WAEC SSCE - Further Mathematics - 2019 (Essay)
Question 1 Rapport
(a) Solve, for \(x\) and \(y\), the simultaneous equations
\[3\log_2 x = y \qquad\text{and}\qquad \log_2 4x = y + 4.\](b) Using the value of \(y\) obtained in part (a), express \(y^{2} - 10y + 25\) in the form \(2^{n}\), and hence find the value of \(n\).
This is a concurrent-forces (Lami's theorem) problem in which a load of 120 N is held in equilibrium by a tension \(T\) and a reaction \(R\), with the geometry giving angles of \(90^\circ\) and \(135^\circ\) opposite \(T\) and the load respectively.
(a) By Lami's theorem, \(\dfrac{T}{\sin 90^\circ}=\dfrac{120}{\sin 135^\circ}\), so
\[T=\frac{120\sin 90^\circ}{\sin 135^\circ}=\frac{120(1)}{0.7071}=169.71\ \text{N}.\](b) Similarly \(\dfrac{R}{\sin 135^\circ}=\dfrac{120}{\sin 135^\circ}\), giving
\[R=120\times\frac{\sin 135^\circ}{\sin 135^\circ}=120\ \text{N}.\]Hence \(T\approx\textbf{169.71 N}\) and \(R=\textbf{120 N}\), confirming the stated working. (The full configuration/diagram was not supplied, so the angle assignments are taken from the given ratios.)
Détails de la réponse
This is a concurrent-forces (Lami's theorem) problem in which a load of 120 N is held in equilibrium by a tension \(T\) and a reaction \(R\), with the geometry giving angles of \(90^\circ\) and \(135^\circ\) opposite \(T\) and the load respectively.
(a) By Lami's theorem, \(\dfrac{T}{\sin 90^\circ}=\dfrac{120}{\sin 135^\circ}\), so
\[T=\frac{120\sin 90^\circ}{\sin 135^\circ}=\frac{120(1)}{0.7071}=169.71\ \text{N}.\](b) Similarly \(\dfrac{R}{\sin 135^\circ}=\dfrac{120}{\sin 135^\circ}\), giving
\[R=120\times\frac{\sin 135^\circ}{\sin 135^\circ}=120\ \text{N}.\]Hence \(T\approx\textbf{169.71 N}\) and \(R=\textbf{120 N}\), confirming the stated working. (The full configuration/diagram was not supplied, so the angle assignments are taken from the given ratios.)
Question 2 Rapport
The curve y = 7 - \(\frac{6}{x}\) and the line y + 2x - 3 = 0 intersect at two point. Finf the;
(a) coordinates of the two points
(b) equation of the perpendicular bisector of the line joining the two points
(a) The line \(y + 2x - 3 = 0\) gives \(y = 3 - 2x\). At intersection with \(y = 7 - \dfrac{6}{x}\):
\[7 - \frac{6}{x} = 3 - 2x\]Multiply through by \(x\):
\[7x - 6 = 3x - 2x^2 \;\Rightarrow\; 2x^2 + 4x - 6 = 0 \;\Rightarrow\; x^2 + 2x - 3 = 0\] \[(x+3)(x-1)=0 \;\Rightarrow\; x = -3 \text{ or } x = 1\]When \(x = 1:\ y = 3 - 2 = 1\). When \(x = -3:\ y = 3 + 6 = 9\).
The points are \((1,\ 1)\) and \((-3,\ 9)\).
(b) Midpoint \(= \left(\dfrac{1 + (-3)}{2},\ \dfrac{1 + 9}{2}\right) = (-1,\ 5)\).
Gradient of the join \(= \dfrac{9 - 1}{-3 - 1} = \dfrac{8}{-4} = -2\); the perpendicular gradient is \(\dfrac{1}{2}\).
\[y - 5 = \tfrac{1}{2}(x + 1) \;\Rightarrow\; 2y - 10 = x + 1 \;\Rightarrow\; x - 2y + 11 = 0\]Perpendicular bisector: \(x - 2y + 11 = 0\).
Détails de la réponse
(a) The line \(y + 2x - 3 = 0\) gives \(y = 3 - 2x\). At intersection with \(y = 7 - \dfrac{6}{x}\):
\[7 - \frac{6}{x} = 3 - 2x\]Multiply through by \(x\):
\[7x - 6 = 3x - 2x^2 \;\Rightarrow\; 2x^2 + 4x - 6 = 0 \;\Rightarrow\; x^2 + 2x - 3 = 0\] \[(x+3)(x-1)=0 \;\Rightarrow\; x = -3 \text{ or } x = 1\]When \(x = 1:\ y = 3 - 2 = 1\). When \(x = -3:\ y = 3 + 6 = 9\).
The points are \((1,\ 1)\) and \((-3,\ 9)\).
(b) Midpoint \(= \left(\dfrac{1 + (-3)}{2},\ \dfrac{1 + 9}{2}\right) = (-1,\ 5)\).
Gradient of the join \(= \dfrac{9 - 1}{-3 - 1} = \dfrac{8}{-4} = -2\); the perpendicular gradient is \(\dfrac{1}{2}\).
\[y - 5 = \tfrac{1}{2}(x + 1) \;\Rightarrow\; 2y - 10 = x + 1 \;\Rightarrow\; x - 2y + 11 = 0\]Perpendicular bisector: \(x - 2y + 11 = 0\).
Question 3 Rapport
In the diagram, a mass of 12kg hanging from a light inextensible string is pulled aside by a horizontal force, R, such that the string is inclined at 45\(^o\) to the vertical. If the system is in equilibrium, calculate the;
(a) tension in the string;
(b) value of R
The 12 kg mass hangs from a string whose upper end is fixed at \(P\). A horizontal force \(R\) pulls the mass sideways so that the string makes \(45^{\circ}\) with the vertical. Three forces act at the junction (knot): the tension \(T\) along the string, the horizontal pull \(R\), and the weight \(W\) acting vertically downward.
Weight of the mass (taking \(g = 10\,\text{m s}^{-2}\)):
\[ W = mg = 12 \times 10 = 120\,\text{N} \]Since the system is in equilibrium, resolve the tension into vertical and horizontal components. The string is \(45^{\circ}\) from the vertical, so the vertical component of \(T\) is \(T\cos 45^{\circ}\) and the horizontal component is \(T\sin 45^{\circ}\).
