WAEC · 2016 · Private · Further Maths · Paper 2

(a)

1. Write $8$ as a power of 2: $8^m = (2^3)^m = 2^{3m}$.
2. So the left side is $2^{3m} + 2^{3m} = 2 \times 2^{3m}$
   $= 2^{3m + 1}$.
3. Write the right side as a power of 2: $\frac14 = 2^{-2}$.
4. Compare the indices: $3m + 1 = -2$, so $3m = -3$ and $m = -1$.

(b)

1. Change the base to 10: $x = \log_4 15 = \dfrac{\log 15}{\log 4}$.
2. $x = \dfrac{1.1761}{0.6021} = 1.953$ to three decimal places.

1. $f(-2) = 0$: $m(-2)^2 + n(-2) + 2 = 0$, so $4m - 2n + 2 = 0$.
2. Divide by 2: $2m - n = -1$.
3. $f(1) = 3$: $m + n + 2 = 3$, so $m + n = 1$.
4. Add the two equations: $3m = 0$, so $m = 0$.
5. Then $n = 1 - 0 = 1$.
6. Check: $f(x) = x + 2$ gives $f(-2) = 0$ ✓ and $f(1) = 3$ ✓.

1. For $P$: factorise $2x^2 - x - 10 = (x + 2)(2x - 5)$.
2. The roots are $x = -2$ and $x = \frac52$.
3. The curve is U-shaped, so it is $\le 0$ between the roots: $P = \{x : -2 \le x \le \frac52\}$.
4. For $Q$: factorise $4x^2 - 1 = (2x - 1)(2x + 1)$.
5. The roots are $x = -\frac12$ and $x = \frac12$.
6. It is $\ge 0$ outside the roots: $x \le -\frac12$ or $x \ge \frac12$.
7. $P \cap Q$ is the part of $P$ that is also in $Q$.
8. On a number line, that is from $-2$ to $-\frac12$ and from $\frac12$ to $\frac52$, with all four ends included.
9. So $P \cap Q = \{x : -2 \le x \le -\frac12\} \cup \{x : \frac12 \le x \le \frac52\}$.

1. $f(x + h) - f(x) = \dfrac{5}{(x + h)^2 + 3} - \dfrac{5}{x^2 + 3}$.
2. Over one denominator, the top is $5(x^2 + 3) - 5[(x + h)^2 + 3] = 5(x^2 - x^2 - 2xh - h^2)$
   $= -10xh - 5h^2$.
3. So $f(x + h) - f(x) = \dfrac{-10xh - 5h^2}{[(x + h)^2 + 3](x^2 + 3)}$.
4. Divide by $h$: $\dfrac{-10x - 5h}{[(x + h)^2 + 3](x^2 + 3)}$.
5. Let $h \to 0$: $f'(x) = -\dfrac{10x}{(x^2 + 3)^2}$.

(a)

1. $\dfrac{{}^nP_5}{{}^nC_4} = \dfrac{n!}{(n - 5)!} \times \dfrac{4!\,(n - 4)!}{n!}$.
2. Cancel $n!$, and write $(n - 4)! = (n - 4)(n - 5)!$: the quotient is $24(n - 4)$.
3. So $24(n - 4) = 24$, which gives $n - 4 = 1$ and $n = 5$.

(b)

1. A number less than 6 has probability $p = \frac56$.
2. With $n = 5$ throws and exactly 3 successes:
3. $P = {}^5C_3\left(\frac56\right)^3\left(\frac16\right)^2$
   $= 10 \times \dfrac{125}{216} \times \dfrac{1}{36}$
   $= \dfrac{1250}{7776}$
   $\approx 0.161$.

