WAEC SSCE - Further Mathematics - 2017 (Essay)

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(a) Pass:fail \(=4:1\), so \(P(\text{pass})=0.8,\ P(\text{fail})=0.2\), with \(n=7\).

(i) At least 3 passed. Let \(X=\) number who pass, \(p=0.8\).

\[P(X\ge3)=1-P(0)-P(1)-P(2)\]

\(P(0)=(0.2)^{7}=0.0000128\), \(P(1)=7(0.8)(0.2)^{6}=0.000358\), \(P(2)=21(0.8)^{2}(0.2)^{5}=0.004301\).

\[P(X\ge3)=1-0.004672=0.995\approx1.00\]

(ii) Between 3 and 6 failed (i.e. \(4\) or \(5\) failed). Let \(Y=\) number who fail, \(p=0.2\).

\(P(Y=4)=\binom{7}{4}(0.2)^{4}(0.8)^{3}=35(0.0016)(0.512)=0.028672\).

\(P(Y=5)=\binom{7}{5}(0.2)^{5}(0.8)^{2}=21(0.00032)(0.64)=0.004301\).

\[P(4\le Y\le5)=0.028672+0.004301=0.032973\approx0.03\]

(b) A fair die thrown 5 times, \(P(\text{six})=\tfrac{1}{6}\). Probability of exactly three sixes:

\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]

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(a) \(g(x)=x+6\), so the inverse is \(g^{-1}(x)=x-6\).

\[f\circ g^{-1}(x)=f(x-6)=\frac{4-5(x-6)}{2}=\frac{4-5x+30}{2}=\frac{34-5x}{2}\]

(b) \(P\) divides \((7,-5)\) and \((-2,7)\) internally in \(5:4\). Using the section formula \(\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\) with \(m:n=5:4\):

\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]

\[P=\left(2,\ \tfrac{5}{3}\right)\]

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In parallelogram \(MNQR\) (vertices in order), the diagonals \(MQ\) and \(NR\) bisect each other, so they share the same midpoint.

Midpoint of \(MQ\): \(\left(\dfrac{4+8}{2},\dfrac{-6+16}{2}\right)=(6,\ 5)\).

Midpoint of \(NR\): \(\left(\dfrac{10+x}{2},\dfrac{2+y}{2}\right)\).

Equate:

\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]

\[R=(2,\ 8)\]

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Binomial with \(n=10\) bottles, defective probability \(p=0.08\), non-defective \(q=0.92\). Let \(X\) be the number of defective bottles.

\[P(X\ge3)=1-P(0)-P(1)-P(2)\]

\(P(0)=(0.92)^{10}=0.43439\).

\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).

\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).

\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]

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(a) Add the two fractions over a common denominator:

\[\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\frac{2}{1-\cos^{2}\theta}\]

Since \(1-\cos^{2}\theta=\sin^{2}\theta\):

\[=\frac{2}{\sin^{2}\theta}\]

(b) \(y=x^{2}(x-3)=x^{3}-3x^{2}\). Then \(\dfrac{dy}{dx}=3x^{2}-6x=3x(x-2)\).

Stationary points where \(\dfrac{dy}{dx}=0\): \(x=0\) or \(x=2\).

At \(x=0,\ y=0\Rightarrow(0,0)\). At \(x=2,\ y=8-12=-4\Rightarrow(2,-4)\).

Line joining them: slope \(=\dfrac{-4-0}{2-0}=-2\), through \((0,0)\):

\[y=-2x\quad\text{or}\quad 2x+y=0\]

Distance:

\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]

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(a)(i) \(f(x)=\dfrac{2x-3}{(x^{2}-1)(x+2)}=\dfrac{2x-3}{(x-1)(x+1)(x+2)}\) is undefined when the denominator is zero:

\[x=1,\quad x=-1,\quad x=-2\]

(ii) Partial fractions. Write

\[\frac{2x-3}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}\]

Cover-up (substitution) gives:

\(x=1:\ A=\dfrac{2(1)-3}{(2)(3)}=-\dfrac{1}{6}\).

