WAEC SSCE - Further Mathematics - 2014 (Essay)

Reading the diagram, the three forces make these angles, measured anticlockwise from the positive \(x\)-axis:

• \(8\sqrt2\ \text{N}\): \(45^\circ\) from the \(y\)-axis in the first quadrant, i.e. at \(45^\circ\)
• \(10\ \text{N}\): \(30^\circ\) from the \(y\)-axis in the second quadrant, i.e. at \(120^\circ\)
• \(20\ \text{N}\): \(60^\circ\) below the negative \(x\)-axis in the third quadrant, i.e. at \(240^\circ\)

Resolve horizontally:

\[\sum F_x=8\sqrt2\cos45^\circ+10\cos120^\circ+20\cos240^\circ.\]

\[\sum F_x=8\sqrt2\cdot\tfrac{\sqrt2}{2}+10(-0.5)+20(-0.5)=8-5-10=-7\ \text{N}.\]

Resolve vertically:

\[\sum F_y=8\sqrt2\sin45^\circ+10\sin120^\circ+20\sin240^\circ.\]

\[\sum F_y=8+8.660+(-17.321)=-0.660\ \text{N}.\]

Both components are negative, so the resultant lies in the third quadrant, just below the negative \(x\)-axis. The acute angle it makes with the negative \(x\)-axis is

\[\alpha=\tan^{-1}\!\left(\frac{|\sum F_y|}{|\sum F_x|}\right)=\tan^{-1}\!\left(\frac{0.660}{7}\right)=\tan^{-1}(0.0943)\approx 5.4^\circ.\]

Measured anticlockwise from the positive \(x\)-axis, the direction of the resultant is

\[\theta=180^\circ+5.4^\circ\approx \mathbf{185.4^\circ}.\]

That is, the resultant points along the negative \(x\)-direction, inclined about \(5.4^\circ\) below it. (For reference its magnitude is \(\sqrt{(-7)^2+(-0.66)^2}\approx 7.03\ \text{N}\).)

(a) With \(x=3i-j\) and \(y=2i+kj\):

\[x\cdot y=(3)(2)+(-1)(k)=6-k,\quad |x|=\sqrt{3^2+(-1)^2}=\sqrt{10},\quad |y|=\sqrt{4+k^2}.\]

Given \(\cos\theta=\dfrac{\sqrt5}{5}=\dfrac{1}{\sqrt5}\):

\[\frac{6-k}{\sqrt{10}\,\sqrt{4+k^2}}=\frac{1}{\sqrt5}.\]

Cross-multiply and square:

\[\sqrt5\,(6-k)=\sqrt{10}\,\sqrt{4+k^2}\;\Rightarrow\;5(6-k)^2=10(4+k^2).\]

\[(6-k)^2=2(4+k^2)\;\Rightarrow\;36-12k+k^2=8+2k^2.\]

\[k^2+12k-28=0\;\Rightarrow\;(k-2)(k+14)=0.\]

So \(k=2\) or \(k=-14\). Both keep \(6-k>0\) (so \(\cos\theta>0\)), hence both are valid: \(\mathbf{k=2}\) or \(\mathbf{k=-14}\).

(b) Take \(A\) as the origin. Then, using the given vectors:

\[\overrightarrow{AB}=\begin{pmatrix}-5\\-1\end{pmatrix}\Rightarrow B=(-5,-1),\qquad \overrightarrow{AC}=\begin{pmatrix}-6\\-9\end{pmatrix}\Rightarrow C=(-6,-9).\]

\[D=B+\overrightarrow{BD}=\begin{pmatrix}-5\\-1\end{pmatrix}+\begin{pmatrix}4\\-7\end{pmatrix}=\begin{pmatrix}-1\\-8\end{pmatrix}\Rightarrow D=(-1,-8).\]

