WAEC · 2010 · Nov/Dec · Further Maths · Paper 2

(a)

1. Work out the bracket first: $2 * 3 = \dfrac{2^2}{2 + 3} = \dfrac45$.
2. Now use $x = \frac45$ and $y = 5$: $\dfrac45 * 5 = \dfrac{\left(\frac45\right)^2}{\frac45 + 5}$
   $= \dfrac{\frac{16}{25}}{\frac{29}{5}}$.
3. Divide the fractions: $\dfrac{16}{25} \times \dfrac{5}{29} = \dfrac{16}{145}$.

(b)

1. Put $x + 1$ for $x$ and $x + 2$ for $y$: $\dfrac{(x + 1)^2}{(x + 1) + (x + 2)} = \dfrac13$.
2. Cross-multiply: $3(x + 1)^2 = 2x + 3$.
3. Expand: $3x^2 + 6x + 3 = 2x + 3$, so $3x^2 + 4x = 0$.
4. Factorise: $x(3x + 4) = 0$.
5. So $x = 0$ or $x = -\dfrac43$.

1. The series is an A.P. with $a = 4$ and $d = 3$.
2. Use $S_n = \dfrac n2[2a + (n - 1)d]$: $\dfrac n2[8 + 3(n - 1)] = 209$.
3. Simplify the bracket: $\dfrac n2(3n + 5) = 209$.
4. Multiply by 2 and rearrange: $3n^2 + 5n - 418 = 0$.
5. Factorise: $(3n + 38)(n - 11) = 0$.
6. $n$ must be a positive whole number, so $n = 11$.

1. The gradient of the tangent is $\tan 135^\circ = -1$.
2. Differentiate: $\dfrac{dy}{dx} = 2x - 3$.
3. Set the gradient equal to $-1$: $2x - 3 = -1$, so $x = 1$.
4. Find $y$ on the curve: $y = 1 - 3 + 4 = 2$, so the point is $(1, 2)$.
5. The tangent: $y - 2 = -1(x - 1)$.
6. Rearrange: $x + y - 3 = 0$.

1. Let the roots of $x^2 - mx + n = 0$ be $\alpha$ and $\beta$: $\alpha + \beta = m$ and $\alpha\beta = n$.
2. The new roots are $\alpha^2$ and $\beta^2$.
3. Their sum: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
   $= m^2 - 2n$.
4. Their product: $\alpha^2\beta^2 = (\alpha\beta)^2 = n^2$.
5. Use $x^2 - (\text{sum})x + \text{product} = 0$: $x^2 - (m^2 - 2n)x + n^2 = 0$.

1. Choosing 4 from $12$ people: ${}^{12}C_4 = 495$ ways.

(a)

1. 2 of the 5 boys and 2 of the 7 girls: ${}^5C_2 \times {}^7C_2 = 10 \times 21$
   $= 210$ ways.
2. Probability $= \dfrac{210}{495} \approx 0.42$.

(b)

1. At least 3 boys means 3 boys and 1 girl, or 4 boys.
2. 3 boys and 1 girl: ${}^5C_3 \times {}^7C_1 = 10 \times 7 = 70$ ways. 4 boys: ${}^5C_4 = 5$ ways.
3. Probability $= \dfrac{70 + 5}{495}$
   $= \dfrac{75}{495}$
   $\approx 0.15$.

(a)

1. Take the class mark (middle value) of each class: $1.41, 1.44, 1.47, 1.50, 1.53, 1.56, 1.59, 1.62$.
2. Multiply each by its frequency: $2.82, 5.76, 27.93, 45.00, 36.72, 21.84, 9.54, 1.62$.
3. Add them: $\sum fx = 151.23$, with $\sum f = 100$.
4. Mean $= \dfrac{151.23}{100} = 1.5123$ m.

(b)

1. The classes from $1.52$ m upwards lie above the mean: $24 + 14 + 6 + 1 = 45$ students.
2. Probability $= \dfrac{45}{100} = 0.45$.

(a)

1. Multiply each vector by its number: $4\mathbf p = \begin{pmatrix} 8 \\ -12 \end{pmatrix}$, $2\mathbf q = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$, $5\mathbf r = \begin{pmatrix} 15 \\ -10 \end{pmatrix}$.
2. Combine the components: $\begin{pmatrix} 8 - 8 + 15 \\ -12 - 4 - 10 \end{pmatrix} = \begin{pmatrix} 15 \\ -26 \end{pmatrix}$.

