WAEC SSCE - Further Mathematics - 2017 (Essay)
Frage 1 Bericht
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.
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.92)^{10}=0.43439\).
\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).
\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).
\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]
Antwortdetails
Binomial with \(n=10\) bottles, defective probability \(p=0.08\), non-defective \(q=0.92\). Let \(X\) be the number of defective bottles.
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.92)^{10}=0.43439\).
\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).
\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).
\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]
Frage 2 Bericht
(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 \(9.6\,\text{kg}\).
Antwortdetails
(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 \(9.6\,\text{kg}\).
Frage 3 Bericht
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:
\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]
\[R=(2,\ 8)\]
Antwortdetails
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:
\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]
\[R=(2,\ 8)\]
Frage 4 Bericht
(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:
\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]
Antwortdetails
(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:
\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]
Frage 5 Bericht
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,
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]
Antwortdetails
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,
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]
Frage 6 Bericht
(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 \(m:n=5:4\):
\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]
\[P=\left(2,\ \tfrac{5}{3}\right)\]
Antwortdetails
(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 \(m:n=5:4\):
\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]
\[P=\left(2,\ \tfrac{5}{3}\right)\]
Frage 7 Bericht
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 \(b=4-5=-1\).
\[a=4,\qquad b=-1\]
Antwortdetails
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 \(b=4-5=-1\).
\[a=4,\qquad b=-1\]
Frage 8 Bericht
(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, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).
Antwortdetails
(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, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).
Frage 9 Bericht
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.
Antwortdetails
(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.
Frage 10 Bericht
(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:
\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]
Antwortdetails
(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:
\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]
Frage 11 Bericht
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
\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]
Antwortdetails
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
\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]
Frage 12 Bericht
(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 \(r=6\,\text{cm}\):
\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]
Antwortdetails
(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 \(r=6\,\text{cm}\):
\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]
Frage 13 Bericht
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 \(3\):
\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]
\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]
Antwortdetails
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 \(3\):
\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]
\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]
Frage 14 Bericht
(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 \(=\sqrt{(3+3)^{2}+(1-1)^{2}}=6\).
\[(x+3)^{2}+(y-1)^{2}=36\]
Antwortdetails
(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 \(=\sqrt{(3+3)^{2}+(1-1)^{2}}=6\).
\[(x+3)^{2}+(y-1)^{2}=36\]
Frage 15 Bericht
(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:
\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]
Antwortdetails
(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:
\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]
