WAEC SSCE - Further Mathematics - 2017 (Essay)
Question 1 Rapport
(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° respectively to the horizontal and the tension in AB is 48N. If the system is in equilibrium :
(i) sketch a diagram to represent the information ; (ii) calculate the tension in DB ;
(a)(i) Diagram
The body \(P\) hangs from the junction \(B\). String \(BA\) rises to the left at \(30^{\circ}\) to the horizontal and carries the tension \(T_{AB}=48\,\text{N}\); string \(BD\) rises to the right at \(60^{\circ}\) to the horizontal and carries the tension \(T_{DB}\); the weight \(W=qg\) acts vertically downward. The three forces meet at \(B\).
(a)(ii) Tension in DB
The three concurrent forces at \(B\) are \(T_{AB}=48\,\text{N}\), \(T_{DB}\) and the weight \(W\). The angles between successive forces are:
| Angle between | Size | Opposite force |
|---|---|---|
| \(T_{AB}\) and \(T_{DB}\) | \(90^{\circ}\) | \(W\) |
| \(T_{DB}\) and \(W\) | \(150^{\circ}\) | \(T_{AB}\) |
| \(W\) and \(T_{AB}\) | \(120^{\circ}\) | \(T_{DB}\) |
Applying Lami's theorem at \(B\):
\[\frac{T_{AB}}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}=\frac{W}{\sin 90^{\circ}}\]From the first two ratios:
\[\frac{48}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}\]\[T_{DB}=\frac{48\sin 120^{\circ}}{\sin 150^{\circ}}=\frac{48\left(\tfrac{\sqrt3}{2}\right)}{\tfrac12}=\frac{48(0.8660)}{0.5}=83.14\,\text{N}\]The tension in DB is \(83.14\,\text{N}\).
(a)(iii) Mass \(q\) of the body
Taking \(g=10\,\text{m s}^{-2}\), the weight is \(W=10q\). Using the first and third ratios of Lami's theorem:
\[\frac{10q}{\sin 90^{\circ}}=\frac{48}{\sin 150^{\circ}}\]\[10q=\frac{48\sin 90^{\circ}}{\sin 150^{\circ}}=\frac{48(1)}{0.5}=96\]\[q=\frac{96}{10}=9.6\,\text{kg}\]The mass of the body is \(9.6\,\text{kg}\).
Détails de la réponse
(a)(i) Diagram
The body \(P\) hangs from the junction \(B\). String \(BA\) rises to the left at \(30^{\circ}\) to the horizontal and carries the tension \(T_{AB}=48\,\text{N}\); string \(BD\) rises to the right at \(60^{\circ}\) to the horizontal and carries the tension \(T_{DB}\); the weight \(W=qg\) acts vertically downward. The three forces meet at \(B\).
(a)(ii) Tension in DB
The three concurrent forces at \(B\) are \(T_{AB}=48\,\text{N}\), \(T_{DB}\) and the weight \(W\). The angles between successive forces are:
| Angle between | Size | Opposite force |
|---|---|---|
| \(T_{AB}\) and \(T_{DB}\) | \(90^{\circ}\) | \(W\) |
| \(T_{DB}\) and \(W\) | \(150^{\circ}\) | \(T_{AB}\) |
| \(W\) and \(T_{AB}\) | \(120^{\circ}\) | \(T_{DB}\) |
Applying Lami's theorem at \(B\):
\[\frac{T_{AB}}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}=\frac{W}{\sin 90^{\circ}}\]From the first two ratios:
\[\frac{48}{\sin 150^{\circ}}=\frac{T_{DB}}{\sin 120^{\circ}}\]\[T_{DB}=\frac{48\sin 120^{\circ}}{\sin 150^{\circ}}=\frac{48\left(\tfrac{\sqrt3}{2}\right)}{\tfrac12}=\frac{48(0.8660)}{0.5}=83.14\,\text{N}\]The tension in DB is \(83.14\,\text{N}\).
