WAEC · 2011 · Nov/Dec · Further Maths · Paper 2

1. Use $\cos 2\theta = 1 - 2\sin^2\theta$, so every term is in $\sin\theta$: $\sin\theta + 1 - 2\sin^2\theta = 0$.
2. Rearrange: $2\sin^2\theta - \sin\theta - 1 = 0$.
3. Factorise: $(2\sin\theta + 1)(\sin\theta - 1) = 0$, so $\sin\theta = 1$ or $\sin\theta = -\frac12$.
4. $\sin\theta = 1$: $\theta = 90^\circ$.
5. $\sin\theta = -\frac12$: the reference angle is $30^\circ$, and sine is negative in the third and fourth quadrants, so $\theta = 210^\circ$ or $330^\circ$.
6. The truth set is $\{90^\circ, 210^\circ, 330^\circ\}$.

1. Rearrange: $2y = -3x + 4$, so $y = -\frac32x + 2$, with gradient $-\frac32$.
2. The perpendicular gradient is $-1 \div \left(-\frac32\right) = \frac23$.
3. $y + 2 = \frac23(x - 3)$.
4. Multiply by 3: $3y + 6 = 2x - 6$.
5. So $2x - 3y - 12 = 0$.

1. First equation: $2\log y = \log y^2$, so the left side is $\log[(x - 1)y^2]$.
2. And $2\log 3 = \log 9$.
3. So $(x - 1)y^2 = 9$.
4. Second equation: $\log x + \log y = \log(xy)$, so $xy = 6$, which gives $x = \frac6y$.
5. Substitute into the first: $\left(\frac6y - 1\right)y^2 = 9$.
6. Multiply out: $6y - y^2 = 9$, so $y^2 - 6y + 9 = 0$.
7. Factorise: $(y - 3)^2 = 0$, so $y = 3$.
8. Then $x = \frac63 = 2$.
9. Check: $\log 1 + 2\log 3 = 2\log 3$ ✓ and $\log 2 + \log 3 = \log 6$ ✓.
10. So $x = 2$, $y = 3$.

1. The common ratio is the second term divided by the first: $r = \dfrac{3 - 2\sqrt2}{\sqrt2 - 1}$.
2. Notice that $(\sqrt2 - 1)^2 = 2 - 2\sqrt2 + 1$
   $= 3 - 2\sqrt2$.
3. So the second term is the first term squared, and $r = \sqrt2 - 1$.
4. The third term is the second term times $r$: $T_3 = (3 - 2\sqrt2)(\sqrt2 - 1)$.
5. Expand: $3\sqrt2 - 3 - 2 \times 2 + 2\sqrt2 = 3\sqrt2 - 3 - 4 + 2\sqrt2$.
6. Collect the terms: $T_3 = 5\sqrt2 - 7$.

(a)

1. Mutually exclusive: $P(X \cup Y) = P(X) + P(Y)$, so $P(Y) = \frac{11}{15} - \frac13$
   $= \frac{6}{15}$
   $= \frac25$.

(b)

1. Independent: $P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)$, so $\frac{11}{15} = \frac13 + P(Y) - \frac13P(Y)$.
2. Collect: $\frac23P(Y) = \frac{11}{15} - \frac{5}{15}$
   $= \frac{6}{15}$.
3. So $P(Y) = \frac{6}{15} \times \frac32$
   $= \frac35$.

(a)

1. Total marks: $12 \times 60 + 20 \times 50 + xy = 720 + 1000 + xy$
   $= 1720 + xy$.
2. Number of students: $12 + 20 + y = 32 + y$.
3. So the mean is $\dfrac{1720 + xy}{32 + y}$.

(b)

1. Red then blue: $\frac47 \times \frac36 = \frac{12}{42}$.
2. Blue then red: $\frac37 \times \frac46 = \frac{12}{42}$.
3. Add: $\dfrac{24}{42} = \dfrac47$.

1. $\mathbf n + 5\mathbf s = \begin{pmatrix} -12 + 5 \\ 5 - 5 \end{pmatrix}$
   $= \begin{pmatrix} -7 \\ 0 \end{pmatrix}$, of length 7.
2. Unit vector: $\frac17\begin{pmatrix} -7 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$.
3. $\mathbf q = 35\begin{pmatrix} -1 \\ 0 \end{pmatrix}$
   $= \begin{pmatrix} -35 \\ 0 \end{pmatrix}$.

(a)

1. Keep the units in hours: $15\text{ min} = \frac14\text{ h}$.
2. $v = u + at$: $50 = 20 + \frac14a$, so $a = 120\text{ km h}^{-2}$.

(b)

1. At the top $v = 0$: $0 = 40^2 - 2(10)s$.
2. $20s = 1600$, so $s = 80\text{ m}$.

(a)

1. Four strips from 1 to 5, so $h = 1$.
2. Work out $y = 2x + \dfrac{8}{x^2}$ at each ordinate.
3. $x = 1$: $10$.
4. $x = 2$: $6$.
5. $x = 3$: $6 + 0.8889 = 6.8889$.
6. $x = 4$: $8.5$.
7. $x = 5$: $10.32$.
8. Trapezium rule: $\frac12h[(\text{first} + \text{last}) + 2(\text{the rest})]$.
9. $\frac12[(10 + 10.32) + 2(6 + 6.8889 + 8.5)] = \frac12[20.32 + 42.7778]$
   $= 31.5489$.
10. So the integral is about $31.55$.