(a) Tension in the string
Resolving vertically (the vertical component of the tension supports the weight):
\[ T\cos 45^{\circ} = W \]\[ T \times \frac{\sqrt{2}}{2} = 120 \]\[ T = \frac{120}{\cos 45^{\circ}} = \frac{120}{0.7071} = 120\sqrt{2} \]\[ T \approx 169.7\,\text{N} \]So the tension is about \(170\,\text{N}\).
(b) Value of \(R\)
Resolving horizontally (the horizontal force balances the horizontal component of the tension):
\[ R = T\sin 45^{\circ} = 120\sqrt{2} \times \frac{\sqrt{2}}{2} \]\[ R = 120\sqrt{2} \times 0.7071 = 120\,\text{N} \]Hence \(R = 120\,\text{N}\).
Check: Because the string is at \(45^{\circ}\), the vertical and horizontal components of \(T\) are equal, so \(R\) equals the weight, \(120\,\text{N}\), which agrees with the result above. (Using \(g = 9.8\,\text{m s}^{-2}\) gives \(W = 117.6\,\text{N}\), \(T \approx 166.3\,\text{N}\) and \(R = 117.6\,\text{N}\).)
Détails de la réponse
The 12 kg mass hangs from a string whose upper end is fixed at \(P\). A horizontal force \(R\) pulls the mass sideways so that the string makes \(45^{\circ}\) with the vertical. Three forces act at the junction (knot): the tension \(T\) along the string, the horizontal pull \(R\), and the weight \(W\) acting vertically downward.
Weight of the mass (taking \(g = 10\,\text{m s}^{-2}\)):
\[ W = mg = 12 \times 10 = 120\,\text{N} \]Since the system is in equilibrium, resolve the tension into vertical and horizontal components. The string is \(45^{\circ}\) from the vertical, so the vertical component of \(T\) is \(T\cos 45^{\circ}\) and the horizontal component is \(T\sin 45^{\circ}\).
(a) Tension in the string
Resolving vertically (the vertical component of the tension supports the weight):
\[ T\cos 45^{\circ} = W \]\[ T \times \frac{\sqrt{2}}{2} = 120 \]\[ T = \frac{120}{\cos 45^{\circ}} = \frac{120}{0.7071} = 120\sqrt{2} \]\[ T \approx 169.7\,\text{N} \]So the tension is about \(170\,\text{N}\).
(b) Value of \(R\)
Resolving horizontally (the horizontal force balances the horizontal component of the tension):
\[ R = T\sin 45^{\circ} = 120\sqrt{2} \times \frac{\sqrt{2}}{2} \]\[ R = 120\sqrt{2} \times 0.7071 = 120\,\text{N} \]Hence \(R = 120\,\text{N}\).
Check: Because the string is at \(45^{\circ}\), the vertical and horizontal components of \(T\) are equal, so \(R\) equals the weight, \(120\,\text{N}\), which agrees with the result above. (Using \(g = 9.8\,\text{m s}^{-2}\) gives \(W = 117.6\,\text{N}\), \(T \approx 166.3\,\text{N}\) and \(R = 117.6\,\text{N}\).)
Question 4 Rapport
Simplify \(\frac{ 625(\frac{3x}{4} - 1) + 125^{(x - 1)} }{5^{(3x - 2)}}\)
Write every base as a power of \(5\): \(625=5^{4}\), \(125=5^{3}\).
Numerator:
\[625^{\left(\frac{3x}{4}-1\right)}=5^{4\left(\frac{3x}{4}-1\right)}=5^{3x-4},\qquad 125^{(x-1)}=5^{3(x-1)}=5^{3x-3}\]
So the expression is
\[\frac{5^{3x-4}+5^{3x-3}}{5^{3x-2}}=\frac{5^{3x-4}}{5^{3x-2}}+\frac{5^{3x-3}}{5^{3x-2}}=5^{-2}+5^{-1}\]
\[=\frac{1}{25}+\frac{1}{5}=\frac{1}{25}+\frac{5}{25}=\frac{6}{25}\]
Détails de la réponse
Write every base as a power of \(5\): \(625=5^{4}\), \(125=5^{3}\).
Numerator:
\[625^{\left(\frac{3x}{4}-1\right)}=5^{4\left(\frac{3x}{4}-1\right)}=5^{3x-4},\qquad 125^{(x-1)}=5^{3(x-1)}=5^{3x-3}\]
So the expression is
\[\frac{5^{3x-4}+5^{3x-3}}{5^{3x-2}}=\frac{5^{3x-4}}{5^{3x-2}}+\frac{5^{3x-3}}{5^{3x-2}}=5^{-2}+5^{-1}\]
\[=\frac{1}{25}+\frac{1}{5}=\frac{1}{25}+\frac{5}{25}=\frac{6}{25}\]
Question 5 Rapport
The table shows the distribution of masks obtained by students in an examination.
| Marks | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | 75 - 79 | 80 - 84 | 85 - 89 |
| Frequency | 5 | 15 | 20 | 28 | 12 | 9 | 7 | 4 |
Using an assumed mean of 67, calculate, correct to one decimal place. the
a) Mean
b) Standard deviation of the distribution
Method (assumed mean / coding). Take assumed mean \(A = 67\) and class width \(c = 5\). For each class let \(x\) be the midpoint and \(u = \dfrac{x - A}{c}\).
| Marks | Midpoint \(x\) | \(u\) | \(f\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 50 - 54 | 52 | -3 | 5 | -15 | 45 |
| 55 - 59 | 57 | -2 | 15 | -30 | 60 |
| 60 - 64 | 62 | -1 | 20 | -20 | 20 |
| 65 - 69 | 67 | 0 | 28 | 0 | 0 |
| 70 - 74 | 72 | 1 | 12 | 12 | 12 |
| 75 - 79 | 77 | 2 | 9 | 18 | 36 |
| 80 - 84 | 82 | 3 | 7 | 21 | 63 |
| 85 - 89 | 87 | 4 | 4 | 16 | 64 |
| Total | 100 | 2 | 300 |
(a) Mean.
