WAEC SSCE - Further Mathematics - 2010 (Essay)
Question 1 Report
The equation of a curve is \(y = x(3 - x^{2})\). Find the equation of its normal of the point where x = 2.
Expand the curve: \(y = x(3 - x^2) = 3x - x^3\).
Differentiate to get the gradient of the tangent:
\[\frac{dy}{dx} = 3 - 3x^2.\]
At \(x = 2\): gradient of tangent \(= 3 - 3(4) = -9\).
The \(y\)-coordinate there is \(y = 2(3 - 4) = -2\), giving the point \((2, -2)\).
The normal is perpendicular to the tangent, so its gradient is
\[m_{\text{normal}} = -\frac{1}{-9} = \frac{1}{9}.\]
Equation of the normal through \((2, -2)\):
\[y - (-2) = \frac{1}{9}(x - 2) \;\Rightarrow\; 9y + 18 = x - 2 \;\Rightarrow\; x - 9y - 20 = 0.\]
Answer Details
Expand the curve: \(y = x(3 - x^2) = 3x - x^3\).
Differentiate to get the gradient of the tangent:
\[\frac{dy}{dx} = 3 - 3x^2.\]
At \(x = 2\): gradient of tangent \(= 3 - 3(4) = -9\).
The \(y\)-coordinate there is \(y = 2(3 - 4) = -2\), giving the point \((2, -2)\).
The normal is perpendicular to the tangent, so its gradient is
\[m_{\text{normal}} = -\frac{1}{-9} = \frac{1}{9}.\]
Equation of the normal through \((2, -2)\):
\[y - (-2) = \frac{1}{9}(x - 2) \;\Rightarrow\; 9y + 18 = x - 2 \;\Rightarrow\; x - 9y - 20 = 0.\]
Question 2 Report
(a) Copy and complete the table for the relation: \(y = 2\cos x + 3\sin x\) for \(0° \leq x \leq 360°\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 2.00 | 3.23 | 1.60 | -3.23 |
(b) Using a scale of 2 cm to 60° on the x- axis and 2 cm to one unit on the y- axis, draw the graph of \(y = 2\cos x + 3\sin x\) for \(0° \leq x \leq 360°\).
(c) From the graph, find the : (i) maximum value of y, correct to two decimal places ; (ii) solution of the equation \(\frac{2}{3}\cos x + \sin x = \frac{5}{6}\).
(a) For each value of \(x\), calculate \(y=2\cos x+3\sin x\), correct to two decimal places.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.23 | 3.60 | 3.00 | 1.60 | -0.23 | -2.00 | -3.23 |
For example,
\[y(60^\circ)=2\cos60^\circ+3\sin60^\circ=2(0.5)+3(0.8660)=3.60.\]
(b) The required graph is shown below. The plotted curve is smooth, with the horizontal scale \(2\text{ cm}\) to \(60^\circ\) and vertical scale \(2\text{ cm}\) to 1 unit.
(c)(i) The maximum ordinate of the curve is
\[\boxed{3.61}\]
correct to two decimal places.
(c)(ii) Since
\[\frac{2}{3}\cos x+\sin x=\frac56,\]
multiplying by 3 gives
\[2\cos x+3\sin x=2.50.\]
Thus, draw the line \(y=2.50\) on the graph and read the abscissae of its intersections with the curve:
\[\boxed{x\approx10^\circ\text{ or }102^\circ}.\]
Answer Details
(a) For each value of \(x\), calculate \(y=2\cos x+3\sin x\), correct to two decimal places.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.23 | 3.60 | 3.00 | 1.60 | -0.23 | -2.00 | -3.23 |
For example,
\[y(60^\circ)=2\cos60^\circ+3\sin60^\circ=2(0.5)+3(0.8660)=3.60.\]
(b) The required graph is shown below. The plotted curve is smooth, with the horizontal scale \(2\text{ cm}\) to \(60^\circ\) and vertical scale \(2\text{ cm}\) to 1 unit.
(c)(i) The maximum ordinate of the curve is
\[\boxed{3.61}\]
correct to two decimal places.
(c)(ii) Since
\[\frac{2}{3}\cos x+\sin x=\frac56,\]
multiplying by 3 gives
\[2\cos x+3\sin x=2.50.\]
Thus, draw the line \(y=2.50\) on the graph and read the abscissae of its intersections with the curve:
\[\boxed{x\approx10^\circ\text{ or }102^\circ}.\]
Question 3 Report
The table gives the distribution of marks of 60 candidates in a test.
| Marks | 23-25 | 26-28 | 29-31 | 32-34 | 35-37 | 38-40 |
| Frequency | 3 | 7 | 15 | 21 | 10 | 4 |
(a) Draw a cumulative frequency curve of the distribution.
(b) From your curve, estimate the : (i) 80th percentile ; (ii) median ; (iii) semi-interquartile range.
Setting up class boundaries and cumulative frequencies
The marks are recorded in whole numbers, so each class boundary lies halfway between the top of one class and the bottom of the next. The lower boundary of the first class is \(22.5\), and we plot the cumulative frequency against each upper class boundary.
| Marks | Frequency | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 23-25 | 3 | 25.5 | 3 |
| 26-28 | 7 | 28.5 | 10 |
| 29-31 | 15 | 31.5 | 25 |
| 32-34 | 21 | 34.5 | 46 |
| 35-37 | 10 | 37.5 | 56 |
| 38-40 | 4 | 40.5 | 60 |
Total frequency \(N = 60\).
(a) The cumulative frequency curve (ogive)
Plot the points \((22.5,\,0)\), \((25.5,\,3)\), \((28.5,\,10)\), \((31.5,\,25)\), \((34.5,\,46)\), \((37.5,\,56)\) and \((40.5,\,60)\), taking the horizontal axis as the marks and the vertical axis as the cumulative frequency. Join the points with a smooth increasing S-shaped curve. The estimates below are read from this curve.
(b) Estimates from the curve
(i) 80th percentile
The position on the vertical axis is
\[ \frac{80}{100}\times 60 = 48. \]
This falls in the class \(35\text{-}37\) (boundaries \(34.5\) to \(37.5\), frequency \(10\), with cumulative frequency \(46\) reached at \(34.5\)). By interpolation,
\[ P_{80} = 34.5 + \left(\frac{48 - 46}{10}\right)\times 3 = 34.5 + 0.6 = 35.1. \]
So the 80th percentile is approximately \(\mathbf{35.1}\) marks.
(ii) Median
The median is at position
\[ \frac{N}{2} = \frac{60}{2} = 30, \]
which lies in the class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) reached at \(31.5\)). Thus
\[ \text{Median} = 31.5 + \left(\frac{30 - 25}{21}\right)\times 3 = 31.5 + \frac{15}{21} \approx 31.5 + 0.71 = 32.2. \]
The median is approximately \(\mathbf{32.2}\) marks.
