WAEC SSCE - Further Mathematics - 2013 (Essay)

Resolve each force into east (x) and north (y) components using its bearing, where east \(= F\sin\theta\), north \(= F\cos\theta\).

\( \mathbf{P} = (4\sin 30^\circ,\ 4\cos 30^\circ) = (2.000,\ 3.464) \).
\( \mathbf{R} = (10\sin 300^\circ,\ 10\cos 300^\circ) = (-8.660,\ 5.000) \).

Resultant: \( \mathbf{P} + \mathbf{R} = (-6.660,\ 8.464) \).

The equilibrant \( \mathbf{E} \) is equal and opposite: \( \mathbf{E} = (6.660,\ -8.464) \).

Magnitude: \[ |\mathbf{E}| = \sqrt{6.660^2 + 8.464^2} = \sqrt{44.36 + 71.64} = \sqrt{116} \approx 10.8\text{ N}. \]

Direction (bearing): the equilibrant points east and south, so \[ \text{bearing} = 180^\circ - \tan^{-1}\!\left(\frac{6.660}{8.464}\right) = 180^\circ - 38.2^\circ \approx 142^\circ. \]

The equilibrating force is about \( 10.8\text{ N} \) on a bearing of \( 142^\circ \).

Take downward as positive with initial speed \(u = 20\text{ ms}^{-1}\), \(g = 10\text{ ms}^{-2}\), height \(s = 45\text{ m}\).

(a) Time to reach the ground. Using \( s = ut + \tfrac12 g t^2 \):

\[ 45 = 20t + 5t^2 \Rightarrow 5t^2 + 20t - 45 = 0 \Rightarrow t^2 + 4t - 9 = 0. \]

\[ t = \frac{-4 + \sqrt{16 + 36}}{2} = \frac{-4 + \sqrt{52}}{2} = \frac{-4 + 7.211}{2} \approx 1.61\text{ s}. \]

(b) Speed on hitting the ground. Using \( v^2 = u^2 + 2gs \):

\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]

Class width \(=3\); mid-values \(13,16,19,22,25\). Code with \(u=\dfrac{x-19}{3}\), \(A=19\).

(a) Mean age.\[\bar{x}=19+\frac{\sum fu}{\sum f}\times 3=19+\frac{-18}{22}\times 3=19-2.4545=16.5\text{ years}\]

(b) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=3\sqrt{\frac{40}{22}-\left(\frac{-18}{22}\right)^2}\]\[=3\sqrt{1.8182-0.6694}=3\sqrt{1.1488}=3(1.0718)=3.22\]

To three significant figures: mean age = 16.5 years and standard deviation = 3.22 years.

(a)–(c) Scatter diagram, mean point and required line of best fit

(b) Calculate the mean coordinates.

The \(x\)-values have total \(360\), so

The heights total \(12.4\), not \(11.4\):

Therefore the mean point to plot is \((45,1.55)\).

(c) Line of best fit. The instruction requires a straight line through both \((45,1.55)\) and \((10,1)\). This is the red line on the graph.

(d)(i) Equation of the line.

Using \((10,1)\):

So an appropriate equation is

(d)(ii) Height after 75 days.

The alternative equation \(y=0.017x+0.78\) is not consistent with the stated construction because it does not pass through \((10,1)\): it gives \(y=0.95\) when \(x=10\). For this question, use the two points explicitly specified for the line.

(a) Winning is binomial with \( p = 0.8,\ q = 0.2,\ n = 4 \).

(i) Wins all races: \( p^4 = 0.8^4 = 0.4096. \)

(ii) Wins no race: \( q^4 = 0.2^4 = 0.0016. \)

(iii) Wins at most 3 races \( = 1 - P(\text{all 4}) = 1 - 0.4096 = 0.5904. \)

(b) Committee of 5 chosen from 5 girls and 10 boys (15 people). Total \( = \binom{15}{5} = 3003 \).

(i) Exactly 3 boys (so 2 girls): \[ \frac{\binom{10}{3}\binom{5}{2}}{3003} = \frac{120 \times 10}{3003} = \frac{1200}{3003} = \frac{400}{1001} \approx 0.400. \]

(ii) At least one girl \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)

(a) Maximum and minimum points of \( y = 2x^{3} - 3x^{2} - 12x + 4 \).

At stationary points \( \dfrac{dy}{dx} = 0 \):

Setting \( \dfrac{dy}{dx} = 0 \):

The nature of each point is tested with the second derivative:

At \( x = -1 \):

Maximum point \( (-1,\ 11) \).

At \( x = 2 \):

Minimum point \( (2,\ -16) \).

(b) Sketch of the curve.

To locate the curve, note the following guide points:

Since the leading coefficient is positive, the curve rises from the bottom-left, climbs to the maximum \((-1,\ 11)\), falls through the \(y\)-intercept \((0,\ 4)\) to the minimum \((2,\ -16)\), then rises again to the top-right, as shown below.

Rectangle with one side three times the other; perimeter rises by 2%.

Let the shorter side be \(x\), so the longer side is \(3x\).

