WAEC · 2011 · May/June · Further Maths · Paper 2

1. $f(-1) = 0$, so by the factor theorem $(x + 1)$ is a factor.
2. Divide by $x + 1$ with synthetic division: write $-1$ on the left and $6, 13, 2, -5$.
3. Bring down 6.
4. Then $6 \times (-1) = -6$ and $13 - 6 = 7$.
5. $7 \times (-1) = -7$ and $2 - 7 = -5$.
6. Last, $-5 \times (-1) = 5$ and $-5 + 5 = 0$: no remainder.
7. So $f(x) = (x + 1)(6x^2 + 7x - 5)$.
8. Factorise the quadratic: two numbers that multiply to $6 \times (-5) = -30$ and add to 7 are 10 and $-3$.
9. So $6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5)$
   $= (2x - 1)(3x + 5)$.
10. The factors are $(x + 1)$, $(2x - 1)$ and $(3x + 5)$: $f(x) = (x + 1)(2x - 1)(3x + 5)$.

1. For $2x^2 - 7x + 4 = 0$: $\alpha + \beta = \frac72$ and $\alpha\beta = \frac42 = 2$.
2. The sum of the new roots: $\dfrac\alpha\beta + \dfrac\beta\alpha = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}$.
3. The top: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
   $= \frac{49}{4} - 4$
   $= \frac{33}{4}$.
4. Divide by $\alpha\beta = 2$: the new sum is $\frac{33}{8}$.
5. The product of the new roots: $\dfrac\alpha\beta \times \dfrac\beta\alpha = 1$.
6. The equation is $x^2 - (\text{sum})x + \text{product} = 0$: $x^2 - \frac{33}{8}x + 1 = 0$.
7. Multiply through by 8: $8x^2 - 33x + 8 = 0$.

(a)

1. Each term is $\binom5r 2^{5 - r}(-x)^r$, with coefficients $1, 5, 10, 10, 5, 1$.
2. $r = 0$: $32$.
3. $r = 1$: $5 \times 16 \times (-x) = -80x$.
4. $r = 2$: $10 \times 8 \times x^2 = 80x^2$.
5. $r = 3$: $10 \times 4 \times (-x^3) = -40x^3$.
6. $r = 4$: $5 \times 2 \times x^4 = 10x^4$.
7. $r = 5$: $-x^5$.
8. So $(2 - x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5$.

(b)

1. $2 - x = 1.98$, so $x = 0.02$.
2. Substitute: $32 - 1.6 + 0.032 - 0.00032 + 0.0000016 - \ldots$
3. That is $30.4316816\ldots$, so $(1.98)^5 = 30.4317$ to four decimal places.

1. Split the index: $2^{2y + 2} = 2^2 \times 2^{2y} = 4(2^y)^2$.
2. So the equation is $4(2^y)^2 - 9(2^y) = -2$.
3. Let $x = 2^y$: $4x^2 - 9x = -2$.
4. Add 2 to both sides: $4x^2 - 9x + 2 = 0$.
5. Factorise: $(4x - 1)(x - 2) = 0$, so $x = \frac14$ or $x = 2$.
6. Back to $y$: $2^y = \frac14 = 2^{-2}$ gives $y = -2$, and $2^y = 2 = 2^1$ gives $y = 1$.

1. For a force $(F, \theta)$ on a bearing, the column vector is $\begin{pmatrix} F\sin\theta \\ F\cos\theta \end{pmatrix}$ (east on top, north below).

(a)

1. $\mathbf r = \begin{pmatrix} 10\sin90^\circ \\ 10\cos90^\circ \end{pmatrix}$
   $= \begin{pmatrix} 10 \\ 0 \end{pmatrix}$.
2. $\mathbf q = \begin{pmatrix} 8\sin135^\circ \\ 8\cos135^\circ \end{pmatrix}$
   $= \begin{pmatrix} 4\sqrt2 \\ -4\sqrt2 \end{pmatrix}$
   $\approx \begin{pmatrix} 5.66 \\ -5.66 \end{pmatrix}$.

(b)

1. $\mathbf r + \mathbf q = \begin{pmatrix} 10 + 4\sqrt2 \\ -4\sqrt2 \end{pmatrix}$
   $\approx \begin{pmatrix} 15.66 \\ -5.66 \end{pmatrix}$.