(iii) Semi-interquartile range
Lower quartile \(Q_1\) is at position \(\dfrac{N}{4} = 15\), lying in class \(29\text{-}31\) (boundaries \(28.5\) to \(31.5\), frequency \(15\), cumulative frequency \(10\) at \(28.5\)):
\[ Q_1 = 28.5 + \left(\frac{15 - 10}{15}\right)\times 3 = 28.5 + 1 = 29.5. \]
Upper quartile \(Q_3\) is at position \(\dfrac{3N}{4} = 45\), lying in class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) at \(31.5\)):
\[ Q_3 = 31.5 + \left(\frac{45 - 25}{21}\right)\times 3 = 31.5 + \frac{60}{21} \approx 31.5 + 2.86 = 34.4. \]
Therefore the semi-interquartile range is
\[ \frac{Q_3 - Q_1}{2} = \frac{34.4 - 29.5}{2} = \frac{4.9}{2} \approx 2.4. \]
The semi-interquartile range is approximately \(\mathbf{2.4}\) marks.
Answer Details
Setting up class boundaries and cumulative frequencies
The marks are recorded in whole numbers, so each class boundary lies halfway between the top of one class and the bottom of the next. The lower boundary of the first class is \(22.5\), and we plot the cumulative frequency against each upper class boundary.
| Marks | Frequency | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 23-25 | 3 | 25.5 | 3 |
| 26-28 | 7 | 28.5 | 10 |
| 29-31 | 15 | 31.5 | 25 |
| 32-34 | 21 | 34.5 | 46 |
| 35-37 | 10 | 37.5 | 56 |
| 38-40 | 4 | 40.5 | 60 |
Total frequency \(N = 60\).
(a) The cumulative frequency curve (ogive)
Plot the points \((22.5,\,0)\), \((25.5,\,3)\), \((28.5,\,10)\), \((31.5,\,25)\), \((34.5,\,46)\), \((37.5,\,56)\) and \((40.5,\,60)\), taking the horizontal axis as the marks and the vertical axis as the cumulative frequency. Join the points with a smooth increasing S-shaped curve. The estimates below are read from this curve.
(b) Estimates from the curve
(i) 80th percentile
The position on the vertical axis is
\[ \frac{80}{100}\times 60 = 48. \]
This falls in the class \(35\text{-}37\) (boundaries \(34.5\) to \(37.5\), frequency \(10\), with cumulative frequency \(46\) reached at \(34.5\)). By interpolation,
\[ P_{80} = 34.5 + \left(\frac{48 - 46}{10}\right)\times 3 = 34.5 + 0.6 = 35.1. \]
So the 80th percentile is approximately \(\mathbf{35.1}\) marks.
(ii) Median
The median is at position
\[ \frac{N}{2} = \frac{60}{2} = 30, \]
which lies in the class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) reached at \(31.5\)). Thus
\[ \text{Median} = 31.5 + \left(\frac{30 - 25}{21}\right)\times 3 = 31.5 + \frac{15}{21} \approx 31.5 + 0.71 = 32.2. \]
The median is approximately \(\mathbf{32.2}\) marks.
(iii) Semi-interquartile range
Lower quartile \(Q_1\) is at position \(\dfrac{N}{4} = 15\), lying in class \(29\text{-}31\) (boundaries \(28.5\) to \(31.5\), frequency \(15\), cumulative frequency \(10\) at \(28.5\)):
\[ Q_1 = 28.5 + \left(\frac{15 - 10}{15}\right)\times 3 = 28.5 + 1 = 29.5. \]
Upper quartile \(Q_3\) is at position \(\dfrac{3N}{4} = 45\), lying in class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) at \(31.5\)):
\[ Q_3 = 31.5 + \left(\frac{45 - 25}{21}\right)\times 3 = 31.5 + \frac{60}{21} \approx 31.5 + 2.86 = 34.4. \]
Therefore the semi-interquartile range is
\[ \frac{Q_3 - Q_1}{2} = \frac{34.4 - 29.5}{2} = \frac{4.9}{2} \approx 2.4. \]
The semi-interquartile range is approximately \(\mathbf{2.4}\) marks.
Question 4 Report
(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability of obtaining :
(i) exactly 8 heads ; (ii) at least 5 heads ; (iii) at most 1 head.
(b) In how many ways can four letters from the word SHEEP be arranged (i) without any restriction ; (ii) with only one E.
(a) Eight coins, \(n = 8\), \(p = 0.5\) for a head, so each outcome count is \(\binom{8}{r}/256\).
(i) Exactly 8 heads: \(\left(\tfrac12\right)^8 = \dfrac{1}{256} \approx 0.004\).
(ii) At least 5 heads \(= \dfrac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{256} = \dfrac{56+28+8+1}{256} = \dfrac{93}{256} \approx 0.363\).
(iii) At most 1 head \(= \dfrac{\binom{8}{0}+\binom{8}{1}}{256} = \dfrac{1+8}{256} = \dfrac{9}{256} \approx 0.035\).
(b) The word SHEEP has letters S, H, E, E, P (the E repeats). Choose and arrange 4 letters.
(i) Without restriction, split by how many E's are used:
- Both E's plus 2 of \(\{S,H,P\}\): \(\binom{3}{2}\times\dfrac{4!}{2!} = 3\times12 = 36\).
- At most one E, i.e. the four distinct letters \(S,H,E,P\): \(4! = 24\).
Total \(= 36 + 24 = 60\) arrangements.
(ii) With only one E, the other three must be \(S, H, P\), giving four distinct letters: \(4! = 24\) arrangements.
Answer Details
(a) Eight coins, \(n = 8\), \(p = 0.5\) for a head, so each outcome count is \(\binom{8}{r}/256\).
(i) Exactly 8 heads: \(\left(\tfrac12\right)^8 = \dfrac{1}{256} \approx 0.004\).
(ii) At least 5 heads \(= \dfrac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{256} = \dfrac{56+28+8+1}{256} = \dfrac{93}{256} \approx 0.363\).
(iii) At most 1 head \(= \dfrac{\binom{8}{0}+\binom{8}{1}}{256} = \dfrac{1+8}{256} = \dfrac{9}{256} \approx 0.035\).
(b) The word SHEEP has letters S, H, E, E, P (the E repeats). Choose and arrange 4 letters.
(i) Without restriction, split by how many E's are used:
- Both E's plus 2 of \(\{S,H,P\}\): \(\binom{3}{2}\times\dfrac{4!}{2!} = 3\times12 = 36\).
- At most one E, i.e. the four distinct letters \(S,H,E,P\): \(4! = 24\).
Total \(= 36 + 24 = 60\) arrangements.