\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 67 + \frac{2}{100}\times 5 = 67 + 0.1 = \mathbf{67.1} \](b) Standard deviation.
\[ \text{SD} = c\sqrt{\frac{\sum fu^2}{\sum f} - \left(\frac{\sum fu}{\sum f}\right)^2} = 5\sqrt{\frac{300}{100} - \left(\frac{2}{100}\right)^2} \] \[ = 5\sqrt{3 - 0.0004} = 5\sqrt{2.9996} = 5 \times 1.7319 \approx \mathbf{8.7} \]Détails de la réponse
Method (assumed mean / coding). Take assumed mean \(A = 67\) and class width \(c = 5\). For each class let \(x\) be the midpoint and \(u = \dfrac{x - A}{c}\).
| Marks | Midpoint \(x\) | \(u\) | \(f\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 50 - 54 | 52 | -3 | 5 | -15 | 45 |
| 55 - 59 | 57 | -2 | 15 | -30 | 60 |
| 60 - 64 | 62 | -1 | 20 | -20 | 20 |
| 65 - 69 | 67 | 0 | 28 | 0 | 0 |
| 70 - 74 | 72 | 1 | 12 | 12 | 12 |
| 75 - 79 | 77 | 2 | 9 | 18 | 36 |
| 80 - 84 | 82 | 3 | 7 | 21 | 63 |
| 85 - 89 | 87 | 4 | 4 | 16 | 64 |
| Total | 100 | 2 | 300 |
(a) Mean.
\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 67 + \frac{2}{100}\times 5 = 67 + 0.1 = \mathbf{67.1} \](b) Standard deviation.
\[ \text{SD} = c\sqrt{\frac{\sum fu^2}{\sum f} - \left(\frac{\sum fu}{\sum f}\right)^2} = 5\sqrt{\frac{300}{100} - \left(\frac{2}{100}\right)^2} \] \[ = 5\sqrt{3 - 0.0004} = 5\sqrt{2.9996} = 5 \times 1.7319 \approx \mathbf{8.7} \]Question 6 Rapport
(a) Given that m = i - i, n = 2i + 3j and 2m + n - r = 0, find |r|
(b) The distance, S metres of a moving particle at any time tseconds is given by
S = 3t - \(\frac{t^3}{3}\) + 9
Find the;
(i) time
(ii) distance travelled
When the particle is momentarily at rest
(a) Taking \(m=i-j\) and \(n=2i+3j\), with \(2m+n-r=0\) we get \(r=2m+n\).
\[r=2(i-j)+(2i+3j)=(2i-2j)+(2i+3j)=4i+j\]
\[|r|=\sqrt{4^{2}+1^{2}}=\sqrt{17}\approx4.12\]
(b) \(S=3t-\dfrac{t^{3}}{3}+9\). Velocity is \(v=\dfrac{dS}{dt}=3-t^{2}\).
(i) Time when momentarily at rest: \(v=0\):
\[3-t^{2}=0\ \Rightarrow\ t^{2}=3\ \Rightarrow\ t=\sqrt{3}\ \text{s}\ (\approx1.73\ \text{s})\]
(ii) Distance at that instant:
\[S=3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3}+9=3\sqrt{3}-\sqrt{3}+9=2\sqrt{3}+9\approx12.46\ \text{m}\]
Détails de la réponse
(a) Taking \(m=i-j\) and \(n=2i+3j\), with \(2m+n-r=0\) we get \(r=2m+n\).
\[r=2(i-j)+(2i+3j)=(2i-2j)+(2i+3j)=4i+j\]
\[|r|=\sqrt{4^{2}+1^{2}}=\sqrt{17}\approx4.12\]
(b) \(S=3t-\dfrac{t^{3}}{3}+9\). Velocity is \(v=\dfrac{dS}{dt}=3-t^{2}\).
(i) Time when momentarily at rest: \(v=0\):
\[3-t^{2}=0\ \Rightarrow\ t^{2}=3\ \Rightarrow\ t=\sqrt{3}\ \text{s}\ (\approx1.73\ \text{s})\]
(ii) Distance at that instant:
\[S=3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3}+9=3\sqrt{3}-\sqrt{3}+9=2\sqrt{3}+9\approx12.46\ \text{m}\]
Question 7 Rapport
Find the angle between \(\over {OP}\) = (\(^{-3}_{-4}\)) and \(\over{OQ}\) = (\(^8_{-15}\))
\(\vec{OP}=\begin{pmatrix}-3\\-4\end{pmatrix},\ \vec{OQ}=\begin{pmatrix}8\\-15\end{pmatrix}\).
Dot product: \(\vec{OP}\cdot\vec{OQ}=(-3)(8)+(-4)(-15)=-24+60=36\).
Magnitudes: \(|\vec{OP}|=\sqrt{9+16}=5\), \(|\vec{OQ}|=\sqrt{64+225}=\sqrt{289}=17\).
\[\cos\theta=\frac{36}{5\times 17}=\frac{36}{85}=0.4235\Rightarrow \theta=\cos^{-1}(0.4235)\approx 65^\circ.\]The angle between the vectors is about \(65^\circ\).
Détails de la réponse
\(\vec{OP}=\begin{pmatrix}-3\\-4\end{pmatrix},\ \vec{OQ}=\begin{pmatrix}8\\-15\end{pmatrix}\).
Dot product: \(\vec{OP}\cdot\vec{OQ}=(-3)(8)+(-4)(-15)=-24+60=36\).
Magnitudes: \(|\vec{OP}|=\sqrt{9+16}=5\), \(|\vec{OQ}|=\sqrt{64+225}=\sqrt{289}=17\).
\[\cos\theta=\frac{36}{5\times 17}=\frac{36}{85}=0.4235\Rightarrow \theta=\cos^{-1}(0.4235)\approx 65^\circ.\]The angle between the vectors is about \(65^\circ\).
Question 8 Rapport
Given that M : (x, y) \(\to\) (7x, 3x - y) and N : (x, y) \(\to\) (2x - y; 5x + 3y)
(a) write down matrices M and N of the linear transformation
(b) find the image of P(2, -3) under the linear transformation N followed by M;
(c) find the coordinates of the point Q whose image is Q(2, 4) under the linear transformation N
(a) Reading the images off column by column:
\[M=\begin{pmatrix}7&0\\3&-1\end{pmatrix},\qquad N=\begin{pmatrix}2&-1\\5&3\end{pmatrix}\]
(b) Image of \(P(2,-3)\) under N then M.
Apply \(N\): \((2x-y,\ 5x+3y)=(2(2)-(-3),\ 5(2)+3(-3))=(7,\ 1)\).