(a)(iii) Mass \(q\) of the body
Taking \(g=10\,\text{m s}^{-2}\), the weight is \(W=10q\). Using the first and third ratios of Lami's theorem:
\[\frac{10q}{\sin 90^{\circ}}=\frac{48}{\sin 150^{\circ}}\]\[10q=\frac{48\sin 90^{\circ}}{\sin 150^{\circ}}=\frac{48(1)}{0.5}=96\]\[q=\frac{96}{10}=9.6\,\text{kg}\]The mass of the body is \(9.6\,\text{kg}\).
Question 2 Rapport
Evaluate : \(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\).
Rewrite the integrand by splitting \(x-1=(x+1)-2\):
\[\frac{x-1}{(x+1)^{2}}=\frac{(x+1)-2}{(x+1)^{2}}=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\]
Integrate term by term:
\[\int\left(\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\right)dx=\ln(x+1)+\frac{2}{x+1}+C\]
Evaluate from \(1\) to \(3\):
\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]
\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]
Détails de la réponse
Rewrite the integrand by splitting \(x-1=(x+1)-2\):
\[\frac{x-1}{(x+1)^{2}}=\frac{(x+1)-2}{(x+1)^{2}}=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\]
Integrate term by term:
\[\int\left(\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\right)dx=\ln(x+1)+\frac{2}{x+1}+C\]
Evaluate from \(1\) to \(3\):
\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]
\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]
Question 3 Rapport
(a) If \(f(x) = \int (4x - x^{2}) \mathrm {d} x\) and f(3) = 21, find f(x).
(b) The second, fourth and eigth terms of an Arithmetic Progression (A.P) form the first three consecutive terms of a Geometric Progression (G.P). The sum of the third and fifth terms of the A.P is 20, find the :
(i) first four terms of the A.P
(ii) sum of the first ten terms of the A.P
(a) \(f(x)=\displaystyle\int(4x-x^{2})\,dx=2x^{2}-\dfrac{x^{3}}{3}+C\).
Use \(f(3)=21\): \(2(9)-\dfrac{27}{3}+C=18-9+C=9+C=21\Rightarrow C=12\).
\[f(x)=2x^{2}-\frac{x^{3}}{3}+12\]
(b) Let the AP have first term \(a\) and common difference \(d\). The 2nd, 4th, 8th terms are \(a+d,\ a+3d,\ a+7d\) and form a GP:
\[(a+3d)^{2}=(a+d)(a+7d)\]
\[a^{2}+6ad+9d^{2}=a^{2}+8ad+7d^{2}\Rightarrow 2d^{2}-2ad=0\Rightarrow 2d(d-a)=0\]
Since \(d\neq0\), \(a=d\).
Sum of 3rd and 5th terms is \(20\): \((a+2d)+(a+4d)=2a+6d=20\Rightarrow a+3d=10\).
With \(a=d\): \(4d=10\Rightarrow d=2.5,\ a=2.5\).
(i) First four terms: \(2.5,\ 5,\ 7.5,\ 10\).
(ii) Sum of first ten terms:
\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]
Détails de la réponse
(a) \(f(x)=\displaystyle\int(4x-x^{2})\,dx=2x^{2}-\dfrac{x^{3}}{3}+C\).
Use \(f(3)=21\): \(2(9)-\dfrac{27}{3}+C=18-9+C=9+C=21\Rightarrow C=12\).
\[f(x)=2x^{2}-\frac{x^{3}}{3}+12\]
(b) Let the AP have first term \(a\) and common difference \(d\). The 2nd, 4th, 8th terms are \(a+d,\ a+3d,\ a+7d\) and form a GP:
\[(a+3d)^{2}=(a+d)(a+7d)\]
\[a^{2}+6ad+9d^{2}=a^{2}+8ad+7d^{2}\Rightarrow 2d^{2}-2ad=0\Rightarrow 2d(d-a)=0\]
Since \(d\neq0\), \(a=d\).
Sum of 3rd and 5th terms is \(20\): \((a+2d)+(a+4d)=2a+6d=20\Rightarrow a+3d=10\).
With \(a=d\): \(4d=10\Rightarrow d=2.5,\ a=2.5\).
(i) First four terms: \(2.5,\ 5,\ 7.5,\ 10\).
