WAEC SSCE - Further Mathematics - 2008 (Essay)

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 \(k=-3.\)

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

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

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

(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.

(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.

(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 \(-24:\)

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

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

(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: \((5,\,0).\)

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.

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 \((r=0,1,2).\)

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

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

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

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

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):

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

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.

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 \(h\to0:\)

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

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 \(x\)

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

Factorising:

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

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

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

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":

Therefore the required probability is:

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.)

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.

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

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

(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 \(5\ \text{kg}\)):

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

(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,

\[k\approx3.8.\]

(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 \(\det(N)=-11.\))