WAEC · 2018 · May/June · Further Maths · Paper 2

1. Expand along the first row, with the signs $+, -, +$: $(x - 3)\begin{vmatrix} 2 & 2 \\ -4 & 6 - x \end{vmatrix} - (-4)\begin{vmatrix} 5 & 2 \\ 2 & 6 - x \end{vmatrix} + 3\begin{vmatrix} 5 & 2 \\ 2 & -4 \end{vmatrix}$.
2. First minor: $2(6 - x) - 2(-4) = 12 - 2x + 8 = 20 - 2x$.
3. Second minor: $5(6 - x) - 2 \times 2 = 30 - 5x - 4$
   $= 26 - 5x$.
4. Third minor: $5(-4) - 2 \times 2 = -24$.
5. So $(x - 3)(20 - 2x) + 4(26 - 5x) + 3(-24) = -24$.
6. Expand: $-2x^2 + 26x - 60 + 104 - 20x - 72 = -24$, so $-2x^2 + 6x - 28 = -24$.
7. Add 24 to both sides: $-2x^2 + 6x - 4 = 0$.
8. Divide by $-2$: $x^2 - 3x + 2 = 0$.
9. Factorise: $(x - 1)(x - 2) = 0$, so $x = 1$ or $x = 2$.

1. Swapping the base and the number turns a log upside down: $\log_x 3 = \dfrac{1}{\log_3 x}$.
2. Let $y = \log_3 x$.
3. The equation becomes $y - \dfrac3y + 2 = 0$.
4. Multiply every term by $y$: $y^2 + 2y - 3 = 0$.
5. Factorise: $(y + 3)(y - 1) = 0$, so $y = 1$ or $y = -3$.
6. Back to $x$: $\log_3 x = 1$ gives $x = 3$, and $\log_3 x = -3$ gives $x = 3^{-3} = \frac{1}{27}$.

(a)

1. $u = x - 2$, so $x = u + 2$.
2. Expand: $x^3 = (u + 2)^3 = u^3 + 6u^2 + 12u + 8$.
3. Add 5: $x^3 + 5 = u^3 + 6u^2 + 12u + 13$.
4. The bottom is $u^4$.
5. So the fraction is $\dfrac{u^3 + 6u^2 + 12u + 13}{u^4} = \dfrac1u + \dfrac{6}{u^2} + \dfrac{12}{u^3} + \dfrac{13}{u^4}$.

(b)

1. Put $u = x - 2$ back: $\dfrac{1}{x - 2} + \dfrac{6}{(x - 2)^2} + \dfrac{12}{(x - 2)^3} + \dfrac{13}{(x - 2)^4}$.

1. The sum: $S_{12} = \dfrac{12}{2}[2a + 11d] = 168$.
2. Divide by 6: $2a + 11d = 28$.
3. The third term: $a + 2d = 7$, so $a = 7 - 2d$.
4. Substitute: $2(7 - 2d) + 11d = 28$, so $14 + 7d = 28$.
5. So $d = 2$, and $a = 7 - 4 = 3$.

1. The judges' values are already ranks.
2. $d = X - Y$: $2, 2, -3, -1, -3, -1, 2, 2$.
3. $\sum d^2 = 4 + 4 + 9 + 1 + 9 + 1 + 4 + 4$
   $= 36$.
4. $\rho = 1 - \dfrac{6 \times 36}{8 \times 63}$
   $= 1 - \dfrac{216}{504}$
   $= \dfrac47$
   $\approx 0.5714$.

(a)

1. Only one solves it: Kunle solves and Tayo doesn't, or Tayo solves and Kunle doesn't.
2. $\dfrac13 \times \dfrac45 + \dfrac23 \times \dfrac15 = \dfrac{4}{15} + \dfrac{2}{15}$
   $= \dfrac{6}{15}$
   $= \dfrac25$.

(b)

1. $\,{}^{10}C_8 = {}^{10}C_2$
   $= \dfrac{10 \times 9}{2}$
   $= 45$ ways.

