WAEC SSCE - Further Mathematics - 2013 (Essay)
Question 1 Report
A stone is dropped vertically downwards from the top of a tower of height 45m with a speed of 20 ms\(^{-1}\). Find the :
(a) time it takes to reach the ground ;
(b) speed with which it hits the ground. [Take \(g = 10 ms^{-2}\)].
Take downward as positive with initial speed \(u = 20\text{ ms}^{-1}\), \(g = 10\text{ ms}^{-2}\), height \(s = 45\text{ m}\).
(a) Time to reach the ground. Using \( s = ut + \tfrac12 g t^2 \):
\[ 45 = 20t + 5t^2 \Rightarrow 5t^2 + 20t - 45 = 0 \Rightarrow t^2 + 4t - 9 = 0. \]
\[ t = \frac{-4 + \sqrt{16 + 36}}{2} = \frac{-4 + \sqrt{52}}{2} = \frac{-4 + 7.211}{2} \approx 1.61\text{ s}. \]
(b) Speed on hitting the ground. Using \( v^2 = u^2 + 2gs \):
\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]
Answer Details
Take downward as positive with initial speed \(u = 20\text{ ms}^{-1}\), \(g = 10\text{ ms}^{-2}\), height \(s = 45\text{ m}\).
(a) Time to reach the ground. Using \( s = ut + \tfrac12 g t^2 \):
\[ 45 = 20t + 5t^2 \Rightarrow 5t^2 + 20t - 45 = 0 \Rightarrow t^2 + 4t - 9 = 0. \]
\[ t = \frac{-4 + \sqrt{16 + 36}}{2} = \frac{-4 + \sqrt{52}}{2} = \frac{-4 + 7.211}{2} \approx 1.61\text{ s}. \]
(b) Speed on hitting the ground. Using \( v^2 = u^2 + 2gs \):
\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]
Question 2 Report
(a) A bag contains 5 blue, 4 green and 3 yellow balls. All the balls are identical except for colour. Three balls are drawn at random without replacement. Find the probability that : (i) all three balls have the same colour ; (ii) two balls have the same colour.
(b) The table shows the ranks of the marks scored by 7 candidates in Physics and Chemistry tests.
| Physics | 6 | 5 | 4 | 3 | 2 | 7 | 1 |
| Chemistry | 7 | 6 | 2 | 4 | 1 | 5 | 3 |
Calculate the Spearman's rank correlation coefficient.
(a) Bag: 5 blue, 4 green, 3 yellow (12 balls), draw 3 without replacement. Total selections \(={}^{12}C_3=220\).
(i) All three the same colour.\[{}^{5}C_3+{}^{4}C_3+{}^{3}C_3=10+4+1=15\]\[P=\frac{15}{220}=\frac{3}{44}\approx 0.068\]
(ii) Exactly two of the same colour (two alike and one different):\[{}^{5}C_2(7)+{}^{4}C_2(8)+{}^{3}C_2(9)=10(7)+6(8)+3(9)=70+48+27=145\]\[P=\frac{145}{220}=\frac{29}{44}\approx 0.659\](Check: all-different \(=5\times4\times3=60\); \(15+145+60=220\).)
(b) Spearman's rank correlation (Physics vs Chemistry ranks). \(n=7\).
| Physics | Chemistry | \(d\) | \(d^2\) |
|---|---|---|---|
| 6 | 7 | -1 | 1 |
| 5 | 6 | -1 | 1 |
| 4 | 2 | 2 | 4 |
| 3 | 4 | -1 | 1 |
| 2 | 1 | 1 | 1 |
| 7 | 5 | 2 | 4 |
| 1 | 3 | -2 | 4 |
| \(\sum d^2\) | 16 | ||
\[r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6(16)}{7(48)}=1-\frac{96}{336}=1-0.286=0.714\]
Answers: (a)(i) \(\tfrac{3}{44}\); (a)(ii) \(\tfrac{29}{44}\); (b) \(r_s\approx 0.714\).
Answer Details
(a) Bag: 5 blue, 4 green, 3 yellow (12 balls), draw 3 without replacement. Total selections \(={}^{12}C_3=220\).
(i) All three the same colour.\[{}^{5}C_3+{}^{4}C_3+{}^{3}C_3=10+4+1=15\]\[P=\frac{15}{220}=\frac{3}{44}\approx 0.068\]
(ii) Exactly two of the same colour (two alike and one different):\[{}^{5}C_2(7)+{}^{4}C_2(8)+{}^{3}C_2(9)=10(7)+6(8)+3(9)=70+48+27=145\]\[P=\frac{145}{220}=\frac{29}{44}\approx 0.659\](Check: all-different \(=5\times4\times3=60\); \(15+145+60=220\).)
(b) Spearman's rank correlation (Physics vs Chemistry ranks). \(n=7\).
| Physics | Chemistry | \(d\) | \(d^2\) |
|---|---|---|---|
| 6 | 7 | -1 | 1 |
| 5 | 6 | -1 | 1 |
| 4 | 2 | 2 | 4 |
| 3 | 4 | -1 | 1 |
| 2 | 1 | 1 | 1 |
| 7 | 5 | 2 | 4 |
| 1 | 3 | -2 | 4 |
| \(\sum d^2\) | 16 | ||
\[r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6(16)}{7(48)}=1-\frac{96}{336}=1-0.286=0.714\]
Answers: (a)(i) \(\tfrac{3}{44}\); (a)(ii) \(\tfrac{29}{44}\); (b) \(r_s\approx 0.714\).
Question 3 Report
A particle is under the action of forces \(P = (4N, 030°)\) and \(R = (10N, 300°)\). Find the force that will keep the particle in equilibrium.
Resolve each force into east (x) and north (y) components using its bearing, where east \(= F\sin\theta\), north \(= F\cos\theta\).
\( \mathbf{P} = (4\sin 30^\circ,\ 4\cos 30^\circ) = (2.000,\ 3.464) \).
\( \mathbf{R} = (10\sin 300^\circ,\ 10\cos 300^\circ) = (-8.660,\ 5.000) \).