The perimeter \(P=8x\) is directly proportional to \(x\), so a \(2\%\) rise in perimeter is a \(2\%\) rise in \(x\). Using differentials for the area:

Alternatively, the new side is \(1.02x\), so the new area is \(3(1.02x)^{2}=1.0404\times 3x^{2}\), an increase of \(4.04\%\), which rounds to about \(4\%\).

The area increases by approximately \(4\%\).

Momentum \( = \text{mass} \times \text{velocity} \), and change in momentum \( = m(\vec{v}_f - \vec{v}_i) \).

Initial velocity: \( \vec{v}_i = 40\mathbf{j} \); final velocity: \( \vec{v}_f = 30\mathbf{i} \); mass \( m = 0.1\text{ kg} \).

\[ \Delta \vec{p} = 0.1(30\mathbf{i} - 40\mathbf{j}) = (3\mathbf{i} - 4\mathbf{j})\text{ kg ms}^{-1}. \]

Magnitude: \[ |\Delta \vec{p}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\text{ kg ms}^{-1}. \]

The change in momentum is \( 3\mathbf{i} - 4\mathbf{j} \) (magnitude \(5\text{ kg ms}^{-1}\)).

(a) Tabulate \( S = t^2 - 6t + 5 \) for \( 0 \le t \le 6 \) and plot S against t.

The graph is a parabola (minimum at \(t = 3\)).

(b)(i) Velocity is zero at the turning point of the graph, i.e. \( \dfrac{dS}{dt} = 2t - 6 = 0 \Rightarrow t = 3\text{ s}. \)

(ii) Average velocity over \( 0 \le t \le 4 \): \[ \frac{S(4) - S(0)}{4 - 0} = \frac{-3 - 5}{4} = -2\text{ ms}^{-1}. \]

(iii) Total distance over \( 0 \le t \le 5 \): the particle moves from \(S=5\) down to the minimum \(S(3) = -4\) (distance \(9\text{ m}\)), then back up to \(S(5) = 0\) (distance \(4\text{ m}\)):

\[ \text{Total distance} = 9 + 4 = 13\text{ m}. \]

(a) Differentiate \( y = \dfrac{x^2 + 1}{(x+1)^2} \) using the quotient rule with \( u = x^2+1,\ v = (x+1)^2 \):

\( u' = 2x,\ v' = 2(x+1) \).

\[ \frac{dy}{dx} = \frac{2x(x+1)^2 - (x^2+1)\,2(x+1)}{(x+1)^4} = \frac{2(x+1)\big[x(x+1) - (x^2+1)\big]}{(x+1)^4}. \]

\[ = \frac{2\big[x^2 + x - x^2 - 1\big]}{(x+1)^3} = \frac{2(x-1)}{(x+1)^3}. \]

(b)(i) Expand the determinant along the first row:

\[ \Delta = 1(3\cdot3 - (-1)\cdot1) - 2(2\cdot3 - (-1)(-1)) + (-1)(2\cdot1 - 3(-1)). \]

\[ \Delta = 1(10) - 2(5) - 1(5) = 10 - 10 - 5 = -5. \]

(ii) The system \( x+2y-z=4,\ 2x+3y-z=2,\ -x+y+3z=-1 \) has coefficient determinant \(\Delta = -5 \neq 0\), so use Cramer's rule.

\( \Delta_x = \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} = 40 - 10 - 5 = 25 \Rightarrow x = \dfrac{25}{-5} = -5. \)

\( \Delta_y = \begin{vmatrix} 1 & 4 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 3 \end{vmatrix} = 5 - 20 - 0 = -15 \Rightarrow y = \dfrac{-15}{-5} = 3. \)

\( \Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 2 \\ -1 & 1 & -1 \end{vmatrix} = -5 - 0 + 20 = 15 \Rightarrow z = \dfrac{15}{-5} = -3. \)

Solution: \( x = -5,\ y = 3,\ z = -3 \).

Choosing 3 prefects from 4 girls and 5 boys (9 people). Total selections: \[ \binom{9}{3} = 84. \]

(a) Only boys chosen: \[ \binom{5}{3} = 10, \qquad P = \frac{10}{84} = \frac{5}{42}. \]

(b) More girls than boys means either 3 girls (0 boys) or 2 girls and 1 boy:

\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).

\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]

(a) First compute the direction vector \( \mathbf{a} + \mathbf{b} - 2\mathbf{c} \):

\[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix}. \]

Its magnitude is \( \sqrt{10^2 + 4^2} = \sqrt{116} = 2\sqrt{29} \). The required vector \( \mathbf{d} \) has magnitude \( \sqrt{41} \) in this direction:

\[ \mathbf{d} = \sqrt{41}\cdot\frac{1}{\sqrt{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \sqrt{\frac{41}{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \approx \begin{pmatrix} 5.94 \\ 2.38 \end{pmatrix}. \]

So \( \mathbf{d} \approx 5.94\mathbf{i} + 2.38\mathbf{j} \) (check: \( 5.94^2 + 2.38^2 \approx 41 \)).