(ii) With only one E, the other three must be \(S, H, P\), giving four distinct letters: \(4! = 24\) arrangements.
Question 5 Report
The position vector of a particle of mass 3 kg moving along a space curve is given by \(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\) at any time t seconds. Find the force acting on it at t = 2 seconds.
The force is found from Newton's second law \(\mathbf{F} = m\mathbf{a}\), where \(\mathbf{a}\) is the second derivative of the position vector.
Given \(\mathbf{r} = (4t^3 - t^2)\mathbf{i} - (2t^2 - t)\mathbf{j}\), differentiate once for velocity:
\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = (12t^2 - 2t)\mathbf{i} - (4t - 1)\mathbf{j}.\]
Differentiate again for acceleration:
\[\mathbf{a} = \frac{d\mathbf{v}}{dt} = (24t - 2)\mathbf{i} - 4\mathbf{j}.\]
At \(t = 2\): \(\mathbf{a} = (24(2) - 2)\mathbf{i} - 4\mathbf{j} = 46\mathbf{i} - 4\mathbf{j}\).
With mass \(m = 3\ \text{kg}\):
\[\mathbf{F} = 3(46\mathbf{i} - 4\mathbf{j}) = 138\mathbf{i} - 12\mathbf{j}\ \text{N}.\]
Its magnitude is \(|\mathbf{F}| = \sqrt{138^2 + 12^2} = \sqrt{19188} \approx 138.5\ \text{N}\).
Answer Details
The force is found from Newton's second law \(\mathbf{F} = m\mathbf{a}\), where \(\mathbf{a}\) is the second derivative of the position vector.
Given \(\mathbf{r} = (4t^3 - t^2)\mathbf{i} - (2t^2 - t)\mathbf{j}\), differentiate once for velocity:
\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = (12t^2 - 2t)\mathbf{i} - (4t - 1)\mathbf{j}.\]
Differentiate again for acceleration:
\[\mathbf{a} = \frac{d\mathbf{v}}{dt} = (24t - 2)\mathbf{i} - 4\mathbf{j}.\]
At \(t = 2\): \(\mathbf{a} = (24(2) - 2)\mathbf{i} - 4\mathbf{j} = 46\mathbf{i} - 4\mathbf{j}\).
With mass \(m = 3\ \text{kg}\):
\[\mathbf{F} = 3(46\mathbf{i} - 4\mathbf{j}) = 138\mathbf{i} - 12\mathbf{j}\ \text{N}.\]
Its magnitude is \(|\mathbf{F}| = \sqrt{138^2 + 12^2} = \sqrt{19188} \approx 138.5\ \text{N}\).
Question 6 Report
Five students are to be selected from a large population. If 60% of them are boys and the rest are girls, find the probability that :
(a) exactly 3 of them are boys;
(b) at least 3 of them are girls.
Selecting \(5\) students, \(P(\text{boy}) = 0.6\) and \(P(\text{girl}) = 0.4\). Use the binomial distribution.
(a) Exactly 3 boys (so 2 girls):
\[\binom{5}{3}(0.6)^3(0.4)^2 = 10(0.216)(0.16) = 0.3456 \approx 0.346.\]
(b) At least 3 girls means \(3\), \(4\) or \(5\) girls, with \(P(\text{girl}) = 0.4\):
\[P(3) = \binom{5}{3}(0.4)^3(0.6)^2 = 10(0.064)(0.36) = 0.2304,\]
\[P(4) = \binom{5}{4}(0.4)^4(0.6) = 5(0.0256)(0.6) = 0.0768,\]
\[P(5) = (0.4)^5 = 0.01024.\]
Sum: \(0.2304 + 0.0768 + 0.01024 = 0.31744 \approx 0.317\).
Answer Details
Selecting \(5\) students, \(P(\text{boy}) = 0.6\) and \(P(\text{girl}) = 0.4\). Use the binomial distribution.
(a) Exactly 3 boys (so 2 girls):
\[\binom{5}{3}(0.6)^3(0.4)^2 = 10(0.216)(0.16) = 0.3456 \approx 0.346.\]
(b) At least 3 girls means \(3\), \(4\) or \(5\) girls, with \(P(\text{girl}) = 0.4\):
\[P(3) = \binom{5}{3}(0.4)^3(0.6)^2 = 10(0.064)(0.36) = 0.2304,\]
\[P(4) = \binom{5}{4}(0.4)^4(0.6) = 5(0.0256)(0.6) = 0.0768,\]
\[P(5) = (0.4)^5 = 0.01024.\]
Sum: \(0.2304 + 0.0768 + 0.01024 = 0.31744 \approx 0.317\).
Question 7 Report
The images of (3, 2) and (-1, 4) under a linear transformation T are (-1, 4) and (7, 11) respectively. P is another transformation where \(P : (x, y) \to (x + y, x + 2y)\).
(a) Find the matrices T and P of the linear transformations T and P;
(b) Calculate TP.
(c) Find the image of the point X(4, 3) under TP.
(a) Let \(T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Using the images \((3,2)\to(-1,4)\) and \((-1,4)\to(7,11)\):
From the \(x\)-outputs: \(3a + 2b = -1\) and \(-a + 4b = 7\). Solving gives \(b = \tfrac{10}{7},\ a = -\tfrac{9}{7}\).
From the \(y\)-outputs: \(3c + 2d = 4\) and \(-c + 4d = 11\). Solving gives \(d = \tfrac{37}{14},\ c = -\tfrac{3}{7}\).
\[T = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}.\]
For \(P:(x,y)\to(x+y,\,x+2y)\), \(P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\).
(b) \(\displaystyle TP = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}.\)
(c) Image of \(X(4,3)\) under \(TP\):
\[\frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 74 \\ 328 \end{pmatrix} = \left(\frac{37}{7},\ \frac{164}{7}\right).\]
Answer Details
(a) Let \(T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Using the images \((3,2)\to(-1,4)\) and \((-1,4)\to(7,11)\):
From the \(x\)-outputs: \(3a + 2b = -1\) and \(-a + 4b = 7\). Solving gives \(b = \tfrac{10}{7},\ a = -\tfrac{9}{7}\).
From the \(y\)-outputs: \(3c + 2d = 4\) and \(-c + 4d = 11\). Solving gives \(d = \tfrac{37}{14},\ c = -\tfrac{3}{7}\).
\[T = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}.\]
For \(P:(x,y)\to(x+y,\,x+2y)\), \(P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\).
(b) \(\displaystyle TP = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}.\)
(c) Image of \(X(4,3)\) under \(TP\):
\[\frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 74 \\ 328 \end{pmatrix} = \left(\frac{37}{7},\ \frac{164}{7}\right).\]
Question 8 Report
If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\) at the point (-2, 1).