1. $4\mathbf m + 2\mathbf n - 3\mathbf p = (12 + 4 + 3)\mathbf i + (-8 + 6 - 18)\mathbf j$.
2. $= 19\mathbf i - 20\mathbf j$.
3. Its magnitude is $\sqrt{361 + 400} = \sqrt{761}$
   $\approx 27.586$.

(a)

1. Same direction: $20(80) + 30(50) = (20 + 30)v$.
2. $1600 + 1500 = 50v$, so $v = 62\text{ m s}^{-1}$.

(b)

1. Opposite directions: $20(80) + 30(-50) = 50v$.
2. $1600 - 1500 = 50v$, so $v = 2\text{ m s}^{-1}$, in the direction the $20\text{ kg}$ body was moving.

1. Substitute each point into $x^2 + y^2 + 2gx + 2fy + k = 0$.
2. $(3, 2)$: $6g + 4f + k = -13$.
3. $(-1, -2)$: $-2g - 4f + k = -5$.
4. $(5, -4)$: $10g - 8f + k = -41$.
5. First minus second: $8g + 8f = -8$, so $g + f = -1$.
6. Third minus first: $4g - 12f = -28$, so $g - 3f = -7$.
7. Subtract: $4f = 6$, so $f = \frac32$ and $g = -\frac52$.
8. Then $k = -13 + 15 - 6 = -4$.

(a)

1. The centre is $(-g, -f) = \left(\frac52, -\frac32\right)$.

(b)

1. $r = \sqrt{g^2 + f^2 - k}$
   $= \sqrt{\frac{25}{4} + \frac94 + 4}$
   $= \sqrt{12.5}$
   $\approx 3.536$.

(c)

1. $x^2 + y^2 - 5x + 3y - 4 = 0$.

(a)

1. Write each power in terms of $2^y$: $2^{3y + 2} = 4(2^y)^3$ and $2^{2y + 2} = 4(2^y)^2$.
2. Let $x = 2^y$: $4x^3 - 28x^2 - 31x - 8 = 0$.
3. Try factors of 8: $x = 8$ gives $2048 - 1792 - 248 - 8 = 0$, so $(x - 8)$ is a factor.
4. Divide it out: $4x^3 - 28x^2 - 31x - 8 = (x - 8)(4x^2 + 4x + 1)$
   $= (x - 8)(2x + 1)^2$.
5. So $x = 8$ or $x = -\frac12$.
6. But $x = 2^y$ is always positive, so reject $-\frac12$.
7. $2^y = 8 = 2^3$, so $y = 3$.

(b)

1. Let $u = x^2 + 1$, so $\dfrac{du}{dx} = 2x$ and $x\,dx = \frac12\,du$.
2. The integral becomes $\displaystyle\int u^{\frac12} \times \frac12\,du = \frac12 \times \frac{u^{\frac32}}{\frac32} + c$
   $= \frac13u^{\frac32} + c$.
3. Put $u$ back: $\dfrac13(x^2 + 1)^{\frac32} + c$.

(a)(i)

1. Each term is $\binom5r 2^{5 - r}\left(-\frac12x\right)^r$ for $r = 0, 1, \ldots, 5$.
2. $r = 0$: $32$.
3. $r = 1$: $5 \times 16 \times \left(-\frac12x\right) = -40x$.
4. $r = 2$: $10 \times 8 \times \frac14x^2 = 20x^2$.
5. $r = 3$: $10 \times 4 \times \left(-\frac18x^3\right) = -5x^3$.
6. $r = 4$: $5 \times 2 \times \frac{1}{16}x^4 = \frac58x^4$.
7. $r = 5$: $-\frac{1}{32}x^5$.
8. So $\left(2 - \frac12x\right)^5 = 32 - 40x + 20x^2 - 5x^3 + \frac58x^4 - \frac{1}{32}x^5$.