(b)(i)

1. $f$ acts first: $g \circ f(x) = g(x^2 + 1)$
   $= \dfrac{3(x^2 + 1) + 1}{(x^2 + 1) - 1}$.
2. Simplify the top and bottom: $\dfrac{3x^2 + 4}{x^2}$.
3. Divide each term by $x^2$: $g \circ f(x) = 3 + \dfrac{4}{x^2}$.

(ii)

1. $g \circ f(2) = 3 + \dfrac44 = 4$.
2. Check: $f(2) = 5$ and $g(5) = \dfrac{16}{4} = 4$ ✓.

(a)

1. Integrate the gradient: $y = x^2 - x^3 + c$.
2. $(1, 2)$ lies on the curve: $2 = 1 - 1 + c$, so $c = 2$ and $y = x^2 - x^3 + 2$.

(b)(i)

1. The curve cuts the $x$-axis where $x^2 - 1 = 0$: $x = -1$ or $x = 1$.
2. The gradient is $\dfrac{dy}{dx} = 2x$.
3. At $(-1, 0)$ the tangent's gradient is $-2$, so the normal's is $\frac12$: $y = \frac12(x + 1)$, that is $x - 2y + 1 = 0$.
4. At $(1, 0)$ the tangent's gradient is $2$, so the normal's is $-\frac12$: $y = -\frac12(x - 1)$, that is $x + 2y - 1 = 0$.

(ii)

1. Add the two equations: $2x = 0$, so $x = 0$.
2. Then $2y = 1$, so $y = \frac12$.
3. They meet at $\left(0, \frac12\right)$.

(a)

1. By Cramer's rule, first find $\Delta = \begin{vmatrix} 3 & 5 & -4 \\ 6 & 3 & -5 \\ -2 & 2 & 1 \end{vmatrix}$.
2. Expand along the top row: $\Delta = 3(3 + 10) - 5(6 - 10) - 4(12 + 6)$
   $= 39 + 20 - 72$
   $= -13$.
3. Replace the $x$ column by the right-hand sides: $\Delta_x = -5(3 + 10) - 5(26 - 55) - 4(52 + 33)$
   $= -65 + 145 - 340$
   $= -260$.
4. Replace the $y$ column: $\Delta_y = 3(26 - 55) + 5(6 - 10) - 4(-66 + 52)$
   $= -87 - 20 + 56$
   $= -51$.
5. Replace the $z$ column: $\Delta_z = 3(-33 - 52) - 5(-66 + 52) - 5(12 + 6)$
   $= -255 + 70 - 90$
   $= -275$.
6. So $x = \dfrac{-260}{-13} = 20$, $y = \dfrac{-51}{-13}$
   $= \dfrac{51}{13}$
   $\approx 3.92$ and $z = \dfrac{-275}{-13}$
   $= \dfrac{275}{13}$
   $\approx 21.15$.
7. Check in the third equation: $-40 + \dfrac{102}{13} + \dfrac{275}{13} = -40 + 29$
   $= -11$ ✓.

(b)

1. Factorise the bottom: $x^2 + x - 6 = (x - 2)(x + 3)$.
2. Write $\dfrac{A}{x - 2} + \dfrac{B}{x + 3}$ and multiply through: $3 - 4x = A(x + 3) + B(x - 2)$.
3. Put $x = 2$: $-5 = 5A$, so $A = -1$.
4. Put $x = -3$: $15 = -5B$, so $B = -3$.
5. So $g(x) = -\dfrac{1}{x - 2} - \dfrac{3}{x + 3}$.

(a)

1. The curve meets the $x$-axis where $x^2 - 4 = 0$: $x = -2$ and $x = 2$.
2. $\displaystyle\int_{-2}^2 (x^2 - 4)\,dx = \left[\frac{x^3}{3} - 4x\right]_{-2}^2$
   $= \left(\frac83 - 8\right) - \left(-\frac83 + 8\right)$
   $= -\frac{32}{3}$.
3. The region is below the axis, so the integral is negative.
4. The area is $\dfrac{32}{3} \approx 10.67$ square units.

(b)

1. $V = \pi\displaystyle\int_{-2}^2 (x^2 - 4)^2\,dx$
   $= \pi\int_{-2}^2 (x^4 - 8x^2 + 16)\,dx$.
2. $= \pi\left[\dfrac{x^5}{5} - \dfrac{8x^3}{3} + 16x\right]_{-2}^2$
   $= 2\pi\left(\dfrac{32}{5} - \dfrac{64}{3} + 32\right)$
   $= \dfrac{512}{15}\pi$.
3. With $\pi = \frac{22}{7}$: $V = \dfrac{512}{15} \times \dfrac{22}{7}$
   $\approx 107.28$ cubic units.