Differentiate \(3x^2 + 2y^2 + xy + x - 7 = 0\) implicitly with respect to \(x\). Use the product rule on \(xy\):
\[6x + 4y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) + 1 = 0.\]
Collect the \(\dfrac{dy}{dx}\) terms:
\[\frac{dy}{dx}(4y + x) = -(6x + y + 1) \;\Rightarrow\; \frac{dy}{dx} = -\frac{6x + y + 1}{4y + x}.\]
At the point \((-2, 1)\):
\[\frac{dy}{dx} = -\frac{6(-2) + 1 + 1}{4(1) + (-2)} = -\frac{-12 + 2}{4 - 2} = -\frac{-10}{2} = 5.\]
Answer: \(\dfrac{dy}{dx} = 5\) at \((-2, 1)\).
Answer Details
Differentiate \(3x^2 + 2y^2 + xy + x - 7 = 0\) implicitly with respect to \(x\). Use the product rule on \(xy\):
\[6x + 4y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) + 1 = 0.\]
Collect the \(\dfrac{dy}{dx}\) terms:
\[\frac{dy}{dx}(4y + x) = -(6x + y + 1) \;\Rightarrow\; \frac{dy}{dx} = -\frac{6x + y + 1}{4y + x}.\]
At the point \((-2, 1)\):
\[\frac{dy}{dx} = -\frac{6(-2) + 1 + 1}{4(1) + (-2)} = -\frac{-12 + 2}{4 - 2} = -\frac{-10}{2} = 5.\]
Answer: \(\dfrac{dy}{dx} = 5\) at \((-2, 1)\).
Question 9 Report
(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6j) and (2i + 3j) and the distance covered by the ships after t seconds are in metres, find (i) MN ; (ii) |MN|, when t = 3 seconds.
(b) A particle is acted upon by forces \(F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j\) and \(F_{4} = -4i + qj\), where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values of p and q.
(a) After time \(t\), position \(=\) initial position \(+\) velocity\(\times t\):
\[M: (3+5t)\mathbf{i} + (7+6t)\mathbf{j},\qquad N: (4+2t)\mathbf{i} + (5+3t)\mathbf{j}.\]
(i) \(\overrightarrow{MN} = \) position of \(N\) minus position of \(M\):
\[\overrightarrow{MN} = (4+2t-3-5t)\mathbf{i} + (5+3t-7-6t)\mathbf{j} = (1 - 3t)\mathbf{i} - (2 + 3t)\mathbf{j}.\]
(ii) At \(t = 3\): \(\overrightarrow{MN} = (1-9)\mathbf{i} - (2+9)\mathbf{j} = -8\mathbf{i} - 11\mathbf{j}\), so
\[|\overrightarrow{MN}| = \sqrt{(-8)^2 + (-11)^2} = \sqrt{64 + 121} = \sqrt{185} \approx 13.6\ \text{m}.\]
(b) For equilibrium the sum of the forces is zero. Adding components:
\(\mathbf{i}\): \(5 + q - 2p - 4 = 0 \Rightarrow q = 2p - 1\).
\(\mathbf{j}\): \(p + 1 + 3 + q = 0 \Rightarrow q = -p - 4\).
Equating: \(2p - 1 = -p - 4 \Rightarrow 3p = -3 \Rightarrow p = -1\), and \(q = -(-1) - 4 = -3\).
Answer: \(p = -1,\; q = -3\).
Answer Details
(a) After time \(t\), position \(=\) initial position \(+\) velocity\(\times t\):
\[M: (3+5t)\mathbf{i} + (7+6t)\mathbf{j},\qquad N: (4+2t)\mathbf{i} + (5+3t)\mathbf{j}.\]
(i) \(\overrightarrow{MN} = \) position of \(N\) minus position of \(M\):
\[\overrightarrow{MN} = (4+2t-3-5t)\mathbf{i} + (5+3t-7-6t)\mathbf{j} = (1 - 3t)\mathbf{i} - (2 + 3t)\mathbf{j}.\]
(ii) At \(t = 3\): \(\overrightarrow{MN} = (1-9)\mathbf{i} - (2+9)\mathbf{j} = -8\mathbf{i} - 11\mathbf{j}\), so
\[|\overrightarrow{MN}| = \sqrt{(-8)^2 + (-11)^2} = \sqrt{64 + 121} = \sqrt{185} \approx 13.6\ \text{m}.\]
(b) For equilibrium the sum of the forces is zero. Adding components:
\(\mathbf{i}\): \(5 + q - 2p - 4 = 0 \Rightarrow q = 2p - 1\).
\(\mathbf{j}\): \(p + 1 + 3 + q = 0 \Rightarrow q = -p - 4\).
Equating: \(2p - 1 = -p - 4 \Rightarrow 3p = -3 \Rightarrow p = -1\), and \(q = -(-1) - 4 = -3\).
Answer: \(p = -1,\; q = -3\).
Question 10 Report
The table shows the marks obtained by a group of students in a class test.
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 |
| No of students |
4 | 9 | 18 | 23 | 10 | 6 |
(a) Draw a histogram for the distribution ;
(b) Use your histogram to estimate the median of the distribution.
(a) Histogram
Convert the class limits to continuous class boundaries. Since every class has width 5, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 40–44 | 39.5–44.5 | 4 | 4 |
| 45–49 | 44.5–49.5 | 9 | 13 |
| 50–54 | 49.5–54.5 | 18 | 31 |
| 55–59 | 54.5–59.5 | 23 | 54 |
| 60–64 | 59.5–64.5 | 10 | 64 |
| 65–69 | 64.5–69.5 | 6 | 70 |
The required histogram is:
(b) Median
The total frequency is
\[N=70,\qquad \frac{N}{2}=35.\]
The 35th observation lies in the class \(55\text{–}59\), since the cumulative frequency before this class is 31 and the cumulative frequency at the end is 54.
Using \(L=54.5\), \(c_f=31\), \(f=23\) and class width \(c=5\),
\[\begin{aligned}\text{Median} &=L+\frac{\left(\frac{N}{2}-c_f\right)}{f}\times c\\&=54.5+\frac{35-31}{23}\times5\\&=54.5+0.87\\&\approx55.4.\end{aligned}\]
Therefore, the estimated median mark is \(55.4\).
Answer Details
(a) Histogram
Convert the class limits to continuous class boundaries. Since every class has width 5, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 40–44 | 39.5–44.5 | 4 | 4 |
| 45–49 | 44.5–49.5 | 9 | 13 |
| 50–54 | 49.5–54.5 | 18 | 31 |
| 55–59 | 54.5–59.5 | 23 | 54 |
| 60–64 | 59.5–64.5 | 10 | 64 |
| 65–69 | 64.5–69.5 | 6 | 70 |
The required histogram is:
(b) Median
The total frequency is
\[N=70,\qquad \frac{N}{2}=35.\]
The 35th observation lies in the class \(55\text{–}59\), since the cumulative frequency before this class is 31 and the cumulative frequency at the end is 54.