(ii) Sum of first ten terms:
\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]
Question 4 Rapport
(a) Given that \(m = (6i + 8j)\) and \(n = (-8i + \frac{7}{3}j)\), find the :
(i) magnitudes and direction of m and n ; (ii) angle between m and n.
(b) The position vectors of points P, Q, R and S are \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}, \begin{pmatrix} 10 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 12 \end{pmatrix}\) and \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\) respectively. Show that \(\overrightarrow{PQ}\) is perpendicular to \(\overrightarrow{RS}\).
(a)(i) Magnitudes and directions.
\(m=6i+8j\): \(|m|=\sqrt{6^{2}+8^{2}}=10\). Direction \(=\tan^{-1}\dfrac{8}{6}=53.1^{\circ}\) above the positive \(x\)-axis.
\(n=-8i+\tfrac{7}{3}j\): \(|n|=\sqrt{(-8)^{2}+\left(\tfrac{7}{3}\right)^{2}}=\sqrt{\dfrac{576+49}{9}}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\approx8.33\). It is in the second quadrant, direction \(=180^{\circ}-\tan^{-1}\dfrac{7/3}{8}=180^{\circ}-16.3^{\circ}=163.7^{\circ}\).
(ii) Angle between m and n.
\[m\cdot n=6(-8)+8\left(\tfrac{7}{3}\right)=-48+\tfrac{56}{3}=-\tfrac{88}{3}\]
\[\cos\theta=\frac{m\cdot n}{|m|\,|n|}=\frac{-88/3}{10\times25/3}=\frac{-88}{250}=-0.352\]
\[\theta=\cos^{-1}(-0.352)\approx110.6^{\circ}\]
(b) \(\overrightarrow{PQ}=Q-P=\binom{10}{4}-\binom{-2}{3}=\binom{12}{1}\); \(\overrightarrow{RS}=S-R=\binom{4}{0}-\binom{3}{12}=\binom{1}{-12}\).
\[\overrightarrow{PQ}\cdot\overrightarrow{RS}=(12)(1)+(1)(-12)=0\]
Since the scalar product is zero, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).
Détails de la réponse
(a)(i) Magnitudes and directions.
\(m=6i+8j\): \(|m|=\sqrt{6^{2}+8^{2}}=10\). Direction \(=\tan^{-1}\dfrac{8}{6}=53.1^{\circ}\) above the positive \(x\)-axis.
\(n=-8i+\tfrac{7}{3}j\): \(|n|=\sqrt{(-8)^{2}+\left(\tfrac{7}{3}\right)^{2}}=\sqrt{\dfrac{576+49}{9}}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\approx8.33\). It is in the second quadrant, direction \(=180^{\circ}-\tan^{-1}\dfrac{7/3}{8}=180^{\circ}-16.3^{\circ}=163.7^{\circ}\).
(ii) Angle between m and n.
\[m\cdot n=6(-8)+8\left(\tfrac{7}{3}\right)=-48+\tfrac{56}{3}=-\tfrac{88}{3}\]
\[\cos\theta=\frac{m\cdot n}{|m|\,|n|}=\frac{-88/3}{10\times25/3}=\frac{-88}{250}=-0.352\]
\[\theta=\cos^{-1}(-0.352)\approx110.6^{\circ}\]
(b) \(\overrightarrow{PQ}=Q-P=\binom{10}{4}-\binom{-2}{3}=\binom{12}{1}\); \(\overrightarrow{RS}=S-R=\binom{4}{0}-\binom{3}{12}=\binom{1}{-12}\).
\[\overrightarrow{PQ}\cdot\overrightarrow{RS}=(12)(1)+(1)(-12)=0\]
Since the scalar product is zero, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).
Question 5 Rapport
Forces \(F_{1} (18N, 330°), F_{2} (10N, 090°)\) and \(F_{3} (25N, 180°)\) act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.
Resolve each force into components (angles measured anticlockwise from the positive \(x\)-axis).
\(F_{1}=18\,\text{N}\) at \(330^{\circ}\): \(x=18\cos330^{\circ}=15.59\), \(y=18\sin330^{\circ}=-9.00\).
\(F_{2}=10\,\text{N}\) at \(090^{\circ}\): \(x=0\), \(y=10\).
