WAEC SSCE - Further Mathematics - 2007 (Essay)

Antwortdetails

Class width \(=5\); mid-values \(19,24,\dots,54\). Code with \(u=\dfrac{x-39}{5}\), \(A=39\).

Age	Mid \(x\)	\(f\)	\(u\)	\(fu\)	\(fu^2\)
17-21	19	12	-4	-48	192
22-26	24	24	-3	-72	216
27-31	29	30	-2	-60	120
32-36	34	37	-1	-37	37
37-41	39	45	0	0	0
42-46	44	25	1	25	25
47-51	49	10	2	20	40
52-56	54	7	3	21	63
Total		190		-151	693

(a)(i) Mean.\[\bar{x}=39+\frac{\sum fu}{\sum f}\times 5=39+\frac{-151}{190}\times 5=39-3.97=35.0\text{ years}\]

(a)(ii) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=5\sqrt{\frac{693}{190}-\left(\frac{-151}{190}\right)^2}\]\[=5\sqrt{3.647-0.632}=5\sqrt{3.015}=5(1.737)=8.68\text{ years}\]

(b) Probability at most 36 years old. Ages \(\le 36\) cover the classes \(17-21,\,22-26,\,27-31,\,32-36\), giving \(12+24+30+37=103\) workers out of 190.\[P(\text{at most }36)=\frac{103}{190}=0.542\]

Mean \(\approx\) 35.0 years, standard deviation \(\approx\) 8.68 years, and \(P=0.542\).

Antwortdetails

(a) For \(x^2 + mx + 11 = 0\) with roots \(\alpha,\beta\): \(\alpha + \beta = -m\) and \(\alpha\beta = 11\).

\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = m^2 - 22\]

Setting this equal to 27:

\[m^2 - 22 = 27 \ \Rightarrow\ m^2 = 49 \ \Rightarrow\ m = \pm 7\]

(b) From the line \(2x + 3y = 1\), \(x = \dfrac{1 - 3y}{2}\). Substitute into the circle \(2x^2 + 2y^2 + 4x + 9y - 9 = 0\) (dividing by 2 first: \(x^2 + y^2 + 2x + 4.5y - 4.5 = 0\)):

\[\frac{(1 - 3y)^2}{4} + y^2 + (1 - 3y) + 4.5y - 4.5 = 0\]

Multiply through by 4 and simplify:

\[13y^2 - 13 = 0 \ \Rightarrow\ y^2 = 1 \ \Rightarrow\ y = \pm 1\]

For \(y = 1\): \(x = \dfrac{1 - 3}{2} = -1\), giving \((-1, 1)\).

For \(y = -1\): \(x = \dfrac{1 + 3}{2} = 2\), giving \((2, -1)\), which lies in the fourth quadrant.

Therefore \(P = (-1, 1)\) and \(Q = (2, -1)\).

Antwortdetails

Differentiate the curve \(y = 2x^2 + x - 3\):

\[\frac{dy}{dx} = 4x + 1\]

At \((2, 7)\) the gradient of the tangent is \(4(2) + 1 = 9\). The normal is perpendicular, so its gradient is:

\[m = -\frac{1}{9}\]

Equation of the normal through \((2, 7)\):

\[y - 7 = -\tfrac{1}{9}(x - 2)\]

The normal meets the \(x\)-axis where \(y = 0\):

\[-7 = -\tfrac{1}{9}(x - 2) \ \Rightarrow\ 63 = x - 2 \ \Rightarrow\ x = 65\]

Therefore \(P = (65,\ 0)\).

Antwortdetails

(a) Solve \(\sqrt{4x-3}-\sqrt{2x-5}=2\).

Isolate one surd and square:

\[\sqrt{4x-3}=2+\sqrt{2x-5}\]\[4x-3=4+4\sqrt{2x-5}+(2x-5)\]\[4x-3=2x-1+4\sqrt{2x-5}\]\[2x-2=4\sqrt{2x-5}\;\Rightarrow\;x-1=2\sqrt{2x-5}\]

Square again:

\[x^{2}-2x+1=4(2x-5)=8x-20\]\[x^{2}-10x+21=0\;\Rightarrow\;(x-3)(x-7)=0\]

So \(x=3\) or \(x=7\). Check both in the original equation:

• \(x=3:\ \sqrt{9}-\sqrt{1}=3-1=2\) ✓
• \(x=7:\ \sqrt{25}-\sqrt{9}=5-3=2\) ✓

Both are valid, so \(x=3\) or \(x=7\).

(b) Area enclosed by \(y^{2}=4x\) and \(y+x=0\).