(b)

1. The point dividing $\mathbf p$ to $\mathbf q$ in the ratio $2 : 3$ has position vector $\dfrac{3\mathbf p + 2\mathbf q}{2 + 3}$.
2. Work out the top: $3\begin{pmatrix} 2 \\ -3 \end{pmatrix} + 2\begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 14 \\ -5 \end{pmatrix}$.
3. Divide by 5: $\dfrac15\begin{pmatrix} 14 \\ -5 \end{pmatrix} = \begin{pmatrix} 2.8 \\ -1 \end{pmatrix}$.

1. The weight is $6 \times 10 = 60$ N downwards.
2. The forces are the weight, the horizontal force $F$ and the tension $T$ along the string, $35^\circ$ to the vertical.

(a)

1. Resolve vertically: $T\cos35^\circ = 60$.
2. Resolve horizontally: $T\sin35^\circ = F$.
3. Divide the two equations: $F = 60\tan35^\circ = 42.0$ N.

(b)

1. From the vertical equation: $T = \dfrac{60}{\cos35^\circ} = 73.2$ N.

(a)

1. Let the centre be $(x, y)$.
2. The centre is the same distance from $A$ and $B$: $(x - 2)^2 + (y - 1)^2 = (x - 4)^2 + (y + 5)^2$.
3. Expand and simplify: $-4x - 2y + 5 = -8x + 10y + 41$, so $x - 3y = 9$.
4. The tangent $2x - y - 13 = 0$ has gradient 2, and the radius to $B$ is perpendicular to it, so the gradient of the radius is $-\frac12$.
5. So $\dfrac{y + 5}{x - 4} = -\dfrac12$, which gives $x + 2y = -6$.
6. Subtract the two equations: $5y = -15$, so $y = -3$, and then $x = 9 + 3(-3) = 0$.
7. The centre is $(0, -3)$.

(b)

1. Radius squared: $(0 - 2)^2 + (-3 - 1)^2 = 20$.
2. The circle: $x^2 + (y + 3)^2 = 20$.
3. Expand: $x^2 + y^2 + 6y - 11 = 0$.

(a)

1. Differentiate each side with respect to $x$, using the product rule on the right: $2x + 2y\dfrac{dy}{dx} = p(1 + x^2)\dfrac{dy}{dx} + 2pxy$.
2. Collect the $\dfrac{dy}{dx}$ terms: $(2y - p - px^2)\dfrac{dy}{dx} = 2pxy - 2x$.
3. Divide: $\dfrac{dy}{dx} = \dfrac{2pxy - 2x}{2y - p - px^2}$.

(b)

1. Expand: $\dfrac{d^2y}{dx^2} = 36 - 24x + 4x^2$.
2. Integrate once: $\dfrac{dy}{dx} = 36x - 12x^2 + \dfrac43x^3 + c$.
3. Integrate again: $y = 18x^2 - 4x^3 + \dfrac13x^4 + cx + k$.
4. At $x = 0$, $y = 0$: $k = 0$.
5. At $x = 3$, $y = 0$: $162 - 108 + 27 + 3c = 0$, so $c = -27$.
6. So $y = \dfrac13x^4 - 4x^3 + 18x^2 - 27x$.

(a)

1. Find where the curve meets the $x$-axis: $3x^2 + 2x - 1 = 0$, i.e. $(3x - 1)(x + 1) = 0$, so $x = -1$ or $x = \frac13$.
2. Integrate between these limits: $\displaystyle\int_{-1}^{1/3}(3x^2 + 2x - 1)\,dx = \big[x^3 + x^2 - x\big]_{-1}^{1/3}$.
3. Upper limit: $\frac1{27} + \frac19 - \frac13 = -\frac5{27}$.
4. Lower limit: $-1 + 1 + 1 = 1$.
5. Subtract: $-\frac5{27} - 1 = -\frac{32}{27}$.
6. The region is below the $x$-axis, so the area is $\dfrac{32}{27}$ square units.