\(F_{3}=25\,\text{N}\) at \(180^{\circ}\): \(x=-25\), \(y=0\).
Sum the components:
\[R_{x}=15.59+0-25=-9.41,\qquad R_{y}=-9+10+0=1.00\]
Magnitude:
\[|R|=\sqrt{(-9.41)^{2}+(1.00)^{2}}=\sqrt{89.55}\approx9.5\ \text{N}\]
Direction: \(R\) lies in the second quadrant. The reference angle is \(\tan^{-1}\dfrac{1}{9.41}=6.1^{\circ}\), so
\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]
Détails de la réponse
Resolve each force into components (angles measured anticlockwise from the positive \(x\)-axis).
\(F_{1}=18\,\text{N}\) at \(330^{\circ}\): \(x=18\cos330^{\circ}=15.59\), \(y=18\sin330^{\circ}=-9.00\).
\(F_{2}=10\,\text{N}\) at \(090^{\circ}\): \(x=0\), \(y=10\).
\(F_{3}=25\,\text{N}\) at \(180^{\circ}\): \(x=-25\), \(y=0\).
Sum the components:
\[R_{x}=15.59+0-25=-9.41,\qquad R_{y}=-9+10+0=1.00\]
Magnitude:
\[|R|=\sqrt{(-9.41)^{2}+(1.00)^{2}}=\sqrt{89.55}\approx9.5\ \text{N}\]
Direction: \(R\) lies in the second quadrant. The reference angle is \(\tan^{-1}\dfrac{1}{9.41}=6.1^{\circ}\), so
\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]
Question 6 Rapport
(a) Given that \(\log_{10} p = a, \log_{10} q = b\) and \(\log_{10} s = c\), express \(\log_{10} (\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\) in terms of a, b and c.
(b) The radius of a circle is 6cm. If the area is increasing at the rate of 20\(cm^{2}s^{-1}\), find, leaving the answer in terms of \(\pi\), the rate at which the radius is increasing.
(a) Using the laws of logarithms with \(\log_{10}p=a,\ \log_{10}q=b,\ \log_{10}s=c\):
\[\log_{10}\!\left(\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\right)=\tfrac{1}{3}\log_{10}p+4\log_{10}q-2\log_{10}s=\frac{a}{3}+4b-2c\]
(b) Area of a circle: \(A=\pi r^{2}\). Differentiate with respect to time:
\[\frac{dA}{dt}=2\pi r\,\frac{dr}{dt}\]
With \(\dfrac{dA}{dt}=20\,\text{cm}^2\text{s}^{-1}\) and \(r=6\,\text{cm}\):
\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]
Détails de la réponse
(a) Using the laws of logarithms with \(\log_{10}p=a,\ \log_{10}q=b,\ \log_{10}s=c\):
\[\log_{10}\!\left(\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\right)=\tfrac{1}{3}\log_{10}p+4\log_{10}q-2\log_{10}s=\frac{a}{3}+4b-2c\]
(b) Area of a circle: \(A=\pi r^{2}\). Differentiate with respect to time:
\[\frac{dA}{dt}=2\pi r\,\frac{dr}{dt}\]
With \(\dfrac{dA}{dt}=20\,\text{cm}^2\text{s}^{-1}\) and \(r=6\,\text{cm}\):
\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]
Question 7 Rapport
(a) If \(f(x) = \frac{4 - 5x}{2}\), and \(g(x) = x + 6, x \in R\), find \(f \circ g^{-1}\).
(b) P(x, y) divides the line joining (7, -5) and (-2, 7) internally in 5 : 4. Find the coordinates of P.