Resultant: \( \mathbf{P} + \mathbf{R} = (-6.660,\ 8.464) \).
The equilibrant \( \mathbf{E} \) is equal and opposite: \( \mathbf{E} = (6.660,\ -8.464) \).
Magnitude: \[ |\mathbf{E}| = \sqrt{6.660^2 + 8.464^2} = \sqrt{44.36 + 71.64} = \sqrt{116} \approx 10.8\text{ N}. \]
Direction (bearing): the equilibrant points east and south, so \[ \text{bearing} = 180^\circ - \tan^{-1}\!\left(\frac{6.660}{8.464}\right) = 180^\circ - 38.2^\circ \approx 142^\circ. \]
The equilibrating force is about \( 10.8\text{ N} \) on a bearing of \( 142^\circ \).
Answer Details
Resolve each force into east (x) and north (y) components using its bearing, where east \(= F\sin\theta\), north \(= F\cos\theta\).
\( \mathbf{P} = (4\sin 30^\circ,\ 4\cos 30^\circ) = (2.000,\ 3.464) \).
\( \mathbf{R} = (10\sin 300^\circ,\ 10\cos 300^\circ) = (-8.660,\ 5.000) \).
Resultant: \( \mathbf{P} + \mathbf{R} = (-6.660,\ 8.464) \).
The equilibrant \( \mathbf{E} \) is equal and opposite: \( \mathbf{E} = (6.660,\ -8.464) \).
Magnitude: \[ |\mathbf{E}| = \sqrt{6.660^2 + 8.464^2} = \sqrt{44.36 + 71.64} = \sqrt{116} \approx 10.8\text{ N}. \]
Direction (bearing): the equilibrant points east and south, so \[ \text{bearing} = 180^\circ - \tan^{-1}\!\left(\frac{6.660}{8.464}\right) = 180^\circ - 38.2^\circ \approx 142^\circ. \]
The equilibrating force is about \( 10.8\text{ N} \) on a bearing of \( 142^\circ \).
Question 4 Report
The table gives the relationship between the height, in metres, of a plant and the number of days it is left to grow.
| Number of days (x) |
10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| Height (y) | 1.0 | 1.1 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 | 2.3 |
(a) Using a scale of 2 cm to represent 0.5 units on the y- axis and 2cm to 10 units on the x- axis, draw a scatter diagram for the information.
(b) Find \(\bar{x}\), the mean of x, and \(\bar{y}\), the mean of y, and plot \((\bar{x}, \bar{y})\) on the diagram.
(c) Draw the line of best fit to pass through \((\bar{x}, \bar{y})\) and \((10, 1)\).
(d) From graph, find the :
(i) equation of the line of best fit ; (ii) height of plant in 75 days.
(a) & (c) Scatter diagram, mean point and line of best fit
Plotting the eight pairs of readings using a scale of 2 cm to 10 units on the x-axis and 2 cm to 0.5 units on the y-axis gives the scatter diagram below. The mean point \((\bar{x},\bar{y})=(45,\,1.55)\) is marked, and the line of best fit is drawn straight through \((10,\,1.0)\) and \((45,\,1.55)\).
(b) Mean of x and mean of y
\[\bar{x}=\frac{10+20+30+40+50+60+70+80}{8}=\frac{360}{8}=45\] \[\bar{y}=\frac{1.0+1.1+1.2+1.4+1.6+1.8+2.0+2.3}{8}=\frac{12.4}{8}=1.55\]Therefore the mean point is \((\bar{x},\bar{y})=(45,\,1.55)\), which is plotted on the diagram above.
(d)(i) Equation of the line of best fit
The line passes through \((\bar{x},\bar{y})=(45,\,1.55)\) and \((10,\,1.0)\). Its gradient is
\[m=\frac{1.55-1.0}{45-10}=\frac{0.55}{35}=0.0157\]Using the point \((10,\,1.0)\):
\[y-1.0=0.0157\,(x-10)\] \[y-1.0=0.0157x-0.157\] \[\boxed{\,y=0.0157x+0.843\,}\]Check: at \(x=45\), \(y=0.0157(45)+0.843=0.707+0.843=1.55=\bar{y}\), so the line indeed passes through the mean point.
(d)(ii) Height of plant after 75 days
Substituting \(x=75\) into the equation of the line of best fit:
\[y=0.0157(75)+0.843=1.179+0.843=2.02\]Reading off the graph at \(x=75\) confirms this value. The height of the plant after 75 days is approximately 2.0 m (2.02 m).
Answer Details
(a) & (c) Scatter diagram, mean point and line of best fit
Plotting the eight pairs of readings using a scale of 2 cm to 10 units on the x-axis and 2 cm to 0.5 units on the y-axis gives the scatter diagram below. The mean point \((\bar{x},\bar{y})=(45,\,1.55)\) is marked, and the line of best fit is drawn straight through \((10,\,1.0)\) and \((45,\,1.55)\).
(b) Mean of x and mean of y
\[\bar{x}=\frac{10+20+30+40+50+60+70+80}{8}=\frac{360}{8}=45\] \[\bar{y}=\frac{1.0+1.1+1.2+1.4+1.6+1.8+2.0+2.3}{8}=\frac{12.4}{8}=1.55\]Therefore the mean point is \((\bar{x},\bar{y})=(45,\,1.55)\), which is plotted on the diagram above.
(d)(i) Equation of the line of best fit
The line passes through \((\bar{x},\bar{y})=(45,\,1.55)\) and \((10,\,1.0)\). Its gradient is
\[m=\frac{1.55-1.0}{45-10}=\frac{0.55}{35}=0.0157\]Using the point \((10,\,1.0)\):
\[y-1.0=0.0157\,(x-10)\] \[y-1.0=0.0157x-0.157\] \[\boxed{\,y=0.0157x+0.843\,}\]Check: at \(x=45\), \(y=0.0157(45)+0.843=0.707+0.843=1.55=\bar{y}\), so the line indeed passes through the mean point.
(d)(ii) Height of plant after 75 days
Substituting \(x=75\) into the equation of the line of best fit:
\[y=0.0157(75)+0.843=1.179+0.843=2.02\]Reading off the graph at \(x=75\) confirms this value. The height of the plant after 75 days is approximately 2.0 m (2.02 m).