(a)

1. Frequencies: score 1: 2, score 2: 9, score 3: 4, score 4: 2, score 5: 2, score 6: 1 (total 20).

(b)

1. $\sum fx = 2 + 18 + 12 + 8 + 10 + 6 = 56$, so the mean is $\dfrac{56}{20} = 2.8$.
2. Distances from 2.8: $1.8, 0.8, 0.2, 1.2, 2.2, 3.2$.
3. $\sum f|x - \bar x| = 3.6 + 7.2 + 0.8 + 2.4 + 4.4 + 3.2$
   $= 21.6$.
4. Mean deviation $= \dfrac{21.6}{20} = 1.08$.

(a)

1. The ball is momentarily at rest at the top: $\dfrac{dh}{dt} = 15.4 - 9.8t = 0$.
2. So $t = \dfrac{15.4}{9.8}$
   $= \dfrac{11}{7}$
   $\approx 1.6$ s.

(b)

1. Put $t = \frac{11}{7}$ into the height: $15.4 \times \frac{11}{7} = 24.2$ and $4.9 \times \left(\frac{11}{7}\right)^2 = 12.1$.
2. So the maximum height is $24.2 - 12.1 = 12.1$ m.

1. The bearings differ by $135^\circ - 45^\circ = 90^\circ$, so the forces are at right angles.
2. By Pythagoras: $50^2 = 40^2 + Q^2$.
3. $Q^2 = 2500 - 1600 = 900$, so $Q = 30\text{ N}$.

(a)

1. Write $\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3}$ and multiply through: $x + 6 = A(x + 1)^2 + B(x + 1) + C$.
2. Put $x = -1$: $5 = C$.
3. The $x^2$ terms: $0 = A$.
4. The $x$ terms: $1 = 2A + B$, so $B = 1$.
5. So the answer is $\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}$.

(b)

1. Integrate each piece: $\displaystyle\int (x + 1)^{-2}\,dx = -\frac{1}{x + 1}$ and $\displaystyle\int 5(x + 1)^{-3}\,dx = -\frac{5}{2(x + 1)^2}$.
2. At $x = 2$: $-\frac13 - \frac{5}{18} = -\frac{11}{18}$.
3. At $x = 1$: $-\frac12 - \frac58 = -\frac98$.
4. Subtract: $-\frac{11}{18} + \frac98 = \frac{-44 + 81}{72}$
   $= \frac{37}{72}$
   $\approx 0.514$.

(a)

1. Factorise: two numbers that multiply to $3 \times (-2) = -6$ and add to 1 are 3 and $-2$.
2. So $3x^2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1)$
   $= (3x - 2)(x + 1)$, and $(3x - 2)(x + 1) \le 0$.
3. The roots are $x = -1$ and $x = \frac23$. "$\le 0$" is between the roots, ends included: $-1 \le x \le \frac23$.

(b)

1. Write everything in terms of $\sin\theta$: $\cos^2\theta = 1 - \sin^2\theta$ and $\cos2\theta = 1 - 2\sin^2\theta$.
2. Substitute: $1 - \sin^2\theta + (1 - 2\sin^2\theta)\sin\theta - 1 = 0$.
3. Simplify: $-\sin^2\theta + \sin\theta - 2\sin^3\theta = 0$.
4. Multiply by $-1$: $2\sin^3\theta + \sin^2\theta - \sin\theta = 0$.
5. Take out $\sin\theta$: $\sin\theta(2\sin^2\theta + \sin\theta - 1) = 0$.
6. Factorise the bracket: $\sin\theta(2\sin\theta - 1)(\sin\theta + 1) = 0$.
7. $\sin\theta = 0$ gives $\theta = 0^\circ, 180^\circ, 360^\circ$.
8. $\sin\theta = \frac12$ gives $\theta = 30^\circ, 150^\circ$.
9. $\sin\theta = -1$ gives $\theta = 270^\circ$.
10. So $\theta = 0^\circ, 30^\circ, 150^\circ, 180^\circ, 270^\circ, 360^\circ$.