(a) \(g(x)=x+6\), so the inverse is \(g^{-1}(x)=x-6\).
\[f\circ g^{-1}(x)=f(x-6)=\frac{4-5(x-6)}{2}=\frac{4-5x+30}{2}=\frac{34-5x}{2}\]
(b) \(P\) divides \((7,-5)\) and \((-2,7)\) internally in \(5:4\). Using the section formula \(\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\) with \(m:n=5:4\):
\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]
\[P=\left(2,\ \tfrac{5}{3}\right)\]
Détails de la réponse
(a) \(g(x)=x+6\), so the inverse is \(g^{-1}(x)=x-6\).
\[f\circ g^{-1}(x)=f(x-6)=\frac{4-5(x-6)}{2}=\frac{4-5x+30}{2}=\frac{34-5x}{2}\]
(b) \(P\) divides \((7,-5)\) and \((-2,7)\) internally in \(5:4\). Using the section formula \(\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\) with \(m:n=5:4\):
\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]
\[P=\left(2,\ \tfrac{5}{3}\right)\]
Question 8 Rapport
(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, correct to two decimal places, the probability that :
(i) at least 3 passed the test ; (ii) between 3 and 6 students failed the test.
(b) A fair die is thrown five times; find the probability of obtaining a six three times.
(a) Pass:fail \(=4:1\), so \(P(\text{pass})=0.8,\ P(\text{fail})=0.2\), with \(n=7\).
(i) At least 3 passed. Let \(X=\) number who pass, \(p=0.8\).
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.2)^{7}=0.0000128\), \(P(1)=7(0.8)(0.2)^{6}=0.000358\), \(P(2)=21(0.8)^{2}(0.2)^{5}=0.004301\).
\[P(X\ge3)=1-0.004672=0.995\approx1.00\]
(ii) Between 3 and 6 failed (i.e. \(4\) or \(5\) failed). Let \(Y=\) number who fail, \(p=0.2\).
\(P(Y=4)=\binom{7}{4}(0.2)^{4}(0.8)^{3}=35(0.0016)(0.512)=0.028672\).
\(P(Y=5)=\binom{7}{5}(0.2)^{5}(0.8)^{2}=21(0.00032)(0.64)=0.004301\).
\[P(4\le Y\le5)=0.028672+0.004301=0.032973\approx0.03\]
(b) A fair die thrown 5 times, \(P(\text{six})=\tfrac{1}{6}\). Probability of exactly three sixes:
\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]
Détails de la réponse
(a) Pass:fail \(=4:1\), so \(P(\text{pass})=0.8,\ P(\text{fail})=0.2\), with \(n=7\).
(i) At least 3 passed. Let \(X=\) number who pass, \(p=0.8\).
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.2)^{7}=0.0000128\), \(P(1)=7(0.8)(0.2)^{6}=0.000358\), \(P(2)=21(0.8)^{2}(0.2)^{5}=0.004301\).
\[P(X\ge3)=1-0.004672=0.995\approx1.00\]
(ii) Between 3 and 6 failed (i.e. \(4\) or \(5\) failed). Let \(Y=\) number who fail, \(p=0.2\).
\(P(Y=4)=\binom{7}{4}(0.2)^{4}(0.8)^{3}=35(0.0016)(0.512)=0.028672\).
\(P(Y=5)=\binom{7}{5}(0.2)^{5}(0.8)^{2}=21(0.00032)(0.64)=0.004301\).
\[P(4\le Y\le5)=0.028672+0.004301=0.032973\approx0.03\]
(b) A fair die thrown 5 times, \(P(\text{six})=\tfrac{1}{6}\). Probability of exactly three sixes:
\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]
Question 9 Rapport
A parallelogram MNQR has vertices M(4, -6), N(10, 2), Q(8, 16) and R(x, y). Find the coordinates of R.
In parallelogram \(MNQR\) (vertices in order), the diagonals \(MQ\) and \(NR\) bisect each other, so they share the same midpoint.
Midpoint of \(MQ\): \(\left(\dfrac{4+8}{2},\dfrac{-6+16}{2}\right)=(6,\ 5)\).
Midpoint of \(NR\): \(\left(\dfrac{10+x}{2},\dfrac{2+y}{2}\right)\).
Equate:
\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]
\[R=(2,\ 8)\]
Détails de la réponse
In parallelogram \(MNQR\) (vertices in order), the diagonals \(MQ\) and \(NR\) bisect each other, so they share the same midpoint.
Midpoint of \(MQ\): \(\left(\dfrac{4+8}{2},\dfrac{-6+16}{2}\right)=(6,\ 5)\).
Midpoint of \(NR\): \(\left(\dfrac{10+x}{2},\dfrac{2+y}{2}\right)\).