Question 5 Report
The initial velocity of a particle of mass 0.1kg is 40 m/s in the direction of the unit vector j. The velocity of the particle changed to 30 m/s in the direction of the unit vector i. Find the change in momentum.
Momentum \( = \text{mass} \times \text{velocity} \), and change in momentum \( = m(\vec{v}_f - \vec{v}_i) \).
Initial velocity: \( \vec{v}_i = 40\mathbf{j} \); final velocity: \( \vec{v}_f = 30\mathbf{i} \); mass \( m = 0.1\text{ kg} \).
\[ \Delta \vec{p} = 0.1(30\mathbf{i} - 40\mathbf{j}) = (3\mathbf{i} - 4\mathbf{j})\text{ kg ms}^{-1}. \]
Magnitude: \[ |\Delta \vec{p}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\text{ kg ms}^{-1}. \]
The change in momentum is \( 3\mathbf{i} - 4\mathbf{j} \) (magnitude \(5\text{ kg ms}^{-1}\)).
Answer Details
Momentum \( = \text{mass} \times \text{velocity} \), and change in momentum \( = m(\vec{v}_f - \vec{v}_i) \).
Initial velocity: \( \vec{v}_i = 40\mathbf{j} \); final velocity: \( \vec{v}_f = 30\mathbf{i} \); mass \( m = 0.1\text{ kg} \).
\[ \Delta \vec{p} = 0.1(30\mathbf{i} - 40\mathbf{j}) = (3\mathbf{i} - 4\mathbf{j})\text{ kg ms}^{-1}. \]
Magnitude: \[ |\Delta \vec{p}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\text{ kg ms}^{-1}. \]
The change in momentum is \( 3\mathbf{i} - 4\mathbf{j} \) (magnitude \(5\text{ kg ms}^{-1}\)).
Question 6 Report
Three school prefects are to be chosen from four girls and five boys. What is the probability that :
(a) only boys will be chosen ;
(b) more girls than boys will be chosen ?
Choosing 3 prefects from 4 girls and 5 boys (9 people). Total selections: \[ \binom{9}{3} = 84. \]
(a) Only boys chosen: \[ \binom{5}{3} = 10, \qquad P = \frac{10}{84} = \frac{5}{42}. \]
(b) More girls than boys means either 3 girls (0 boys) or 2 girls and 1 boy:
\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).
\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]
Answer Details
Choosing 3 prefects from 4 girls and 5 boys (9 people). Total selections: \[ \binom{9}{3} = 84. \]
(a) Only boys chosen: \[ \binom{5}{3} = 10, \qquad P = \frac{10}{84} = \frac{5}{42}. \]
(b) More girls than boys means either 3 girls (0 boys) or 2 girls and 1 boy:
\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).
\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]
Question 7 Report
The displacement S metres of a particle from a fixed point O at time t seconds is given by \(S = t^{2} - 6t + 5\).
(a) On a graph sheet, draw a displacement- time graph for the interval \(0 \leq x \leq 6\).
(b) From the graph, find the : (i) time at which the velocity is zero ; (ii) average velocity over the interval \(0 \leq x \leq 4\) ; (iii) total distance covered in the interval \(0 \leq x \leq 5\).
(a) Displacement–time graph
The displacement is \(S = t^{2} - 6t + 5\). Compute \(S\) for each whole second in the interval \(0 \le t \le 6\):
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| S (m) | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
Plotting these points and joining them with a smooth curve gives the parabola below (scale on the time axis: 2 cm to 1 s; on the displacement axis: 2 cm to 1 m).
(b) Reading the graph
(i) Time at which the velocity is zero.
The velocity is the gradient of the displacement–time graph, \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t}\). The velocity is zero where the tangent to the curve is horizontal, i.e. at the lowest point (turning point) of the parabola. From the graph this occurs at
(Check: \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t} = 2t - 6 = 0 \Rightarrow t = 3\ \text{s}.\))
(ii) Average velocity over the interval \(0 \le t \le 4\).
Average velocity is the change in displacement divided by the time taken (the gradient of the chord joining the two end points). From the graph, \(S = 5\) m at \(t = 0\) and \(S = -3\) m at \(t = 4\):
The average velocity is \(-2\ \text{m/s}\); the negative sign shows the net motion is directed back towards, and past, \(O\). (For interest, the total path length in this interval is \(9 + 1 = 10\) m, so the average speed is \(10/4 = 2.5\) m/s.)
(iii) Total distance covered in the interval \(0 \le t \le 5\).
Distance is the actual length of path travelled, so the two legs of the journey are added as positive lengths. From the graph the particle first moves from \(S = 5\) m down to the turning point \(S = -4\) m at \(t = 3\) s, then rises back to \(S = 0\) m at \(t = 5\) s.
Answer Details
(a) Displacement–time graph
The displacement is \(S = t^{2} - 6t + 5\). Compute \(S\) for each whole second in the interval \(0 \le t \le 6\):
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| S (m) | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
Plotting these points and joining them with a smooth curve gives the parabola below (scale on the time axis: 2 cm to 1 s; on the displacement axis: 2 cm to 1 m).
(b) Reading the graph
(i) Time at which the velocity is zero.
The velocity is the gradient of the displacement–time graph, \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t}\). The velocity is zero where the tangent to the curve is horizontal, i.e. at the lowest point (turning point) of the parabola. From the graph this occurs at
(Check: \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t} = 2t - 6 = 0 \Rightarrow t = 3\ \text{s}.\))
(ii) Average velocity over the interval \(0 \le t \le 4\).
Average velocity is the change in displacement divided by the time taken (the gradient of the chord joining the two end points). From the graph, \(S = 5\) m at \(t = 0\) and \(S = -3\) m at \(t = 4\):
The average velocity is \(-2\ \text{m/s}\); the negative sign shows the net motion is directed back towards, and past, \(O\). (For interest, the total path length in this interval is \(9 + 1 = 10\) m, so the average speed is \(10/4 = 2.5\) m/s.)