Using \(L=54.5\), \(c_f=31\), \(f=23\) and class width \(c=5\),
\[\begin{aligned}\text{Median} &=L+\frac{\left(\frac{N}{2}-c_f\right)}{f}\times c\\&=54.5+\frac{35-31}{23}\times5\\&=54.5+0.87\\&\approx55.4.\end{aligned}\]
Therefore, the estimated median mark is \(55.4\).
Question 11 Report
(a) The position vectors of the points X and Y are \(x = (-2i + 5j)\) and \(y = (i - 7j)\) respectively. Find :
(i) (3x + 2y) ; (ii) \(|(y - 2x)|\) ; (iii) the angle between x and y ; (iv) the unit vector in the direction of \((x + y)\).
(b) A bullet of mass 0.084kg is fired horizontally into a 20 kg block of wood at rest on a smooth floor. If they both move at a velocity of \(0.24 ms^{-1}\) after impact; Calculate, correct to two decimal places, the initial velocity of the bullet.
(a) With \(\mathbf{x} = -2\mathbf{i} + 5\mathbf{j}\) and \(\mathbf{y} = \mathbf{i} - 7\mathbf{j}\):
(i) \(3\mathbf{x} + 2\mathbf{y} = (-6\mathbf{i} + 15\mathbf{j}) + (2\mathbf{i} - 14\mathbf{j}) = -4\mathbf{i} + \mathbf{j}\).
(ii) \(\mathbf{y} - 2\mathbf{x} = (\mathbf{i} - 7\mathbf{j}) - (-4\mathbf{i} + 10\mathbf{j}) = 5\mathbf{i} - 17\mathbf{j}\), so \(|\mathbf{y} - 2\mathbf{x}| = \sqrt{25 + 289} = \sqrt{314} \approx 17.7\).
(iii) \(\cos\theta = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}||\mathbf{y}|}\), where \(\mathbf{x}\cdot\mathbf{y} = (-2)(1) + (5)(-7) = -37\), \(|\mathbf{x}| = \sqrt{29}\), \(|\mathbf{y}| = \sqrt{50}\):
\[\cos\theta = \frac{-37}{\sqrt{29}\,\sqrt{50}} = \frac{-37}{38.08} = -0.9717 \Rightarrow \theta \approx 166.3^\circ.\]
(iv) \(\mathbf{x} + \mathbf{y} = -\mathbf{i} - 2\mathbf{j}\), \(|\mathbf{x} + \mathbf{y}| = \sqrt{5}\). Unit vector \(= \dfrac{-\mathbf{i} - 2\mathbf{j}}{\sqrt{5}} \approx -0.45\mathbf{i} - 0.89\mathbf{j}\).
(b) Conservation of momentum (block initially at rest). Let \(u\) be the bullet's initial speed:
\[0.084\,u = (0.084 + 20)(0.24) \Rightarrow u = \frac{20.084 \times 0.24}{0.084} = \frac{4.82016}{0.084} \approx 57.38\ \text{m s}^{-1}.\]
Answer Details
(a) With \(\mathbf{x} = -2\mathbf{i} + 5\mathbf{j}\) and \(\mathbf{y} = \mathbf{i} - 7\mathbf{j}\):
(i) \(3\mathbf{x} + 2\mathbf{y} = (-6\mathbf{i} + 15\mathbf{j}) + (2\mathbf{i} - 14\mathbf{j}) = -4\mathbf{i} + \mathbf{j}\).
(ii) \(\mathbf{y} - 2\mathbf{x} = (\mathbf{i} - 7\mathbf{j}) - (-4\mathbf{i} + 10\mathbf{j}) = 5\mathbf{i} - 17\mathbf{j}\), so \(|\mathbf{y} - 2\mathbf{x}| = \sqrt{25 + 289} = \sqrt{314} \approx 17.7\).
(iii) \(\cos\theta = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}||\mathbf{y}|}\), where \(\mathbf{x}\cdot\mathbf{y} = (-2)(1) + (5)(-7) = -37\), \(|\mathbf{x}| = \sqrt{29}\), \(|\mathbf{y}| = \sqrt{50}\):
\[\cos\theta = \frac{-37}{\sqrt{29}\,\sqrt{50}} = \frac{-37}{38.08} = -0.9717 \Rightarrow \theta \approx 166.3^\circ.\]
(iv) \(\mathbf{x} + \mathbf{y} = -\mathbf{i} - 2\mathbf{j}\), \(|\mathbf{x} + \mathbf{y}| = \sqrt{5}\). Unit vector \(= \dfrac{-\mathbf{i} - 2\mathbf{j}}{\sqrt{5}} \approx -0.45\mathbf{i} - 0.89\mathbf{j}\).
(b) Conservation of momentum (block initially at rest). Let \(u\) be the bullet's initial speed:
\[0.084\,u = (0.084 + 20)(0.24) \Rightarrow u = \frac{20.084 \times 0.24}{0.084} = \frac{4.82016}{0.084} \approx 57.38\ \text{m s}^{-1}.\]
Question 12 Report
(a) Express \(\frac{2\sqrt{2}}{\sqrt{48} - \sqrt{8} - \sqrt{27}}\) in the form \(p + q\sqrt{r}\), where p, q and r are rational numbers.
(b) If \(V = A\log_{10} (M + N)\), express N in terms of M, V and A.
(a) Simplify the surds: \(\sqrt{48} = 4\sqrt{3},\ \sqrt{8} = 2\sqrt{2},\ \sqrt{27} = 3\sqrt{3}\).
\[\sqrt{48} - \sqrt{8} - \sqrt{27} = 4\sqrt{3} - 2\sqrt{2} - 3\sqrt{3} = \sqrt{3} - 2\sqrt{2}\]So the expression is \(\dfrac{2\sqrt{2}}{\sqrt{3} - 2\sqrt{2}}\). Rationalise using \((\sqrt{3} + 2\sqrt{2})\):
\[\frac{2\sqrt{2}(\sqrt{3} + 2\sqrt{2})}{(\sqrt{3})^2 - (2\sqrt{2})^2} = \frac{2\sqrt{6} + 8}{3 - 8} = \frac{2\sqrt{6} + 8}{-5}\] \[= -\frac{8}{5} - \frac{2}{5}\sqrt{6}\]This is of the form \(p + q\sqrt{r}\) with \(p = -\tfrac{8}{5},\ q = -\tfrac{2}{5},\ r = 6\).