The line is \(x=-y\). Substitute into the parabola:

\[y^{2}=4(-y)\;\Rightarrow\;y^{2}+4y=0\;\Rightarrow\;y(y+4)=0\]

So \(y=0\) (giving \((0,0)\)) and \(y=-4\) (giving \((4,-4)\)). Integrating with respect to \(y\), the line \(x=-y\) lies to the right of the parabola \(x=\tfrac{y^{2}}{4}\) on \(-4\le y\le 0\):

\[A=\int_{-4}^{0}\left(-y-\frac{y^{2}}{4}\right)\,dy=\left[-\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{-4}^{0}\]\[=0-\left(-\frac{16}{2}-\frac{-64}{12}\right)=0-\left(-8+\frac{16}{3}\right)=\frac{8}{3}\]

The finite area is \(\dfrac{8}{3}\approx 2.67\) square units.

Antwortdetails

Let the winning probabilities be \(p_1 = \tfrac{1}{5},\ p_2 = \tfrac{1}{4},\ p_3 = \tfrac{1}{3},\ p_4 = \tfrac{1}{2}\), with losing probabilities \(q_1 = \tfrac{4}{5},\ q_2 = \tfrac{3}{4},\ q_3 = \tfrac{2}{3},\ q_4 = \tfrac{1}{2}\).

"At most two win" is easier via the complement: \(1 - P(3\text{ win}) - P(4\text{ win})\).

All four win:

\[P(4) = \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} = \frac{1}{120}\]

Exactly three win (one loses at a time, summed):

\[\tfrac{4}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{3}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2}\] \[= \frac{4 + 3 + 2 + 1}{120} = \frac{10}{120}\]

So \(P(3\text{ or }4) = \dfrac{10}{120} + \dfrac{1}{120} = \dfrac{11}{120}\), and:

\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]

Antwortdetails

Height: \(h = 20t - \tfrac{3}{2}t^2 - \tfrac{2}{3}t^3\). Velocity is \(v = \dfrac{dh}{dt}\):

\[v = 20 - 3t - 2t^2\]

(a) Momentarily at rest when \(v = 0\):

\[2t^2 + 3t - 20 = 0 \ \Rightarrow\ t = \frac{-3 \pm \sqrt{9 + 160}}{4} = \frac{-3 \pm 13}{4}\]

Taking the positive root, \(t = \dfrac{10}{4} = 2.5\,\text{s}\).

(b) Maximum height at \(t = 2.5\,\text{s}\):

\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]

Antwortdetails

(a) Trapezium rule, seven ordinates, \(\displaystyle\int_{0}^{3}\frac{dx}{x^{2}+1}\).

Seven ordinates means six strips, so \(h=\dfrac{3-0}{6}=0.5\). With \(y=\dfrac{1}{x^{2}+1}\):

x	0	0.5	1	1.5	2	2.5	3
y	1.0000	0.8000	0.5000	0.3077	0.2000	0.1379	0.1000

\[\int\approx\frac{h}{2}\Big[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)\Big]\]\[=\frac{0.5}{2}\Big[(1.1000)+2(1.9456)\Big]=0.25(1.1000+3.8912)\]\[=0.25(4.9912)=1.2478\approx 1.25\]

(b) Matrix method: \(-2x+y=3,\ -x+4y=1\).

\[\begin{pmatrix}-2&1\\-1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}\]

Determinant \(=(-2)(4)-(1)(-1)=-8+1=-7\). Inverse:

\[\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}4&-1\\1&-2\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}12-1\\3-2\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}11\\1\end{pmatrix}\]\[x=-\frac{11}{7},\qquad y=-\frac{1}{7}\]

Antwortdetails

Five different digits are available: \(4, 5, 6, 7, 8\).

(a)(i) Repetition allowed: each of the 5 positions can be any of the 5 digits:

\[5^5 = 3125\]

(ii) Repetition not allowed: arrange all 5 distinct digits:

\[5! = 120\]

(b) A number is odd if its last digit is odd. Among the digits, the odd ones are \(5\) and \(7\), giving 2 choices for the units position. With no repetition, the remaining 4 positions are filled by the remaining 4 digits in \(4!\) ways:

\[2 \times 4! = 2 \times 24 = 48\]

So 48 of the numbers are odd.

Antwortdetails

Let the first term be \(a\) and common ratio \(r\), with \(|r| < 1\).

Second term: \(ar = 3\), so \(a = \dfrac{3}{r}\).