Apply \(M\) to \((7,1)\): \((7x,\ 3x-y)=(7(7),\ 3(7)-1)=(49,\ 20)\).
Image is \((49,\ 20)\).
(c) Point whose image under N is \((2,4)\). Solve \(N(x,y)=(2,4)\):
\[2x-y=2,\qquad 5x+3y=4\]
From the first, \(y=2x-2\). Substituting: \(5x+3(2x-2)=4\Rightarrow11x-6=4\Rightarrow x=\dfrac{10}{11}\).
\[y=2\left(\tfrac{10}{11}\right)-2=-\tfrac{2}{11}\]
The point is \(\left(\dfrac{10}{11},\ -\dfrac{2}{11}\right)\).
Détails de la réponse
(a) Reading the images off column by column:
\[M=\begin{pmatrix}7&0\\3&-1\end{pmatrix},\qquad N=\begin{pmatrix}2&-1\\5&3\end{pmatrix}\]
(b) Image of \(P(2,-3)\) under N then M.
Apply \(N\): \((2x-y,\ 5x+3y)=(2(2)-(-3),\ 5(2)+3(-3))=(7,\ 1)\).
Apply \(M\) to \((7,1)\): \((7x,\ 3x-y)=(7(7),\ 3(7)-1)=(49,\ 20)\).
Image is \((49,\ 20)\).
(c) Point whose image under N is \((2,4)\). Solve \(N(x,y)=(2,4)\):
\[2x-y=2,\qquad 5x+3y=4\]
From the first, \(y=2x-2\). Substituting: \(5x+3(2x-2)=4\Rightarrow11x-6=4\Rightarrow x=\dfrac{10}{11}\).
\[y=2\left(\tfrac{10}{11}\right)-2=-\tfrac{2}{11}\]
The point is \(\left(\dfrac{10}{11},\ -\dfrac{2}{11}\right)\).
Question 9 Rapport
Three soldies, X, Y and Z have probabilities \(\frac{1}{3}, \frac{1}{5}\) and \(\frac{1}{4}\) respectively of hitting a target. If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target
Hit probabilities: \(P(X)=\tfrac13,\ P(Y)=\tfrac15,\ P(Z)=\tfrac14\); miss probabilities: \(\tfrac23,\tfrac45,\tfrac34\). "Only one hits" means exactly one succeeds while the other two miss (events independent):
\[P(\text{only }X)=\frac13\cdot\frac45\cdot\frac34=\frac{12}{60}=\frac15,\] \[P(\text{only }Y)=\frac23\cdot\frac15\cdot\frac34=\frac{6}{60}=\frac{1}{10},\] \[P(\text{only }Z)=\frac23\cdot\frac45\cdot\frac14=\frac{8}{60}=\frac{2}{15}.\]Adding (LCD 30):
\[P(\text{exactly one})=\frac{6}{30}+\frac{3}{30}+\frac{4}{30}=\frac{13}{30}\approx 0.43.\]Détails de la réponse
Hit probabilities: \(P(X)=\tfrac13,\ P(Y)=\tfrac15,\ P(Z)=\tfrac14\); miss probabilities: \(\tfrac23,\tfrac45,\tfrac34\). "Only one hits" means exactly one succeeds while the other two miss (events independent):
\[P(\text{only }X)=\frac13\cdot\frac45\cdot\frac34=\frac{12}{60}=\frac15,\] \[P(\text{only }Y)=\frac23\cdot\frac15\cdot\frac34=\frac{6}{60}=\frac{1}{10},\] \[P(\text{only }Z)=\frac23\cdot\frac45\cdot\frac14=\frac{8}{60}=\frac{2}{15}.\]Adding (LCD 30):
\[P(\text{exactly one})=\frac{6}{30}+\frac{3}{30}+\frac{4}{30}=\frac{13}{30}\approx 0.43.\]Question 10 Rapport
A uniform beam, WX, of length 90 cm and weight 50N is suspended on a pivot, 35 cm from W. It is kept in equilibrum by a means of forces T and 20N applied at Y and Z respectively. |WY| = 10cm and |XZ| = 10cm. Find the value of T
Set distances from end \(W\). The beam is \(90\,\text{cm}\) long.
- \(Y\): \(|WY| = 10\,\text{cm}\), so \(Y\) is at \(10\,\text{cm}\).
- Pivot: \(35\,\text{cm}\) from \(W\).
- Weight \(50\,\text{N}\) acts at the centre (uniform beam): \(45\,\text{cm}\).
- \(Z\): \(|XZ| = 10\,\text{cm}\), so \(Z\) is at \(90 - 10 = 80\,\text{cm}\).
Taking moments about the pivot (which removes the pivot reaction). Distances from the pivot:
- \(T\) at \(Y\): \(35 - 10 = 25\,\text{cm}\) on the \(W\) side.
- Weight \(50\,\text{N}\): \(45 - 35 = 10\,\text{cm}\) on the \(X\) side.
- \(20\,\text{N}\) at \(Z\): \(80 - 35 = 45\,\text{cm}\) on the \(X\) side.
For equilibrium, the anticlockwise moment of \(T\) balances the clockwise moments of the weight and the \(20\,\text{N}\) force:
\[T \times 25 = 50 \times 10 + 20 \times 45\] \[25\,T = 500 + 900 = 1400\] \[T = \frac{1400}{25} = 56\,\text{N}\]\(T = 56\,\text{N}\).
Détails de la réponse
Set distances from end \(W\). The beam is \(90\,\text{cm}\) long.
- \(Y\): \(|WY| = 10\,\text{cm}\), so \(Y\) is at \(10\,\text{cm}\).
- Pivot: \(35\,\text{cm}\) from \(W\).
- Weight \(50\,\text{N}\) acts at the centre (uniform beam): \(45\,\text{cm}\).
- \(Z\): \(|XZ| = 10\,\text{cm}\), so \(Z\) is at \(90 - 10 = 80\,\text{cm}\).
Taking moments about the pivot (which removes the pivot reaction). Distances from the pivot:
- \(T\) at \(Y\): \(35 - 10 = 25\,\text{cm}\) on the \(W\) side.
- Weight \(50\,\text{N}\): \(45 - 35 = 10\,\text{cm}\) on the \(X\) side.
- \(20\,\text{N}\) at \(Z\): \(80 - 35 = 45\,\text{cm}\) on the \(X\) side.