(b) From \(V = A\log_{10}(M + N)\):
\[\frac{V}{A} = \log_{10}(M + N) \ \Rightarrow\ M + N = 10^{V/A}\] \[N = 10^{V/A} - M\]Answer Details
(a) Simplify the surds: \(\sqrt{48} = 4\sqrt{3},\ \sqrt{8} = 2\sqrt{2},\ \sqrt{27} = 3\sqrt{3}\).
\[\sqrt{48} - \sqrt{8} - \sqrt{27} = 4\sqrt{3} - 2\sqrt{2} - 3\sqrt{3} = \sqrt{3} - 2\sqrt{2}\]So the expression is \(\dfrac{2\sqrt{2}}{\sqrt{3} - 2\sqrt{2}}\). Rationalise using \((\sqrt{3} + 2\sqrt{2})\):
\[\frac{2\sqrt{2}(\sqrt{3} + 2\sqrt{2})}{(\sqrt{3})^2 - (2\sqrt{2})^2} = \frac{2\sqrt{6} + 8}{3 - 8} = \frac{2\sqrt{6} + 8}{-5}\] \[= -\frac{8}{5} - \frac{2}{5}\sqrt{6}\]This is of the form \(p + q\sqrt{r}\) with \(p = -\tfrac{8}{5},\ q = -\tfrac{2}{5},\ r = 6\).
(b) From \(V = A\log_{10}(M + N)\):
\[\frac{V}{A} = \log_{10}(M + N) \ \Rightarrow\ M + N = 10^{V/A}\] \[N = 10^{V/A} - M\]Question 13 Report
(a) A manufacturer produces light bulbs which are tested in the following way. A batch is accepted in either of the following cases:
(i) a first sample of 5 bulbs contains no faulty bulbs ; (ii) a first sample of 5 bulbs contains at least one faulty bulb but a second sample of size 5 has no faulty bulb. If 10% of the bulbs are faulty, what is the probability that the batch is accepted?
(b) A bag contains 15 identical marbles of which 3 are black, Keshi picks a marble at random from the bag and replaces it. If this is repeated 10 times; what is the probability that he :
(i) did not pick a black ball? (ii) picked a black ball at most three times?
(a) With \(10\%\) faulty, \(P(\text{a bulb is good}) = 0.9\). For a sample of 5, \(P(\text{no faulty}) = (0.9)^5 = 0.59049\).
The batch is accepted if the first sample has no faulty, OR the first sample has at least one faulty but the second sample has none:
\[P(\text{accept}) = (0.9)^5 + \big[1 - (0.9)^5\big](0.9)^5 = 0.59049 + (0.40951)(0.59049).\]
\[= 0.59049 + 0.241813 = 0.832303 \approx 0.832.\]
(b) Bag of 15 marbles, 3 black, so \(P(\text{black}) = \tfrac{3}{15} = 0.2\), \(P(\text{not black}) = 0.8\), with replacement, \(n = 10\).
(i) Did not pick a black at all: \((0.8)^{10} = 0.1073742 \approx 0.107\).
(ii) Picked black at most three times \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.107374,\ P(1)=\binom{10}{1}(0.2)(0.8)^9=0.268435,\]
\[P(2)=\binom{10}{2}(0.2)^2(0.8)^8=0.301990,\ P(3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201327.\]
Sum \(= 0.879\) (to three decimal places).
Answer Details
(a) With \(10\%\) faulty, \(P(\text{a bulb is good}) = 0.9\). For a sample of 5, \(P(\text{no faulty}) = (0.9)^5 = 0.59049\).
The batch is accepted if the first sample has no faulty, OR the first sample has at least one faulty but the second sample has none:
\[P(\text{accept}) = (0.9)^5 + \big[1 - (0.9)^5\big](0.9)^5 = 0.59049 + (0.40951)(0.59049).\]
\[= 0.59049 + 0.241813 = 0.832303 \approx 0.832.\]
(b) Bag of 15 marbles, 3 black, so \(P(\text{black}) = \tfrac{3}{15} = 0.2\), \(P(\text{not black}) = 0.8\), with replacement, \(n = 10\).
(i) Did not pick a black at all: \((0.8)^{10} = 0.1073742 \approx 0.107\).
(ii) Picked black at most three times \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.107374,\ P(1)=\binom{10}{1}(0.2)(0.8)^9=0.268435,\]
\[P(2)=\binom{10}{2}(0.2)^2(0.8)^8=0.301990,\ P(3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201327.\]
Sum \(= 0.879\) (to three decimal places).
Question 14 Report
A particle moves from point O along a straight line such that its acceleration at any time, t seconds is \(a = (4 - 2t) ms^{-2}\). At t = 0, its distance from O is 18 metres while its velocity is \(5 ms^{-1}\).
(a) At what time will the velocity be greatest?
(b) Calculate the : (i) time ; (ii) distance of the particle from O when the particle is momentarily at rest.
Integrate the acceleration to get velocity, then position, applying the initial conditions.
\[v = \int (4 - 2t)\,dt = 4t - t^2 + C.\]
At \(t = 0,\ v = 5\), so \(C = 5\) and \(v = 4t - t^2 + 5\).
(a) Velocity is greatest when \(\dfrac{dv}{dt} = a = 0\):
\[4 - 2t = 0 \Rightarrow t = 2\ \text{s}.\]
(b) The particle is momentarily at rest when \(v = 0\):
\[4t - t^2 + 5 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t = 5\ \text{s}\ (t>0).\]
(i) Time \(= 5\ \text{s}\).
Now integrate for displacement: \(s = \int(4t - t^2 + 5)\,dt = 2t^2 - \dfrac{t^3}{3} + 5t + D\). At \(t = 0,\ s = 18\), so \(D = 18\).
(ii) At \(t = 5\):
\[s = 2(25) - \frac{125}{3} + 25 + 18 = 93 - \frac{125}{3} = \frac{154}{3} \approx 51.3\ \text{m}.\]
The particle is \(\dfrac{154}{3}\ \text{m} \approx 51.3\ \text{m}\) from \(O\).
Answer Details
Integrate the acceleration to get velocity, then position, applying the initial conditions.
\[v = \int (4 - 2t)\,dt = 4t - t^2 + C.\]
At \(t = 0,\ v = 5\), so \(C = 5\) and \(v = 4t - t^2 + 5\).
(a) Velocity is greatest when \(\dfrac{dv}{dt} = a = 0\):
\[4 - 2t = 0 \Rightarrow t = 2\ \text{s}.\]
(b) The particle is momentarily at rest when \(v = 0\):
\[4t - t^2 + 5 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t = 5\ \text{s}\ (t>0).\]
(i) Time \(= 5\ \text{s}\).
Now integrate for displacement: \(s = \int(4t - t^2 + 5)\,dt = 2t^2 - \dfrac{t^3}{3} + 5t + D\). At \(t = 0,\ s = 18\), so \(D = 18\).