(iii) Total distance covered in the interval \(0 \le t \le 5\).
Distance is the actual length of path travelled, so the two legs of the journey are added as positive lengths. From the graph the particle first moves from \(S = 5\) m down to the turning point \(S = -4\) m at \(t = 3\) s, then rises back to \(S = 0\) m at \(t = 5\) s.
Question 8 Report
(a) Find the maximum and minimum points of the curve \(y = 2x^{3} - 3x^{2} - 12x + 4\).
(b) Sketch the curve in (a) above.
(a) Maximum and minimum points of \( y = 2x^{3} - 3x^{2} - 12x + 4 \).
At stationary points \( \dfrac{dy}{dx} = 0 \):
\[ \frac{dy}{dx} = 6x^{2} - 6x - 12 = 6\left(x^{2} - x - 2\right) = 6(x-2)(x+1). \]Setting \( \dfrac{dy}{dx} = 0 \):
\[ 6(x-2)(x+1) = 0 \implies x = 2 \ \text{ or } \ x = -1. \]The nature of each point is tested with the second derivative:
\[ \frac{d^{2}y}{dx^{2}} = 12x - 6. \]At \( x = -1 \):
\[ \frac{d^{2}y}{dx^{2}} = 12(-1) - 6 = -18 < 0 \quad \Rightarrow \ \text{maximum}. \]\[ y = 2(-1)^{3} - 3(-1)^{2} - 12(-1) + 4 = -2 - 3 + 12 + 4 = 11. \]Maximum point \( (-1,\ 11) \).
At \( x = 2 \):
\[ \frac{d^{2}y}{dx^{2}} = 12(2) - 6 = 18 > 0 \quad \Rightarrow \ \text{minimum}. \]\[ y = 2(2)^{3} - 3(2)^{2} - 12(2) + 4 = 16 - 12 - 24 + 4 = -16. \]Minimum point \( (2,\ -16) \).
(b) Sketch of the curve.
To locate the curve, note the following guide points:
| Feature | Point |
| Maximum turning point | \((-1,\ 11)\) |
| Minimum turning point | \((2,\ -16)\) |
| \(y\)-intercept \((x=0)\) | \((0,\ 4)\) |
| An \(x\)-intercept \((y=0)\) | \((-2,\ 0)\) |
Since the leading coefficient is positive, the curve rises from the bottom-left, climbs to the maximum \((-1,\ 11)\), falls through the \(y\)-intercept \((0,\ 4)\) to the minimum \((2,\ -16)\), then rises again to the top-right, as shown below.
Answer Details
(a) Maximum and minimum points of \( y = 2x^{3} - 3x^{2} - 12x + 4 \).
At stationary points \( \dfrac{dy}{dx} = 0 \):
\[ \frac{dy}{dx} = 6x^{2} - 6x - 12 = 6\left(x^{2} - x - 2\right) = 6(x-2)(x+1). \]Setting \( \dfrac{dy}{dx} = 0 \):
\[ 6(x-2)(x+1) = 0 \implies x = 2 \ \text{ or } \ x = -1. \]The nature of each point is tested with the second derivative:
\[ \frac{d^{2}y}{dx^{2}} = 12x - 6. \]At \( x = -1 \):
\[ \frac{d^{2}y}{dx^{2}} = 12(-1) - 6 = -18 < 0 \quad \Rightarrow \ \text{maximum}. \]\[ y = 2(-1)^{3} - 3(-1)^{2} - 12(-1) + 4 = -2 - 3 + 12 + 4 = 11. \]Maximum point \( (-1,\ 11) \).
At \( x = 2 \):
\[ \frac{d^{2}y}{dx^{2}} = 12(2) - 6 = 18 > 0 \quad \Rightarrow \ \text{minimum}. \]\[ y = 2(2)^{3} - 3(2)^{2} - 12(2) + 4 = 16 - 12 - 24 + 4 = -16. \]Minimum point \( (2,\ -16) \).
(b) Sketch of the curve.
To locate the curve, note the following guide points:
| Feature | Point |
| Maximum turning point | \((-1,\ 11)\) |
| Minimum turning point | \((2,\ -16)\) |
| \(y\)-intercept \((x=0)\) | \((0,\ 4)\) |
| An \(x\)-intercept \((y=0)\) | \((-2,\ 0)\) |
Since the leading coefficient is positive, the curve rises from the bottom-left, climbs to the maximum \((-1,\ 11)\), falls through the \(y\)-intercept \((0,\ 4)\) to the minimum \((2,\ -16)\), then rises again to the top-right, as shown below.
Question 9 Report
A side of a rectangle is three times the other. If the perimeter increases by 2%, find the percentage increase in the area of the rectangle.
Rectangle with one side three times the other; perimeter rises by 2%.
Let the shorter side be \(x\), so the longer side is \(3x\).
\[P=2(x+3x)=8x,\qquad A=x(3x)=3x^{2}\]The perimeter \(P=8x\) is directly proportional to \(x\), so a \(2\%\) rise in perimeter is a \(2\%\) rise in \(x\). Using differentials for the area:
\[A=3x^{2}\;\Rightarrow\;\frac{dA}{A}=2\cdot\frac{dx}{x}\]\[\frac{\Delta A}{A}\times 100\%=2\times 2\%=4\%\]Alternatively, the new side is \(1.02x\), so the new area is \(3(1.02x)^{2}=1.0404\times 3x^{2}\), an increase of \(4.04\%\), which rounds to about \(4\%\).
The area increases by approximately \(4\%\).
Answer Details
Rectangle with one side three times the other; perimeter rises by 2%.
Let the shorter side be \(x\), so the longer side is \(3x\).
\[P=2(x+3x)=8x,\qquad A=x(3x)=3x^{2}\]The perimeter \(P=8x\) is directly proportional to \(x\), so a \(2\%\) rise in perimeter is a \(2\%\) rise in \(x\). Using differentials for the area:
\[A=3x^{2}\;\Rightarrow\;\frac{dA}{A}=2\cdot\frac{dx}{x}\]\[\frac{\Delta A}{A}\times 100\%=2\times 2\%=4\%\]Alternatively, the new side is \(1.02x\), so the new area is \(3(1.02x)^{2}=1.0404\times 3x^{2}\), an increase of \(4.04\%\), which rounds to about \(4\%\).
