WAEC · 2019 · Private · Further Maths · Paper 2

1. Write each term out: $\dbinom{2x + 1}{3} = \dfrac{(2x + 1)(2x)(2x - 1)}{6}$ and $\dbinom{2x - 1}{3} = \dfrac{(2x - 1)(2x - 2)(2x - 3)}{6}$.
2. $2\dbinom x2 = 2 \times \dfrac{x(x - 1)}{2}$
   $= x(x - 1)$.
3. Multiply the whole expression by 6.
4. The first two terms share $(2x - 1)$: $(2x - 1)[2x(2x + 1) - (2x - 2)(2x - 3)]$.
5. The bracket: $4x^2 + 2x - (4x^2 - 10x + 6) = 12x - 6$
   $= 6(2x - 1)$, so the two terms give $6(2x - 1)^2 = 24x^2 - 24x + 6$.
6. Take away $6x(x - 1) = 6x^2 - 6x$: $18x^2 - 18x + 6$.
7. Divide by 6: the expression is $3x^2 - 3x + 1$.

1. $f(x + h) = 3(x + h)^2 + 2(x + h) - 1$
   $= 3x^2 + 6xh + 3h^2 + 2x + 2h - 1$.
2. Take away $f(x) = 3x^2 + 2x - 1$: $f(x + h) - f(x) = 6xh + 3h^2 + 2h$.
3. Divide by $h$: $6x + 3h + 2$.
4. Let $h \to 0$: $f'(x) = 6x + 2$.

1. The $y$-intercept is where $x = 0$: $-2y + 6 = 0$, so $y = 3$: the point $(0, 3)$.
2. $r^2 = (0 - 2)^2 + (3 - 3)^2 = 4$.
3. $(x - 2)^2 + (y - 3)^2 = 4$.
4. Expand: $x^2 - 4x + 4 + y^2 - 6y + 9 = 4$, so $x^2 + y^2 - 4x - 6y + 9 = 0$.

1. For $3x^2 + 4x - 5 = 0$: $\alpha + \beta = -\frac43$ and $\alpha\beta = -\frac53$.
2. Square the difference: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
3. Substitute: $\frac{16}{9} - 4\left(-\frac53\right) = \frac{16}{9} + \frac{60}{9}$
   $= \frac{76}{9}$.
4. Take the square root: $\alpha - \beta = \pm\dfrac{\sqrt{76}}{3}$.
5. Simplify the surd: $\sqrt{76} = \sqrt{4 \times 19}$
   $= 2\sqrt{19}$, so $\alpha - \beta = \pm\dfrac{2\sqrt{19}}{3}$.

1. The miss chances are $\frac23$, $\frac45$ and $\frac34$.
2. Only $X$: $\frac13 \cdot \frac45 \cdot \frac34 = \frac{12}{60}$.
3. Only $Y$: $\frac23 \cdot \frac15 \cdot \frac34 = \frac{6}{60}$.
4. Only $Z$: $\frac23 \cdot \frac45 \cdot \frac14 = \frac{8}{60}$.
5. Add: $\frac{26}{60} = \frac{13}{30}$
   $\approx 0.43$.

1. The class widths are unequal: $4, 10, 20, 3, 2$ (from the boundaries).
2. Frequency densities: $\frac24 = 0.5$, $\frac{6}{10} = 0.6$, $\frac{18}{20} = 0.9$, $\frac23 \approx 0.67$, $\frac12 = 0.5$.
3. Draw each bar over its class boundaries with these heights, so that each bar's area is its frequency.

1. $\overrightarrow{OP} \cdot \overrightarrow{OQ} = (-3)(8) + (-4)(-15)$
   $= -24 + 60$
   $= 36$.
2. $|\overrightarrow{OP}| = 5$ and $|\overrightarrow{OQ}| = \sqrt{64 + 225}$
   $= 17$.
3. $\cos\theta = \dfrac{36}{85}$
   $\approx 0.4235$, so $\theta \approx 64.94^\circ$.

1. The weight is $12 \times 10 = 120\text{ N}$.
2. The string is at $45^\circ$ to the vertical.

(a)

1. Up: $T\cos45^\circ = 120$, so $T = 120\sqrt2 \approx 169.71\text{ N}$.

(b)

1. Across: $R = T\sin45^\circ = 120\text{ N}$.

(a)

1. Change each equation to index form: $3\log_2 x = y$ gives $2^y = x^3$, and $\log_2 4x = y + 4$ gives $2^{y + 4} = 4x$.
2. $2^{y + 4} = 16 \times 2^y = 16x^3$, so $16x^3 = 4x$.
3. Rearrange: $16x^3 - 4x = 0$, so $4x(4x^2 - 1) = 0$ and $x = 0$ or $x = \pm\frac12$.
4. Only $x = \frac12$ has a logarithm ($\log_2 0$ and $\log_2(-\frac12)$ don't exist).
5. Then $y = 3\log_2 \frac12 = 3(-1) = -3$.
6. So $x = \frac12$, $y = -3$.

(b)

1. $(-3) * 5 = (-3)^2 - 2(-3)(5) + 5^2$
   $= 9 + 30 + 25$
   $= 64$.
2. $64 = 2^6$, so $n = 6$.

