WAEC SSCE - Further Mathematics - 2011 (Essay)

(a) Histogram

Since all the class intervals have the same width, namely 10 (in units of 100 km), the bar heights are proportional to the frequencies. Use continuous class boundaries as follows:

(b) Estimate of the mode

The modal class is \(30\text{–}39\), with boundaries \(29.5\text{–}39.5\). From the intersection of the diagonals drawn in the modal rectangle on the histogram, the modal value is approximately \(33.5\).

Equivalently,

\[\text{Mode}=29.5+\frac{93-69}{(93-69)+(93-57)}\times 10\]

\[=29.5+\frac{24}{60}\times10=\boxed{33.5}\]

Thus, the estimated modal tyre life span is \(\boxed{33.5\times100=3350\text{ km}}\).

Since \(f(-1) = 0\), \((x + 1)\) is a factor of \(f(x) = 6x^3 + 13x^2 + 2x - 5\).

Divide \(f(x)\) by \((x + 1)\):

\[6x^3 + 13x^2 + 2x - 5 = (x + 1)(6x^2 + 7x - 5).\]

Now factor the quadratic \(6x^2 + 7x - 5\). We need two numbers multiplying to \(6\times(-5) = -30\) and adding to \(7\): these are \(10\) and \(-3\):

\[6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5) = (3x + 5)(2x - 1).\]

Therefore

\[f(x) = (x + 1)(2x - 1)(3x + 5).\]

(a) A force \((F,\theta)\) resolves into the column vector \(\begin{pmatrix}F\cos\theta\\ F\sin\theta\end{pmatrix}\), where \(\theta\) is measured anticlockwise from the positive x-axis.

For \(r=(10\text{N},090^{\circ})\): \(\cos90^{\circ}=0,\ \sin90^{\circ}=1\).

\[r=\begin{pmatrix}0\\ 10\end{pmatrix}\]

For \(q=(8\text{N},135^{\circ})\): \(\cos135^{\circ}=-\frac{\sqrt2}{2},\ \sin135^{\circ}=\frac{\sqrt2}{2}\).

\[q=\begin{pmatrix}8(-\frac{\sqrt2}{2})\\ 8(\frac{\sqrt2}{2})\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 4\sqrt2\end{pmatrix}\]

(b) Add corresponding components:

\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]

(a) Differentiate \(\dfrac{x^2}{4}+y^2=1\) implicitly:

\[\frac{2x}{4}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}\]

At \(\left(1,\tfrac{\sqrt3}{2}\right)\): \(\dfrac{dy}{dx}=-\dfrac{1}{4\cdot\frac{\sqrt3}{2}}=-\dfrac{1}{2\sqrt3}=-\dfrac{\sqrt3}{6}\).

Tangent: \(y-\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3}{6}(x-1)\). Multiplying through by 6 and simplifying:

\[\sqrt3\,x+6y=4\sqrt3\quad\text{or}\quad x+2\sqrt3\,y=4\]

(b) Factorise the denominator: \(x^2+x-2=(x+2)(x-1)\). Write

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

Put \(x=1\): \(5=3B\Rightarrow B=\tfrac53\). Put \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).

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

(a) First principles: \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\) with \(f(x)=(2x+3)^2\).

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

\[=2(2x+3)(2h)+(2h)^2=4h(2x+3)+4h^2\]

Divide by \(h\): \(4(2x+3)+4h\). Taking the limit as \(h\to0\):

\[f'(x)=4(2x+3)=8x+12\]

(b) First expand the numerator: \((x+1)(x^2-2x+2)=x^3-x^2+2\). Then

\[\frac{x^3-x^2+2}{x^2}=x-1+\frac{2}{x^2}=x-1+2x^{-2}\]

\[\int_{1}^{2}\left(x-1+2x^{-2}\right)dx=\left[\frac{x^2}{2}-x-\frac{2}{x}\right]_{1}^{2}\]

At \(x=2\): \(2-2-1=-1\). At \(x=1\): \(\tfrac12-1-2=-2.5\).

\[=-1-(-2.5)=\tfrac32\]

Resolve each force into components \((F\cos\theta,\ F\sin\theta)\).