(ii) At \(t = 5\):
\[s = 2(25) - \frac{125}{3} + 25 + 18 = 93 - \frac{125}{3} = \frac{154}{3} \approx 51.3\ \text{m}.\]
The particle is \(\dfrac{154}{3}\ \text{m} \approx 51.3\ \text{m}\) from \(O\).
Question 15 Report
If the quadratic equation \((2x - 1) - p(x^{2} + 2) = 0\), where p is a constant, has real roots :
(a) show that \(2p^{2} + p - 1 < 0\);
(b) find the values of p.
(a) Expand the equation \((2x - 1) - p(x^2 + 2) = 0\) into standard quadratic form:
\[-px^2 + 2x - 1 - 2p = 0 \;\Rightarrow\; px^2 - 2x + (1 + 2p) = 0.\]
Here \(a = p,\ b = -2,\ c = 1 + 2p\). For real roots the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-2)^2 - 4p(1 + 2p) \ge 0 \;\Rightarrow\; 4 - 4p - 8p^2 \ge 0.\]
Divide through by \(4\): \(1 - p - 2p^2 \ge 0\). Multiplying by \(-1\) (reversing the inequality):
\[2p^2 + p - 1 \le 0.\]
(For two distinct real roots the inequality is strict, \(2p^2 + p - 1 < 0\), as stated.)
(b) Solve \(2p^2 + p - 1 = 0\): factorising, \((2p - 1)(p + 1) = 0\), so \(p = \tfrac12\) or \(p = -1\).
Since the coefficient of \(p^2\) is positive, \(2p^2 + p - 1 \le 0\) between the roots:
\[-1 \le p \le \tfrac{1}{2}.\]
Answer Details
(a) Expand the equation \((2x - 1) - p(x^2 + 2) = 0\) into standard quadratic form:
\[-px^2 + 2x - 1 - 2p = 0 \;\Rightarrow\; px^2 - 2x + (1 + 2p) = 0.\]
Here \(a = p,\ b = -2,\ c = 1 + 2p\). For real roots the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-2)^2 - 4p(1 + 2p) \ge 0 \;\Rightarrow\; 4 - 4p - 8p^2 \ge 0.\]
Divide through by \(4\): \(1 - p - 2p^2 \ge 0\). Multiplying by \(-1\) (reversing the inequality):
\[2p^2 + p - 1 \le 0.\]
(For two distinct real roots the inequality is strict, \(2p^2 + p - 1 < 0\), as stated.)
(b) Solve \(2p^2 + p - 1 = 0\): factorising, \((2p - 1)(p + 1) = 0\), so \(p = \tfrac12\) or \(p = -1\).
Since the coefficient of \(p^2\) is positive, \(2p^2 + p - 1 \le 0\) between the roots:
\[-1 \le p \le \tfrac{1}{2}.\]
Question 16 Report
(a) If \(y = (2x + 3)^{7} + \frac{x + 1}{2x - 1}\), find the value of \(\frac{\mathrm d y}{\mathrm d x}\) at x = -1.
(b) Using the substitution, \(u = x + 2\), evaluate \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\).
(a) Differentiate \(y = (2x+3)^7 + \dfrac{x+1}{2x-1}\). For the first term use the chain rule; for the second use the quotient rule:
\[\frac{dy}{dx} = 7(2x+3)^6(2) + \frac{(2x-1)(1) - (x+1)(2)}{(2x-1)^2} = 14(2x+3)^6 + \frac{-3}{(2x-1)^2}.\]
At \(x = -1\): \(2x+3 = 1\) so \((2x+3)^6 = 1\), and \(2x-1 = -3\) so \((2x-1)^2 = 9\):
\[\frac{dy}{dx} = 14(1) - \frac{3}{9} = 14 - \frac{1}{3} = \frac{41}{3} = 13\tfrac{2}{3}.\]
(b) With \(u = x + 2\), we have \(x = u - 2\), \(x - 1 = u - 3\), \(dx = du\). Limits: \(x=1\Rightarrow u=3\), \(x=2\Rightarrow u=4\).
\[\int_1^2 \frac{x-1}{(x+2)^4}\,dx = \int_3^4 \frac{u-3}{u^4}\,du = \int_3^4 \left(u^{-3} - 3u^{-4}\right)du.\]
\[= \left[-\frac{1}{2u^2} + \frac{1}{u^3}\right]_3^4 = \left(-\frac{1}{32} + \frac{1}{64}\right) - \left(-\frac{1}{18} + \frac{1}{27}\right) = -\frac{1}{64} + \frac{1}{54} = \frac{5}{1728} \approx 0.0029.\]
Answer Details
(a) Differentiate \(y = (2x+3)^7 + \dfrac{x+1}{2x-1}\). For the first term use the chain rule; for the second use the quotient rule:
\[\frac{dy}{dx} = 7(2x+3)^6(2) + \frac{(2x-1)(1) - (x+1)(2)}{(2x-1)^2} = 14(2x+3)^6 + \frac{-3}{(2x-1)^2}.\]
At \(x = -1\): \(2x+3 = 1\) so \((2x+3)^6 = 1\), and \(2x-1 = -3\) so \((2x-1)^2 = 9\):
\[\frac{dy}{dx} = 14(1) - \frac{3}{9} = 14 - \frac{1}{3} = \frac{41}{3} = 13\tfrac{2}{3}.\]
(b) With \(u = x + 2\), we have \(x = u - 2\), \(x - 1 = u - 3\), \(dx = du\). Limits: \(x=1\Rightarrow u=3\), \(x=2\Rightarrow u=4\).
\[\int_1^2 \frac{x-1}{(x+2)^4}\,dx = \int_3^4 \frac{u-3}{u^4}\,du = \int_3^4 \left(u^{-3} - 3u^{-4}\right)du.\]
\[= \left[-\frac{1}{2u^2} + \frac{1}{u^3}\right]_3^4 = \left(-\frac{1}{32} + \frac{1}{64}\right) - \left(-\frac{1}{18} + \frac{1}{27}\right) = -\frac{1}{64} + \frac{1}{54} = \frac{5}{1728} \approx 0.0029.\]
Question 17 Report
(a) The nth term of a sequence is given by \(T_{n} = 4T_{n - 1} - 3\). If twice the third term is five times the second term, find the first three terms of the sequence.
(b) Given that \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\), find the values of the constants k, m and r.