The area increases by approximately \(4\%\).
Question 10 Report
(a) Three vectors a, b and c are \(\begin{pmatrix} 8 \\ 3 \end{pmatrix}, \begin{pmatrix} 6 \\ -5 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) respectively. Find the vector d such that \(|d| = \sqrt{41}\) and d is in the direction of \(a + b - 2c\).
(b) The coordinates of A and B are (3, 4) and (3, n) respectively. If AOB = 30°, find, correct to 2 decimal places, the values of n.
(a) First compute the direction vector \( \mathbf{a} + \mathbf{b} - 2\mathbf{c} \):
\[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix}. \]
Its magnitude is \( \sqrt{10^2 + 4^2} = \sqrt{116} = 2\sqrt{29} \). The required vector \( \mathbf{d} \) has magnitude \( \sqrt{41} \) in this direction:
\[ \mathbf{d} = \sqrt{41}\cdot\frac{1}{\sqrt{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \sqrt{\frac{41}{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \approx \begin{pmatrix} 5.94 \\ 2.38 \end{pmatrix}. \]
So \( \mathbf{d} \approx 5.94\mathbf{i} + 2.38\mathbf{j} \) (check: \( 5.94^2 + 2.38^2 \approx 41 \)).
(b) With \( O \) the origin, \( \vec{OA} = (3,4) \) so \( |OA| = 5 \), and \( \vec{OB} = (3,n) \) with \( |OB| = \sqrt{9+n^2} \). The angle \( AOB = 30^\circ \):
\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]
\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]
\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]
\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]
\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]
Answer Details
(a) First compute the direction vector \( \mathbf{a} + \mathbf{b} - 2\mathbf{c} \):
\[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix}. \]
Its magnitude is \( \sqrt{10^2 + 4^2} = \sqrt{116} = 2\sqrt{29} \). The required vector \( \mathbf{d} \) has magnitude \( \sqrt{41} \) in this direction:
\[ \mathbf{d} = \sqrt{41}\cdot\frac{1}{\sqrt{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \sqrt{\frac{41}{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \approx \begin{pmatrix} 5.94 \\ 2.38 \end{pmatrix}. \]
So \( \mathbf{d} \approx 5.94\mathbf{i} + 2.38\mathbf{j} \) (check: \( 5.94^2 + 2.38^2 \approx 41 \)).
(b) With \( O \) the origin, \( \vec{OA} = (3,4) \) so \( |OA| = 5 \), and \( \vec{OB} = (3,n) \) with \( |OB| = \sqrt{9+n^2} \). The angle \( AOB = 30^\circ \):
\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]
\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]
\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]
\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]
\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]
Question 11 Report
(a) The sum of the first three terms of a decreasing exponential sequence (G.P) is equal to 7 and the product of these three is equal to 8. Find the :
(i) common ratio ; (ii) first three terms of the sequence.
(b) Using the trapezium rule with the ordinates at x = 1, 2, 3, 4 and 5, calculate, correct to two decimal places, the value of \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\).
(a) Let the three terms be \( \dfrac{a}{r},\ a,\ ar \).
Product: \( \dfrac{a}{r}\cdot a \cdot ar = a^3 = 8 \Rightarrow a = 2. \)
Sum: \( \dfrac{2}{r} + 2 + 2r = 7 \Rightarrow \dfrac{2}{r} + 2r = 5 \Rightarrow 2r^2 - 5r + 2 = 0. \)
\[ (2r - 1)(r - 2) = 0 \Rightarrow r = \tfrac12 \text{ or } r = 2. \]
(i) The sequence is decreasing, so \( r = \dfrac12 \).
(ii) The three terms are \( \dfrac{a}{r} = 4,\ a = 2,\ ar = 1 \), i.e. \( 4,\ 2,\ 1 \).
(b) With \( f(x) = x + \dfrac{2}{x^2} \) and ordinates at \( x = 1,2,3,4,5 \) (\(h = 1\)):
\( f(1)=3,\ f(2)=2.5,\ f(3)=3.2222,\ f(4)=4.125,\ f(5)=5.08. \)
Trapezium rule: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]
\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]
Answer Details
(a) Let the three terms be \( \dfrac{a}{r},\ a,\ ar \).
Product: \( \dfrac{a}{r}\cdot a \cdot ar = a^3 = 8 \Rightarrow a = 2. \)
Sum: \( \dfrac{2}{r} + 2 + 2r = 7 \Rightarrow \dfrac{2}{r} + 2r = 5 \Rightarrow 2r^2 - 5r + 2 = 0. \)
\[ (2r - 1)(r - 2) = 0 \Rightarrow r = \tfrac12 \text{ or } r = 2. \]
(i) The sequence is decreasing, so \( r = \dfrac12 \).
(ii) The three terms are \( \dfrac{a}{r} = 4,\ a = 2,\ ar = 1 \), i.e. \( 4,\ 2,\ 1 \).
(b) With \( f(x) = x + \dfrac{2}{x^2} \) and ordinates at \( x = 1,2,3,4,5 \) (\(h = 1\)):
\( f(1)=3,\ f(2)=2.5,\ f(3)=3.2222,\ f(4)=4.125,\ f(5)=5.08. \)
Trapezium rule: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]
\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]
Question 12 Report
The table shows the distribution of ages of 22 students in a school.
| Age (years) | 12-14 | 15-17 | 18-20 | 21-23 | 24-26 |
| Frequency | 6 | 10 | 3 | 2 | 1 |
Using an assumed mean of 19, calculate, correct to three significant figures, the :
(a) mean age ; (b) standard deviation ; of the distribution.