Equate:
\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]
\[R=(2,\ 8)\]
Question 10 Rapport
Bottles of the same sizes produced in a factory are packed in boxes. Each box contains 10 bottles. If 8% of the bottles are defective, find, correct to two decimal places, the probability that box chosen at random contains at least 3 defective bottles.
Binomial with \(n=10\) bottles, defective probability \(p=0.08\), non-defective \(q=0.92\). Let \(X\) be the number of defective bottles.
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.92)^{10}=0.43439\).
\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).
\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).
\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]
Détails de la réponse
Binomial with \(n=10\) bottles, defective probability \(p=0.08\), non-defective \(q=0.92\). Let \(X\) be the number of defective bottles.
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.92)^{10}=0.43439\).
\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).
\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).
\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]
Question 11 Rapport
The table shows the heights in cm of some seedlings in a certain garden.
| Height (cm) | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 |
| Frequency | 3 | 9 | 21 | 12 | 5 |
(a) Draw the cumulative frequency curve for the distribution.
(b) Using the curve in (a), find thesemi-interquartile range.
The continuous class boundaries and cumulative frequencies are:
| Height (cm) | Class boundaries (cm) | Frequency | Cumulative frequency |
|---|---|---|---|
| 36–40 | 35.5–40.5 | 3 | 3 |
| 41–45 | 40.5–45.5 | 9 | 12 |
| 46–50 | 45.5–50.5 | 21 | 33 |
| 51–55 | 50.5–55.5 | 12 | 45 |
| 56–60 | 55.5–60.5 | 5 | 50 |
(a) The cumulative frequency curve is shown below. The point [35.5, 0[0m is included before plotting the upper class boundaries against their cumulative frequencies.
(b) Total frequency, \(N=50\).
On the curve, the lower quartile corresponds to cumulative frequency
\[\frac{N}{4}=\frac{50}{4}=12.5.\]
Reading across from \(12.5\) to the curve and down to the height axis gives
\[Q_1\approx 45.6\text{ cm}.\]
The upper quartile corresponds to cumulative frequency
\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]
Reading across from \(37.5\) to the curve and down to the height axis gives
\[Q_3\approx 52.0\text{ cm}.\]
Hence,
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]
Détails de la réponse
The continuous class boundaries and cumulative frequencies are:
| Height (cm) | Class boundaries (cm) | Frequency | Cumulative frequency |
|---|---|---|---|
| 36–40 | 35.5–40.5 | 3 | 3 |
| 41–45 | 40.5–45.5 | 9 | 12 |
| 46–50 | 45.5–50.5 | 21 | 33 |
| 51–55 | 50.5–55.5 | 12 | 45 |
| 56–60 | 55.5–60.5 | 5 | 50 |
(a) The cumulative frequency curve is shown below. The point [35.5, 0[0m is included before plotting the upper class boundaries against their cumulative frequencies.
(b) Total frequency, \(N=50\).
On the curve, the lower quartile corresponds to cumulative frequency
\[\frac{N}{4}=\frac{50}{4}=12.5.\]
Reading across from \(12.5\) to the curve and down to the height axis gives
\[Q_1\approx 45.6\text{ cm}.\]
The upper quartile corresponds to cumulative frequency
\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]
Reading across from \(37.5\) to the curve and down to the height axis gives
\[Q_3\approx 52.0\text{ cm}.\]
Hence,
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]
Question 12 Rapport
If (x + 1) and (x - 2) are factors of the polynomial \(g(x) = x^{4} + ax^{3} + bx^{2} - 16x - 12\), find the values of a and b.
By the Factor Theorem, \(g(-1)=0\) and \(g(2)=0\).
\(g(x)=x^{4}+ax^{3}+bx^{2}-16x-12\).
Using \(g(-1)=0\):
\[1-a+b+16-12=0\ \Rightarrow\ -a+b+5=0\ \Rightarrow\ b=a-5\quad(1)\]
Using \(g(2)=0\):
\[16+8a+4b-32-12=0\ \Rightarrow\ 8a+4b-28=0\ \Rightarrow\ 2a+b=7\quad(2)\]
Substitute (1) into (2): \(2a+(a-5)=7\Rightarrow3a=12\Rightarrow a=4\).
