WAEC · 2019 · May/June · Further Maths · Paper 2

1. Write the numbers as powers of 5: $625^{\frac{3x}{4} - 1} = 5^{4\left(\frac{3x}{4} - 1\right)}$
   $= 5^{3x - 4}$ and $125^{x - 1} = 5^{3x - 3}$.
2. The top is $5^{3x - 4} + 5^{3x - 3}$.
3. Take out the smaller power: $5^{3x - 4}(1 + 5) = 6 \times 5^{3x - 4}$.
4. Divide by the bottom, subtracting the indices: $\dfrac{6 \times 5^{3x - 4}}{5^{3x - 2}} = 6 \times 5^{-2}$.
5. $5^{-2} = \frac{1}{25}$, so the answer is $\dfrac{6}{25}$.

(a)

1. For external division in the ratio $m : n = 3 : 2$, use $\left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right)$, with $(x_1, y_1) = (7, -5)$ and $(x_2, y_2) = (-2, 7)$.
2. The $x$-coordinate: $\dfrac{3(-2) - 2(7)}{3 - 2} = -6 - 14$
   $= -20$.
3. The $y$-coordinate: $\dfrac{3(7) - 2(-5)}{3 - 2} = 21 + 10$
   $= 31$.
4. The point is $(-20, 31)$.

(b)

1. Put the two fractions over the common denominator $(1 - \sqrt2)(2 + \sqrt2)$: the top is $2(2 + \sqrt2) - 2(1 - \sqrt2) = 2 + 4\sqrt2$.
2. The bottom is $(1 - \sqrt2)(2 + \sqrt2) = 2 + \sqrt2 - 2\sqrt2 - 2$
   $= -\sqrt2$.
3. So the value is $\dfrac{2 + 4\sqrt2}{-\sqrt2}$.
4. Multiply top and bottom by $\sqrt2$: $\dfrac{2\sqrt2 + 8}{-2} = -4 - \sqrt2$.

1. The first term is $a = -3$ and the common difference is $d = -1 - (-3) = 2$.
2. $S_n = \dfrac n2[2(-3) + (n - 1)(2)]$
   $= \dfrac n2(2n - 8)$
   $= n(n - 4)$.
3. Set $n(n - 4) = 165$: $n^2 - 4n - 165 = 0$.
4. Factorise: $(n - 15)(n + 11) = 0$, so $n = 15$ or $n = -11$.
5. A number of terms can't be negative, so 15 terms add up to 165.

1. Draw a right-angled triangle with opposite side $p - q$ and hypotenuse $p + q$.
2. Adjacent side: $\sqrt{(p + q)^2 - (p - q)^2} = \sqrt{4pq}$
   $= 2\sqrt{pq}$.
3. So $\tan X = \dfrac{p - q}{2\sqrt{pq}}$ and $\tan^2 X = \dfrac{p^2 - 2pq + q^2}{4pq}$.
4. $1 - \tan^2 X = \dfrac{4pq - p^2 + 2pq - q^2}{4pq}$
   $= \dfrac{6pq - p^2 - q^2}{4pq}$.

1. The 10 N and 16 N forces have a resultant of 14 N, balancing the third force.
2. $14^2 = 10^2 + 16^2 + 2(10)(16)\cos\theta$, so $196 = 356 + 320\cos\theta$.
3. $\cos\theta = -\dfrac{160}{320}$
   $= -\dfrac12$, so $\theta = 120^\circ$.

1. From the pivot (35 cm from $W$): $T$ at $Y$ is $25$ cm on the $W$ side.
2. The weight, at the middle (45 cm from $W$), is $10$ cm on the $X$ side.
3. The 20 N at $Z$ (80 cm from $W$) is $45$ cm on the $X$ side.
4. Moments about the pivot: $25T = 50 \times 10 + 20 \times 45 = 1400$, so $T = 56\text{ N}$.

1. Class marks $3, 8, 13, 18, 23, 28$ and $d = x - 13$: $-10, -5, 0, 5, 10, 15$.
2. $\sum fd = -180 - 60 + 0 + 75 + 200 + 150$
   $= 185$ and $\sum f = 100$.
3. Mean $= 13 + \dfrac{185}{100} = 14.85$ years.

1. With two dice there are 36 equally likely outcomes.
2. A sum of 7 happens 6 ways, so $P = \frac{6}{36}$.
3. A sum of 4 happens 3 ways: $(1, 3), (2, 2), (3, 1)$, so $P = \frac{3}{36}$.
4. The two throws are independent: $\frac{6}{36} \times \frac{3}{36} = \frac{18}{1296}$
   $= \frac{1}{72}$.

(a)

1. From the line, $y = 3 - 2x$.
2. So $7 - \dfrac6x = 3 - 2x$.
3. Multiply by $x$: $7x - 6 = 3x - 2x^2$, so $2x^2 + 4x - 6 = 0$, that is $x^2 + 2x - 3 = 0$.
4. $(x + 3)(x - 1) = 0$: the points are $(1, 1)$ and $(-3, 9)$.