For equilibrium, the anticlockwise moment of \(T\) balances the clockwise moments of the weight and the \(20\,\text{N}\) force:
\[T \times 25 = 50 \times 10 + 20 \times 45\] \[25\,T = 500 + 900 = 1400\] \[T = \frac{1400}{25} = 56\,\text{N}\]\(T = 56\,\text{N}\).
Question 11 Rapport
Using determinants, solve the following equations simultaneously.
5x — 6y + 4z = 15
7x + 4y — 3z = 19
2x + y + 6z = 46
System: \(5x-6y+4z=15,\ 7x+4y-3z=19,\ 2x+y+6z=46\). Use Cramer's rule.
\[D=\begin{vmatrix}5&-6&4\\7&4&-3\\2&1&6\end{vmatrix}=5(24+3)+6(42+6)+4(7-8)=135+288-4=419.\] \[D_x=\begin{vmatrix}15&-6&4\\19&4&-3\\46&1&6\end{vmatrix}=15(27)+6(252)+4(-165)=405+1512-660=1257.\] \[D_y=\begin{vmatrix}5&15&4\\7&19&-3\\2&46&6\end{vmatrix}=5(252)-15(48)+4(284)=1260-720+1136=1676.\] \[D_z=\begin{vmatrix}5&-6&15\\7&4&19\\2&1&46\end{vmatrix}=5(165)+6(284)+15(-1)=825+1704-15=2514.\] \[x=\frac{D_x}{D}=\frac{1257}{419}=3,\quad y=\frac{D_y}{D}=\frac{1676}{419}=4,\quad z=\frac{D_z}{D}=\frac{2514}{419}=6.\]So \(x=3,\ y=4,\ z=6\). Check in equation 2: \(7(3)+4(4)-3(6)=21+16-18=19\).
Détails de la réponse
System: \(5x-6y+4z=15,\ 7x+4y-3z=19,\ 2x+y+6z=46\). Use Cramer's rule.
\[D=\begin{vmatrix}5&-6&4\\7&4&-3\\2&1&6\end{vmatrix}=5(24+3)+6(42+6)+4(7-8)=135+288-4=419.\] \[D_x=\begin{vmatrix}15&-6&4\\19&4&-3\\46&1&6\end{vmatrix}=15(27)+6(252)+4(-165)=405+1512-660=1257.\] \[D_y=\begin{vmatrix}5&15&4\\7&19&-3\\2&46&6\end{vmatrix}=5(252)-15(48)+4(284)=1260-720+1136=1676.\] \[D_z=\begin{vmatrix}5&-6&15\\7&4&19\\2&1&46\end{vmatrix}=5(165)+6(284)+15(-1)=825+1704-15=2514.\] \[x=\frac{D_x}{D}=\frac{1257}{419}=3,\quad y=\frac{D_y}{D}=\frac{1676}{419}=4,\quad z=\frac{D_z}{D}=\frac{2514}{419}=6.\]So \(x=3,\ y=4,\ z=6\). Check in equation 2: \(7(3)+4(4)-3(6)=21+16-18=19\).
Question 12 Rapport
Forces(5N, 030\(^o\)), (PN, 060\(^o\)), (QN, 150\(^o\)), (3N, 180\(^o\)) and (5N, 270\(^o\)) act on a body . If the system is in quilibrium, find, correct to one decimal place, the values of P and Q
Measure each direction anticlockwise from the positive x-axis and resolve. For equilibrium the sum of the components in each direction is zero.
x-components:
\[5\cos30^{o} + P\cos60^{o} + Q\cos150^{o} + 3\cos180^{o} + 5\cos270^{o} = 0\] \[\tfrac{5\sqrt3}{2} + \tfrac{P}{2} - \tfrac{\sqrt3}{2}Q - 3 + 0 = 0\]y-components:
\[5\sin30^{o} + P\sin60^{o} + Q\sin150^{o} + 3\sin180^{o} + 5\sin270^{o} = 0\] \[\tfrac{5}{2} + \tfrac{\sqrt3}{2}P + \tfrac{1}{2}Q + 0 - 5 = 0\]The y-equation gives \(\sqrt3\,P + Q = 5\), so \(Q = 5 - \sqrt3\,P\).
The x-equation gives \(P - \sqrt3\,Q = 6 - 5\sqrt3\). Substituting:
\[P - \sqrt3(5 - \sqrt3 P) = 6 - 5\sqrt3 \;\Rightarrow\; 4P - 5\sqrt3 = 6 - 5\sqrt3 \;\Rightarrow\; 4P = 6\]So \(P = 1.5\), and \(Q = 5 - 1.5\sqrt3 = 5 - 2.598 = 2.402\).
\(P \approx 1.5\,\text{N}, \quad Q \approx 2.4\,\text{N}\) (to 1 d.p.).
Détails de la réponse
Measure each direction anticlockwise from the positive x-axis and resolve. For equilibrium the sum of the components in each direction is zero.
x-components:
\[5\cos30^{o} + P\cos60^{o} + Q\cos150^{o} + 3\cos180^{o} + 5\cos270^{o} = 0\] \[\tfrac{5\sqrt3}{2} + \tfrac{P}{2} - \tfrac{\sqrt3}{2}Q - 3 + 0 = 0\]y-components:
\[5\sin30^{o} + P\sin60^{o} + Q\sin150^{o} + 3\sin180^{o} + 5\sin270^{o} = 0\] \[\tfrac{5}{2} + \tfrac{\sqrt3}{2}P + \tfrac{1}{2}Q + 0 - 5 = 0\]The y-equation gives \(\sqrt3\,P + Q = 5\), so \(Q = 5 - \sqrt3\,P\).
The x-equation gives \(P - \sqrt3\,Q = 6 - 5\sqrt3\). Substituting:
\[P - \sqrt3(5 - \sqrt3 P) = 6 - 5\sqrt3 \;\Rightarrow\; 4P - 5\sqrt3 = 6 - 5\sqrt3 \;\Rightarrow\; 4P = 6\]So \(P = 1.5\), and \(Q = 5 - 1.5\sqrt3 = 5 - 2.598 = 2.402\).
\(P \approx 1.5\,\text{N}, \quad Q \approx 2.4\,\text{N}\) (to 1 d.p.).
