WAEC SSCE - Further Mathematics - 2006 (Essay)

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Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)

First equation. Since \(32=2^5,\) equate indices:

\[2x-3y=5.\quad(1)\]

Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)

Substitute (2) into (1):

\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]

Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)

A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence

\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]

Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.

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(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).

Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):

\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]

So \(a=9,\ b=-4,\ c=3\).

(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).

\((7,3):\ 58+14g+6f+c=0\)  (i)
\((-3,3):\ 18-6g+6f+c=0\)  (ii)
\((2,8):\ 68+4g+16f+c=0\)  (iii)

(i) - (ii): \(40+20g=0\Rightarrow g=-2\).

(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\)  (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\)  (v).

(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).

(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).

(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).

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(a) Choosing 2 from 4 boys and 5 girls (9 pupils):

\[P(\text{both boys})=\frac{\binom{4}{2}}{\binom{9}{2}}=\frac{6}{36}=\frac16.\]

(b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).

(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).

(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).

(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).

\[=1-0.000064-0.001536=0.9984.\]

(iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).

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Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)

(a) Values of m and n.

\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)

So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)

\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)

Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)

\[m=12,\qquad n=-19.\]

(b) Remainder when f(x) is divided by \((x+1).\)

By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)

\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]

Remainder \(=14.\)

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Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).

(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):

\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]

The maximum height reached is 320 m.

(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).

\[T = 2t = 2\times 8 = 16\ \text{s}.\]

The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)

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(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:

\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]

At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).

So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).

(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):

\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]

Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.

(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:

\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]

So \(\tan 15^\circ=2-\sqrt3\).

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(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).

(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).

(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).

(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).

Take moments about A (clockwise = anticlockwise):

\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\]

(ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).

\[R_A=210-100=\textbf{110 N}.\]

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A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.

Resultant force.

\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]

Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)

\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]

(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)

\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]

(b) Magnitude of the velocity after 4 s.

\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]

(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))

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(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:

\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]

So 20 triangles can be formed.

(b) Seating 6 people in a row with father and mother NOT together.

Total arrangements of 6 people in a row: \(6!=720.\)

Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:

\[5!\times2!=120\times2=240.\]

Therefore arrangements with them NOT together:

\[720-240=480.\]

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(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:

\[m_P(2u)+m_Q(0)=m_P\!\left(-\tfrac12 u\right)+m_Q(u).\] \[2m_P u+\tfrac12 m_P u=m_Q u\Rightarrow \tfrac52 m_P=m_Q\Rightarrow \frac{m_P}{m_Q}=\frac{2}{5}.\]

Ratio \(m_P:m_Q=\textbf{2:5}\).

(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:

\[R^2=P^2+Q^2+2PQ\cos\theta\Rightarrow 25=9+49+2(3)(7)\cos\theta.\] \[42\cos\theta=25-58=-33\Rightarrow \cos\theta=-0.7857\Rightarrow \theta\approx 141.8^\circ.\]

(c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then

\[\vec{CV}=\vec{CB}+\vec{BV}=\begin{pmatrix}2\\-3\end{pmatrix}+\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}.\]

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Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)

Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)

\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]

Gradient at the point. At \(x=1:\)

\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]

Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)

\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]

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(a) P owns \(5R,3B,2W\) (total 10); Q owns \(3R,2B,4W\) (total 9). One bus of P meets one bus of Q. First find P(same colour):

\[P(\text{both R})=\tfrac{5}{10}\cdot\tfrac{3}{9}=\tfrac{15}{90},\quad P(\text{both B})=\tfrac{3}{10}\cdot\tfrac{2}{9}=\tfrac{6}{90},\quad P(\text{both W})=\tfrac{2}{10}\cdot\tfrac{4}{9}=\tfrac{8}{90}.\] \[P(\text{same})=\frac{15+6+8}{90}=\frac{29}{90}.\] \[P(\text{not same colour})=1-\frac{29}{90}=\frac{61}{90}.\]

(b) Outward journey: air Nigeria \(\to\) Ghana in 6 ways; ship Ghana \(\to\) Liberia in 4 ways. On the return he must not repeat the airline or shipping line, so: ship Liberia \(\to\) Ghana in \(4-1=3\) ways; air Ghana \(\to\) Nigeria in \(6-1=5\) ways.

\[\text{Total}=6\times 4\times 3\times 5=360\ \text{ways}.\]

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(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).

\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]

The value of the integral is 1.

(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).

Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.

So \(p=3,\ q=7\).

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(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:

\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]

Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).

(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):

\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\]

(ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.

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The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.

Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).

Ages	Frequency \(f\)	Mid-value \(x\)	\(d=x-24.5\)	\(fd\)
15 - 18	40	16.5	-8	-320
19 - 22	33	20.5	-4	-132
23 - 26	25	24.5	0	0
27 - 30	10	28.5	4	40
31 - 34	8	32.5	8	64
35 - 38	4	36.5	12	48
Total	120			-300

\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]

The mean age of the distribution is 22 years.

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Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)

Set up the equations. Sum of 2nd and 5th terms is 42:

\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]

Difference between 6th and 3rd terms is 12:

\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]

(i) Common difference. \(3d=12\Rightarrow d=4.\)

(ii) First term. Substitute into \(2a+5d=42:\)

\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]

(iii) 20th term.

\[T_{20}=a+19d=11+19(4)=11+76=87.\]