Sum to infinity: \(\dfrac{a}{1 - r} = \dfrac{25}{2}\). Substitute \(a\):

\[\frac{3/r}{1 - r} = \frac{25}{2} \ \Rightarrow\ \frac{3}{r(1 - r)} = \frac{25}{2}\] \[6 = 25r(1 - r) \ \Rightarrow\ 25r^2 - 25r + 6 = 0\]

Solve:

\[r = \frac{25 \pm \sqrt{625 - 600}}{50} = \frac{25 \pm 5}{50}\] \[r = \frac{3}{5} \quad\text{or}\quad r = \frac{2}{5}\]

Both satisfy \(|r| < 1\), so the common ratio is \(r = \dfrac{3}{5}\) or \(r = \dfrac{2}{5}\).

Antwortdetails

(a) Find k given the determinant equals \(-207\).

Expand along the first row:

\[5\big(k(k+2)-54\big)-2\big(-(k+2)-18\big)+(-3)\big(-9-3k\big)\]\[=5(k^{2}+2k-54)-2(-k-20)-3(-9-3k)\]\[=5k^{2}+10k-270+2k+40+27+9k\]\[=5k^{2}+21k-203\]

Set equal to \(-207\):

\[5k^{2}+21k-203=-207\;\Rightarrow\;5k^{2}+21k+4=0\]\[k=\frac{-21\pm\sqrt{441-80}}{10}=\frac{-21\pm 19}{10}\]

So \(k=-\dfrac{1}{5}\) or \(k=-4\).

(b) Curve \(x(y^{2}+1)-y(x^{2}+1)+4=0\).

Expand: \(xy^{2}+x-x^{2}y-y+4=0\). Differentiate implicitly:

\[y^{2}+2xy\,y'+1-2xy-x^{2}y'-y'=0\]\[y'(2xy-x^{2}-1)=-(y^{2}-2xy+1)\]

(i) Gradient at any point:

\[\frac{dy}{dx}=\frac{y^{2}-2xy+1}{x^{2}-2xy+1}\]

(ii) At \((-1,-3)\):

\[\frac{dy}{dx}=\frac{9-2(-1)(-3)+1}{1-2(-1)(-3)+1}=\frac{9-6+1}{1-6+1}=\frac{4}{-4}=-1\]

Tangent through \((-1,-3)\):

\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]

Antwortdetails

Resolve each force into components with bearings measured clockwise from north, using \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).

• \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
• \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
• \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)

Sum the components:

\[\sum x = 10 - 10 - 2 = -2,\qquad \sum y = 0 - 17.32 + 3.46 = -13.86\]

Magnitude of the resultant:

\[R = \sqrt{(-2)^2 + (-13.86)^2} = \sqrt{4 + 192.0} = \sqrt{196.0} = 14.0\,\text{N}\]

The magnitude of the resultant force is \(14.0\,\text{N}\) (to 1 decimal place).

Antwortdetails

(a) Angle between \(p=\binom{3}{-1}\) and \(q=\binom{3}{4}\).

\[\cos\theta=\frac{p\cdot q}{|p|\,|q|}=\frac{(3)(3)+(-1)(4)}{\sqrt{3^{2}+(-1)^{2}}\,\sqrt{3^{2}+4^{2}}}=\frac{9-4}{\sqrt{10}\cdot 5}=\frac{5}{5\sqrt{10}}=\frac{1}{\sqrt{10}}\]\[\theta=\cos^{-1}(0.3162)=71.6^{\circ}\ (\text{1 d.p.})\]

(b) Forces on square A(0,0), B(2,0), C(2,2), D(0,2).

Write each force as its magnitude times the unit vector of its line of action:

• \(10\,\text{N}\) along \(\overrightarrow{BA}=(-1,0):\ (-10,0)\)
• \(15\,\text{N}\) along \(\overrightarrow{BC}=(0,1):\ (0,15)\)
• \(20\,\text{N}\) along \(\overrightarrow{DC}=(1,0):\ (20,0)\)
• \(5\,\text{N}\) along \(\overrightarrow{AD}=(0,1):\ (0,5)\)

Resultant components:

\[R_x=-10+20=10,\qquad R_y=15+5=20\]

(i) Magnitude:

\[|R|=\sqrt{10^{2}+20^{2}}=\sqrt{500}=10\sqrt{5}\approx 22.4\,\text{N}\]

(ii) Direction (measured from the positive x-axis, i.e. along AB):

\[\tan\alpha=\frac{20}{10}=2\;\Rightarrow\;\alpha=63.4^{\circ}\]

The resultant is about \(22.4\,\text{N}\) at \(63.4^{\circ}\) to AB.