(b) With \( O \) the origin, \( \vec{OA} = (3,4) \) so \( |OA| = 5 \), and \( \vec{OB} = (3,n) \) with \( |OB| = \sqrt{9+n^2} \). The angle \( AOB = 30^\circ \):

\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]

\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]

\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]

\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]

\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]

Line \(2y=x+3\) meets circle \(x^{2}+y^{2}-2x+6y-15=0\).

From the line, \(x=2y-3\). Substitute into the circle:

So \(y=0\) or \(y=2\).

• \(y=0:\ x=2(0)-3=-3\), giving \((-3,0)\).
• \(y=2:\ x=2(2)-3=1\), giving \((1,2)\).

N is in the first quadrant, so \(N=(1,2)\) and \(M=(-3,0)\).

(a) Equal coefficients of \(x^{2}\) and \(x^{3}\) in \((p+qx)^{7}\).

The general term is \({}^{7}C_r\,p^{7-r}(qx)^{r}\).

Set them equal:

(b) Weekly contributions 180, 260, 340, ...

This is an A.P. with \(a=180\), common difference \(d=80\), and \(n=16\).

He would have contributed \(\text{N}12{,}480.00\) in 16 weeks.

Gradient of \(x^{3}+y^{3}-2xy=11\) at \((2,-1)\).

Differentiate implicitly with respect to \(x\), using the product rule on \(2xy\):

Collect the derivative terms:

At \((2,-1)\):

The gradient of the curve at \((2,-1)\) is \(14\).

(a) Let the three terms be \( \dfrac{a}{r},\ a,\ ar \).

Product: \( \dfrac{a}{r}\cdot a \cdot ar = a^3 = 8 \Rightarrow a = 2. \)

Sum: \( \dfrac{2}{r} + 2 + 2r = 7 \Rightarrow \dfrac{2}{r} + 2r = 5 \Rightarrow 2r^2 - 5r + 2 = 0. \)

\[ (2r - 1)(r - 2) = 0 \Rightarrow r = \tfrac12 \text{ or } r = 2. \]

(i) The sequence is decreasing, so \( r = \dfrac12 \).

(ii) The three terms are \( \dfrac{a}{r} = 4,\ a = 2,\ ar = 1 \), i.e. \( 4,\ 2,\ 1 \).

(b) With \( f(x) = x + \dfrac{2}{x^2} \) and ordinates at \( x = 1,2,3,4,5 \) (\(h = 1\)):

\( f(1)=3,\ f(2)=2.5,\ f(3)=3.2222,\ f(4)=4.125,\ f(5)=5.08. \)

Trapezium rule: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]

\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]

(a) Prepare a table of values for \(y = \sin 2x\) and \(y = \cos x\) at intervals of \(30^\circ\) for \(0^\circ \le x \le 210^\circ\). For example, at \(x = 30^\circ\): \(\sin 2x = \sin 60^\circ = 0.87\) and \(\cos x = \cos 30^\circ = 0.87\); at \(x = 120^\circ\): \(\sin 2x = \sin 240^\circ = -0.87\) and \(\cos x = \cos 120^\circ = -0.50\). The complete table is:

Plotting these points, using a scale of 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 0.2 units on the \(y\)-axis, gives the two smooth curves below:

(b)(i) The truth set of \(\sin 2x = 0\) is read where the curve \(y = \sin 2x\) crosses the \(x\)-axis (i.e. where \(y = 0\)). From the graph these crossings occur at \(x = 0^\circ,\ 90^\circ\) and \(180^\circ\):

(b)(ii) The truth set of \(\sin 2x - \cos x = 0\), i.e. \(\sin 2x = \cos x\), is read where the two curves intersect. From the graph the curves cross at \(x = 30^\circ,\ 90^\circ\) and \(150^\circ\).

This agrees with the algebra: \(\sin 2x = \cos x \Rightarrow 2\sin x\cos x = \cos x \Rightarrow \cos x\,(2\sin x - 1) = 0\), so \(\cos x = 0 \Rightarrow x = 90^\circ\), or \(\sin x = \tfrac{1}{2} \Rightarrow x = 30^\circ,\ 150^\circ\). Hence the truth set is:

(a) Bag: 5 blue, 4 green, 3 yellow (12 balls), draw 3 without replacement. Total selections \(={}^{12}C_3=220\).

(i) All three the same colour.\[{}^{5}C_3+{}^{4}C_3+{}^{3}C_3=10+4+1=15\]\[P=\frac{15}{220}=\frac{3}{44}\approx 0.068\]

(ii) Exactly two of the same colour (two alike and one different):\[{}^{5}C_2(7)+{}^{4}C_2(8)+{}^{3}C_2(9)=10(7)+6(8)+3(9)=70+48+27=145\]\[P=\frac{145}{220}=\frac{29}{44}\approx 0.659\](Check: all-different \(=5\times4\times3=60\); \(15+145+60=220\).)

(b) Spearman's rank correlation (Physics vs Chemistry ranks). \(n=7\).

\[r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6(16)}{7(48)}=1-\frac{96}{336}=1-0.286=0.714\]

Answers: (a)(i) \(\tfrac{3}{44}\); (a)(ii) \(\tfrac{29}{44}\); (b) \(r_s\approx 0.714\).