Then \(b=4-5=-1\).
\[a=4,\qquad b=-1\]
Détails de la réponse
By the Factor Theorem, \(g(-1)=0\) and \(g(2)=0\).
\(g(x)=x^{4}+ax^{3}+bx^{2}-16x-12\).
Using \(g(-1)=0\):
\[1-a+b+16-12=0\ \Rightarrow\ -a+b+5=0\ \Rightarrow\ b=a-5\quad(1)\]
Using \(g(2)=0\):
\[16+8a+4b-32-12=0\ \Rightarrow\ 8a+4b-28=0\ \Rightarrow\ 2a+b=7\quad(2)\]
Substitute (1) into (2): \(2a+(a-5)=7\Rightarrow3a=12\Rightarrow a=4\).
Then \(b=4-5=-1\).
\[a=4,\qquad b=-1\]
Question 13 Rapport
(a) Simplify : \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\) and leave your answer in terms of \(\sin \theta\).
(b) Find the equation of the line joining the stationary points of \(y = x^{2} (x - 3)\) and the distance between them.
(a) Add the two fractions over a common denominator:
\[\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\frac{2}{1-\cos^{2}\theta}\]
Since \(1-\cos^{2}\theta=\sin^{2}\theta\):
\[=\frac{2}{\sin^{2}\theta}\]
(b) \(y=x^{2}(x-3)=x^{3}-3x^{2}\). Then \(\dfrac{dy}{dx}=3x^{2}-6x=3x(x-2)\).
Stationary points where \(\dfrac{dy}{dx}=0\): \(x=0\) or \(x=2\).
At \(x=0,\ y=0\Rightarrow(0,0)\). At \(x=2,\ y=8-12=-4\Rightarrow(2,-4)\).
Line joining them: slope \(=\dfrac{-4-0}{2-0}=-2\), through \((0,0)\):
\[y=-2x\quad\text{or}\quad 2x+y=0\]
Distance:
\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]
Détails de la réponse
(a) Add the two fractions over a common denominator:
\[\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\frac{2}{1-\cos^{2}\theta}\]
Since \(1-\cos^{2}\theta=\sin^{2}\theta\):
\[=\frac{2}{\sin^{2}\theta}\]
(b) \(y=x^{2}(x-3)=x^{3}-3x^{2}\). Then \(\dfrac{dy}{dx}=3x^{2}-6x=3x(x-2)\).
Stationary points where \(\dfrac{dy}{dx}=0\): \(x=0\) or \(x=2\).
At \(x=0,\ y=0\Rightarrow(0,0)\). At \(x=2,\ y=8-12=-4\Rightarrow(2,-4)\).
Line joining them: slope \(=\dfrac{-4-0}{2-0}=-2\), through \((0,0)\):
\[y=-2x\quad\text{or}\quad 2x+y=0\]
Distance:
\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]
Question 14 Rapport
(a) If \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\)
(i) find the values of x for which f(x) is undefined.
(ii) express f(x) in partial fractions.
(b) A circle with centre (-3, 1) passes through the point (3, 1). Find its equation.
(a)(i) \(f(x)=\dfrac{2x-3}{(x^{2}-1)(x+2)}=\dfrac{2x-3}{(x-1)(x+1)(x+2)}\) is undefined when the denominator is zero:
\[x=1,\quad x=-1,\quad x=-2\]
(ii) Partial fractions. Write
\[\frac{2x-3}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}\]
Cover-up (substitution) gives:
\(x=1:\ A=\dfrac{2(1)-3}{(2)(3)}=-\dfrac{1}{6}\).
\(x=-1:\ B=\dfrac{2(-1)-3}{(-2)(1)}=\dfrac{-5}{-2}=\dfrac{5}{2}\).