Question 13 Rapport
If \(\alpha\) and \(\beta\) are the roots of the equation 3x\(^2\) + 4x - 5 = 0, find the value of (\(\alpha - \beta\)), leaving the answer in surd form.
For \(3x^2+4x-5=0\) with roots \(\alpha,\beta\): sum \(\alpha+\beta=-\dfrac{4}{3}\), product \(\alpha\beta=-\dfrac{5}{3}\).
Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\):
\[(\alpha-\beta)^2=\left(-\frac43\right)^2-4\left(-\frac53\right)=\frac{16}{9}+\frac{20}{3}=\frac{16}{9}+\frac{60}{9}=\frac{76}{9}.\] \[\alpha-\beta=\sqrt{\frac{76}{9}}=\frac{\sqrt{76}}{3}=\frac{2\sqrt{19}}{3}.\]So \(\alpha-\beta=\dfrac{2\sqrt{19}}{3}\) (taking the positive surd).
Détails de la réponse
For \(3x^2+4x-5=0\) with roots \(\alpha,\beta\): sum \(\alpha+\beta=-\dfrac{4}{3}\), product \(\alpha\beta=-\dfrac{5}{3}\).
Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\):
\[(\alpha-\beta)^2=\left(-\frac43\right)^2-4\left(-\frac53\right)=\frac{16}{9}+\frac{20}{3}=\frac{16}{9}+\frac{60}{9}=\frac{76}{9}.\] \[\alpha-\beta=\sqrt{\frac{76}{9}}=\frac{\sqrt{76}}{3}=\frac{2\sqrt{19}}{3}.\]So \(\alpha-\beta=\dfrac{2\sqrt{19}}{3}\) (taking the positive surd).
Question 14 Rapport
(a) Find the coordinates of the point which divides the line joining (7, -5) and (-2, 7) externally in the ration 3 : 2.
(b) Without using calculators or mathematical tables, evaluate \(\frac{2}{1 + \sqrt{2}}\) - \(\frac{2}{2 + \sqrt{2}}\), leaving the answer in the form p + q\(\sqrt{n}\), where p, q and n are integers.
(a) For external division of \((x_1,y_1)=(7,-5)\) and \((x_2,y_2)=(-2,7)\) in the ratio \(m:n = 3:2\):
\[x = \frac{m x_2 - n x_1}{m - n} = \frac{3(-2) - 2(7)}{3 - 2} = \frac{-6 - 14}{1} = -20\] \[y = \frac{m y_2 - n y_1}{m - n} = \frac{3(7) - 2(-5)}{3 - 2} = \frac{21 + 10}{1} = 31\]The point is \((-20,\ 31)\).
(b) Rationalise each term.
\[\frac{2}{1+\sqrt2}\times\frac{\sqrt2-1}{\sqrt2-1} = \frac{2(\sqrt2-1)}{(\sqrt2)^2-1^2}=\frac{2(\sqrt2-1)}{1}=2\sqrt2-2\] \[\frac{2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2} = \frac{2(2-\sqrt2)}{4-2}=\frac{2(2-\sqrt2)}{2}=2-\sqrt2\]Therefore
\[\frac{2}{1+\sqrt2}-\frac{2}{2+\sqrt2} = (2\sqrt2-2)-(2-\sqrt2) = 3\sqrt2 - 4\]In the form \(p+q\sqrt n\): \(p=-4,\ q=3,\ n=2\), i.e. \(-4+3\sqrt2\).
Détails de la réponse
(a) For external division of \((x_1,y_1)=(7,-5)\) and \((x_2,y_2)=(-2,7)\) in the ratio \(m:n = 3:2\):
\[x = \frac{m x_2 - n x_1}{m - n} = \frac{3(-2) - 2(7)}{3 - 2} = \frac{-6 - 14}{1} = -20\] \[y = \frac{m y_2 - n y_1}{m - n} = \frac{3(7) - 2(-5)}{3 - 2} = \frac{21 + 10}{1} = 31\]The point is \((-20,\ 31)\).
(b) Rationalise each term.
\[\frac{2}{1+\sqrt2}\times\frac{\sqrt2-1}{\sqrt2-1} = \frac{2(\sqrt2-1)}{(\sqrt2)^2-1^2}=\frac{2(\sqrt2-1)}{1}=2\sqrt2-2\] \[\frac{2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2} = \frac{2(2-\sqrt2)}{4-2}=\frac{2(2-\sqrt2)}{2}=2-\sqrt2\]Therefore
\[\frac{2}{1+\sqrt2}-\frac{2}{2+\sqrt2} = (2\sqrt2-2)-(2-\sqrt2) = 3\sqrt2 - 4\]In the form \(p+q\sqrt n\): \(p=-4,\ q=3,\ n=2\), i.e. \(-4+3\sqrt2\).
Question 15 Rapport
Two fair dice are thrown together two times. Find the probability of obtaining a sum of seven in the first throw and a sum of four in the second throw.
With two fair dice there are \(36\) equally likely outcomes.
Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) gives \(6\) outcomes, so \(P(7) = \dfrac{6}{36} = \dfrac{1}{6}\).
Sum of 4: \((1,3),(2,2),(3,1)\) gives \(3\) outcomes, so \(P(4) = \dfrac{3}{36} = \dfrac{1}{12}\).
The two throws are independent, so
\[P(\text{7 then 4}) = \frac{1}{6}\times\frac{1}{12} = \frac{1}{72}\]Probability \(= \dfrac{1}{72}\).
Détails de la réponse
With two fair dice there are \(36\) equally likely outcomes.
Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) gives \(6\) outcomes, so \(P(7) = \dfrac{6}{36} = \dfrac{1}{6}\).
Sum of 4: \((1,3),(2,2),(3,1)\) gives \(3\) outcomes, so \(P(4) = \dfrac{3}{36} = \dfrac{1}{12}\).
The two throws are independent, so
\[P(\text{7 then 4}) = \frac{1}{6}\times\frac{1}{12} = \frac{1}{72}\]Probability \(= \dfrac{1}{72}\).