(b)

1. Midpoint: $(-1, 5)$.
2. Gradient of the chord: $\dfrac{9 - 1}{-3 - 1} = -2$.
3. The bisector has gradient $\frac12$: $y - 5 = \frac12(x + 1)$, so $2y - x - 11 = 0$.

(a)

1. Read off the coefficients: $a = 4$, $b = -p$, $c = 1$.
2. Real roots need $b^2 - 4ac \ge 0$: $p^2 - 16 \ge 0$.
3. Factorise: $(p - 4)(p + 4) \ge 0$, with roots $p = -4$ and $p = 4$.
4. "$\ge 0$" is outside the roots: $p \le -4$ or $p \ge 4$.

(b)(i)

1. Each term is $\binom6r(3x)^r$: the coefficients are $1$, $6 \times 3$, $15 \times 9$, $20 \times 27$, $15 \times 81$, $6 \times 243$, $729$.
2. So $(1 + 3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6$.

(ii)

1. Find $x$: $1 + 3x = 1.03$ gives $x = 0.01$.
2. Substitute: $1 + 0.18 + 0.0135 + 0.00054 + 0.00001215 + \ldots = 1.19405\ldots$
3. So $(1.03)^6 = 1.194$ to 4 significant figures.

(a)

1. Read the coefficients: $M = \begin{pmatrix} 0 & 7 \\ 3 & -1 \end{pmatrix}$ and $N = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$.

(b)

1. $N$ followed by $M$ is $MN = \begin{pmatrix} 0 + 35 & 0 + 21 \\ 6 - 5 & -3 - 3 \end{pmatrix}$
   $= \begin{pmatrix} 35 & 21 \\ 1 & -6 \end{pmatrix}$.
2. $MN\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 70 - 63 \\ 2 + 18 \end{pmatrix}$
   $= \begin{pmatrix} 7 \\ 20 \end{pmatrix}$: the image is $(7, 20)$.

(c)

1. Solve $N\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$: $2x - y = 2$ and $5x + 3y = 4$.
2. Multiply the first by 3 and add: $11x = 10$, so $x = \frac{10}{11}$ and $y = 2x - 2 = -\frac{2}{11}$.
3. So $Q\left(\frac{10}{11}, -\frac{2}{11}\right)$.

1. Let $F$ be the number who failed: $F \sim B(10, 0.4)$.

(a)

1. At least two failed $= 1 - [0.6^{10} + 10(0.4)(0.6)^9]$
   $= 1 - 0.0464$
   $= 0.9536$.

(b)

1. Half passed means 5 passed: $\binom{10}{5}(0.6)^5(0.4)^5 = 252 \times 0.07776 \times 0.01024$
   $\approx 0.2007$.

(c)

1. At most two failed: $0.0060 + 0.0403 + 45(0.4)^2(0.6)^8 = 0.0060 + 0.0403 + 0.1209$
   $= 0.1673$.

(a)

1. Rank the entrance marks (1 for the highest): A 7, B 6, C 9, D 5, E 10, F 3, G 8, H 4, I 1, J 2.
2. Rank the school marks: A 9, B 4, C 5, D 7, E 10, F 6, G 8, H 1, I 3, J 2.
3. $d$: $-2, 2, 4, -2, 0, -3, 0, 3, -2, 0$, so $\sum d^2 = 50$.
4. $\rho = 1 - \dfrac{6 \times 50}{10 \times 99}$
   $= 1 - 0.3030$
   $\approx 0.697$.

(b)

1. There is a fairly strong positive correlation: students who do well in the entrance examination tend to do well in school.

(a)

1. After 4 s: $v = 20 + 2.5 \times 4$
   $= 30\text{ m s}^{-1}$.
2. The graph starts at $20$ (not 0), rises to $30$ at $t = 4$, stays level until $t = 12$, then falls to 0.

(b)

1. $2.5 : r = 3 : 4$, so the retardation is $r = \dfrac{4 \times 2.5}{3}$
   $= \dfrac{10}{3}\text{ m s}^{-2}$.
2. Time to stop: $t = 30 \div \frac{10}{3} = 9\text{ s}$.

(c)

1. Distance = area: $\frac12(20 + 30)(4) + 30 \times 8 + \frac12(30)(9)$.
2. $= 100 + 240 + 135 = 475\text{ m}$.

(a)

1. $\mathbf r = 2\mathbf m + \mathbf n$
   $= 2(\mathbf i - \mathbf j) + (2\mathbf i + 3\mathbf j)$
   $= 4\mathbf i + \mathbf j$.
2. $|\mathbf r| = \sqrt{4^2 + 1^2}$
   $= \sqrt{17}$
   $\approx 4.123$.

(b)(i)

1. Momentarily at rest when $\dfrac{dS}{dt} = 3 - t^2 = 0$, so $t = \sqrt3 \approx 1.732$ s (time is positive).

(ii)

1. $S = 3\sqrt3 - \dfrac{(\sqrt3)^3}{3} + 9$
   $= 3\sqrt3 - \sqrt3 + 9$
   $= 9 + 2\sqrt3$
   $\approx 12.46$ m.