}$.

(b)(i)

1. Take $P$'s original direction as positive.
2. Momentum before: $60(5) + 50(-16) = 300 - 800 = -500$.
3. Momentum after: $60(4) + 50v = 240 + 50v$.
4. Momentum is conserved: $240 + 50v = -500$, so $50v = -740$ and $v = -14.8$.
5. So $Q$ moves at $14.8\text{ m s}^{-1}$, still in its original direction.

(ii)

1. After the collision $P$ has $u = 4$ and $a = -0.25$.
2. It stops when $v = 0$: $0 = 4 - 0.25t$.
3. So $t = \frac{4}{0.25} = 16$ seconds.

(a)

1. $\overrightarrow{AB} = \mathbf b - \mathbf a$
   $= -2\mathbf i - 3\mathbf j$ and $\overrightarrow{BC} = \mathbf c - \mathbf b$
   $= 4\mathbf i - \mathbf j$.
2. Dot product: $(-2)(4) + (-3)(-1) = -8 + 3 = -5$.
3. Magnitudes: $|\overrightarrow{AB}| = \sqrt{13}$ and $|\overrightarrow{BC}| = \sqrt{17}$.
4. $\cos\theta = \dfrac{-5}{\sqrt{13}\sqrt{17}}$
   $\approx -0.3363$, so $\theta \approx 109.7^\circ$.
5. The acute angle is $180^\circ - 109.7^\circ = 70.3^\circ$.

(b)

1. The weight is $5 \times 10 = 50\text{ N}$.
2. Let $T_1$ be the tension in the string at $30^\circ$ and $T_2$ the one at $45^\circ$.
3. Horizontal components balance: $T_1\cos30^\circ = T_2\cos45^\circ$, so $T_2 = \dfrac{T_1\cos30^\circ}{\cos45^\circ}$.
4. Vertical components balance the weight: $T_1\sin30^\circ + T_2\sin45^\circ = 50$.
5. Substitute, using $\cos45^\circ = \sin45^\circ$: $T_1(\sin30^\circ + \cos30^\circ) = 50$, so $T_1 = \dfrac{50}{0.5 + 0.8660}$
   $\approx 36.6\text{ N}$.
6. Then $T_2 = \dfrac{36.60 \times 0.8660}{0.7071}$
   $\approx 44.8\text{ N}$.