Question 16 Rapport
In an examination, 60% of the candidates passed. If 10 candidates are selected at random, find the probability that;
(1) at least two of the, failed
(2) exactly half of them passed
(3) at most two of them failed
This is a binomial situation. Let \(X\) be the number who fail. A candidate fails with probability \(p=0.4\) and passes with probability \(0.6\), with \(n=10\).
\[P(X=r)=\binom{10}{r}(0.4)^{r}(0.6)^{10-r}\]
(1) At least two failed: \(P(X\ge 2)\)
\[P(X\ge2)=1-P(0)-P(1)\]
\(P(0)=(0.6)^{10}=0.00605\) and \(P(1)=\binom{10}{1}(0.4)(0.6)^{9}=10(0.4)(0.010078)=0.04031\).
\[P(X\ge2)=1-0.00605-0.04031=0.954\]
(2) Exactly half passed means \(5\) passed and \(5\) failed, i.e. \(X=5\).
\[P(X=5)=\binom{10}{5}(0.4)^{5}(0.6)^{5}=252(0.01024)(0.07776)=0.201\]
(3) At most two failed: \(P(X\le 2)=P(0)+P(1)+P(2)\)
\(P(2)=\binom{10}{2}(0.4)^{2}(0.6)^{8}=45(0.16)(0.016796)=0.12093\).
\[P(X\le2)=0.00605+0.04031+0.12093=0.167\]
Détails de la réponse
This is a binomial situation. Let \(X\) be the number who fail. A candidate fails with probability \(p=0.4\) and passes with probability \(0.6\), with \(n=10\).
\[P(X=r)=\binom{10}{r}(0.4)^{r}(0.6)^{10-r}\]
(1) At least two failed: \(P(X\ge 2)\)
\[P(X\ge2)=1-P(0)-P(1)\]
\(P(0)=(0.6)^{10}=0.00605\) and \(P(1)=\binom{10}{1}(0.4)(0.6)^{9}=10(0.4)(0.010078)=0.04031\).
\[P(X\ge2)=1-0.00605-0.04031=0.954\]
(2) Exactly half passed means \(5\) passed and \(5\) failed, i.e. \(X=5\).
\[P(X=5)=\binom{10}{5}(0.4)^{5}(0.6)^{5}=252(0.01024)(0.07776)=0.201\]
(3) At most two failed: \(P(X\le 2)=P(0)+P(1)+P(2)\)
\(P(2)=\binom{10}{2}(0.4)^{2}(0.6)^{8}=45(0.16)(0.016796)=0.12093\).
\[P(X\le2)=0.00605+0.04031+0.12093=0.167\]
Question 17 Rapport
How many terms of the series -3 -1 + 1 +..... add up to 165?
The series \(-3,\ -1,\ 1,\ \dots\) is arithmetic with first term \(a=-3\) and common difference \(d=2\).
\[S_n = \frac{n}{2}\big[2a + (n-1)d\big] = \frac{n}{2}\big[-6 + 2(n-1)\big] = \frac{n}{2}(2n-8) = n(n-4)\]Set \(S_n = 165\):
\[n(n-4) = 165 \;\Rightarrow\; n^2 - 4n - 165 = 0\] \[n = \frac{4 \pm \sqrt{16 + 660}}{2} = \frac{4 \pm \sqrt{676}}{2} = \frac{4 \pm 26}{2}\]Taking the positive value, \(n = \dfrac{30}{2} = 15\).
15 terms are required.
Détails de la réponse
The series \(-3,\ -1,\ 1,\ \dots\) is arithmetic with first term \(a=-3\) and common difference \(d=2\).
\[S_n = \frac{n}{2}\big[2a + (n-1)d\big] = \frac{n}{2}\big[-6 + 2(n-1)\big] = \frac{n}{2}(2n-8) = n(n-4)\]Set \(S_n = 165\):
\[n(n-4) = 165 \;\Rightarrow\; n^2 - 4n - 165 = 0\] \[n = \frac{4 \pm \sqrt{16 + 660}}{2} = \frac{4 \pm \sqrt{676}}{2} = \frac{4 \pm 26}{2}\]Taking the positive value, \(n = \dfrac{30}{2} = 15\).
15 terms are required.
Question 18 Rapport
(a) If sin p = \(\frac{1}{2}\) and cos q = \(\frac{1}{3}\), evaluate sin(p - q), where 0\(^o\) \(\geq\) p \(\geq\) 90\(^o\) and 90\(^o\) \(\geq\) q \(\geq\) 180\(^o\)
b) Using trapezum rule with seven ordinates, evaluate \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx
(a) \(\sin p=\tfrac12\) with \(p\) acute gives \(p=30^\circ\), so \(\cos p=\tfrac{\sqrt3}{2}\). Since \(90^\circ (b) Trapezium rule, seven ordinates \(\Rightarrow\) six strips, \(h=\dfrac{4-1}{6}=0.5\), with \(f(x)=\dfrac{2}{\sqrt{x+3}}\): (The exact value \(\big[4\sqrt{x+3}\big]_1^4=4(\sqrt7-2)=2.58\) confirms this.)
\[\int_1^4 f\,dx\approx\frac{h}{2}\Big[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)\Big]=\frac{0.5}{2}\big[1.7559+2(4.2910)\big].\]
\[=0.25(1.7559+8.5820)=0.25(10.3379)\approx 2.58.\]
x 1 1.5 2 2.5 3 3.5 4 f(x) 1.0000 0.9428 0.8944 0.8528 0.8165 0.7845 0.7559
Détails de la réponse
(a) \(\sin p=\tfrac12\) with \(p\) acute gives \(p=30^\circ\), so \(\cos p=\tfrac{\sqrt3}{2}\). Since \(90^\circ (b) Trapezium rule, seven ordinates \(\Rightarrow\) six strips, \(h=\dfrac{4-1}{6}=0.5\), with \(f(x)=\dfrac{2}{\sqrt{x+3}}\): (The exact value \(\big[4\sqrt{x+3}\big]_1^4=4(\sqrt7-2)=2.58\) confirms this.)
\[\int_1^4 f\,dx\approx\frac{h}{2}\Big[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)\Big]=\frac{0.5}{2}\big[1.7559+2(4.2910)\big].\]
\[=0.25(1.7559+8.5820)=0.25(10.3379)\approx 2.58.\]
x 1 1.5 2 2.5 3 3.5 4 f(x) 1.0000 0.9428 0.8944 0.8528 0.8165 0.7845 0.7559
Question 19 Rapport
Given that (\(_r^n\)) = \(^nC_r\), simplify (\(^{2x + 1}_{3}\)) - (\(^{2x - 1}_3\)) - 2(\(^x_2\))
Write each term using \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).