\(x=-2:\ C=\dfrac{2(-2)-3}{(-3)(-1)}=\dfrac{-7}{3}\).
\[f(x)=-\frac{1}{6(x-1)}+\frac{5}{2(x+1)}-\frac{7}{3(x+2)}\]
(b) Circle with centre \((-3,1)\) through \((3,1)\): radius \(=\sqrt{(3+3)^{2}+(1-1)^{2}}=6\).
\[(x+3)^{2}+(y-1)^{2}=36\]
Détails de la réponse
(a)(i) \(f(x)=\dfrac{2x-3}{(x^{2}-1)(x+2)}=\dfrac{2x-3}{(x-1)(x+1)(x+2)}\) is undefined when the denominator is zero:
\[x=1,\quad x=-1,\quad x=-2\]
(ii) Partial fractions. Write
\[\frac{2x-3}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}\]
Cover-up (substitution) gives:
\(x=1:\ A=\dfrac{2(1)-3}{(2)(3)}=-\dfrac{1}{6}\).
\(x=-1:\ B=\dfrac{2(-1)-3}{(-2)(1)}=\dfrac{-5}{-2}=\dfrac{5}{2}\).
\(x=-2:\ C=\dfrac{2(-2)-3}{(-3)(-1)}=\dfrac{-7}{3}\).
\[f(x)=-\frac{1}{6(x-1)}+\frac{5}{2(x+1)}-\frac{7}{3(x+2)}\]
(b) Circle with centre \((-3,1)\) through \((3,1)\): radius \(=\sqrt{(3+3)^{2}+(1-1)^{2}}=6\).
\[(x+3)^{2}+(y-1)^{2}=36\]
Question 15 Rapport
The table shows the frequency distribution of the ages of patients in a clinic.
| Ages (years) | 17 - 19 | 20 - 22 | 23 - 28 | 29 - 34 | 35 - 43 |
| No. of patients | 6 | 9 | 12 | 18 | 18 |
(a) Draw a histogram for the distribution
(b) Find, correct to two decimal places, the mean age of the patients.
(a) Histogram
Since the class intervals have unequal widths, plot frequency density against age.
| Age class (years) | Class boundaries | Class width | Frequency | Frequency density \(=\frac{f}{\text{class width}}\) |
|---|---|---|---|---|
| 17 - 19 | 16.5 - 19.5 | 3 | 6 | 2 |
| 20 - 22 | 19.5 - 22.5 | 3 | 9 | 3 |
| 23 - 28 | 22.5 - 28.5 | 6 | 12 | 2 |
| 29 - 34 | 28.5 - 34.5 | 6 | 18 | 3 |
| 35 - 43 | 34.5 - 43.5 | 9 | 18 | 2 |
(b) Mean age
| Age class | Midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 17 - 19 | 18.0 | 6 | 108 |
| 20 - 22 | 21.0 | 9 | 189 |
| 23 - 28 | 25.5 | 12 | 306 |
| 29 - 34 | 31.5 | 18 | 567 |
| 35 - 43 | 39.0 | 18 | 702 |
| Total | 63 | 1872 | |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1872}{63}=29.714\ldots\]
Therefore, the mean age of the patients is \(29.71\text{ years}\), correct to two decimal places.
Détails de la réponse
(a) Histogram
Since the class intervals have unequal widths, plot frequency density against age.
| Age class (years) | Class boundaries | Class width | Frequency | Frequency density \(=\frac{f}{\text{class width}}\) |
|---|---|---|---|---|
| 17 - 19 | 16.5 - 19.5 | 3 | 6 | 2 |
| 20 - 22 | 19.5 - 22.5 | 3 | 9 | 3 |
| 23 - 28 | 22.5 - 28.5 | 6 | 12 | 2 |
| 29 - 34 | 28.5 - 34.5 | 6 | 18 | 3 |
| 35 - 43 | 34.5 - 43.5 | 9 | 18 | 2 |
(b) Mean age
| Age class | Midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 17 - 19 | 18.0 | 6 | 108 |
| 20 - 22 | 21.0 | 9 | 189 |
| 23 - 28 | 25.5 | 12 | 306 |
| 29 - 34 | 31.5 | 18 | 567 |
| 35 - 43 | 39.0 | 18 | 702 |
| Total | 63 | 1872 | |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1872}{63}=29.714\ldots\]
Therefore, the mean age of the patients is \(29.71\text{ years}\), correct to two decimal places.
