WAEC SSCE - Further Mathematics - 2015 (Essay)

Given \(f''(x)=2,\ f'(1)=0,\ f(0)=-8\), find \(f(x)\).

Integrate \(f''(x)=2\) once:

\[f'(x)=2x+C_1\]

Use \(f'(1)=0\):

\[2(1)+C_1=0\;\Rightarrow\;C_1=-2,\qquad f'(x)=2x-2\]

Integrate again:

\[f(x)=x^{2}-2x+C_2\]

Use \(f(0)=-8\):

\[0-0+C_2=-8\;\Rightarrow\;C_2=-8\]

Therefore

\[f(x)=x^{2}-2x-8\]

(a) \(\overrightarrow{AB}=\binom{4}{5},\ \overrightarrow{BC}=\binom{-3}{5}\).

First, \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\binom{1}{10}\).

(i) Angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\[\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}|\,|\overrightarrow{AC}|}=\frac{(4)(1)+(5)(10)}{\sqrt{41}\,\sqrt{101}}=\frac{54}{64.35}=0.8392\]\[\theta=\cos^{-1}(0.8392)\approx 33.0^{\circ}\]

(ii) Unit vector along \(\overrightarrow{AB}-\overrightarrow{BC}\):

\[\overrightarrow{AB}-\overrightarrow{BC}=\binom{4-(-3)}{5-5}=\binom{7}{0},\quad|\cdot|=7\]\[\text{unit vector}=\frac{1}{7}\binom{7}{0}=\binom{1}{0}=i\]

(b) \(\overrightarrow{PQ}=2i+8j,\ \overrightarrow{PR}=11i-12j\); M divides QR in the ratio \(3:7\).

Using the section formula from P (with \(QM:MR=3:7\)):

\[\overrightarrow{PM}=\frac{7\,\overrightarrow{PQ}+3\,\overrightarrow{PR}}{10}=\frac{7(2i+8j)+3(11i-12j)}{10}\]\[=\frac{(14i+56j)+(33i-36j)}{10}=\frac{47i+20j}{10}=4.7i+2j\]

(a) \(f:x\to x^{2}+1\) and \(g:x\to 5-3x\) on \(\mathbb{R}\).

(i) Domain of \(f^{-1}\). The domain of the inverse equals the range of \(f\). Since \(x^{2}\ge 0\), \(f(x)=x^{2}+1\ge 1\). Hence

\[\text{Domain of }f^{-1}=\{x:x\ge 1\}=[1,\infty)\]

(ii) \(g^{-1}(2)\). Let \(y=5-3x\Rightarrow x=\dfrac{5-y}{3}\), so \(g^{-1}(x)=\dfrac{5-x}{3}\).

\[g^{-1}(2)=\frac{5-2}{3}=1\]

(b) Evaluate \(\displaystyle\int\frac{x+3}{x^{2}+6x+9}\,dx\).

The denominator is a perfect square: \(x^{2}+6x+9=(x+3)^{2}\).

\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]

\(a\,\Delta\,b=a^{3}-b^{3}\); evaluate \((\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)\).

By the definition,

\[(\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)=(\sqrt3+\sqrt2)^{3}-(\sqrt3-\sqrt2)^{3}\]

Expand each cube:

\[(\sqrt3+\sqrt2)^{3}=3\sqrt3+3(3)\sqrt2+3\sqrt3(2)+2\sqrt2=9\sqrt3+11\sqrt2\]\[(\sqrt3-\sqrt2)^{3}=3\sqrt3-3(3)\sqrt2+3\sqrt3(2)-2\sqrt2=9\sqrt3-11\sqrt2\]

Subtract:

\[(9\sqrt3+11\sqrt2)-(9\sqrt3-11\sqrt2)=22\sqrt2\]

The value is \(22\sqrt2\).

Solve \(\tan(2x-15)^{\circ}-1=0\) for \(0^{\circ}\le x\le 360^{\circ}\).

\[\tan(2x-15)^{\circ}=1\]

The tangent is \(1\) at \(45^{\circ}\) and every \(180^{\circ}\) thereafter, so

\[2x-15=45+180n\quad(n\in\mathbb{Z})\]\[2x=60+180n\;\Rightarrow\;x=30+90n\]

Take values of \(x\) in \([0^{\circ},360^{\circ}]\):

• \(n=0:\ x=30^{\circ}\)
• \(n=1:\ x=120^{\circ}\)
• \(n=2:\ x=210^{\circ}\)
• \(n=3:\ x=300^{\circ}\)

Hence \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).

Car: initial velocity \(u\), \(a=3\,\text{ms}^{-2}\), reaches \(v=33\,\text{ms}^{-1}\) after \(t=6\,\text{s}\).

First find \(u\) using \(v=u+at\):

\[33=u+3(6)\;\Rightarrow\;u=33-18=15\,\text{ms}^{-1}\]

Distance travelled, using \(s=ut+\tfrac12 at^{2}\):

\[s=15(6)+\tfrac12(3)(6)^{2}=90+54=144\,\text{m}\]

(Check with \(s=\tfrac{u+v}{2}t=\tfrac{15+33}{2}\times 6=24\times 6=144\,\text{m}\).)