Class width \(=3\); mid-values \(13,16,19,22,25\). Code with \(u=\dfrac{x-19}{3}\), \(A=19\).
| Age | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 12-14 | 13 | 6 | -2 | -12 | 24 |
| 15-17 | 16 | 10 | -1 | -10 | 10 |
| 18-20 | 19 | 3 | 0 | 0 | 0 |
| 21-23 | 22 | 2 | 1 | 2 | 2 |
| 24-26 | 25 | 1 | 2 | 2 | 4 |
| Total | 22 | -18 | 40 |
(a) Mean age.\[\bar{x}=19+\frac{\sum fu}{\sum f}\times 3=19+\frac{-18}{22}\times 3=19-2.4545=16.5\text{ years}\]
(b) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=3\sqrt{\frac{40}{22}-\left(\frac{-18}{22}\right)^2}\]\[=3\sqrt{1.8182-0.6694}=3\sqrt{1.1488}=3(1.0718)=3.22\]
To three significant figures: mean age = 16.5 years and standard deviation = 3.22 years.
Answer Details
Class width \(=3\); mid-values \(13,16,19,22,25\). Code with \(u=\dfrac{x-19}{3}\), \(A=19\).
| Age | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 12-14 | 13 | 6 | -2 | -12 | 24 |
| 15-17 | 16 | 10 | -1 | -10 | 10 |
| 18-20 | 19 | 3 | 0 | 0 | 0 |
| 21-23 | 22 | 2 | 1 | 2 | 2 |
| 24-26 | 25 | 1 | 2 | 2 | 4 |
| Total | 22 | -18 | 40 |
(a) Mean age.\[\bar{x}=19+\frac{\sum fu}{\sum f}\times 3=19+\frac{-18}{22}\times 3=19-2.4545=16.5\text{ years}\]
(b) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=3\sqrt{\frac{40}{22}-\left(\frac{-18}{22}\right)^2}\]\[=3\sqrt{1.8182-0.6694}=3\sqrt{1.1488}=3(1.0718)=3.22\]
To three significant figures: mean age = 16.5 years and standard deviation = 3.22 years.
Question 13 Report
(a) Using a scale of 2 cm to 30° on the x- axis, 2 cm to 0.2 units on the y- axis, on the same graph sheet, draw the graphs of \(y = \sin 2x\) and \(y = \cos x\) for \(0° \leq x \leq 210°\) at intervals of 30°.
(b) Using the graphs in (a), find the truth set of :
(i) \(\sin 2x = 0\) ; (ii) \(\sin 2x - \cos x = 0\).
(a) Prepare a table of values for \(y = \sin 2x\) and \(y = \cos x\) at intervals of \(30^\circ\) for \(0^\circ \le x \le 210^\circ\). For example, at \(x = 30^\circ\): \(\sin 2x = \sin 60^\circ = 0.87\) and \(\cos x = \cos 30^\circ = 0.87\); at \(x = 120^\circ\): \(\sin 2x = \sin 240^\circ = -0.87\) and \(\cos x = \cos 120^\circ = -0.50\). The complete table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) |
| \(y = \sin 2x\) | 0 | 0.87 | 0.87 | 0 | -0.87 | -0.87 | 0 | 0.87 |
| \(y = \cos x\) | 1 | 0.87 | 0.50 | 0 | -0.50 | -0.87 | -1 | -0.87 |
Plotting these points, using a scale of 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 0.2 units on the \(y\)-axis, gives the two smooth curves below:
(b)(i) The truth set of \(\sin 2x = 0\) is read where the curve \(y = \sin 2x\) crosses the \(x\)-axis (i.e. where \(y = 0\)). From the graph these crossings occur at \(x = 0^\circ,\ 90^\circ\) and \(180^\circ\):
\[ \{\,0^\circ,\ 90^\circ,\ 180^\circ\,\}. \](b)(ii) The truth set of \(\sin 2x - \cos x = 0\), i.e. \(\sin 2x = \cos x\), is read where the two curves intersect. From the graph the curves cross at \(x = 30^\circ,\ 90^\circ\) and \(150^\circ\).
This agrees with the algebra: \(\sin 2x = \cos x \Rightarrow 2\sin x\cos x = \cos x \Rightarrow \cos x\,(2\sin x - 1) = 0\), so \(\cos x = 0 \Rightarrow x = 90^\circ\), or \(\sin x = \tfrac{1}{2} \Rightarrow x = 30^\circ,\ 150^\circ\). Hence the truth set is:
\[ \{\,30^\circ,\ 90^\circ,\ 150^\circ\,\}. \]Answer Details
(a) Prepare a table of values for \(y = \sin 2x\) and \(y = \cos x\) at intervals of \(30^\circ\) for \(0^\circ \le x \le 210^\circ\). For example, at \(x = 30^\circ\): \(\sin 2x = \sin 60^\circ = 0.87\) and \(\cos x = \cos 30^\circ = 0.87\); at \(x = 120^\circ\): \(\sin 2x = \sin 240^\circ = -0.87\) and \(\cos x = \cos 120^\circ = -0.50\). The complete table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) |
| \(y = \sin 2x\) | 0 | 0.87 | 0.87 | 0 | -0.87 | -0.87 | 0 | 0.87 |
| \(y = \cos x\) | 1 | 0.87 | 0.50 | 0 | -0.50 | -0.87 | -1 | -0.87 |
Plotting these points, using a scale of 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 0.2 units on the \(y\)-axis, gives the two smooth curves below:
(b)(i) The truth set of \(\sin 2x = 0\) is read where the curve \(y = \sin 2x\) crosses the \(x\)-axis (i.e. where \(y = 0\)). From the graph these crossings occur at \(x = 0^\circ,\ 90^\circ\) and \(180^\circ\):
\[ \{\,0^\circ,\ 90^\circ,\ 180^\circ\,\}. \](b)(ii) The truth set of \(\sin 2x - \cos x = 0\), i.e. \(\sin 2x = \cos x\), is read where the two curves intersect. From the graph the curves cross at \(x = 30^\circ,\ 90^\circ\) and \(150^\circ\).
