WAEC · 2008 · May/June · Further Maths · Paper 2

1. Put $x = -2$ into $f(x) = px^2 + qx + 2$ and use $f(-2) = 0$: $4p - 2q + 2 = 0$.
2. Divide by 2: $2p - q = -1$.
3. Put $x = 1$ and use $f(1) = -3$: $p + q + 2 = -3$, so $p + q = -5$.
4. Add the two equations to remove $q$: $3p = -6$, so $p = -2$.
5. Then $q = -5 - p$, which gives $q = -3$.
6. So $f(x) = -2x^2 - 3x + 2$.
7. Put $x = -4$: $-2(16) - 3(-4) + 2 = -32 + 12 + 2$.
8. $f(-4) = -18$.

1. Use $\sin^2 A + \cos^2 A = 1$: $\cos^2 A = 1 - \frac{9}{25}$
   $= \frac{16}{25}$.
2. $A$ is obtuse, so its cosine is negative: $\cos A = -\frac45$.
3. For $B$: $\sin^2 B = 1 - \frac{225}{289}$
   $= \frac{64}{289}$.
4. $B$ is acute, so $\sin B = \frac{8}{17}$.
5. Use $\cos(A + B) = \cos A \cos B - \sin A \sin B$.
6. Substitute: $\left(-\frac45\right)\left(\frac{15}{17}\right) - \left(\frac35\right)\left(\frac{8}{17}\right) = -\frac{60}{85} - \frac{24}{85}$.
7. $\cos(A + B) = -\frac{84}{85}$.

1. Let $y = x^3 + 2x$.
2. Increase $x$ by a small amount $\Delta x$: $y + \Delta y = (x + \Delta x)^3 + 2(x + \Delta x)$.
3. Expand: $y + \Delta y = x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 + 2x + 2\Delta x$.
4. Subtract $y = x^3 + 2x$: $\Delta y = 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 + 2\Delta x$.
5. Divide by $\Delta x$: $\dfrac{\Delta y}{\Delta x} = 3x^2 + 3x\Delta x + (\Delta x)^2 + 2$.
6. Let $\Delta x \to 0$: the terms with $\Delta x$ vanish, so $\dfrac{dy}{dx} = 3x^2 + 2$.

1. Gradient of $PR$: $\dfrac{5 - 3}{k - (-1)} = \dfrac{2}{k + 1}$.
2. Gradient of $QR$: $\dfrac{5 - 4}{k - (-4)} = \dfrac{1}{k + 4}$.
3. $\angle PRQ = 90^\circ$, so the lines are perpendicular and the product of the gradients is $-1$: $\dfrac{2}{(k + 1)(k + 4)} = -1$.
4. Multiply out: $2 = -(k^2 + 5k + 4)$, so $k^2 + 5k + 6 = 0$.
5. Factorise: $(k + 2)(k + 3) = 0$.
6. So $k = -2$ or $k = -3$.

(a)

1. Every guest chose yam, plantain or both, so the three regions add up to 25: $(2x + 1) + x + (x - 2)^2 = 25$.
2. Expand $(x - 2)^2 = x^2 - 4x + 4$: $x^2 - x + 5 = 25$.
3. Rearrange: $x^2 - x - 20 = 0$, which factorises as $(x - 5)(x + 4) = 0$.
4. A number of guests cannot be negative, so $x = 5$.

(b)

1. Only plantain: $2(5) + 1 = 11$ guests.
2. Only yam: $(5 - 2)^2 = 9$ guests.
3. So $11 + 9 = 20$ guests chose only one.
4. Probability $= \frac{20}{25} = \frac45$.

1. Find the class boundaries: $9.5, 14.5, 24.5, 29.5, 39.5, 44.5, 49.5$.
2. Find the class widths: $5, 10, 5, 10, 5, 5$.
3. They are not all equal.
4. With unequal widths, the height of a bar is the frequency density.
5. Take 5 marks as the standard width: height $= \dfrac{\text{frequency}}{\text{width}} \times 5$.
6. For 15–24: $\frac{30}{10} \times 5 = 15$.
7. For 30–39: $\frac{18}{10} \times 5 = 9$.
8. The other classes already have width 5, so their heights are their frequencies: 14, 22, 12 and 4.
9. Draw bars on the boundaries with heights 14, 15, 22, 9, 12, 4, no gaps, and label the axes "Marks" and "Frequency per 5 marks".

1. Use $s = ut + \frac12 at^2$ with $u = 50$, $a = 6$ and $s = 20$.
2. Substitute: $20 = 50t + 3t^2$.
3. Rearrange into a quadratic: $3t^2 + 50t - 20 = 0$.
4. Use the formula: $t = \dfrac{-50 \pm \sqrt{50^2 + 4(3)(20)}}{2(3)}$
   $= \dfrac{-50 \pm \sqrt{2740}}{6}$.
5. $\sqrt{2740} \approx 52.345$.
6. Time cannot be negative, so $t = \dfrac{2.345}{6} \approx 0.391$.
7. The car took about $0.4$ hours (about 23 minutes).