\(F_{1}=(35,210^{\circ})\): \(\cos210^{\circ}=-\frac{\sqrt3}{2},\ \sin210^{\circ}=-\frac12\), so \(F_{1}=\left(-\tfrac{35\sqrt3}{2},\,-\tfrac{35}{2}\right)\approx(-30.31,\,-17.5)\).

\(F_{2}=(35\sqrt3,300^{\circ})\): \(\cos300^{\circ}=\frac12,\ \sin300^{\circ}=-\frac{\sqrt3}{2}\), so \(F_{2}=\left(\tfrac{35\sqrt3}{2},\,-\tfrac{35\cdot3}{2}\right)\approx(30.31,\,-52.5)\).

(a) Resultant of \(F_{1}\) and \(F_{2}\):

\[R=\big(-30.31+30.31,\ -17.5-52.5\big)=(0,\,-70)\]

Magnitude \(=\sqrt{0^{2}+(-70)^{2}}=70\text{N}\), directed vertically downward (along \(270^{\circ}\)).

(b) The 40N resistance opposes the motion, i.e. it acts opposite to this 70N resultant. The net (resultant) force acting on the particle is therefore

\[70-40=30\text{N}\]

directed the same way as the \(70\)N resultant (\(270^{\circ}\)). Its magnitude is \(\boxed{30\text{N}}\).

Let \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The two mappings give:

\(A\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\Rightarrow 2a-3b=3,\ 2c-3d=4\)

\(A\begin{pmatrix}4\\5\end{pmatrix}=\begin{pmatrix}5\\6\end{pmatrix}\Rightarrow 4a+5b=5,\ 4c+5d=6\)

(a) Solve for \(a,b\): from \(2a-3b=3\) and \(4a+5b=5\), eliminating \(a\) gives \(11b=-1\Rightarrow b=-\tfrac{1}{11}\), then \(a=\tfrac{15}{11}\). Solve for \(c,d\): \(11d=-2\Rightarrow d=-\tfrac{2}{11}\), then \(c=\tfrac{19}{11}\).

\[A=\frac{1}{11}\begin{pmatrix}15&-1\\19&-2\end{pmatrix}\]

(b) \(\det A=\dfrac{(15)(-2)-(-1)(19)}{11^2}=\dfrac{-11}{121}=-\dfrac{1}{11}\).

\[A^{-1}=\frac{1}{\det A}\cdot\frac{1}{11}\begin{pmatrix}-2&1\\-19&15\end{pmatrix}=\begin{pmatrix}2&-1\\19&-15\end{pmatrix}\]

(c) The point mapping to \((-1,1)\) is \(A^{-1}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}2(-1)-1(1)\\19(-1)-15(1)\end{pmatrix}=\begin{pmatrix}-3\\-34\end{pmatrix}\).

So the point is \((-3,-34)\).

Total marbles \(=4R+6B+8G=18\).

(a) Without replacement.

(i) all green:

(ii) all the same colour = all red + all blue + all green:

(b) With replacement (each draw from all 18):

(a) Evaluate each term.

\({}^{9}P_{3}=9\times8\times7=504\), and \({}^{15}C_{3}=\dfrac{15\times14\times13}{3!}=455\), so \(\dfrac{{}^{9}P_{3}}{{}^{15}C_{3}}=\dfrac{504}{455}=1.1077\).

\({}^{5}C_{3}=10\), and \({}^{3}P_{2}=3\times2=6\), so \(\dfrac{{}^{5}C_{3}}{{}^{3}P_{2}}=\dfrac{10}{6}=1.6667\).

Sum \(=1.1077+1.6667=2.7744\approx 2.77\) (2 d.p.).

(b) We form a committee of 2 tutors and 5 pupils from 6 tutors and 10 pupils.

Tutors: one particular tutor is already chosen, so pick 1 more tutor from the remaining 5: \({}^{5}C_{1}=5\).

Pupils: two particular pupils are excluded, so choose 5 pupils from the remaining 8: \({}^{8}C_{5}=56\).

By the multiplication principle:

\[5\times56=280\ \text{ways}\]

(a) Using \((a+b)^5=\sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\) with \(a=2\) and \(b=-x\):

\[(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5\]

The coefficients come from \(\binom{5}{k}2^{5-k}(-1)^k\): \(32,\,-80,\,80,\,-40,\,10,\,-1\).