(a)

1. Write 2 as a log: $2 = \log_{10} 100$.
2. The left side is then $\log_{10}\dfrac{100x}{20} = \log_{10} 5x$.
3. Drop the logs: $5x = x^2 + 4$, so $x^2 - 5x + 4 = 0$.
4. Factorise: $(x - 1)(x - 4) = 0$, so $x = 1$ or $x = 4$.
5. Both make every log positive.

(b)

1. Complete the squares: $(x + 1)^2 + (y - 3)^2 = 1 + 9 - n$.
2. The radius squared is $10 - n = 2^2 = 4$, so $n = 6$.

(c)

1. At $x = 1$: $y = 7 - 4 = 3$.
2. The gradient is $\dfrac{dy}{dx} = 7 - 8x$, which is $-1$ at $x = 1$.
3. The tangent is $y - 3 = -1(x - 1)$, so $y = -x + 4$, that is $x + y - 4 = 0$.

1. $X \sim B(7, 0.65)$, with $q = 0.35$.

(a)

1. $P(X = 5)$
   $= \binom75(0.65)^5(0.35)^2$
   $\approx 0.298$.

(b)

1. $P(3) + P(4) = 0.1442 + 0.2679$
   $\approx 0.412$.

(c)

1. At most 2: $P(0) + P(1) + P(2) = 0.0006 + 0.0084 + 0.0466$
   $\approx 0.056$.

(a)

1. Take $d = x - 40$: the deviations add to $\sum d = -1$ and their squares to $\sum d^2 = 527$.
2. $\sigma = \sqrt{\dfrac{\sum d^2}{n} - \left(\dfrac{\sum d}{n}\right)^2}$
   $= \sqrt{52.7 - 0.01}$
   $\approx 7.26$.

(b)

1. With replacement, $p(\text{red}) = \frac{15}{50} = 0.3$ on every draw, with $n = 6$.

(i)

1. $P(4 \text{ red}) = \binom64(0.3)^4(0.7)^2$
   $\approx 0.06$.

(ii)

1. Equal numbers means 3 red: $\binom63(0.3)^3(0.7)^3 \approx 0.19$.

(iii)

1. At least 2 green means at most 4 red: $1 - [P(5 \text{ red}) + P(6 \text{ red})] = 1 - (0.0102 + 0.0007)$
   $\approx 0.99$.

(a)

1. The lift accelerates upwards: $R - mg = ma$.
2. $R - 5 = 0.5 \times 2$, so $R = 6\text{ N}$.

(b)

1. From $X$: $A$ at $0.4$, $P$ at $1.1$, the centre at $2.0$, $B$ at $2.8$ and $Q$ at $3.7$ m.
2. Moments about $B$: $90(2.4) + 100(0.9) = 80(1.7) + 10M(0.8)$.
3. $216 + 90 = 136 + 8M$, so $M = 21.25 \approx 21.3\text{ kg}$.
4. Up = down: $90 + R_B = 80 + 212.5 + 100$, so $R_B = 302.5\text{ N}$.

(a)

1. $|\mathbf m| = \sqrt{36 + 64} = 10.00$.
2. $\mathbf m$ points 6 east and 8 north: $\tan^{-1}\frac68 = 36.87^\circ$ from north, a bearing of $036.87^\circ$.
3. $|\mathbf n| = \sqrt{64 + \frac{49}{9}}$
   $= \sqrt{\frac{625}{9}}$
   $= \frac{25}{3}$
   $\approx 8.33$.
4. $\mathbf n$ points 8 west and $\frac73$ north: $\tan^{-1}\frac{8}{7/3} = 73.74^\circ$ west of north.
5. So the bearing is $360^\circ - 73.74^\circ = 286.26^\circ$.

(b)(i)

1. $v^2 = u^2 + 2as$: $0 = 900 + 100a$, so $a = -9\text{ m s}^{-2}$.

(ii)

1. $v = u + at$: $0 = 30 - 9t$, so $t = \frac{30}{9} \approx 3.33\text{ s}$.