(a)

1. Work out each symbol: $\,{}^9P_3 = 9 \times 8 \times 7 = 504$ and $\,{}^{15}C_3 = \dfrac{15 \times 14 \times 13}{6}$
   $= 455$.
2. $\,{}^5C_3 = 10$ and $\,{}^3P_2 = 3 \times 2 = 6$.
3. So the value is $\dfrac{504}{455} + \dfrac{10}{6} = 1.1077 + 1.6667$
   $= 2.7744$, which is $2.77$ to two decimal places.

(b)

1. Put the particular tutor on.
2. Choose the second tutor from the other 5: $\,{}^5C_1 = 5$.
3. Leave out the two particular pupils.
4. Choose 5 pupils from the other 8: $\,{}^8C_5 = 56$.
5. So $5 \times 56 = 280$ ways.

(a)

1. The classes have equal widths, so draw bars of heights $30, 69, 93, 57, 36, 15$ over the boundaries $9.5, 19.5, \ldots, 69.5$.

(b)

1. On the tallest bar (30–39), join each top corner to the top corner of the opposite neighbour.
2. The lines cross over the mode.
3. By calculation: $29.5 + \dfrac{93 - 69}{(93 - 69) + (93 - 57)} \times 10 = 29.5 + \dfrac{24}{60} \times 10$
   $= 33.5$ thousand km.

(i)

1. $F_1$: $35\sin210^\circ = -17.5$ east and $35\cos210^\circ = -30.31$ north.
2. $F_2$: $35\sqrt3\sin300^\circ = -52.5$ east and $35\sqrt3\cos300^\circ = 30.31$ north.
3. Add: $F_1 + F_2 = -70\mathbf i + 0\mathbf j$, so the resultant is $70\text{ N}$ (due west).

(ii)

1. The resistance acts against the motion, so the resultant force is $70 - 40 = 30\text{ N}$.

(a)

1. Expand: $f(x) = (2x + 3)^2 = 4x^2 + 12x + 9$.
2. $f(x + h) = 4(x + h)^2 + 12(x + h) + 9$
   $= 4x^2 + 8xh + 4h^2 + 12x + 12h + 9$.
3. Take away $f(x)$: $f(x + h) - f(x) = 8xh + 4h^2 + 12h$.
4. Divide by $h$: $8x + 4h + 12$.
5. Let $h \to 0$: $f'(x) = 8x + 12$.

(b)

1. Multiply out the top: $(x + 1)(x^2 - 2x + 2) = x^3 - x^2 + 2$.
2. Divide each term by $x^2$: the integrand is $x - 1 + 2x^{-2}$.
3. Integrate: $\left[\dfrac{x^2}{2} - x - \dfrac2x\right]_1^2$.
4. At $x = 2$: $2 - 2 - 1 = -1$.
5. At $x = 1$: $\frac12 - 1 - 2 = -\frac52$.
6. Subtract: $-1 - \left(-\frac52\right) = \dfrac32$.

(a)

1. Multiply in the order asked, $\mathbf{BA}$: row of $\mathbf B$ times column of $\mathbf A$.
2. Top row: $3(-2) + 1(4) = -2$ and $3(5) + 1(3) = 18$.
3. Bottom row: $2(-2) + 3(4) = 8$ and $2(5) + 3(3) = 19$.
4. So $\mathbf{BA} = \begin{pmatrix} -2 & 18 \\ 8 & 19 \end{pmatrix}$.
5. The right side: $\begin{pmatrix} 6 & 14 \\ -4 & 2x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix} = \begin{pmatrix} 6 + y & 18 \\ 8 & 2x - 3 \end{pmatrix}$.
6. Match the top-left entries: $6 + y = -2$, so $y = -8$.
7. Match the bottom-right entries: $2x - 3 = 19$, so $x = 11$.

(b)(i)

1. Write $y = \dfrac{5x - 1}{3}$.
2. Multiply by 3: $3y = 5x - 1$.
3. Add 1: $3y + 1 = 5x$.
4. Divide by 5: $x = \dfrac{3y + 1}{5}$, so $g^{-1}(x) = \dfrac{3x + 1}{5}$.