For ABCD (vertices in order) to be a parallelogram, opposite sides must be equal, i.e. \(\overrightarrow{AB}=\overrightarrow{DC}\). Now

\[\overrightarrow{DC}=C-D=\begin{pmatrix}-6-(-1)\\-9-(-8)\end{pmatrix}=\begin{pmatrix}-5\\-1\end{pmatrix}=\overrightarrow{AB}.\]

Since \(\overrightarrow{AB}=\overrightarrow{DC}\), the side AB is equal and parallel to DC. As a check, \(\overrightarrow{AD}=D-A=\begin{pmatrix}-1\\-8\end{pmatrix}\) and \(\overrightarrow{BC}=C-B=\begin{pmatrix}-1\\-8\end{pmatrix}\), so \(\overrightarrow{AD}=\overrightarrow{BC}\) as well. (Equivalently, the diagonals bisect each other: midpoint of \(AC=(-3,-4.5)=\) midpoint of \(BD\).)

Therefore ABCD is a parallelogram.

(a) Direction of the resultant of the two forces.

\(\overrightarrow{AB}=B-A=(3-(-1),\,5-2)=(4,3),\) with \(|\overrightarrow{AB}|=\sqrt{16+9}=5.\) A force of 5 N along \(\overrightarrow{AB}\) is \(\mathbf{F_1}=5\cdot\dfrac{(4,3)}{5}=(4,3).\)

\(\overrightarrow{CB}=B-C=(3-4,\,5-8)=(-1,-3),\) with \(|\overrightarrow{CB}|=\sqrt{1+9}=\sqrt{10}.\) A force of \(3\sqrt{10}\) N along \(\overrightarrow{CB}\) is \(\mathbf{F_2}=3\sqrt{10}\cdot\dfrac{(-1,-3)}{\sqrt{10}}=(-3,-9).\)

Resultant \(\mathbf{R}=\mathbf{F_1}+\mathbf{F_2}=(4-3,\,3-9)=(1,-6).\)

The resultant lies in the fourth quadrant. Its direction with the positive x-axis:

\[\theta=\tan^{-1}\!\left(\frac{-6}{1}\right),\qquad \tan^{-1}6=80.5^\circ.\]

So the resultant acts at \(80.5^\circ\) below the positive x-axis (equivalently a direction of \(-80.5^\circ,\) or \(279.5^\circ\) measured anticlockwise). Its magnitude, for reference, is \(\sqrt{1^2+6^2}=\sqrt{37}\approx6.08\text{ N}.\)

(b) Straight-line motion from rest.

(i) Acceleration. \(u=0,\ v=20\text{ m/s},\ s=8\text{ m}.\) Using \(v^2=u^2+2as:\)

\[400=0+2a(8)\Rightarrow a=\frac{400}{16}=25\text{ m/s}^2.\]

(ii) Time to cover 40 m. Using \(s=ut+\tfrac12 at^2\) with \(u=0:\)

\[40=\tfrac12(25)t^2\Rightarrow t^2=\frac{40}{12.5}=3.2\Rightarrow t=\sqrt{3.2}\approx1.79\text{ s}.\]

The panel has \(5\) men and \(3\) women, i.e. \(8\) people, from which a committee of \(3\) is chosen. Order does not matter, so we use combinations.

(a) Number of ways of forming the committee

\[\binom{8}{3}=\frac{8\times 7\times 6}{3\times 2\times 1}=56.\]

There are \(56\) possible committees.

(b) Probability that at least one woman is on the committee

Use the complement "no woman", i.e. an all-men committee chosen from the \(5\) men:

\[\binom{5}{3}=\frac{5\times 4\times 3}{3\times 2\times 1}=10.\]

\[P(\text{no woman})=\frac{10}{56}=\frac{5}{28}.\]

Therefore

\[P(\text{at least one woman})=1-\frac{5}{28}=\frac{23}{28}.\]

Differentiate \(xy^{2}+x^{2}y=4xy\) implicitly with respect to \(x\), treating \(y\) as a function of \(x\) and using the product rule on each term.