\(x=-1:\ B=\dfrac{2(-1)-3}{(-2)(1)}=\dfrac{-5}{-2}=\dfrac{5}{2}\).

\(x=-2:\ C=\dfrac{2(-2)-3}{(-3)(-1)}=\dfrac{-7}{3}\).

\[f(x)=-\frac{1}{6(x-1)}+\frac{5}{2(x+1)}-\frac{7}{3(x+2)}\]

(b) Circle with centre \((-3,1)\) through \((3,1)\): radius \(=\sqrt{(3+3)^{2}+(1-1)^{2}}=6\).

\[(x+3)^{2}+(y-1)^{2}=36\]

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By the Factor Theorem, \(g(-1)=0\) and \(g(2)=0\).

\(g(x)=x^{4}+ax^{3}+bx^{2}-16x-12\).

Using \(g(-1)=0\):

\[1-a+b+16-12=0\ \Rightarrow\ -a+b+5=0\ \Rightarrow\ b=a-5\quad(1)\]

Using \(g(2)=0\):

\[16+8a+4b-32-12=0\ \Rightarrow\ 8a+4b-28=0\ \Rightarrow\ 2a+b=7\quad(2)\]

Substitute (1) into (2): \(2a+(a-5)=7\Rightarrow3a=12\Rightarrow a=4\).

Then \(b=4-5=-1\).

\[a=4,\qquad b=-1\]

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Rewrite the integrand by splitting \(x-1=(x+1)-2\):

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

Integrate term by term:

\[\int\left(\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\right)dx=\ln(x+1)+\frac{2}{x+1}+C\]

Evaluate from \(1\) to \(3\):

\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]

\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]

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(a)(i) Diagram

The body \(P\) hangs from the junction \(B\). String \(BA\) rises to the left at \(30^{\circ}\) to the horizontal and carries the tension \(T_{AB}=48\,\text{N}\); string \(BD\) rises to the right at \(60^{\circ}\) to the horizontal and carries the tension \(T_{DB}\); the weight \(W=qg\) acts vertically downward. The three forces meet at \(B\).

(a)(ii) Tension in DB

The three concurrent forces at \(B\) are \(T_{AB}=48\,\text{N}\), \(T_{DB}\) and the weight \(W\). The angles between successive forces are:

Angle between	Size	Opposite force
\(T_{AB}\) and \(T_{DB}\)	\(90^{\circ}\)	\(W\)
\(T_{DB}\) and \(W\)	\(150^{\circ}\)	\(T_{AB}\)
\(W\) and \(T_{AB}\)	\(120^{\circ}\)	\(T_{DB}\)

Applying Lami's theorem at \(B\):

\[\frac{T_{AB}}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}=\frac{W}{\sin 90^{\circ}}\]

From the first two ratios:

\[\frac{48}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}\]\[T_{DB}=\frac{48\sin 120^{\circ}}{\sin 150^{\circ}}=\frac{48\left(\tfrac{\sqrt3}{2}\right)}{\tfrac12}=\frac{48(0.8660)}{0.5}=83.14\,\text{N}\]

The tension in DB is \(83.14\,\text{N}\).

(a)(iii) Mass \(q\) of the body

Taking \(g=10\,\text{m s}^{-2}\), the weight is \(W=10q\). Using the first and third ratios of Lami's theorem:

\[\frac{10q}{\sin 90^{\circ}}=\frac{48}{\sin 150^{\circ}}\]\[10q=\frac{48\sin 90^{\circ}}{\sin 150^{\circ}}=\frac{48(1)}{0.5}=96\]\[q=\frac{96}{10}=9.6\,\text{kg}\]

The mass of the body is \(9.6\,\text{kg}\).