(a)

1. The magnitude is $\sqrt{x^2 + 15^2}$, so $x^2 + 225 = 17^2 = 289$.
2. Subtract 225: $x^2 = 64$.
3. Take both square roots: $x = 8$ or $x = -8$.

(b)

1. For $8\mathbf i + 15\mathbf j$: the force points up (north) and to the right (east).
2. Its angle from north is $\tan^{-1}\frac{8}{15}$.
3. $\tan^{-1}\frac{8}{15} \approx 28.07^\circ$, so the direction is $\text{N}28^\circ\text{E}$, a bearing of $028^\circ$.
4. For $-8\mathbf i + 15\mathbf j$: the force points north and to the left (west), at the same $28^\circ$ from north.
5. Its direction is $\text{N}28^\circ\text{W}$, a bearing of $360^\circ - 28^\circ = 332^\circ$.

(a)

1. $S_1 = \frac52 + \frac52 = 5$, so $T_1 = 5$.
2. $S_2 = 10 + 5 = 15$, so $T_2 = S_2 - S_1 = 10$.
3. $S_3 = 22.5 + 7.5 = 30$, so $T_3 = 30 - 15 = 15$.
4. $S_4 = 40 + 10 = 50$, so $T_4 = 50 - 30 = 20$.
5. The first four terms are $5, 10, 15, 20$: an AP with $a = 5$ and $d = 5$.
6. $n$th term: $T_n = a + (n - 1)d = 5 + 5(n - 1)$, which simplifies to $T_n = 5n$.

(b)(i)

1. Put $y = 0$ (the $x$-axis): $x^2 - 10x + 25 = 0$.
2. This is $(x - 5)^2 = 0$, a repeated root, so the circle meets the $x$-axis at one point only: it touches it.

(ii)

1. The repeated root is $x = 5$, so the point of contact is $(5, 0)$.

1. Factorise the denominator: $x^3 - 4x^2 + 3x = x(x^2 - 4x + 3)$
   $= x(x - 1)(x - 3)$.
2. Write $\dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x - 3}$.
3. Multiply through by the denominator: $2x^2 - 5x + 1 = A(x - 1)(x - 3) + Bx(x - 3) + Cx(x - 1)$.
4. Put $x = 0$: $1 = 3A$, so $A = \frac13$.
5. Put $x = 1$: $2 - 5 + 1 = B(1)(-2)$, so $-2 = -2B$ and $B = 1$.
6. Put $x = 3$: $18 - 15 + 1 = C(3)(2)$, so $4 = 6C$ and $C = \frac23$.
7. So the expression is $\dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)}$.

(a)(i)

1. An anticlockwise rotation through $\theta$ about the origin has matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.
2. With $\theta = 60^\circ$: $\cos 60^\circ = \frac12$ and $\sin 60^\circ = \frac{\sqrt3}{2}$, giving $\begin{pmatrix} \frac12 & -\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac12 \end{pmatrix}$.

(ii)

1. Multiply the matrix by $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$.
2. Top row: $\frac32 + \frac{5\sqrt3}{2} = \frac{3 + 5\sqrt3}{2}$.
3. Bottom row: $\frac{3\sqrt3}{2} - \frac52 = \frac{3\sqrt3 - 5}{2}$.
4. So $P_1\left(\frac{3 + 5\sqrt3}{2}, \frac{3\sqrt3 - 5}{2}\right)$, about $(5.83, 0.10)$.

(b)(i)

1. Read the coefficients of $x$ and $y$ row by row: $N = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix}$.

(ii)

1. $N^2 = N \times N$.
2. Top row: $(4 + 9,\ 6 - 3) = (13, 3)$.
3. Bottom row: $(6 - 3,\ 9 + 1) = (3, 10)$.
4. So $\begin{pmatrix} 13 & 3 \\ 3 & 10 \end{pmatrix} + a\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.
5. Top-right entries: $3 + 3a = 0$, so $a = -1$.
6. Top-left entries: $13 + 2a + b = 0$, so $13 - 2 + b = 0$ and $b = -11$.
7. Check with the bottom-right entries: $10 - a + b = 10 + 1 - 11 = 0$ ✓.
8. So $a = -1$ and $b = -11$.

(a)

1. Add the frequencies as you go: $50, 100, 140, 200, 300, 400, 450, 475, 490, 500$.
2. There are 500 students.
3. Pair each total with its upper class boundary: $9.5, 19.5, 29.5, \ldots, 99.5$.

(b)

1. Plot the points $(9.5, 50), (19.5, 100), \ldots, (99.5, 500)$, starting from $(-0.5, 0)$, and join them with a smooth curve.