(a)

1. $p$ is acute, so $\cos p = \sqrt{1 - \frac14}$
   $= \dfrac{\sqrt3}{2}$.
2. $q$ is obtuse, so $\sin q$ is positive: $\sin q = \sqrt{1 - \frac19}$
   $= \dfrac{2\sqrt2}{3}$.
3. $\sin(p - q) = \sin p\cos q - \cos p\sin q$
   $= \dfrac12\left(-\dfrac13\right) - \dfrac{\sqrt3}{2} \times \dfrac{2\sqrt2}{3}$.
4. $= -\dfrac16 - \dfrac{2\sqrt6}{6}$
   $= \dfrac{-1 - 2\sqrt6}{6}$.

(b)

1. Seven ordinates means six strips, $h = 0.5$.
2. The ordinates of $\dfrac{2}{\sqrt{x + 3}}$ at $x = 1, 1.5, \ldots, 4$ are $1,\ 0.9428,\ 0.8944,\ 0.8528,\ 0.8165,\ 0.7845,\ 0.7559$.
3. First and last: $1.7559$.
4. Twice the rest: $2 \times 4.2910 = 8.5820$.
5. Trapezium rule: $\dfrac{0.5}{2}(1.7559 + 8.5820) = 2.5845$, about $2.58$.

1. $\Delta = \begin{vmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{vmatrix}$
   $= 5(24 + 3) + 6(42 + 6) + 4(7 - 8)$
   $= 135 + 288 - 4$
   $= 419$.
2. Replace the $x$ column by $15, 19, 46$: $\Delta_x = 15(27) + 6(114 + 138) + 4(19 - 184)$
   $= 1257$.
3. Replace the $y$ column: $\Delta_y = 5(114 + 138) - 15(48) + 4(322 - 38)$
   $= 1676$.
4. Replace the $z$ column: $\Delta_z = 5(184 - 19) + 6(322 - 38) + 15(7 - 8)$
   $= 2514$.
5. So $x = \dfrac{1257}{419} = 3$, $y = \dfrac{1676}{419} = 4$ and $z = \dfrac{2514}{419} = 6$.
6. Check: $15 - 24 + 24 = 15$ ✓, $21 + 16 - 18 = 19$ ✓, $6 + 4 + 36 = 46$ ✓.

1. Class marks $52, 57, \ldots, 87$ and $d = x - 67$: $-15, -10, -5, 0, 5, 10, 15, 20$.
2. $\sum f = 100$, $\sum fd = 10$ and $\sum fd^2 = 7500$.

(a)

1. Mean $= 67 + \dfrac{10}{100} = 67.1$.

(b)

1. $\sigma = \sqrt{\dfrac{7500}{100} - 0.1^2}$
   $= \sqrt{74.99}$
   $\approx 8.7$.

(a)

1. $X \sim B(12, 0.3)$.

(i)

1. $P(X = 6)$
   $= {}^{12}C_6(0.3)^6(0.7)^6$
   $\approx 0.0792$.

(ii)

1. $P(X \ge 4) = 1 - [P(0) + P(1) + P(2) + P(3)]$
   $= 1 - (0.0138 + 0.0712 + 0.1678 + 0.2397)$
   $\approx 0.5075$.

(iii)

1. $P(X = 0) = 0.7^{12} \approx 0.0138$.

(b)

1. There are $\,{}^{13}C_7 = 1716$ committees. 4 boys and 3 girls: $\,{}^8C_4 \times {}^5C_3 = 70 \times 10$
   $= 700$.
2. So $P = \dfrac{700}{1716}$
   $= \dfrac{175}{429}$
   $\approx 0.4079$.

(a)

1. In $PQRS$, $\overrightarrow{QR} = \overrightarrow{PS}$
   $= \begin{pmatrix} -2 + 1 \\ 3 - 4 \end{pmatrix}$
   $= \begin{pmatrix} -1 \\ -1 \end{pmatrix}$.
2. So $R = (2 - 1, 3 - 1) = (1, 2)$: $x = 1$ and $y = 2$.

(b)(i)

1. $v^2 = u^2 + 2as$: $400 = 0 + 16a$, so $a = 25\text{ m s}^{-2}$.

(ii)

1. $s = ut + \frac12at^2$: $40 = \frac{25}{2}t^2$, so $t^2 = 3.2$ and $t \approx 1.79\text{ s}$.

1. East: $5\sin30^\circ + P\sin60^\circ + Q\sin150^\circ + 3\sin180^\circ + 5\sin270^\circ = 0$.
2. So $2.5 + 0.866P + 0.5Q - 5 = 0$, which gives $0.866P + 0.5Q = 2.5$.
3. North: $5\cos30^\circ + P\cos60^\circ + Q\cos150^\circ + 3\cos180^\circ + 5\cos270^\circ = 0$.
4. So $4.330 + 0.5P - 0.866Q - 3 = 0$, which gives $0.5P - 0.866Q = -1.330$.
5. Multiply the first by 0.866 and the second by 0.5, then add: $P = 2.165 - 0.665 = 1.5\text{ N}$.
6. Then $0.5Q = 2.5 - 1.299$, so $Q \approx 2.4\text{ N}$.