(b) Choose \(x\) so that \(2-x=1.98\), i.e. \(x=0.02\). Substitute:

• \(32\)
• \(-80(0.02)=-1.6\)
• \(+80(0.02)^2=+0.032\)
• \(-40(0.02)^3=-0.00032\)
• \(+10(0.02)^4=+0.0000016\)
• \(-(0.02)^5\approx 0\)

Adding: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).

\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]

Four out of twenty lack a valid licence, so \(P(\text{no licence})=\tfrac{4}{20}=0.2\) and \(P(\text{valid})=0.8\). With \(n=6\), \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\) (\(X\) = number with valid licences).

(a) exactly 2 valid:

(b) more than 3 valid \(=P(4)+P(5)+P(6)\):

(c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).

(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):

From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).

(b) \(A(-1,3),B(2,-1),C(5,3)\). \(\vec{AB}=\begin{pmatrix}3\\-4\end{pmatrix},\ |\vec{AB}|=5\); \(\vec{AC}=\begin{pmatrix}6\\0\end{pmatrix},\ |\vec{AC}|=6\).

(i) Unit vectors: \(\hat{AB}=\begin{pmatrix}3/5\\-4/5\end{pmatrix},\quad \hat{AC}=\begin{pmatrix}1\\0\end{pmatrix}.\)

(ii) \(\vec{AB}\cdot\vec{AC}=3(6)+(-4)(0)=18\), so

(a) Weight \(W=6.5\times 10=65\ \text{N}\). Let string \(AP\) (\(50^\circ\) to the downward vertical) carry tension \(T_1\) up to the left and \(BP\) (\(60^\circ\)) carry \(T_2\) up to the right.

(i) Component form:

(ii) Equilibrium of P: \(\vec{T_1}+\vec{T_2}+\vec{W}=\vec{0}\).

(iii) Horizontal: \(T_1\sin 50^\circ=T_2\sin 60^\circ\). Using Lami's theorem (angles \(110^\circ,120^\circ,130^\circ\)):

Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).

(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).

(a) Compute \(BA\):

\[BA=\begin{pmatrix}3&1\\2&3\end{pmatrix}\begin{pmatrix}-2&5\\4&3\end{pmatrix}=\begin{pmatrix}-6+4&15+3\\-4+12&10+9\end{pmatrix}=\begin{pmatrix}-2&18\\8&19\end{pmatrix}\]

Now the right-hand side:

\[2\begin{pmatrix}3&7\\-2&x\end{pmatrix}+\begin{pmatrix}y&4\\12&-3\end{pmatrix}=\begin{pmatrix}6+y&18\\8&2x-3\end{pmatrix}\]

Equating entries: top-left \(-2=6+y\Rightarrow y=-8\); bottom-right \(19=2x-3\Rightarrow x=11\). (The other entries \(18=18,\ 8=8\) check out.)

\[\boxed{x=11,\ y=-8}\]

(b)(i) Let \(y=\dfrac{5x-1}{3}\). Then \(3y=5x-1\Rightarrow x=\dfrac{3y+1}{5}\), so

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

(ii) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)

(a) Resolve the polar forces into components.

Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).

(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).

(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).

(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).

The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.

Write everything in terms of \(u=2^{y}\). Note \(2^{2y+2}=2^{2}\cdot 2^{2y}=4(2^{y})^{2}=4u^{2}\).

The equation \(2^{2y+2}-9(2^{y})=-2\) becomes:

\[4u^{2}-9u+2=0\]

Factorising (or using the formula \(u=\frac{9\pm\sqrt{81-32}}{8}=\frac{9\pm 7}{8}\)):

\[(4u-1)(u-2)=0\Rightarrow u=\tfrac14\ \text{or}\ u=2\]

Since \(u=2^{y}\):

• \(2^{y}=2\Rightarrow y=1\)
• \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)

\[\boxed{y=1\ \text{or}\ y=-2}\]

For \(2x^2 - 7x + 4 = 0\), the sum and product of the roots \(\alpha,\beta\) are

\[\alpha + \beta = \frac{7}{2},\qquad \alpha\beta = \frac{4}{2} = 2.\]

Sum of the new roots \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\). Using \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\):

\[\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2(2) = \frac{49}{4} - 4 = \frac{33}{4}.\]

\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{33/4}{2} = \frac{33}{8}.\]

Product of the new roots \(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1\).

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

\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]