\[\underbrace{y^{2}+2xy\frac{dy}{dx}}_{\frac{d}{dx}(xy^{2})}+\underbrace{2xy+x^{2}\frac{dy}{dx}}_{\frac{d}{dx}(x^{2}y)}=\underbrace{4y+4x\frac{dy}{dx}}_{\frac{d}{dx}(4xy)}.\]

Collect the \(\dfrac{dy}{dx}\) terms on one side:

\[(2xy+x^{2}-4x)\frac{dy}{dx}=4y-y^{2}-2xy,\]

\[\frac{dy}{dx}=\frac{4y-y^{2}-2xy}{2xy+x^{2}-4x}.\]

Substitute the point \((1,3)\):

\[\text{Numerator}=4(3)-3^{2}-2(1)(3)=12-9-6=-3,\]

\[\text{Denominator}=2(1)(3)+1^{2}-4(1)=6+1-4=3.\]

\[\frac{dy}{dx}=\frac{-3}{3}=-1.\]

The gradient at \((1,3)\) is \(-1\).

The position vectors give \(P(11,1)\), \(Q\left(5,\tfrac{13}{3}\right)\), \(R(2,6)\).

(a) Show P, Q, R are collinear

Find the displacement vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{QR}\):

\[\overrightarrow{PQ}=Q-P=\left(5-11,\;\tfrac{13}{3}-1\right)=\left(-6,\;\tfrac{10}{3}\right),\]

\[\overrightarrow{QR}=R-Q=\left(2-5,\;6-\tfrac{13}{3}\right)=\left(-3,\;\tfrac{5}{3}\right).\]

Compare them: \(\overrightarrow{PQ}=2\left(-3,\tfrac{5}{3}\right)=2\,\overrightarrow{QR}\). Since \(\overrightarrow{PQ}\) is a scalar multiple of \(\overrightarrow{QR}\), the two vectors are parallel; and because they share the common point \(Q\), the points \(P\), \(Q\) and \(R\) lie on one straight line.

(b) Ratio \(|\overrightarrow{PQ}|:|\overrightarrow{QR}|\)

From \(\overrightarrow{PQ}=2\,\overrightarrow{QR}\), the magnitudes satisfy \(|\overrightarrow{PQ}|=2\,|\overrightarrow{QR}|\). Hence

\[|\overrightarrow{PQ}|:|\overrightarrow{QR}|=2:1.\]

Let the passing probabilities be \(P(K)=\dfrac{9}{10},\ P(W)=\dfrac{4}{5},\ P(A)=x,\) with failing probabilities \(\dfrac{1}{10},\ \dfrac{1}{5},\ (1-x)\) respectively.

(a) Value of x. "Only one passes" means exactly one of the three succeeds while the other two fail:

\[P(\text{only one})=\underbrace{\tfrac{9}{10}\cdot\tfrac{1}{5}\cdot(1-x)}_{K\text{ only}}+\underbrace{\tfrac{1}{10}\cdot\tfrac{4}{5}\cdot(1-x)}_{W\text{ only}}+\underbrace{\tfrac{1}{10}\cdot\tfrac{1}{5}\cdot x}_{A\text{ only}}.\]

\[=\frac{9}{50}(1-x)+\frac{4}{50}(1-x)+\frac{1}{50}x=\frac{13(1-x)+x}{50}=\frac{13-12x}{50}.\]

Set equal to \(\dfrac{9}{50}:\)

\[13-12x=9\Rightarrow 12x=4\Rightarrow x=\frac{1}{3}.\]

(b) Probability that at least one passes. Use the complement (none pass). With \(P(A)=\dfrac13,\) failing \(A=\dfrac23:\)

\[P(\text{none})=\frac{1}{10}\cdot\frac{1}{5}\cdot\frac{2}{3}=\frac{2}{150}=\frac{1}{75}.\]

\[P(\text{at least one})=1-\frac{1}{75}=\frac{74}{75}.\]

The operation is \(m\ast n=\dfrac{m^{2}-n^{2}}{2mn}\).