(ii)

1. $f$ acts first: $g^{-1} \circ f(x) = g^{-1}\left(\frac12x + 1\right)$
   $= \dfrac{3\left(\frac12x + 1\right) + 1}{5}$.
2. Multiply out the top: $\dfrac{\frac32x + 4}{5}$.
3. Multiply top and bottom by 2: $\dfrac{3x + 8}{10}$.

(a)

1. Let $\mathbf A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
2. Its images of $(2, 3)$ and $(4, 5)$ give four equations.
3. First rows: $2a + 3b = 3$ and $4a + 5b = 5$.
4. Double the first and subtract: $b = 1$, so $a = 0$.
5. Second rows: $2c + 3d = 4$ and $4c + 5d = 6$.
6. Likewise $d = 2$, so $c = -1$.
7. So $\mathbf A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}$.

(b)

1. $|\mathbf A| = 0(2) - 1(-1) = 1$, so $\mathbf A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$.

(c)

1. $\mathbf A^{-1}\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 - 1 \\ -1 + 0 \end{pmatrix}$
   $= \begin{pmatrix} -3 \\ -1 \end{pmatrix}$: the point $(-3, -1)$.

(a)

1. Differentiate each term with respect to $x$.
2. $y^2$ gives $2y\dfrac{dy}{dx}$: $\dfrac{x}{2} + 2y\dfrac{dy}{dx} = 0$.
3. Make $\dfrac{dy}{dx}$ the subject: $\dfrac{dy}{dx} = -\dfrac{x}{4y}$.
4. At $\left(1, \frac{\sqrt3}{2}\right)$: $\dfrac{dy}{dx} = -\dfrac{1}{4 \times \frac{\sqrt3}{2}}$
   $= -\dfrac{1}{2\sqrt3}$.
5. The tangent: $y - \dfrac{\sqrt3}{2} = -\dfrac{1}{2\sqrt3}(x - 1)$.
6. Multiply by $2\sqrt3$: $2\sqrt3y - 3 = -x + 1$.
7. Rearrange: $x + 2\sqrt3y - 4 = 0$.

(b)

1. Factorise the bottom: $x^2 + x - 2 = (x + 2)(x - 1)$.
2. Write $\dfrac{A}{x + 2} + \dfrac{B}{x - 1}$ and multiply through: $3x + 2 = A(x - 1) + B(x + 2)$.
3. Put $x = 1$: $5 = 3B$, so $B = \frac53$.
4. Put $x = -2$: $-4 = -3A$, so $A = \frac43$.
5. So the answer is $\dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)}$.

1. A success is a driver with a valid licence: $q = \frac{4}{20} = 0.2$, so $p = 0.8$, with $n = 6$.

(a)

1. $P(X = 2)$
   $= \binom62(0.8)^2(0.2)^4$
   $= 15 \times 0.64 \times 0.0016$
   $= 0.01536$
   $\approx 0.015$.

(b)

1. More than 3 means 4, 5 or 6: $P(4) = 0.24576$, $P(5) = 0.393216$ and $P(6) = 0.262144$.
2. Add: $0.90112 \approx 0.901$.

(c)

1. At least 5: $P(5) + P(6) = 0.393216 + 0.262144$
   $= 0.65536$
   $\approx 0.655$.

(a)

1. Upper class boundaries $9.5, 19.5, \ldots, 99.5$ with cumulative frequencies $2, 7, 15, 33, 53, 68, 73, 77, 79, 80$.
2. Plot each against its upper boundary and join with a smooth curve.

(b)(i)

1. $N = 80$.
2. Read across from 20 and 60: $Q_1 \approx 32.3$ and $Q_3 \approx 54.2$.
3. Semi-interquartile range $= \frac12(54.2 - 32.3) \approx 10.9$.

(ii)

1. Read up from 72: about 74 candidates scored less than 72, so about $80 - 74 = 6$ passed with distinction: $\frac{6}{80} \times 100 = 7.5\%$.

1. There are 18 marbles: 4 red, 6 blue and 8 green.

(a)(i)

1. Without replacement: $\dfrac{8}{18} \times \dfrac{7}{17} \times \dfrac{6}{16} = \dfrac{336}{4896}$
   $= \dfrac{7}{102}$.

(ii)

1. All red: $\dfrac{4 \times 3 \times 2}{4896} = \dfrac{24}{4896}$.
2. All blue: $\dfrac{6 \times 5 \times 4}{4896} = \dfrac{120}{4896}$.
3. Add the three: $\dfrac{24 + 120 + 336}{4896} = \dfrac{480}{4896}$
   $= \dfrac{5}{51}$
   $\approx 0.098$.