The car travels \(144\,\text{m}\).

(a) Simplify \(\dfrac{\sqrt5+4}{3-2\sqrt5}-\dfrac{2+\sqrt5}{4-2\sqrt5}=a+b\sqrt5\).

Rationalise the first fraction by \(3+2\sqrt5\) (denominator \(9-20=-11\)):

\[\frac{(\sqrt5+4)(3+2\sqrt5)}{-11}=\frac{22+11\sqrt5}{-11}=-2-\sqrt5\]

Rationalise the second by \(4+2\sqrt5\) (denominator \(16-20=-4\)):

\[\frac{(2+\sqrt5)(4+2\sqrt5)}{-4}=\frac{18+8\sqrt5}{-4}=-\frac{9}{2}-2\sqrt5\]

Subtract:

\[(-2-\sqrt5)-\left(-\frac{9}{2}-2\sqrt5\right)=-2+\frac{9}{2}+(-\sqrt5+2\sqrt5)=\frac{5}{2}+\sqrt5\]

Hence \(a=\dfrac{5}{2},\ b=1\).

(b)(i) Evaluate the determinant.

\[\begin{vmatrix}2&-1&2\\1&3&4\\1&2&1\end{vmatrix}=2(3-8)-(-1)(1-4)+2(2-3)=-10-3-2=-15\]

(ii) Writing the system as \(2x-y+2z=-5,\ x+3y+4z=1,\ x+2y+z=-2\), the coefficient determinant is \(-15\). By Cramer's rule, replace the x-column with the constants:

\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]

(a) Edem and his wife join a family of 5, so 7 people sit in a row with Edem next to his wife.

Tie Edem and his wife into a single block. Then there are \(6\) units to arrange in a row:

\[6!=720\]

Within the block, Edem and his wife can swap in \(2!=2\) ways. Total:

\[720\times 2=1440\ \text{ways}\]

(b) Bag of 4 red and 5 black (total 9), 5 draws with replacement, so \(P(R)=\dfrac49,\ P(B)=\dfrac59\).

(i) Red picked exactly 3 times:

\[{}^{5}C_3\left(\tfrac49\right)^{3}\left(\tfrac59\right)^{2}=10\cdot\frac{64}{729}\cdot\frac{25}{81}=\frac{16000}{59049}\approx 0.2710\]

(ii) Black picked at most 2 times (0, 1 or 2), with \(P(B)=\tfrac59\):

\[P(0)=\left(\tfrac49\right)^{5}=\frac{1024}{59049}\]\[P(1)={}^{5}C_1\left(\tfrac59\right)\left(\tfrac49\right)^{4}=\frac{6400}{59049}\]\[P(2)={}^{5}C_2\left(\tfrac59\right)^{2}\left(\tfrac49\right)^{3}=\frac{16000}{59049}\]\[P(\le 2)=\frac{1024+6400+16000}{59049}=\frac{23424}{59049}\approx 0.3967\]

(a) Histogram. The classes have unequal widths, so the bars must be drawn using frequency density, not raw frequency:

\[\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\]

First close the gaps between successive classes to obtain the true class boundaries, then compute the width and frequency density of each class.

Height (m)	Class boundaries	Width	Frequency	Freq. density
0.4 - 0.5	0.35 - 0.55	0.2	2	10
0.6 - 0.9	0.55 - 0.95	0.4	8	20
1.0 - 1.2	0.95 - 1.25	0.3	12	40
1.3 - 1.4	1.25 - 1.45	0.2	6	30
1.5 - 1.7	1.45 - 1.75	0.3	6	20

Plot frequency density on the vertical axis against height (class boundaries) on the horizontal axis, drawing the bars with no gaps between them:

(b) Mean height (assumed-mean method, \(A=1.1\) m). Use the class mid-values \(x\) and the deviation \(d=x-A\):

\[\bar{x}=A+\frac{\sum fd}{\sum f}\]

Height (m)	Mid-value \(x\)	Frequency \(f\)	\(d=x-1.1\)	\(fd\)
0.4 - 0.5	0.45	2	-0.65	-1.30
0.6 - 0.9	0.75	8	-0.35	-2.80
1.0 - 1.2	1.10	12	0.00	0.00
1.3 - 1.4	1.35	6	0.25	1.50
1.5 - 1.7	1.60	6	0.50	3.00
Total		34		0.40

Therefore

\[\bar{x}=1.1+\frac{0.40}{34}=1.1+0.0118=1.1118\text{ m}\] \[\boxed{\bar{x}\approx 1.1\text{ m (to 1 decimal place)}}\]

The mean height of the group is 1.1 m.

8 boys and 6 girls; two students chosen at random.

Total ways to choose 2 from 14:

\[{}^{14}C_2=\frac{14\times 13}{2}=91\]

(a) Same sex.

\[{}^{8}C_2+{}^{6}C_2=\frac{8\times7}{2}+\frac{6\times5}{2}=28+15=43\]\[P(\text{same sex})=\frac{43}{91}\]

(b) Different sex.

\[{}^{8}C_1\times{}^{6}C_1=8\times6=48\]\[P(\text{different sex})=\frac{48}{91}\]

(Check: \(\dfrac{43}{91}+\dfrac{48}{91}=1\).)