(a) Evaluate \(-3\ast 2\)

\[-3\ast 2=\frac{(-3)^{2}-2^{2}}{2(-3)(2)}=\frac{9-4}{-12}=\frac{5}{-12}=-\frac{5}{12}.\]

(b) Is \(\ast\) associative?

The operation is associative only if \((m\ast n)\ast p=m\ast(n\ast p)\) for all admissible values. Test the counterexample \(m=1,\;n=2,\;p=3\).

First \(m\ast n=\dfrac{1-4}{2(1)(2)}=-\dfrac{3}{4}\). Then

\[(m\ast n)\ast p=\left(-\tfrac{3}{4}\right)\ast 3=\frac{\left(-\frac{3}{4}\right)^{2}-3^{2}}{2\left(-\frac{3}{4}\right)(3)}=\frac{\frac{9}{16}-9}{-\frac{9}{2}}=\frac{-\frac{135}{16}}{-\frac{9}{2}}=\frac{135}{16}\times\frac{2}{9}=\frac{15}{8}.\]

Next \(n\ast p=\dfrac{4-9}{2(2)(3)}=-\dfrac{5}{12}\). Then

\[m\ast(n\ast p)=1\ast\left(-\tfrac{5}{12}\right)=\frac{1-\frac{25}{144}}{2(1)\left(-\frac{5}{12}\right)}=\frac{\frac{119}{144}}{-\frac{5}{6}}=\frac{119}{144}\times\left(-\frac{6}{5}\right)=-\frac{119}{120}.\]

Since \(\dfrac{15}{8}\neq -\dfrac{119}{120}\), we have \((m\ast n)\ast p\neq m\ast(n\ast p)\). Therefore \(\ast\) is not associative.

For \(3x^{2}+5x+1=0\) with roots \(\alpha,\beta\), the sum and product of roots are

\[\alpha+\beta=-\frac{5}{3},\qquad \alpha\beta=\frac{1}{3}.\]

Use the identity for the sum of cubes:

\[\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta).\]

Substitute the values:

\[\alpha^{3}+\beta^{3}=\left(-\frac{5}{3}\right)^{3}-3\left(\frac{1}{3}\right)\left(-\frac{5}{3}\right)=-\frac{125}{27}+\frac{5}{3}.\]

Writing over a common denominator of \(27\):

\[\alpha^{3}+\beta^{3}=-\frac{125}{27}+\frac{45}{27}=-\frac{80}{27}.\]

Therefore

\[27(\alpha^{3}+\beta^{3})=27\times\left(-\frac{80}{27}\right)=-80.\]

(a) Differentiate \((x-3)(x^2+5)\).

Expanding first: \((x-3)(x^2+5)=x^3+5x-3x^2-15=x^3-3x^2+5x-15.\)

Differentiating term by term:

\[\frac{dy}{dx}=3x^2-6x+5.\]

(The product rule gives the same result: \((1)(x^2+5)+(x-3)(2x)=x^2+5+2x^2-6x=3x^2-6x+5.\))

(b) \((x+1)^2\) is a factor of \(f(x)=x^3+ax^2+bx+3\).

(i) Values of a and b. If \((x+1)^2\) divides \(f(x)\), then \(x=-1\) is a repeated root, so both \(f(-1)=0\) and \(f'(-1)=0\).

\(f(-1)=-1+a-b+3=0\Rightarrow a-b=-2.\)

\(f'(x)=3x^2+2ax+b,\) so \(f'(-1)=3-2a+b=0\Rightarrow 2a-b=3.\)

Subtracting the first from the second: \(a=5.\) Then \(b=a+2=7.\)

\[a=5,\quad b=7.\]

(ii) Zeros of f(x). Now \(f(x)=x^3+5x^2+7x+3.\) Since \((x+1)^2=x^2+2x+1\) is a factor, divide:

\[x^3+5x^2+7x+3=(x+1)^2(x+3).\]

Therefore the zeros are \(x=-1\) (a double/repeated root) and \(x=-3.\)

Method. Assumed mean \(A = 1.45\), class width \(c = 0.2\), \(u = \dfrac{x - 1.45}{0.2}\) where \(x\) is the class midpoint.