This agrees with the algebra: \(\sin 2x = \cos x \Rightarrow 2\sin x\cos x = \cos x \Rightarrow \cos x\,(2\sin x - 1) = 0\), so \(\cos x = 0 \Rightarrow x = 90^\circ\), or \(\sin x = \tfrac{1}{2} \Rightarrow x = 30^\circ,\ 150^\circ\). Hence the truth set is:
\[ \{\,30^\circ,\ 90^\circ,\ 150^\circ\,\}. \]Question 14 Report
The line \(2y = x + 3\) meets the circle \(x^{2} + y^{2} - 2x + 6y - 15 = 0\) at points M and N, where N is in the first quadrant. Find the coordinates of M and N.
Line \(2y=x+3\) meets circle \(x^{2}+y^{2}-2x+6y-15=0\).
From the line, \(x=2y-3\). Substitute into the circle:
\[(2y-3)^{2}+y^{2}-2(2y-3)+6y-15=0\]\[(4y^{2}-12y+9)+y^{2}-4y+6+6y-15=0\]\[5y^{2}-10y+0=0\;\Rightarrow\;5y(y-2)=0\]So \(y=0\) or \(y=2\).
- \(y=0:\ x=2(0)-3=-3\), giving \((-3,0)\).
- \(y=2:\ x=2(2)-3=1\), giving \((1,2)\).
N is in the first quadrant, so \(N=(1,2)\) and \(M=(-3,0)\).
Answer Details
Line \(2y=x+3\) meets circle \(x^{2}+y^{2}-2x+6y-15=0\).
From the line, \(x=2y-3\). Substitute into the circle:
\[(2y-3)^{2}+y^{2}-2(2y-3)+6y-15=0\]\[(4y^{2}-12y+9)+y^{2}-4y+6+6y-15=0\]\[5y^{2}-10y+0=0\;\Rightarrow\;5y(y-2)=0\]So \(y=0\) or \(y=2\).
- \(y=0:\ x=2(0)-3=-3\), giving \((-3,0)\).
- \(y=2:\ x=2(2)-3=1\), giving \((1,2)\).
N is in the first quadrant, so \(N=(1,2)\) and \(M=(-3,0)\).
Question 15 Report
(a) If the coefficient of \(x^{2}\) and \(x^{3}\) in the expansion of \((p + qx)^{7}\) are equal, express q in terms of p.
(b) A man makes a weekly contribution into a fund. In the first week, he paid N180.00, second week N260.00, third week N340.00 and so on. How much would he have contributed in 16 weeks?
(a) Equal coefficients of \(x^{2}\) and \(x^{3}\) in \((p+qx)^{7}\).
The general term is \({}^{7}C_r\,p^{7-r}(qx)^{r}\).
\[\text{Coeff of }x^{2}:\ {}^{7}C_2\,p^{5}q^{2}=21p^{5}q^{2}\]\[\text{Coeff of }x^{3}:\ {}^{7}C_3\,p^{4}q^{3}=35p^{4}q^{3}\]Set them equal:
\[21p^{5}q^{2}=35p^{4}q^{3}\;\Rightarrow\;21p=35q\;\Rightarrow\;q=\frac{3p}{5}\](b) Weekly contributions 180, 260, 340, ...
This is an A.P. with \(a=180\), common difference \(d=80\), and \(n=16\).
\[S_{16}=\frac{n}{2}\big[2a+(n-1)d\big]=\frac{16}{2}\big[2(180)+15(80)\big]\]\[=8\,[360+1200]=8(1560)=12480\]He would have contributed \(\text{N}12{,}480.00\) in 16 weeks.
Answer Details
(a) Equal coefficients of \(x^{2}\) and \(x^{3}\) in \((p+qx)^{7}\).
The general term is \({}^{7}C_r\,p^{7-r}(qx)^{r}\).
\[\text{Coeff of }x^{2}:\ {}^{7}C_2\,p^{5}q^{2}=21p^{5}q^{2}\]\[\text{Coeff of }x^{3}:\ {}^{7}C_3\,p^{4}q^{3}=35p^{4}q^{3}\]Set them equal:
\[21p^{5}q^{2}=35p^{4}q^{3}\;\Rightarrow\;21p=35q\;\Rightarrow\;q=\frac{3p}{5}\](b) Weekly contributions 180, 260, 340, ...
This is an A.P. with \(a=180\), common difference \(d=80\), and \(n=16\).
\[S_{16}=\frac{n}{2}\big[2a+(n-1)d\big]=\frac{16}{2}\big[2(180)+15(80)\big]\]\[=8\,[360+1200]=8(1560)=12480\]He would have contributed \(\text{N}12{,}480.00\) in 16 weeks.
Question 16 Report
(a) The probability that a man wins a race is 0.8. In four different races, what is the probability that he wins : (i) all races ; (ii) no race ; (iii) at most 3 races ?
(b) A class consists of 5 girls and 10 boys. If a committee of 5 is chosen at random from the class, find the probability that :
(i) 3 boys are selected ; (ii) at least one girl is selected.
(a) Winning is binomial with \( p = 0.8,\ q = 0.2,\ n = 4 \).
(i) Wins all races: \( p^4 = 0.8^4 = 0.4096. \)
(ii) Wins no race: \( q^4 = 0.2^4 = 0.0016. \)
(iii) Wins at most 3 races \( = 1 - P(\text{all 4}) = 1 - 0.4096 = 0.5904. \)
(b) Committee of 5 chosen from 5 girls and 10 boys (15 people). Total \( = \binom{15}{5} = 3003 \).
(i) Exactly 3 boys (so 2 girls): \[ \frac{\binom{10}{3}\binom{5}{2}}{3003} = \frac{120 \times 10}{3003} = \frac{1200}{3003} = \frac{400}{1001} \approx 0.400. \]
(ii) At least one girl \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)
Answer Details
(a) Winning is binomial with \( p = 0.8,\ q = 0.2,\ n = 4 \).
(i) Wins all races: \( p^4 = 0.8^4 = 0.4096. \)
(ii) Wins no race: \( q^4 = 0.2^4 = 0.0016. \)
(iii) Wins at most 3 races \( = 1 - P(\text{all 4}) = 1 - 0.4096 = 0.5904. \)
(b) Committee of 5 chosen from 5 girls and 10 boys (15 people). Total \( = \binom{15}{5} = 3003 \).