(a) The recurrence is \(T_n = 4T_{n-1} - 3\). Express the terms through \(T_1\):
\[T_2 = 4T_1 - 3,\qquad T_3 = 4T_2 - 3 = 4(4T_1 - 3) - 3 = 16T_1 - 15.\]
Given \(2T_3 = 5T_2\):
\[2(16T_1 - 15) = 5(4T_1 - 3) \Rightarrow 32T_1 - 30 = 20T_1 - 15 \Rightarrow 12T_1 = 15 \Rightarrow T_1 = \tfrac{5}{4}.\]
Then \(T_2 = 4(\tfrac54) - 3 = 2\) and \(T_3 = 4(2) - 3 = 5\).
The first three terms are \(\tfrac{5}{4},\ 2,\ 5\).
(b) Multiplying the matrices row by row:
Row 1: \(2(1) + 0\cdot m + 1\cdot r = 2 + r = k\).
Row 2: \(5(1) - 3m + r = 2 \Rightarrow -3m + r = -3\).
Row 3: \(0 + 4m + 6r = 26\).
From Row 2, \(r = 3m - 3\). Substitute into Row 3:
\[4m + 6(3m - 3) = 26 \Rightarrow 22m - 18 = 26 \Rightarrow m = 2,\]
so \(r = 3(2) - 3 = 3\) and \(k = 2 + r = 5\).
Answer: \(k = 5,\; m = 2,\; r = 3\).
Answer Details
(a) The recurrence is \(T_n = 4T_{n-1} - 3\). Express the terms through \(T_1\):
\[T_2 = 4T_1 - 3,\qquad T_3 = 4T_2 - 3 = 4(4T_1 - 3) - 3 = 16T_1 - 15.\]
Given \(2T_3 = 5T_2\):
\[2(16T_1 - 15) = 5(4T_1 - 3) \Rightarrow 32T_1 - 30 = 20T_1 - 15 \Rightarrow 12T_1 = 15 \Rightarrow T_1 = \tfrac{5}{4}.\]
Then \(T_2 = 4(\tfrac54) - 3 = 2\) and \(T_3 = 4(2) - 3 = 5\).
The first three terms are \(\tfrac{5}{4},\ 2,\ 5\).
(b) Multiplying the matrices row by row:
Row 1: \(2(1) + 0\cdot m + 1\cdot r = 2 + r = k\).
Row 2: \(5(1) - 3m + r = 2 \Rightarrow -3m + r = -3\).
Row 3: \(0 + 4m + 6r = 26\).
From Row 2, \(r = 3m - 3\). Substitute into Row 3:
\[4m + 6(3m - 3) = 26 \Rightarrow 22m - 18 = 26 \Rightarrow m = 2,\]
so \(r = 3(2) - 3 = 3\) and \(k = 2 + r = 5\).
Answer: \(k = 5,\; m = 2,\; r = 3\).
Question 18 Report
A uniform plank PQ of length 8m and mass 10kg is supported horizontally at the end P and at point R, 3 metres from Q. A boy of mass 20 kg walks along the plank starting from P. If the plank is in equilibrium, calculate the
(a) reactions at P and R when he walked 1.5 metres;
(b) distance he had walked when the two reactions are equal;
(c) distance he walked before the plank tips over.
The plank \(PQ\) is \(8\ \text{m}\) long, mass \(10\ \text{kg}\), supported at \(P\) (left end) and at \(R\), which is \(3\ \text{m}\) from \(Q\), i.e. \(5\ \text{m}\) from \(P\). The plank weight \(10g\) acts at the midpoint, \(4\ \text{m}\) from \(P\). A boy of weight \(20g\) stands at distance \(d\) from \(P\). (Take \(g = 10\ \text{m s}^{-2}\), so weights are \(100\ \text{N}\) and \(200\ \text{N}\).)
(a) When \(d = 1.5\ \text{m}\). Taking moments about \(P\) (so \(R_P\) has no moment):
\[R_R \times 5 = 10g(4) + 20g(1.5) = 40g + 30g = 70g \Rightarrow R_R = 14g = 140\ \text{N}.\]
Vertical equilibrium: \(R_P + R_R = 30g = 300\ \text{N} \Rightarrow R_P = 160\ \text{N}\).
So \(R_P = 160\ \text{N},\ R_R = 140\ \text{N}\).
(b) When the reactions are equal. Each reaction \(= \tfrac12(30g) = 15g = 150\ \text{N}\). Moments about \(P\):
\[15g \times 5 = 40g + 20g\,d \Rightarrow 75g = 40g + 20g\,d \Rightarrow d = \frac{35}{20} = 1.75\ \text{m}.\]
(c) When the plank tips. The plank tips about \(R\) when \(R_P = 0\). Taking moments about \(R\), the boy (on the \(Q\) side, \(d - 5\) from \(R\)) balances the plank weight (\(5 - 4 = 1\ \text{m}\) on the \(P\) side):
\[20g(d - 5) = 10g(1) \Rightarrow d - 5 = 0.5 \Rightarrow d = 5.5\ \text{m}.\]
The boy can walk \(5.5\ \text{m}\) from \(P\) before the plank tips over.
Answer Details
The plank \(PQ\) is \(8\ \text{m}\) long, mass \(10\ \text{kg}\), supported at \(P\) (left end) and at \(R\), which is \(3\ \text{m}\) from \(Q\), i.e. \(5\ \text{m}\) from \(P\). The plank weight \(10g\) acts at the midpoint, \(4\ \text{m}\) from \(P\). A boy of weight \(20g\) stands at distance \(d\) from \(P\). (Take \(g = 10\ \text{m s}^{-2}\), so weights are \(100\ \text{N}\) and \(200\ \text{N}\).)
(a) When \(d = 1.5\ \text{m}\). Taking moments about \(P\) (so \(R_P\) has no moment):
\[R_R \times 5 = 10g(4) + 20g(1.5) = 40g + 30g = 70g \Rightarrow R_R = 14g = 140\ \text{N}.\]
Vertical equilibrium: \(R_P + R_R = 30g = 300\ \text{N} \Rightarrow R_P = 160\ \text{N}\).
So \(R_P = 160\ \text{N},\ R_R = 140\ \text{N}\).
(b) When the reactions are equal. Each reaction \(= \tfrac12(30g) = 15g = 150\ \text{N}\). Moments about \(P\):
\[15g \times 5 = 40g + 20g\,d \Rightarrow 75g = 40g + 20g\,d \Rightarrow d = \frac{35}{20} = 1.75\ \text{m}.\]
(c) When the plank tips. The plank tips about \(R\) when \(R_P = 0\). Taking moments about \(R\), the boy (on the \(Q\) side, \(d - 5\) from \(R\)) balances the plank weight (\(5 - 4 = 1\ \text{m}\) on the \(P\) side):
\[20g(d - 5) = 10g(1) \Rightarrow d - 5 = 0.5 \Rightarrow d = 5.5\ \text{m}.\]
The boy can walk \(5.5\ \text{m}\) from \(P\) before the plank tips over.