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(a) Using the laws of logarithms with \(\log_{10}p=a,\ \log_{10}q=b,\ \log_{10}s=c\):

\[\log_{10}\!\left(\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\right)=\tfrac{1}{3}\log_{10}p+4\log_{10}q-2\log_{10}s=\frac{a}{3}+4b-2c\]

(b) Area of a circle: \(A=\pi r^{2}\). Differentiate with respect to time:

\[\frac{dA}{dt}=2\pi r\,\frac{dr}{dt}\]

With \(\dfrac{dA}{dt}=20\,\text{cm}^2\text{s}^{-1}\) and \(r=6\,\text{cm}\):

\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]

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(a) \(f(x)=\displaystyle\int(4x-x^{2})\,dx=2x^{2}-\dfrac{x^{3}}{3}+C\).

Use \(f(3)=21\): \(2(9)-\dfrac{27}{3}+C=18-9+C=9+C=21\Rightarrow C=12\).

\[f(x)=2x^{2}-\frac{x^{3}}{3}+12\]

(b) Let the AP have first term \(a\) and common difference \(d\). The 2nd, 4th, 8th terms are \(a+d,\ a+3d,\ a+7d\) and form a GP:

\[(a+3d)^{2}=(a+d)(a+7d)\]

\[a^{2}+6ad+9d^{2}=a^{2}+8ad+7d^{2}\Rightarrow 2d^{2}-2ad=0\Rightarrow 2d(d-a)=0\]

Since \(d\neq0\), \(a=d\).

Sum of 3rd and 5th terms is \(20\): \((a+2d)+(a+4d)=2a+6d=20\Rightarrow a+3d=10\).

With \(a=d\): \(4d=10\Rightarrow d=2.5,\ a=2.5\).

(i) First four terms: \(2.5,\ 5,\ 7.5,\ 10\).

(ii) Sum of first ten terms:

\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]

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(a)(i) Magnitudes and directions.

\(m=6i+8j\): \(|m|=\sqrt{6^{2}+8^{2}}=10\). Direction \(=\tan^{-1}\dfrac{8}{6}=53.1^{\circ}\) above the positive \(x\)-axis.

\(n=-8i+\tfrac{7}{3}j\): \(|n|=\sqrt{(-8)^{2}+\left(\tfrac{7}{3}\right)^{2}}=\sqrt{\dfrac{576+49}{9}}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\approx8.33\). It is in the second quadrant, direction \(=180^{\circ}-\tan^{-1}\dfrac{7/3}{8}=180^{\circ}-16.3^{\circ}=163.7^{\circ}\).

(ii) Angle between m and n.

\[m\cdot n=6(-8)+8\left(\tfrac{7}{3}\right)=-48+\tfrac{56}{3}=-\tfrac{88}{3}\]

\[\cos\theta=\frac{m\cdot n}{|m|\,|n|}=\frac{-88/3}{10\times25/3}=\frac{-88}{250}=-0.352\]

\[\theta=\cos^{-1}(-0.352)\approx110.6^{\circ}\]

(b) \(\overrightarrow{PQ}=Q-P=\binom{10}{4}-\binom{-2}{3}=\binom{12}{1}\); \(\overrightarrow{RS}=S-R=\binom{4}{0}-\binom{3}{12}=\binom{1}{-12}\).

\[\overrightarrow{PQ}\cdot\overrightarrow{RS}=(12)(1)+(1)(-12)=0\]

Since the scalar product is zero, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).

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Resolve each force into components (angles measured anticlockwise from the positive \(x\)-axis).

\(F_{1}=18\,\text{N}\) at \(330^{\circ}\): \(x=18\cos330^{\circ}=15.59\), \(y=18\sin330^{\circ}=-9.00\).

\(F_{2}=10\,\text{N}\) at \(090^{\circ}\): \(x=0\), \(y=10\).

\(F_{3}=25\,\text{N}\) at \(180^{\circ}\): \(x=-25\), \(y=0\).

Sum the components:

\[R_{x}=15.59+0-25=-9.41,\qquad R_{y}=-9+10+0=1.00\]

Magnitude:

\[|R|=\sqrt{(-9.41)^{2}+(1.00)^{2}}=\sqrt{89.55}\approx9.5\ \text{N}\]

Direction: \(R\) lies in the second quadrant. The reference angle is \(\tan^{-1}\dfrac{1}{9.41}=6.1^{\circ}\), so

\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]