WAEC SSCE - Further Mathematics - 2008 (Essay)

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(a) New equation whose roots are \((\alpha-2)\) and \((\beta-2).\)

For \(2x^2+5x-6=0,\) sum and product of roots:

\[\alpha+\beta=-\frac{5}{2},\qquad \alpha\beta=\frac{-6}{2}=-3.\]

Sum of new roots:

\[(\alpha-2)+(\beta-2)=(\alpha+\beta)-4=-\frac{5}{2}-4=-\frac{13}{2}.\]

Product of new roots:

\[(\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=-3-2\!\left(-\frac{5}{2}\right)+4=-3+5+4=6.\]

Required equation \(x^2-(\text{sum})x+(\text{product})=0:\)

\[x^2+\frac{13}{2}x+6=0\ \Rightarrow\ 2x^2+13x+12=0.\]

(b) Solve \(\displaystyle\int_0^k(x^2-2x)\,dx=4.\)

\[\int_0^k(x^2-2x)\,dx=\left[\frac{x^3}{3}-x^2\right]_0^k=\frac{k^3}{3}-k^2.\]

Set equal to 4 and multiply through by 3:

\[\frac{k^3}{3}-k^2=4\ \Rightarrow\ k^3-3k^2-12=0.\]

This cubic has a single real root. Testing shows \(f(3)=27-27-12=-12<0\) and \(f(4)=64-48-12=4>0,\) so the root lies between 3 and 4. Solving numerically,

\[k\approx3.8.\]

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Given \(f(x)=px^2+qx+2.\)

From \(f(-2)=0:\ p(-2)^2+q(-2)+2=0\Rightarrow 4p-2q+2=0\Rightarrow 2p-q=-1.\)

From \(f(1)=3:\ p+q+2=3\Rightarrow p+q=1.\)

Add the two results: \((2p-q)+(p+q)=-1+1\Rightarrow 3p=0\Rightarrow p=0.\) Then \(q=1.\)

So \(f(x)=x+2.\) Therefore

\[f(-4)=-4+2=-2.\]

Check: \(f(-2)=-2+2=0\) and \(f(1)=1+2=3,\) as required.

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Take consistent units of kilometres and hours. Initial velocity \(u=50\text{ km h}^{-1},\) distance \(s=20\text{ km},\) acceleration \(a=6\text{ km h}^{-2}.\)

Using \(s=ut+\tfrac12 at^2:\)

\[20=50t+\tfrac12(6)t^2\Rightarrow 20=50t+3t^2.\]

Rearrange into a standard quadratic in \(t:\)

\[3t^2+50t-20=0.\]

Apply the quadratic formula \(t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) with \(a=3,\ b=50,\ c=-20:\)

\[t=\frac{-50\pm\sqrt{2500+240}}{6}=\frac{-50\pm\sqrt{2740}}{6}.\]

Since \(\sqrt{2740}\approx52.35,\) the positive (physical) root is

\[t=\frac{-50+52.35}{6}=\frac{2.35}{6}\approx0.39\text{ h}.\]

\[t\approx0.4\text{ hours (to one decimal place)}.\]

(The negative root is rejected as time cannot be negative.)

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Differentiate \(f(x)=x^3+2x\) from first principles.

By definition,

\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.\]

Compute \(f(x+h):\)

\[f(x+h)=(x+h)^3+2(x+h)=x^3+3x^2h+3xh^2+h^3+2x+2h.\]

Then

\[f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h.\]

Divide by \(h\) (for \(h\neq0\)):

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

Now let \(h\to0:\)

\[f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2+2\right)=3x^2+2.\]

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From the Venn diagram the three disjoint regions are:

• Plantain only (\(P\) region outside the overlap): \(2x+1\)
• Both yam and plantain (the overlap): \(x\)
• Yam only (\(Y\) region outside the overlap): \((x-2)^2\)

Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.

(a) Finding the value of \(x\)

\[ (2x+1) + x + (x-2)^2 = 25 \]

Expand \((x-2)^2 = x^2 - 4x + 4\):

\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]

Factorising:

\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]

A number of guests cannot be negative, so we reject \(x = -4\).

\[ \boxed{x = 5} \]

(b) Probability that a guest chose only one of the two

Substitute \(x = 5\) into each region:

\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]

Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.

Guests who chose only one of the two are those in "plantain only" or "yam only":

\[ 11 + 9 = 20 \]

Therefore the required probability is:

\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]

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The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.