Antwortdetails

(a) If \({}^{18}C_r={}^{18}C_{r+2}\), find \({}^{r}C_5\).

Since \({}^{n}C_a={}^{n}C_b\) with \(a\ne b\) requires \(a+b=n\):

\[r+(r+2)=18\;\Rightarrow\;2r+2=18\;\Rightarrow\;r=8\]\[{}^{r}C_5={}^{8}C_5=\frac{8!}{5!\,3!}=56\]

(b) Binomial with \(p=0.1,\ q=0.9,\ n=6\).

(i) Exactly 5:

\[{}^{6}C_5(0.1)^{5}(0.9)^{1}=6(0.00001)(0.9)=0.000054\approx 0.0001\]

(ii) None:

\[(0.9)^{6}=0.531441\approx 0.5314\]

(iii) At most 2 (0, 1 or 2):

\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]

Antwortdetails

(a)(i) Scatter diagram. Using a scale of 2 cm to 10 marks on each axis, the nine pairs \((43,54),(46,53),(48,63),(39,30),(30,44),(60,75),(8,20),(45,33),(40,49)\) are plotted with Mathematics \((x)\) on the horizontal axis and Physics \((y)\) on the vertical axis. The line of best fit required in part (a)(iii) is drawn on the same diagram.

(a)(ii) The means.

\[\bar{x}=\frac{43+46+48+39+30+60+8+45+40}{9}=\frac{359}{9}=39.9\] \[\bar{y}=\frac{54+53+63+30+44+75+20+33+49}{9}=\frac{421}{9}=46.8\]

So the mean point is \((\bar{x},\bar{y})=(39.9,\,46.8)\).

(a)(iii) Line of best fit. A single straight line is drawn through the mean point \((39.9,46.8)\), oriented so that the points are balanced roughly equally above and below it (shown as the straight line on the scatter diagram above).

(b) Equation of the line. Set out the sums needed for the least-squares gradient:

\(x\)	\(y\)	\(xy\)	\(x^{2}\)
43	54	2322	1849
46	53	2438	2116
48	63	3024	2304
39	30	1170	1521
30	44	1320	900
60	75	4500	3600
8	20	160	64
45	33	1485	2025
40	49	1960	1600
\(\sum x=359\)	\(\sum y=421\)	\(\sum xy=18379\)	\(\sum x^{2}=15979\)

With \(n=9\), the gradient is

\[b=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}=\frac{9(18379)-359(421)}{9(15979)-359^{2}}=\frac{165411-151139}{143811-128881}=\frac{14272}{14930}=0.96.\]

The line passes through \((\bar{x},\bar{y})\), so using \(y-\bar{y}=b(x-\bar{x})\):

\[y-46.8=0.96(x-39.9)\] \[y=0.96x-38.3+46.8\] \[y=0.96x+8.6\]

(c) Physics mark when Mathematics \(=28\).

\[y=0.96(28)+8.6=26.9+8.6=35.5\]

The predicted mark in Physics for a candidate who scored 28 in Mathematics is 35.5 (to one decimal place).

Antwortdetails

Let \(u = x^{1/3}\), so \(x^{2/3} = u^2\). The equation becomes a quadratic:

\[u^2 - 5u + 6 = 0 \ \Rightarrow\ (u - 2)(u - 3) = 0\] \[u = 2 \quad\text{or}\quad u = 3\]

Return to \(x\):

\[x^{1/3} = 2 \ \Rightarrow\ x = 2^3 = 8\] \[x^{1/3} = 3 \ \Rightarrow\ x = 3^3 = 27\]

So \(x = 8\) or \(x = 27\).

Antwortdetails

(a) Expand \(\left(2 - \tfrac{x}{2}\right)^5\) in ascending powers of \(x\), using \(\binom{5}{k}2^{5-k}\left(-\tfrac{x}{2}\right)^k\):

• \(k = 0:\ 2^5 = 32\)
• \(k = 1:\ 5 \cdot 2^4 \cdot \left(-\tfrac{x}{2}\right) = -40x\)
• \(k = 2:\ 10 \cdot 2^3 \cdot \tfrac{x^2}{4} = 20x^2\)
• \(k = 3:\ 10 \cdot 2^2 \cdot \left(-\tfrac{x^3}{8}\right) = -5x^3\)

First four terms: \(32 - 40x + 20x^2 - 5x^3\).

(b) Since \(1.99 = 2 - 0.01\), set \(\tfrac{x}{2} = 0.01\), i.e. \(x = 0.02\):

\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]