(c)(i)

1. $Q_1$ is at the $\frac{500}{4} = 125$th student.
2. Read across from 125 to the curve and down: $Q_1 \approx 25.8$.
3. $Q_3$ is at the $375$th student.
4. Read across from 375: $Q_3 \approx 57.0$.
5. Semi-interquartile range $= \frac12(Q_3 - Q_1)$
   $= \frac12(57.0 - 25.8)$
   $\approx 15.6$.

(ii)

1. Students who failed scored below 37.
2. Read up from 37 to the curve and across: about 185 students failed.

(iii)

1. Read up from 60: about 402 students scored below 60.
2. Read up from 20: about 102 scored below 20.
3. So about $402 - 102 = 300$ scored between 20 and 60.
4. Probability $\approx \frac{300}{500} = 0.6$.

(a)(i)

1. Choose any 4 of the 12 teachers: $^{12}C_4 = \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$
   $= 495$.

(ii)

1. "At least 2 females" means 2, 3 or 4 females.
2. 2 females and 2 males: $^5C_2 \times {^7C_2} = 10 \times 21 = 210$.
3. 3 females and 1 male: $^5C_3 \times {^7C_1} = 10 \times 7 = 70$. 4 females: $^5C_4 = 5$.
4. Add: $210 + 70 + 5 = 285$ ways.

(b)(i)

1. Differences $d = X - Y$: $-2, 1, -1, 0, -1, 4, -1$.
2. Squares: $4, 1, 1, 0, 1, 16, 1$, so $\sum d^2 = 24$.
3. Use $\rho = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)}$ with $n = 7$: $\rho = 1 - \dfrac{6 \times 24}{7 \times 48}$
   $= 1 - \dfrac{144}{336}$.
4. $\rho \approx 1 - 0.429 = 0.571$, which is $0.6$ to one decimal place.

(ii)

1. $\rho = 0.6$ is positive and fairly close to 1: there is a fairly strong positive correlation, so the two judges broadly agree in their rankings.

(a)

1. In parallelogram $PQRM$, $\overrightarrow{MP} = \overrightarrow{RQ}$, so $\mathbf p - \mathbf m = \mathbf q - \mathbf r$.
2. Rearrange: $\mathbf m = \mathbf p + \mathbf r - \mathbf q$.
3. $\mathbf m = (4 - 5 - 1)\mathbf i + (-5 + 2 - 3)\mathbf j$
   $= -2\mathbf i - 6\mathbf j$.

(b)

1. $\overrightarrow{PM} = \mathbf m - \mathbf p$
   $= -6\mathbf i - \mathbf j$, so $|\overrightarrow{PM}| = \sqrt{36 + 1}$
   $= \sqrt{37}$.
2. $\overrightarrow{PQ} = \mathbf q - \mathbf p$
   $= -3\mathbf i + 8\mathbf j$, so $|\overrightarrow{PQ}| = \sqrt{9 + 64}$
   $= \sqrt{73}$.

(c)

1. Dot product: $\overrightarrow{PM} \cdot \overrightarrow{PQ} = (-6)(-3) + (-1)(8)$
   $= 10$.
2. $\cos\theta = \dfrac{10}{\sqrt{37}\sqrt{73}}$
   $\approx 0.1924$, so $\theta \approx 78.9^\circ$.

(d)

1. Area of a parallelogram $= |\overrightarrow{PM}||\overrightarrow{PQ}|\sin\theta$.
2. $\sqrt{37}\sqrt{73}\sin\theta = \sqrt{2701} \times 0.9813$
   $\approx 51$ square units.
3. (Exactly: $|(-6)(8) - (-1)(-3)| = 51$.)

(a)

1. On a smooth plane the only force along the slope is the component of the weight, $mg\sin30^\circ$, acting down the slope.
2. So the deceleration is $g\sin 30^\circ = 10 \times \frac12$
   $= 5\text{ m s}^{-2}$.
3. The object stops at the top, so work backwards from rest: $s = \frac12 at^2$ gives $15 = \frac12(5)t^2$.
4. $t^2 = 6$, so $t = \sqrt6 \approx 2.45$ seconds.

(b)(i)

1. Take the 10 N force along the $x$-axis.
2. The others make $60^\circ$, $30^\circ$, $-30^\circ$ and $-60^\circ$ with it.
3. Vertical components: $5\sin60^\circ + 5\sqrt3\sin30^\circ - 5\sqrt3\sin30^\circ - 5\sin60^\circ = 0$.
4. They cancel.
5. Horizontal components: $2(5\cos60^\circ) + 2(5\sqrt3\cos30^\circ) + 10 = 5 + 15 + 10$
   $= 30$.
6. The resultant is $30\text{ N}$ along the direction of the 10 N force.

(ii)

1. Use $F = ma$: $a = \frac{30}{5} = 6\text{ m s}^{-2}$.