(b)(i)

1. Take out 2 from the $x$ terms: $2x^2 - x - 3 = 2\left(x^2 - \frac12x\right) - 3$.
2. Complete the square: $2\left[\left(x - \frac14\right)^2 - \frac1{16}\right] - 3 = 2\left(x - \frac14\right)^2 - \frac{25}{8}$.
3. Compare: $P = 2$, $Q = -\frac14$ and $R = -\frac{25}{8}$.

(ii)

1. The square is never negative, so the least value is when $x = \frac14$: minimum $= -\dfrac{25}{8}$.

(a)

1. Integrate: $\left[\dfrac{x^3}{3} - x^2 + x\right]_1^m = \dfrac{m^3}{3} - m^2 + m - \dfrac13$.
2. Set this equal to $\frac13$: $\dfrac{m^3}{3} - m^2 + m - \dfrac23 = 0$.
3. Multiply by 3: $m^3 - 3m^2 + 3m - 2 = 0$.
4. Factorise ($m = 2$ makes it zero): $(m - 2)(m^2 - m + 1) = 0$.
5. $m^2 - m + 1 = 0$ has no real roots, so $m = 2$.

(b)

1. Differentiate each side with respect to $x$: $2(x - y)\left(1 - \dfrac{dy}{dx}\right) = 3y + 3x\dfrac{dy}{dx}$.
2. Put $x = 1$, $y = 0$: $2\left(1 - \dfrac{dy}{dx}\right) = 3\dfrac{dy}{dx}$.
3. So $2 = 5\dfrac{dy}{dx}$ and the gradient is $\dfrac25 = 0.4$.

1. The class marks are $5.5, 15.5, 25.5, \dots, 95.5$.
2. Deviations from $A = 45.5$: $d = -40, -30, -20, -10, 0, 10, 20, 30, 40, 50$.
3. Multiply by the frequencies: $fd = -80, -120, -80, -60, 0, 60, 100, 60, 80, 50$, so $\sum fd = 10$.

(i)

1. Mean $= A + \dfrac{\sum fd}{\sum f}$
   $= 45.5 + \dfrac{10}{40}$
   $= 45.75$.

(ii)

1. Square the deviations and multiply by $f$: $fd^2 = 3200, 3600, 1600, 600, 0, 600, 2000, 1800, 3200, 2500$, so $\sum fd^2 = 19\,100$.
2. Variance $= \dfrac{\sum fd^2}{\sum f} - \left(\dfrac{\sum fd}{\sum f}\right)^2$
   $= \dfrac{19\,100}{40} - \left(\dfrac{10}{40}\right)^2$.
3. Work it out: $477.5 - 0.0625 = 477.44$ (2 d.p.).

(a)

1. After the transfer, $Q$ holds 11 balls.
2. If a green ball moves ($\frac58$), $Q$ has 5 green: probability $\frac58 \times \frac5{11} = \frac{25}{88}$.
3. If a white ball moves ($\frac38$), $Q$ has 4 green: probability $\frac38 \times \frac4{11} = \frac{12}{88}$.
4. Add: $\dfrac{25 + 12}{88} = \dfrac{37}{88}$
   $\approx 0.42$.

(b)

1. The number of sixes is binomial with $n = 5$, $p = \frac16$, $q = \frac56$.
2. $P(0) = \left(\frac56\right)^5$
   $= \frac{3125}{7776}$ and $P(1) = 5 \times \frac16 \times \left(\frac56\right)^4$
   $= \frac{3125}{7776}$.
3. $P(\text{at least } 2) = 1 - \dfrac{6250}{7776}$
   $= \dfrac{1526}{7776}$
   $\approx 0.196$.

(c)

1. ${}^xC_3 = \dfrac{x!}{3!(x - 3)!}$ and ${}^xP_2 = \dfrac{x!}{(x - 2)!}$.
2. Divide: $\dfrac{{}^xC_3}{{}^xP_2} = \dfrac{(x - 2)!}{3!(x - 3)!}$
   $= \dfrac{x - 2}{6}$.
3. Solve $\dfrac{x - 2}{6} = \dfrac23$: $x - 2 = 4$, so $x = 6$.

(a)(i)

1. Choose 3 from 9 students: ${}^9C_3 = 84$ ways.

(ii)

1. 2 girls and 1 boy: ${}^4C_2 \times {}^5C_1 = 6 \times 5 = 30$ ways. 3 girls: ${}^4C_3 = 4$ ways.
2. Probability $= \dfrac{30 + 4}{84}$
   $= \dfrac{34}{84}$
   $\approx 0.405$.