\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]

\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]

Set the dot product to zero:

\[(-1-k)(-4-k)+(-2)(-1)=0.\]

Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so

\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]

Factorise:

\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]

Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)

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This is a binomial situation with \(n=7\) families, probability of owning a car \(p=0.65,\) and \(q=1-p=0.35.\) The probability of exactly \(r\) successes is \(P(r)=\binom{7}{r}p^r q^{7-r}.\)

(a) Exactly 5 have cars.

\[P(5)=\binom{7}{5}(0.65)^5(0.35)^2=21\times0.116029\times0.1225\approx0.298.\]

(b) 3 or 4 have cars.

\[P(3)=\binom{7}{3}(0.65)^3(0.35)^4=35\times0.274625\times0.015006\approx0.144.\]

\[P(4)=\binom{7}{4}(0.65)^4(0.35)^3=35\times0.178506\times0.042875\approx0.268.\]

\[P(3\text{ or }4)=0.144+0.268\approx0.412.\]

(c) At most 2 have cars \((r=0,1,2).\)

\[P(0)=(0.35)^7\approx0.000643,\]

\[P(1)=\binom{7}{1}(0.65)(0.35)^6\approx0.008364,\]

\[P(2)=\binom{7}{2}(0.65)^2(0.35)^5\approx0.046590.\]

\[P(\text{at most }2)=0.000643+0.008364+0.046590\approx0.056.\]

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(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation

\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]

(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):

\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]

(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):

\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]

(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):

• \(5\ \text{N}\) at \(60^\circ\)
• \(5\sqrt3\ \text{N}\) at \(30^\circ\)
• \(10\ \text{N}\) at \(0^\circ\)
• \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
• \(5\ \text{N}\) at \(-60^\circ\)

(i) Magnitude of the resultant. Resolve horizontally:

\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]

\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]

Vertically the pairs cancel by symmetry:

\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]

Hence the resultant is

\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]

(ii) Acceleration of the body (mass \(5\ \text{kg}\)):

\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]

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Given \(\sin A=\dfrac{3}{5}\) with A obtuse, and \(\cos B=\dfrac{15}{17}\) with B acute.

Find cos A. A is obtuse (second quadrant), so cosine is negative. Using \(\cos^2A=1-\sin^2A:\)

\[\cos A=-\sqrt{1-\tfrac{9}{25}}=-\sqrt{\tfrac{16}{25}}=-\frac{4}{5}.\]

Find sin B. B is acute, so sine is positive:

\[\sin B=\sqrt{1-\tfrac{225}{289}}=\sqrt{\tfrac{64}{289}}=\frac{8}{17}.\]

Apply the compound-angle formula.

\[\cos(A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\!\left(\frac{15}{17}\right)-\left(\frac{3}{5}\right)\!\left(\frac{8}{17}\right).\]

\[=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}.\]

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Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)

(a) Position vector of M.

\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]

(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)

\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)

\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)

(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)

\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]

\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]

(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):

\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]

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(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.

(i) No restriction. Choose any 4 of the 12:

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \]

(ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):

\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \]

(b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.

Contestant	\(R_X\)	\(R_Y\)	\(d\)	\(d^2\)
P	2	5	-3	9
Q	7	7	0	0
R	1	3	-2	4
S	3	4	-1	1
T	6	1.5	4.5	20.25
U	5	1.5	3.5	12.25
V	4	6	-2	4
Total \(\sum d^2\)				50.5

\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 50.5}{7(49 - 1)} = 1 - \frac{303}{336} \approx 1 - 0.902 = \mathbf{0.1} \]

(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.

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The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.

(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)

\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]

So \(x=8\) or \(x=-8.\)

(b) Directions (to the nearest degree), measured from the positive x-axis.

When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.

\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]

When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.

\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]

So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.

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(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)

Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)

Write

\[\frac{2x^2-5x+1}{x(x-1)(x-3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-3}.\]

Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)

\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)

\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)

\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)

\[\frac{2x^2-5x+1}{x^3-4x^2+3x}=\frac{1}{3x}+\frac{1}{x-1}+\frac{2}{3(x-3)}.\]

(b) Solve \(\begin{vmatrix}x-3&-4&3\\5&2&2\\2&-4&6-x\end{vmatrix}=-24.\)

Expand along the first row:

\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]

\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]

Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is

\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]

Set equal to \(-24:\)

\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]

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

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(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)

Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:

\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)

Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)

First four terms: 5, 10, 15, 20.

General term:

\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]

So \(T_n=5n.\)

(b) Circle \(x^2+y^2-10x-8y+25=0.\)

Complete the square:

\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]

\[(x-5)^2+(y-4)^2=16.\]

Centre \((5,4),\) radius \(r=4.\)

(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.

(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)

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(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.

\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]

(ii) Image of \(P(3,-5).\)

\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]

Numerically \(P'\approx(5.83,\ 0.10).\)

(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)

(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)

(ii) Find a and b in \(N^2+aN+bI=0.\)

\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]

Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)

Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)

Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)

Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.

\[a=-1,\qquad b=-11.\]

(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))

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