WAEC SSCE - Further Mathematics - 2013 (Essay)
Ibeere 1 Ìròyìn
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 required line of best fit
(b) Calculate the mean coordinates.
The \(x\)-values have total \(360\), so
\[ \bar{x}=\frac{360}{8}=45 \]The heights total \(12.4\), not \(11.4\):
\[ \bar{y}=\frac{1.0+1.1+1.2+1.4+1.6+1.8+2.0+2.3}{8} =\frac{12.4}{8}=1.55 \]Therefore the mean point to plot is \((45,1.55)\).
(c) Line of best fit. The instruction requires a straight line through both \((45,1.55)\) and \((10,1)\). This is the red line on the graph.
(d)(i) Equation of the line.
\[ \text{gradient}=\frac{1.55-1}{45-10} =\frac{0.55}{35} =\frac{11}{700} \approx0.0157 \]Using \((10,1)\):
\[ y-1=\frac{11}{700}(x-10) \] \[ y=\frac{11}{700}x+\frac{59}{70} \]So an appropriate equation is
\[ \boxed{y\approx0.0157x+0.843} \](d)(ii) Height after 75 days.
\[ y=0.0157(75)+0.843\approx2.02 \] \[ \boxed{\text{The plant's height is approximately }2.0\text{ m after 75 days.}} \]The alternative equation \(y=0.017x+0.78\) is not consistent with the stated construction because it does not pass through \((10,1)\): it gives \(y=0.95\) when \(x=10\). For this question, use the two points explicitly specified for the line.
Awọn alaye Idahun
(a)–(c) Scatter diagram, mean point and required line of best fit
(b) Calculate the mean coordinates.
The \(x\)-values have total \(360\), so
\[ \bar{x}=\frac{360}{8}=45 \]The heights total \(12.4\), not \(11.4\):
\[ \bar{y}=\frac{1.0+1.1+1.2+1.4+1.6+1.8+2.0+2.3}{8} =\frac{12.4}{8}=1.55 \]Therefore the mean point to plot is \((45,1.55)\).
(c) Line of best fit. The instruction requires a straight line through both \((45,1.55)\) and \((10,1)\). This is the red line on the graph.
(d)(i) Equation of the line.
\[ \text{gradient}=\frac{1.55-1}{45-10} =\frac{0.55}{35} =\frac{11}{700} \approx0.0157 \]Using \((10,1)\):
\[ y-1=\frac{11}{700}(x-10) \] \[ y=\frac{11}{700}x+\frac{59}{70} \]So an appropriate equation is
\[ \boxed{y\approx0.0157x+0.843} \](d)(ii) Height after 75 days.
\[ y=0.0157(75)+0.843\approx2.02 \] \[ \boxed{\text{The plant's height is approximately }2.0\text{ m after 75 days.}} \]The alternative equation \(y=0.017x+0.78\) is not consistent with the stated construction because it does not pass through \((10,1)\): it gives \(y=0.95\) when \(x=10\). For this question, use the two points explicitly specified for the line.
Ibeere 2 Ìròyìn
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 \( 142^\circ \).
Awọn alaye Idahun
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 \( 142^\circ \).
Ibeere 3 Ìròyìn
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 \(14\).
Awọn alaye Idahun
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 \(14\).
Ibeere 4 Ìròyìn
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) Tabulate \( S = t^2 - 6t + 5 \) for \( 0 \le t \le 6 \) and plot S against t.
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| S (m) | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
The graph is a parabola (minimum at \(t = 3\)).
(b)(i) Velocity is zero at the turning point of the graph, i.e. \( \dfrac{dS}{dt} = 2t - 6 = 0 \Rightarrow t = 3\text{ s}. \)
(ii) Average velocity over \( 0 \le t \le 4 \): \[ \frac{S(4) - S(0)}{4 - 0} = \frac{-3 - 5}{4} = -2\text{ ms}^{-1}. \]
(iii) Total distance over \( 0 \le t \le 5 \): the particle moves from \(S=5\) down to the minimum \(S(3) = -4\) (distance \(9\text{ m}\)), then back up to \(S(5) = 0\) (distance \(4\text{ m}\)):
\[ \text{Total distance} = 9 + 4 = 13\text{ m}. \]
Awọn alaye Idahun
(a) Tabulate \( S = t^2 - 6t + 5 \) for \( 0 \le t \le 6 \) and plot S against t.
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| S (m) | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
The graph is a parabola (minimum at \(t = 3\)).
(b)(i) Velocity is zero at the turning point of the graph, i.e. \( \dfrac{dS}{dt} = 2t - 6 = 0 \Rightarrow t = 3\text{ s}. \)
(ii) Average velocity over \( 0 \le t \le 4 \): \[ \frac{S(4) - S(0)}{4 - 0} = \frac{-3 - 5}{4} = -2\text{ ms}^{-1}. \]
(iii) Total distance over \( 0 \le t \le 5 \): the particle moves from \(S=5\) down to the minimum \(S(3) = -4\) (distance \(9\text{ m}\)), then back up to \(S(5) = 0\) (distance \(4\text{ m}\)):
\[ \text{Total distance} = 9 + 4 = 13\text{ m}. \]
Ibeere 5 Ìròyìn
(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 \( AOB = 30^\circ \):
\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]
\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]
\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]
\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]
\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]
Awọn alaye Idahun
(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 \( AOB = 30^\circ \):
\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]
\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]
\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]
\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]
\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]
Ibeere 6 Ìròyìn
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 \(4\%\).
Awọn alaye Idahun
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 \(4\%\).
Ibeere 7 Ìròyìn
(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) \(r_s\approx 0.714\).
Awọn alaye Idahun
(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) \(r_s\approx 0.714\).
Ibeere 8 Ìròyìn
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.
Awọn alaye Idahun
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.
Ibeere 9 Ìròyìn
(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.
Awọn alaye Idahun
(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.
Ibeere 10 Ìròyìn
(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: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]
\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]
Awọn alaye Idahun
(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: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]
\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]
Ibeere 11 Ìròyìn
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 \(M=(-3,0)\).
Awọn alaye Idahun
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 \(M=(-3,0)\).
Ibeere 12 Ìròyìn
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:
\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).
\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]
Awọn alaye Idahun
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:
\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).
\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]
Ibeere 13 Ìròyìn
(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.
Awọn alaye Idahun
(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.
Ibeere 14 Ìròyìn
(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 \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)
Awọn alaye Idahun
(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 \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)
Ibeere 15 Ìròyìn
(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: \( x = -5,\ y = 3,\ z = -3 \).
Awọn alaye Idahun
(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: \( x = -5,\ y = 3,\ z = -3 \).
Ibeere 16 Ìròyìn
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 \( v^2 = u^2 + 2gs \):
\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]
Awọn alaye Idahun
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 \( v^2 = u^2 + 2gs \):
\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]
Ibeere 17 Ìròyìn
(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:
\[ \{\,30^\circ,\ 90^\circ,\ 150^\circ\,\}. \]Awọn alaye Idahun
(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:
\[ \{\,30^\circ,\ 90^\circ,\ 150^\circ\,\}. \]Ibeere 18 Ìròyìn
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 \(5\text{ kg ms}^{-1}\)).
Awọn alaye Idahun
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 \(5\text{ kg ms}^{-1}\)).