(b)

1. The number of goals is binomial with $n = 5$, $p = 0.8$, $q = 0.2$.

(i)

1. $P(0) = (0.2)^5 = 0.00032 \approx 0.0003$.

(ii)

1. $P(1) = 5(0.8)(0.2)^4 = 0.0064$ and $P(2) = 10(0.8)^2(0.2)^3 = 0.0512$.
2. $P(\text{at most } 2) = 0.00032 + 0.0064 + 0.0512$
   $= 0.05792$
   $\approx 0.0579$.

(a)

1. $6\mathbf p - 5\mathbf q = (30\mathbf i + 18\mathbf j) - (20\mathbf i + 35\mathbf j)$
   $= 10\mathbf i - 17\mathbf j$.
2. Magnitude: $\sqrt{10^2 + (-17)^2} = \sqrt{389}$
   $\approx 19.723$.

(b)

1. $m(5\mathbf i + 3\mathbf j) + n(4\mathbf i + 7\mathbf j) = 14\mathbf i + 13\mathbf j$.
2. Compare the $\mathbf i$ and $\mathbf j$ parts: $5m + 4n = 14$ and $3m + 7n = 13$.
3. Solve: multiply the first by 3 and the second by 5 and subtract: $23n = 23$, so $n = 1$ and $m = 2$.

(c)

1. $\mathbf p \cdot \mathbf q = 5 \times 4 + 3 \times 7$
   $= 41$, $|\mathbf p| = \sqrt{34}$, $|\mathbf q| = \sqrt{65}$.
2. $\cos\theta = \dfrac{41}{\sqrt{34}\sqrt{65}}$
   $= 0.8710$.
3. So $\theta = 29.29^\circ$.

(a)

1. Resolve horizontally (to the right): $40\cos30^\circ + 15\sqrt3 = 20\sqrt3 + 15\sqrt3$
   $= 35\sqrt3$ N.
2. Resolve vertically (upwards): $40\sin30^\circ - 20 = 20 - 20 = 0$.
3. So the three forces add up to $35\sqrt3$ N horizontally to the right.
4. The force that stops $Q$ moving is equal and opposite: $35\sqrt3 \approx 60.6$ N horizontally to the left (bearing $270^\circ$).

(b)

1. Resolve east using $F\sin(\text{bearing})$: $10\sin90^\circ + 20\sin210^\circ + 4\sin330^\circ = 10 - 10 - 2$
   $= -2$.
2. Resolve north using $F\cos(\text{bearing})$: $0 + 20\cos210^\circ + 4\cos330^\circ = -17.321 + 3.464$
   $= -13.856$.
3. Magnitude: $\sqrt{(-2)^2 + (-13.856)^2} = \sqrt{196}$
   $= 14.0$ N.
4. Both parts are negative, so the resultant points south-west of south: angle west of south $= \tan^{-1}\dfrac{2}{13.856} = 8.2^\circ$.
5. Bearing $= 180^\circ + 8.2^\circ = 188.2^\circ$.

(a)(i)

1. Add the forces: $(3 - 2)\mathbf i + (5 + 3)\mathbf j = \mathbf i + 8\mathbf j$.
2. Magnitude: $\sqrt{1^2 + 8^2} = \sqrt{65} \approx 8.06$ N.

(ii)

1. $F = ma$: $a = \dfrac{\sqrt{65}}{2} \approx 4.03$ m s$^{-2}$.

(iii)

1. Change in velocity $= at$
   $= \dfrac{\sqrt{65}}{2} \times 5$
   $\approx 20.16$ m s$^{-1}$.

(b)

1. The weight is $1.5 \times 10 = 15$ N at the middle, 50 cm from the end: that is $50 - 20 = 30$ cm from the pivot.
2. $T$ acts 95 cm from the first end: $95 - 20 = 75$ cm from the pivot, on the same side as the weight.

(i)

1. Take moments about the pivot: $T \times 75 = 15 \times 30$, so $T = 6$ N.

(ii)

1. Upward forces equal downward forces: $R + 6 = 15$, so $R = 9$ N.