Mean.

(a) Simplify \(\dfrac{5+\sqrt2}{3-\sqrt2}-\dfrac{5-\sqrt2}{3+\sqrt2}\).

Rationalise each fraction. For the first, multiply by \(\dfrac{3+\sqrt2}{3+\sqrt2}\):

\[\frac{(5+\sqrt2)(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)}=\frac{15+5\sqrt2+3\sqrt2+2}{9-2}=\frac{17+8\sqrt2}{7}.\]

For the second, multiply by \(\dfrac{3-\sqrt2}{3-\sqrt2}\):

\[\frac{(5-\sqrt2)(3-\sqrt2)}{9-2}=\frac{15-5\sqrt2-3\sqrt2+2}{7}=\frac{17-8\sqrt2}{7}.\]

Subtracting:

\[\frac{17+8\sqrt2}{7}-\frac{17-8\sqrt2}{7}=\frac{16\sqrt2}{7}.\]

So in the form \(a+b\sqrt2\): \(a=0,\ b=\dfrac{16}{7},\) i.e. \(\dfrac{16}{7}\sqrt2.\)

(b) Solve by the determinant (Cramer's) method:

\(3x-y-z=-2,\quad x+5y+2z=5,\quad 2x+3y+z=0.\)

Main determinant:

\[D=\begin{vmatrix}3&-1&-1\\1&5&2\\2&3&1\end{vmatrix}=3(5-6)+1(1-4)-1(3-10)=-3-3+7=1.\]

\[D_x=\begin{vmatrix}-2&-1&-1\\5&5&2\\0&3&1\end{vmatrix}=-2(5-6)+1(5-0)-1(15-0)=2+5-15=-8.\]

\[D_y=\begin{vmatrix}3&-2&-1\\1&5&2\\2&0&1\end{vmatrix}=3(5-0)+2(1-4)-1(0-10)=15-6+10=19.\]

\[D_z=\begin{vmatrix}3&-1&-2\\1&5&5\\2&3&0\end{vmatrix}=3(0-15)+1(0-10)-2(3-10)=-45-10+14=-41.\]

Therefore

\[x=\frac{D_x}{D}=-8,\quad y=\frac{D_y}{D}=19,\quad z=\frac{D_z}{D}=-41.\]

Check in equation (1): \(3(-8)-19-(-41)=-24-19+41=-2.\) Correct.

Let \(u=\log_{2}m\). Using the power law of logarithms, \(\log_{2}m^{3}=3\log_{2}m=3u\). The equation becomes a quadratic in \(u\):

\[u^{2}-3u=10\;\Rightarrow\; u^{2}-3u-10=0.\]

Factorising,

\[(u-5)(u+2)=0\;\Rightarrow\; u=5\quad\text{or}\quad u=-2.\]

Now convert back to \(m\) using \(u=\log_{2}m\Rightarrow m=2^{u}\):

• \(u=5:\quad m=2^{5}=32.\)
• \(u=-2:\quad m=2^{-2}=\dfrac{1}{4}.\)

Both values are positive, so both are valid solutions:

\[m=32\quad\text{or}\quad m=\tfrac{1}{4}.\]

(a) Frequency distribution table

(b) Cumulative frequency curve

Plot the cumulative frequencies against the corresponding upper class boundaries to obtain the less-than ogive shown below.

(i) Median

\[\frac{N}{2}=\frac{60}{2}=30.\]

From cumulative frequency 30, the score read from the ogive is

\[\boxed{\text{Median}=48.4}\]

(ii) Quartile deviation

\[Q_1=\frac{N}{4}=15\text{th score},\qquad Q_3=\frac{3N}{4}=45\text{th score}.\]

From the ogive, \(Q_1=35.5\) and \(Q_3=57.5\).

\[\text{Quartile deviation}=\frac{Q_3-Q_1}{2}=\frac{57.5-35.5}{2}=\boxed{11.0}.\]