(a)(i)

1. There are $6 + 8 + 5 = 19$ people: $\,{}^{19}C_5 = 11\,628$ committees.

(ii)

1. At most 2 Ghanaians means 0, 1 or 2.
2. The rest come from the 13 others.
3. $\,{}^6C_0 \times {}^{13}C_5 = 1287$, $\,{}^6C_1 \times {}^{13}C_4 = 6 \times 715$
   $= 4290$ and $\,{}^6C_2 \times {}^{13}C_3 = 15 \times 286$
   $= 4290$.
4. Add: $1287 + 4290 + 4290 = 9867$.

(iii)

1. 1 Nigerian from 8, and the other 4 from the 11 non-Nigerians: $8 \times 330 = 2640$.

(b)

1. There are 7 good articles.
2. Without replacement: $\dfrac{7}{12} \times \dfrac{6}{11} \times \dfrac{5}{10} = \dfrac{210}{1320}$
   $= \dfrac{7}{44}$
   $\approx 0.159$.

(a)

1. Plot the eight points $(3, 143), (5, 95), \ldots, (26, 69)$.

(b)

1. $\bar x = \dfrac{110}{8} = 13.75$ and $\bar y = \dfrac{797}{8} = 99.625$.
2. Draw the line of best fit through $(13.75, 99.6)$, following the downward trend of the points.

(c)(i)

1. Read across from $y = 80$ to the line and down: about day 22.

(ii)

1. Read up from $x = 25$ to the line and across: about 73 cars.

(a)

1. $X \sim B(6, 0.25)$, with $q = 0.75$.

(i)

1. More than three: $P(4) + P(5) + P(6) = 0.032959 + 0.004395 + 0.000244$
   $= 0.0376$.

(ii)

1. At most 2: $P(0) + P(1) + P(2) = 0.177979 + 0.355957 + 0.296631$
   $= 0.8306$.

(b)

1. There are $\binom52 = 10$ pairs of distinct numbers.

(i)

1. A sum of 8: $\{2, 6\}$ and $\{3, 5\}$, so $P = \frac{2}{10} = \frac15$.

(ii)

1. One divides the other: $\{2, 4\}$, $\{2, 6\}$ and $\{3, 6\}$, so $P = \frac{3}{10}$.

(a)(i)

1. $\overrightarrow{PQ} = (8 - 5)\mathbf i + (-1 - 3)\mathbf j$
   $= 3\mathbf i - 4\mathbf j$.
2. $\overrightarrow{QR} = (11 - 8)\mathbf i + (-5 + 1)\mathbf j$
   $= 3\mathbf i - 4\mathbf j$.
3. They are equal, so parallel, and they share $Q$: $P$, $Q$ and $R$ are collinear.

(ii)

1. Match the parts: $5k_1 + 11k_2 = 37$ and $3k_1 - 5k_2 = -1$.
2. From the second, $k_1 = \dfrac{5k_2 - 1}{3}$.
3. Substitute: $\dfrac{25k_2 - 5}{3} + 11k_2 = 37$.
4. Multiply by 3: $58k_2 = 116$, so $k_2 = 2$ and $k_1 = \frac{10 - 1}{3} = 3$.

(b)

1. $\mathbf m \cdot \mathbf n = 18 - 16 = 2$, $|\mathbf m| = 5$ and $|\mathbf n| = \sqrt{52}$.
2. $\cos\theta = \dfrac{2}{5\sqrt{52}}$
   $\approx 0.0555$, so $\theta \approx 87^\circ$.

(a)(i)

1. $s = ut - \frac12gt^2 = 25(4) - 5(16) = 20$, so it is $20\text{ m}$ above the ground.

(ii)

1. At the top $v = 0$: $0 = 25^2 - 20s$, so $s = \dfrac{625}{20} = 31.25\text{ m}$.

(iii)

1. $0 = 25 - 10t$, so $t = 2.5\text{ s}$.

(iv)

1. $30 = 25t - 5t^2$, so $t^2 - 5t + 6 = 0$.
2. $(t - 2)(t - 3) = 0$: at $t = 2\text{ s}$ (going up) and $t = 3\text{ s}$ (coming down).

(b)

1. $Ft = m(v - u)$: $0.5F = 3(4.5 - 2.5) = 6$, so $F = 12\text{ N}$.

(a)

1. Measured from $D$: $P$ is 45 cm, $C$ is 35 cm and the centre is 15 cm on one side.
2. $Q$ is 15 cm on the other.
3. Moments about $D$: $35R_C + 15 \times 15 = 20 \times 15 + 10 \times 45$.
4. $35R_C = 300 + 450 - 225 = 525$, so $R_C = 15\text{ N}$.
5. Up = down: $R_C + R_D = 10 + 20 + 15 = 45$, so $R_D = 30\text{ N}$.

(b)

1. The forces are at right angles: $|\mathbf F| = \sqrt{3^2 + 4^2}$
   $= 5\text{ N}$.
2. $a = \dfrac{F}{m}$
   $= \dfrac55$
   $= 1\text{ m s}^{-2}$.