WAEC · 2016 · May/June · Further Maths · Paper 2

(a)

1. The identity $e$ satisfies $m * e = m$: $m + e + 2 = m$, so $e = -2$.

(b)

1. The inverse satisfies $n * n^{-1} = e$: $n + n^{-1} + 2 = -2$.
2. So $n^{-1} = -4 - n$.

1. On one line, the gradients between pairs of points are equal.
2. From $(2, 1)$ to $(5, 2)$: $\dfrac{2 - 1}{5 - 2} = \dfrac13$.
3. From $(-4, k)$ to $(5, 2)$: $\dfrac{2 - k}{5 + 4} = \dfrac13$, so $2 - k = 3$ and $k = -1$.

(a)

1. Find the $x$ that makes the bracket 5: $x + 2 = 5$, so $x = 3$.
2. Put $x = 3$ into the right-hand side: $f(5) = 6(3)^2 + 5(3) - 8$
   $= 54 + 15 - 8$
   $= 61$.

(b)

1. Multiply the top and the bottom by the conjugate of the bottom, $4\sqrt2 + 2\sqrt3$.
2. The bottom: $(4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20$.
3. The top: $(7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18$
   $= 74 + 26\sqrt6$.
4. Divide each term by 20: $\dfrac{74}{20} + \dfrac{26}{20}\sqrt6 = \dfrac{37}{10} + \dfrac{13}{10}\sqrt6$.
5. So $p = \frac{37}{10}$, $q = \frac{13}{10}$, $r = 6$.

1. The polynomial is the divisor times the quotient, plus the remainder: $f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16)$.
2. Multiply out term by term: $x^2(2x - 5) = 2x^3 - 5x^2$, $5x(2x - 5) = 10x^2 - 25x$ and $1(2x - 5) = 2x - 5$.
3. Add them: $2x^3 + 5x^2 - 23x - 5$.
4. Add the remainder: $2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11$.
5. Compare with $2x^3 + mx^2 + nx + 11$: $m = 5$ and $n = 7$.
6. The constants, 11, agree.

1. The chances of not gaining admission are $\frac15$, $\frac14$ and $\frac13$.

(a)

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

(b)

1. Only Ago and Sulley: they succeed and Musa fails: $\frac45 \times \frac34 \times \frac13 = \frac{1}{5}$.

(a)

1. The five numbers add to $2k + 20$, so their mean is $\dfrac{2k + 20}{5} = k + 1$.
2. Multiply by 5: $2k + 20 = 5k + 5$, so $3k = 15$ and $k = 5$.

(b)

1. The numbers are $1, 4, 5, 9, 11$, with mean 6.
2. The deviations are $-5, -2, -1, 3, 5$.
3. $\sum d^2 = 25 + 4 + 1 + 9 + 25 = 64$, so $\sigma = \sqrt{\dfrac{64}{5}}$
   $= \dfrac{8}{\sqrt5}$
   $= \dfrac{8\sqrt5}{5}$
   $\approx 3.578$.

(a)

1. Momentum before: $3(8) + 5m = 24 + 5m$.
2. After, moving together: $6(3 + m) = 18 + 6m$.
3. Momentum is conserved: $24 + 5m = 18 + 6m$, so $m = 6\text{ kg}$.

(b)

1. Now the $6\text{ kg}$ body moves the other way: momentum before $= 3(8) + 6(-5) = -6$.
2. After: $(3 + 6)v = 9v$, so $v = -\dfrac69 = -\dfrac23$.
3. The common velocity is $0.67\text{ m s}^{-1}$, in the direction the $6\text{ kg}$ body was moving.

1. Match the parts of $\mathbf r = \alpha\mathbf p + \beta\mathbf q$: $5\alpha - \beta = 17$ and $3\alpha + 2\beta = 5$.
2. From the first, $\beta = 5\alpha - 17$.
3. Substitute: $3\alpha + 10\alpha - 34 = 5$, so $\alpha = 3$.
4. Then $\beta = 15 - 17 = -2$, so $\mathbf r = 3\mathbf p - 2\mathbf q$.
5. Rearrange: $2\mathbf q = 3\mathbf p - \mathbf r$, so $\mathbf q = \frac32\mathbf p - \frac12\mathbf r$.

(a)

1. Write the numbers as prime powers: $27 = 3^3$ and $5\sqrt5 = 5^{\frac32}$.
2. So $\frac32\log 27 = \frac32 \times 3\log 3$
   $= \frac92\log 3$ and $3\log 5\sqrt5 = 3 \times \frac32\log 5$
   $= \frac92\log 5$.
3. The top is $\frac92(\log 3 - \log 5)$.
4. Write the bottom the same way: $0.6 = \frac35$, so $\log 0.6 = \log 3 - \log 5$.
5. The bracket cancels, so the value is $\dfrac92 = 4\frac12$.