(b)

1. With replacement: $\left(\dfrac{4}{18}\right)^3 + \left(\dfrac{6}{18}\right)^3 + \left(\dfrac{8}{18}\right)^3 = \dfrac{64 + 216 + 512}{5832}$
   $= \dfrac{11}{81}$
   $\approx 0.136$.

(a)(i)

1. The strings make $40^\circ$ ($AP$) and $30^\circ$ ($BP$) with the horizontal.
2. The weight is $6.5 \times 10 = 65\text{ N}$.
3. $\mathbf T_1 = \begin{pmatrix} -T_1\cos40^\circ \\ T_1\sin40^\circ \end{pmatrix}$, $\mathbf T_2 = \begin{pmatrix} T_2\cos30^\circ \\ T_2\sin30^\circ \end{pmatrix}$ and $\mathbf W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}$.

(ii)

1. At rest: $\mathbf T_1 + \mathbf T_2 + \mathbf W = \mathbf 0$.

(iii)

1. Across: $T_1\cos40^\circ = T_2\cos30^\circ$, so $T_2 = 0.8846\,T_1$.
2. Up: $T_1\sin40^\circ + T_2\sin30^\circ = 65$, so $T_1(0.6428 + 0.4423) = 65$.
3. $T_1 = \dfrac{65}{1.0851}$
   $\approx 59.9\text{ N}$ and $T_2 \approx 53.0\text{ N}$.

(b)

1. Change in momentum $= m(\mathbf v - \mathbf u)$
   $= 5\begin{pmatrix} 3 + 1 \\ 4 - 2 \end{pmatrix}$
   $= \begin{pmatrix} 20 \\ 10 \end{pmatrix}$.
2. Its magnitude is $\sqrt{400 + 100} = 10\sqrt5$
   $\approx 22.36\text{ kg m s}^{-1}$.

(a)

1. Match the top parts: $2m - n = 5$.
2. Match the bottom parts: $m + 2n = -4$.
3. From the first, $n = 2m - 5$.
4. Substitute: $m + 4m - 10 = -4$, so $m = \frac65$.
5. Then $n = \frac{12}{5} - 5 = -\frac{13}{5}$.
6. $m + n = \frac65 - \frac{13}{5}$
   $= -\frac75$
   $= -1\frac25$.

(b)(i)

1. $\overrightarrow{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$, of length 5.
2. $\overrightarrow{AC} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$, of length 6.
3. Unit vectors: $\frac15\begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

(ii)

1. $\overrightarrow{AB} \cdot \overrightarrow{AC} = 18 + 0$
   $= 18$.
2. $\cos B\hat AC = \dfrac{18}{5 \times 6}$
   $= 0.6$, so $B\hat AC \approx 53^\circ$.

(a)(i)

1. $F_1$: $3\sin210^\circ = -1.5$ east and $3\cos210^\circ = -2.598$ north.
2. $F_2$: $4\sin120^\circ = 3.464$ east and $4\cos120^\circ = -2$ north.
3. Resultant: $1.964\mathbf i - 4.598\mathbf j$, of size $\sqrt{3.857 + 21.143} = \sqrt{25}$
   $= 5\text{ N}$.
4. (A check: the bearings differ by $90^\circ$, so $\sqrt{3^2 + 4^2} = 5$.)
5. $a = \dfrac{F}{m}$
   $= \dfrac57$
   $\approx 0.71\text{ m s}^{-2}$.

(ii)

1. From rest, $v = at$
   $= \frac57 \times 3$
   $= \frac{15}{7}$
   $\approx 2.14\text{ m s}^{-1}$.

(b)

1. $F_1 + F_2 + F_3 = (2 + 0 + 6)\mathbf i + (3 - 5 - 4)\mathbf j$
   $= 8\mathbf i - 6\mathbf j$.
2. The fourth force cancels it: $-8\mathbf i + 6\mathbf j$.
3. Its size is $\sqrt{64 + 36} = 10\text{ N}$.
4. It points 8 west and 6 north: $\tan^{-1}\frac86 \approx 53^\circ$ west of north, a bearing of about $307^\circ$.