(i) Exactly 3 boys (so 2 girls): \[ \frac{\binom{10}{3}\binom{5}{2}}{3003} = \frac{120 \times 10}{3003} = \frac{1200}{3003} = \frac{400}{1001} \approx 0.400. \]
(ii) At least one girl \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)
Question 17 Report
Calculate the gradient of the curve \(x^{3} + y^{3} - 2xy = 11\) at (2, -1).
Gradient of \(x^{3}+y^{3}-2xy=11\) at \((2,-1)\).
Differentiate implicitly with respect to \(x\), using the product rule on \(2xy\):
\[3x^{2}+3y^{2}\frac{dy}{dx}-2\left(y+x\frac{dy}{dx}\right)=0\]\[3x^{2}+3y^{2}\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0\]Collect the derivative terms:
\[\frac{dy}{dx}\left(3y^{2}-2x\right)=2y-3x^{2}\]\[\frac{dy}{dx}=\frac{2y-3x^{2}}{3y^{2}-2x}\]At \((2,-1)\):
\[\frac{dy}{dx}=\frac{2(-1)-3(2)^{2}}{3(-1)^{2}-2(2)}=\frac{-2-12}{3-4}=\frac{-14}{-1}=14\]The gradient of the curve at \((2,-1)\) is \(14\).
Answer Details
Gradient of \(x^{3}+y^{3}-2xy=11\) at \((2,-1)\).
Differentiate implicitly with respect to \(x\), using the product rule on \(2xy\):
\[3x^{2}+3y^{2}\frac{dy}{dx}-2\left(y+x\frac{dy}{dx}\right)=0\]\[3x^{2}+3y^{2}\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0\]Collect the derivative terms:
\[\frac{dy}{dx}\left(3y^{2}-2x\right)=2y-3x^{2}\]\[\frac{dy}{dx}=\frac{2y-3x^{2}}{3y^{2}-2x}\]At \((2,-1)\):
\[\frac{dy}{dx}=\frac{2(-1)-3(2)^{2}}{3(-1)^{2}-2(2)}=\frac{-2-12}{3-4}=\frac{-14}{-1}=14\]The gradient of the curve at \((2,-1)\) is \(14\).
Question 18 Report
(a) Differentiate \(\frac{x^{2} + 1}{(x + 1)^{2}}\) with respect to x.
(b)(i) Evaluate \(\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\).
(ii) Using the answer in (b)(i), solve the system of equations.
\(x + 2y - z = 4\)
\(2x + 3y - z = 2\)
\(-x + y + 3z = -1\).
(a) Differentiate \( y = \dfrac{x^2 + 1}{(x+1)^2} \) using the quotient rule with \( u = x^2+1,\ v = (x+1)^2 \):
\( u' = 2x,\ v' = 2(x+1) \).
\[ \frac{dy}{dx} = \frac{2x(x+1)^2 - (x^2+1)\,2(x+1)}{(x+1)^4} = \frac{2(x+1)\big[x(x+1) - (x^2+1)\big]}{(x+1)^4}. \]
\[ = \frac{2\big[x^2 + x - x^2 - 1\big]}{(x+1)^3} = \frac{2(x-1)}{(x+1)^3}. \]
(b)(i) Expand the determinant along the first row:
\[ \Delta = 1(3\cdot3 - (-1)\cdot1) - 2(2\cdot3 - (-1)(-1)) + (-1)(2\cdot1 - 3(-1)). \]
\[ \Delta = 1(10) - 2(5) - 1(5) = 10 - 10 - 5 = -5. \]
(ii) The system \( x+2y-z=4,\ 2x+3y-z=2,\ -x+y+3z=-1 \) has coefficient determinant \(\Delta = -5 \neq 0\), so use Cramer's rule.
\( \Delta_x = \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} = 40 - 10 - 5 = 25 \Rightarrow x = \dfrac{25}{-5} = -5. \)
\( \Delta_y = \begin{vmatrix} 1 & 4 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 3 \end{vmatrix} = 5 - 20 - 0 = -15 \Rightarrow y = \dfrac{-15}{-5} = 3. \)
\( \Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 2 \\ -1 & 1 & -1 \end{vmatrix} = -5 - 0 + 20 = 15 \Rightarrow z = \dfrac{15}{-5} = -3. \)
Solution: \( x = -5,\ y = 3,\ z = -3 \).
Answer Details
(a) Differentiate \( y = \dfrac{x^2 + 1}{(x+1)^2} \) using the quotient rule with \( u = x^2+1,\ v = (x+1)^2 \):
\( u' = 2x,\ v' = 2(x+1) \).
\[ \frac{dy}{dx} = \frac{2x(x+1)^2 - (x^2+1)\,2(x+1)}{(x+1)^4} = \frac{2(x+1)\big[x(x+1) - (x^2+1)\big]}{(x+1)^4}. \]
\[ = \frac{2\big[x^2 + x - x^2 - 1\big]}{(x+1)^3} = \frac{2(x-1)}{(x+1)^3}. \]
(b)(i) Expand the determinant along the first row:
\[ \Delta = 1(3\cdot3 - (-1)\cdot1) - 2(2\cdot3 - (-1)(-1)) + (-1)(2\cdot1 - 3(-1)). \]
\[ \Delta = 1(10) - 2(5) - 1(5) = 10 - 10 - 5 = -5. \]
(ii) The system \( x+2y-z=4,\ 2x+3y-z=2,\ -x+y+3z=-1 \) has coefficient determinant \(\Delta = -5 \neq 0\), so use Cramer's rule.
\( \Delta_x = \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} = 40 - 10 - 5 = 25 \Rightarrow x = \dfrac{25}{-5} = -5. \)
\( \Delta_y = \begin{vmatrix} 1 & 4 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 3 \end{vmatrix} = 5 - 20 - 0 = -15 \Rightarrow y = \dfrac{-15}{-5} = 3. \)
\( \Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 2 \\ -1 & 1 & -1 \end{vmatrix} = -5 - 0 + 20 = 15 \Rightarrow z = \dfrac{15}{-5} = -3. \)
Solution: \( x = -5,\ y = 3,\ z = -3 \).