(b)(i)

1. Read the coefficients off each rule: $A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$.

(ii)

1. $A$ followed by $B$ is the matrix $BA$ (the first transformation goes on the right): $BA = \begin{pmatrix} 2 - 3 & 4 + 3 \\ 1 - 2 & 2 + 2 \end{pmatrix}$
   $= \begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}$.
2. Apply it to $P(-2, 2)$: $\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 + 14 \\ 2 + 8 \end{pmatrix}$
   $= \begin{pmatrix} 16 \\ 10 \end{pmatrix}$.
3. The image is $(16, 10)$.

(a)(i)

1. The coefficients are row 7 of Pascal's triangle: $(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7$.

(ii)

1. The fifth, sixth and seventh terms are $35x^4$, $21x^5$ and $7x^6$, so the coefficients are $35, 21, 7$.
2. They go down by 14 each time: $21 - 35 = -14$ and $7 - 21 = -14$.
3. The common difference is $-14$.

(b)

1. $h = 1$.
2. $y = \sqrt{2x + 8x^{-2}}$ gives $3.1623,\ 2.4495,\ 2.6247,\ 2.9155,\ 3.2125$ at $x = 1, \ldots, 5$.
3. $\frac12[(3.1623 + 3.2125) + 2(2.4495 + 2.6247 + 2.9155)] = \frac12[6.3748 + 15.9794]$
   $= 11.1771$, about $11.18$.

(a)

1. $\,{}^kP_2 = k(k - 1) = 72$, so $k^2 - k - 72 = 0$.
2. Factorise: $(k - 9)(k + 8) = 0$.
3. $k$ must be a positive whole number, so $k = 9$.

(b)

1. Let $c = \cos\theta$: $2c^2 - 5c - 3 = 0$, so $(2c + 1)(c - 3) = 0$.
2. $\cos\theta = 3$ is impossible, so $\cos\theta = -\frac12$.
3. Cosine is negative in the second and third quadrants: $\theta = 180^\circ - 60^\circ = 120^\circ$ or $\theta = 180^\circ + 60^\circ = 240^\circ$.

(a)(i)

1. Cumulative frequencies $3, 20, 61, 146, 243, 358, 459, 523, 544, 550$ at the upper boundaries $10.5, 20.5, \ldots, 100.5$.

(ii)

1. Plot these points, starting from $(0.5, 0)$, and join them with a smooth curve.

(b)(i)

1. Read up from 32: about 74 students scored less.
2. Read up from 74: about 481.
3. So about $481 - 74 \approx 408$ students scored between 32 and 74.

(ii)

1. 18% of 550 is 99.
2. Read across from 99: the pass mark is about 35.

(iii)

1. 8% of 550 is 44, so read across from $550 - 44 = 506$: the lowest distinction mark is about 78.

(a)(i)

1. Glue each subject into a block.
2. The 3 blocks can be ordered in $3! = 6$ ways.
3. Inside the blocks: $2! = 2$, $5! = 120$ and $3! = 6$.
4. So $6 \times 2 \times 120 \times 6 = 8640$.

(ii)

1. Glue only the 5 Physics books: 1 block and 5 other books make 6 units, $6! = 720$ orders.
2. Inside the block, $5! = 120$.
3. So $720 \times 120 = 86\,400$.

(b)

1. $p = \frac{13}{20} = 0.65$, $n = 8$.
2. At least 3 is $1 - [P(0) + P(1) + P(2)]$.
3. $P(0) = 0.35^8 = 0.000225$, $P(1) = 8(0.65)(0.35)^7 = 0.003346$ and $P(2) = 28(0.65)^2(0.35)^6 = 0.021747$.
4. So $1 - 0.025318 = 0.9747$, which is $0.975$ to three significant figures.

(a)

1. $\mathbf m - \mathbf n = 2\mathbf i + \mathbf j$, so $\mathbf r = k(2\mathbf i + \mathbf j)$ for some $k > 0$ (the parts are positive).
2. $|\mathbf s| = \sqrt{4 + 1} = \sqrt5$, so $|\mathbf r| = 3\sqrt5$.
3. $|\mathbf r| = k\sqrt5 = 3\sqrt5$, so $k = 3$: $\alpha = 6$ and $\beta = 3$.

(b)

1. $\mathbf r - \mathbf s = (6 - 2)\mathbf i + (3 + 1)\mathbf j$
   $= 4\mathbf i + 4\mathbf j$.
2. Its magnitude is $\sqrt{32} = 4\sqrt2 \approx 5.66$.
3. Equal east and north parts: the bearing is $045^\circ$.

(a)(i)

1. At the top $v = 0$: $0 = 900 - 20s$, so $s = 45\text{ m}$.

(ii)

1. Up: $0 = 30 - 10t$, so $t = 3\text{ s}$.
2. Coming down takes as long, so it lands after $6\text{ s}$.

(iii)

1. $40 = 30t - 5t^2$, so $t^2 - 6t + 8 = 0$.
2. $(t - 2)(t - 4) = 0$: at $t = 2\text{ s}$ and $t = 4\text{ s}$.