Forces \(3\,\text{N},\ 4\,\text{N},\ 2\,\text{N}\) along \(j,\ -i+j,\ i+j\) respectively.

Convert each to components using the unit vector of its direction.

\[3\,\text{N along }j=(0,1):\quad (0,3)\]\[4\,\text{N along }-i+j=\frac{(-1,1)}{\sqrt2}:\quad \left(-\frac{4}{\sqrt2},\frac{4}{\sqrt2}\right)=(-2\sqrt2,\,2\sqrt2)\]\[2\,\text{N along }i+j=\frac{(1,1)}{\sqrt2}:\quad \left(\frac{2}{\sqrt2},\frac{2}{\sqrt2}\right)=(\sqrt2,\,\sqrt2)\]

Resultant components:

\[R_x=0-2\sqrt2+\sqrt2=-\sqrt2\approx-1.414\]\[R_y=3+2\sqrt2+\sqrt2=3+3\sqrt2\approx 7.243\]

Magnitude:

\[|R|=\sqrt{(-\sqrt2)^{2}+(3+3\sqrt2)^{2}}=\sqrt{2+(27+18\sqrt2)}=\sqrt{54.46}\approx 7.4\,\text{N}\]

The magnitude of the resultant is about \(7.4\,\text{N}\).

(a)(i) Binomial expansion of \((1+x)^{4}\):

\[(1+x)^{4}=1+4x+6x^{2}+4x^{3}+x^{4}\]

(ii) Evaluate \(\left(\dfrac54\right)^{4}=\left(1+\dfrac14\right)^{4}\) with \(x=\dfrac14\):

\[=1+4\left(\tfrac14\right)+6\left(\tfrac14\right)^{2}+4\left(\tfrac14\right)^{3}+\left(\tfrac14\right)^{4}\]\[=1+1+0.375+0.0625+0.003906=2.441406\approx 2.441\]

(b) A.P. with first term 7; its 1st, 2nd and 5th terms form a G.P.

The three A.P. terms are \(7,\ 7+d,\ 7+4d\). For a G.P. the middle term squared equals the product of the outer terms:

\[(7+d)^{2}=7(7+4d)\]\[49+14d+d^{2}=49+28d\]\[d^{2}-14d=0\;\Rightarrow\;d(d-14)=0\]

Taking the non-trivial value, the common difference is \(d=14\).

(a) Bucket of mass \(8\,\text{kg}\) pulled up, rope tension \(150\,\text{N}\), \(g=10\,\text{ms}^{-2}\).

Newton's second law (upward positive):

\[T-mg=ma\;\Rightarrow\;150-8(10)=8a\]\[70=8a\;\Rightarrow\;a=8.75\,\text{ms}^{-2}\ \text{(upward)}\]

(b) Mass \(12\,\text{kg}\), speed rises \(15\to 25\,\text{m/s}\) over \(50\,\text{m}\).

(i) Find the acceleration from \(v^{2}=u^{2}+2as\):

\[25^{2}=15^{2}+2a(50)\;\Rightarrow\;625=225+100a\;\Rightarrow\;a=4\,\text{ms}^{-2}\]\[F=ma=12\times 4=48\,\text{N}\]

(ii) Distance from the start (at \(15\,\text{m/s}\)) until the speed is \(35\,\text{m/s}\):

\[35^{2}=15^{2}+2(4)s\;\Rightarrow\;1225=225+8s\;\Rightarrow\;8s=1000\]\[s=125\,\text{m}\]

Spearman's rank correlation coefficient is \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\), where \(d\) is the difference between paired ranks and \(n\) is the number of individuals. Here \(n=5\).

Finalist	Judge X	Judge Y	\(d=X-Y\)	\(d^2\)
Anne	1	3	-2	4
Linda	4	2	2	4
Susan	3	4	-1	1
Rose	5	5	0	0
Erica	2	1	1	1
\(\sum d^2\)				10

\[r_s=1-\frac{6\times 10}{5(5^2-1)}=1-\frac{60}{5\times 24}=1-\frac{60}{120}=1-0.5=0.5\]

The Spearman's rank correlation coefficient is 0.5, indicating a moderate positive agreement between the two judges.

Square inscribed in a circle with opposite vertices \((2,1)\) and \((6,7)\).

Opposite vertices of the square are the ends of a diagonal, which is also a diameter of the circle.

(a) Centre = midpoint of the diagonal:

\[\left(\frac{2+6}{2},\frac{1+7}{2}\right)=(4,4)\]

(b) Equation. The diameter length is

\[d=\sqrt{(6-2)^{2}+(7-1)^{2}}=\sqrt{16+36}=\sqrt{52}\]

so the radius is \(r=\dfrac{\sqrt{52}}{2}=\sqrt{13}\) and \(r^{2}=13\). The equation of the circle is

\[(x-4)^{2}+(y-4)^{2}=13\]