(ii)

1. Find $x$: $2 - \frac12x = 1.99$ gives $\frac12x = 0.01$, so $x = 0.02$.
2. Substitute: $32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3 + \ldots = 32 - 0.8 + 0.008 - 0.00004 + \ldots$
3. That is $31.20796\ldots$, so $(1.99)^5 = 31.21$ to 2 decimal places.

(b)

1. $(x + 1)$ is a factor, so $f(-1) = 0$: $-1 + q - r + 9 = 0$, which gives $q - r = -8$.
2. The remainder on dividing by $(x + 2)$ is $-17$, so $f(-2) = -17$: $-8 + 4q - 2r + 9 = -17$, which gives $2q - r = -9$.
3. Take the first equation from the second: $q = -1$.
4. Then $r = q + 8 = 7$.

(a)

1. $\sum f = 268$ and $\sum fx = 1333$, so the mean is $\dfrac{1333}{268} \approx 4.974$.

(b)

1. Even numbers of heads are 0, 2, 4, 6, 8 and 10: $2 + 23 + 11 + 100 + 8 + 3 = 147$, so $P = \dfrac{147}{268} \approx 0.549$.

(c)

1. Odd: $7 + 36 + 61 + 12 + 5 = 121$, so $P = \dfrac{121}{268} \approx 0.451$.

1. The chances of failing are $\frac13$, $\frac14$ and $\frac15$.

(a)

1. Only Katty and Baba: Ali fails, they succeed: $\frac13 \times \frac34 \times \frac45 = \frac15$.

(b)

1. None: $\frac13 \times \frac14 \times \frac15 = \frac{1}{60}$.

(c)

1. At most two is everything except all three: $1 - \frac23 \times \frac34 \times \frac45 = 1 - \frac25$
   $= \frac35$.

(a)

1. In $ABCN$, $\overrightarrow{AB} = \overrightarrow{NC}$, so $\mathbf n = \mathbf a + \mathbf c - \mathbf b$.
2. $\mathbf n = (8 - 10 - 2)\mathbf i + (-10 + 4 - 6)\mathbf j$
   $= -4\mathbf i - 12\mathbf j$.

(b)

1. $\overrightarrow{AN} = \mathbf n - \mathbf a$
   $= -12\mathbf i - 2\mathbf j$, so $|\overrightarrow{AN}| = \sqrt{148}$
   $\approx 12.166$.
2. $\overrightarrow{AB} = \mathbf b - \mathbf a$
   $= -6\mathbf i + 16\mathbf j$, so $|\overrightarrow{AB}| = \sqrt{292}$
   $\approx 17.088$.

(c)

1. $\overrightarrow{AN} \cdot \overrightarrow{AB} = 72 - 32$
   $= 40$.
2. $\cos\theta = \dfrac{40}{\sqrt{148}\sqrt{292}}$
   $\approx 0.1924$, so $\theta \approx 78.91^\circ$.

(a)

1. From $X$: $P$ at $0.8$ m, the centre at $2$ m, the $1000\text{ N}$ weight at $2.4$ m and $Q$ at $3$ m.
2. Moments about $P$: $2.2R_Q + 80(0.8) = 350(1.2) + 1000(1.6)$.
3. $2.2R_Q = 420 + 1600 - 64 = 1956$, so $R_Q \approx 889.09\text{ N}$.
4. Up = down: $R_P = 80 + 350 + 1000 - 889.09$
   $\approx 540.91\text{ N}$.

(b)

1. Put the $1200\text{ N}$ weight $x$ m from $Q$, so $2.2 - x$ m from $P$.
2. Moments about $P$: $2.2R_Q + 64 = 420 + 1200(2.2 - x)$, so $R_Q = \dfrac{2996 - 1200x}{2.2}$.
3. Up = down: $R_P = 1630 - R_Q$
   $= \dfrac{590 + 1200x}{2.2}$.
4. Both reactions stay positive for any point between $P$ and $Q$, so the beam balances wherever the weight goes there.