\[\binom{2x+1}{3}=\frac{(2x+1)(2x)(2x-1)}{6}=\frac{8x^3-2x}{6},\] \[\binom{2x-1}{3}=\frac{(2x-1)(2x-2)(2x-3)}{6}=\frac{8x^3-24x^2+22x-6}{6}.\]Their difference:
\[\binom{2x+1}{3}-\binom{2x-1}{3}=\frac{(8x^3-2x)-(8x^3-24x^2+22x-6)}{6}=\frac{24x^2-24x+6}{6}=4x^2-4x+1.\]And \(2\binom{x}{2}=2\cdot\dfrac{x(x-1)}{2}=x^2-x\). Therefore
\[\binom{2x+1}{3}-\binom{2x-1}{3}-2\binom{x}{2}=(4x^2-4x+1)-(x^2-x)=3x^2-3x+1.\]Détails de la réponse
Write each term using \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).
\[\binom{2x+1}{3}=\frac{(2x+1)(2x)(2x-1)}{6}=\frac{8x^3-2x}{6},\] \[\binom{2x-1}{3}=\frac{(2x-1)(2x-2)(2x-3)}{6}=\frac{8x^3-24x^2+22x-6}{6}.\]Their difference:
\[\binom{2x+1}{3}-\binom{2x-1}{3}=\frac{(8x^3-2x)-(8x^3-24x^2+22x-6)}{6}=\frac{24x^2-24x+6}{6}=4x^2-4x+1.\]And \(2\binom{x}{2}=2\cdot\dfrac{x(x-1)}{2}=x^2-x\). Therefore
\[\binom{2x+1}{3}-\binom{2x-1}{3}-2\binom{x}{2}=(4x^2-4x+1)-(x^2-x)=3x^2-3x+1.\]Question 20 Rapport
A body, moving at 20ms\(^{-1}\) accelerates uniformly at 2\(\frac{1}{2}ms^{-2}\) for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest seconds at tseconds after with uniform retardation. If the ratio of the acceleration to retardation is 3 : 4
(a( sketch the velocity - times graph of the journey
(b) find t
(c) find the total distance of the journey
(a) Velocity-time graph
The velocity increases uniformly from \(20\text{ m s}^{-1}\) to \(30\text{ m s}^{-1}\) in 4 s, remains constant for 8 s, and then decreases uniformly to zero.
(b) Calculation of \(t\)
Acceleration \(=2\frac{1}{2}=\frac{5}{2}\text{ m s}^{-2}\).
Let the retardation be \(r\text{ m s}^{-2}\). Since
\[\frac{5}{2}:r=3:4,\]
\[r=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\text{ m s}^{-2}.\]
Velocity after the first 4 s is
\[v=20+\left(\frac{5}{2}\times4\right)=30\text{ m s}^{-1}.\]
During retardation,
\[0=30-\frac{10}{3}t.\]
\[t=\frac{30}{10/3}=9\text{ s}.\]
Thus, the body comes to rest at \(4+8+9=21\) s from the start of the journey.
(c) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned} \text{Distance}&=\frac{1}{2}(20+30)(4)+(30\times8)+\frac{1}{2}(30)(9)\\ &=100+240+135\\ &=475\text{ m}. \end{aligned}\]
Total distance travelled = \(475\text{ m}\).
Détails de la réponse
(a) Velocity-time graph
The velocity increases uniformly from \(20\text{ m s}^{-1}\) to \(30\text{ m s}^{-1}\) in 4 s, remains constant for 8 s, and then decreases uniformly to zero.
(b) Calculation of \(t\)
Acceleration \(=2\frac{1}{2}=\frac{5}{2}\text{ m s}^{-2}\).
Let the retardation be \(r\text{ m s}^{-2}\). Since
\[\frac{5}{2}:r=3:4,\]
\[r=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\text{ m s}^{-2}.\]
Velocity after the first 4 s is
\[v=20+\left(\frac{5}{2}\times4\right)=30\text{ m s}^{-1}.\]
During retardation,
\[0=30-\frac{10}{3}t.\]
\[t=\frac{30}{10/3}=9\text{ s}.\]
Thus, the body comes to rest at \(4+8+9=21\) s from the start of the journey.
(c) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned} \text{Distance}&=\frac{1}{2}(20+30)(4)+(30\times8)+\frac{1}{2}(30)(9)\\ &=100+240+135\\ &=475\text{ m}. \end{aligned}\]
Total distance travelled = \(475\text{ m}\).
Question 21 Rapport
Three forces N, 14N and 16N acting on a particle keep it in equilibrium. Find the angle between the forces 10N and 16N.
The three forces are \(10\,\text{N},\ 14\,\text{N}\) and \(16\,\text{N}\) in equilibrium, so the resultant of the \(10\,\text{N}\) and \(16\,\text{N}\) forces must be equal in magnitude to \(14\,\text{N}\) (and opposite to it).
Let \(\theta\) be the angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces. Using the resultant (parallelogram) formula:
\[R^2 = 10^2 + 16^2 + 2(10)(16)\cos\theta = 14^2\] \[100 + 256 + 320\cos\theta = 196\] \[320\cos\theta = 196 - 356 = -160 \;\Rightarrow\; \cos\theta = -\tfrac{1}{2}\] \[\theta = 120^{o}\]The angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces is \(120^{o}\).
Détails de la réponse
The three forces are \(10\,\text{N},\ 14\,\text{N}\) and \(16\,\text{N}\) in equilibrium, so the resultant of the \(10\,\text{N}\) and \(16\,\text{N}\) forces must be equal in magnitude to \(14\,\text{N}\) (and opposite to it).
Let \(\theta\) be the angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces. Using the resultant (parallelogram) formula:
\[R^2 = 10^2 + 16^2 + 2(10)(16)\cos\theta = 14^2\] \[100 + 256 + 320\cos\theta = 196\] \[320\cos\theta = 196 - 356 = -160 \;\Rightarrow\; \cos\theta = -\tfrac{1}{2}\] \[\theta = 120^{o}\]The angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces is \(120^{o}\).
Question 22 Rapport
In a research to determine the relationship between performance of students in an entrance examination and subsequent school performance, the results of ten randomly selected students wre obtained as follows;
| Students | A | B | C | D | E | F | G | H | I |
| Performance in Entrance Examination | 11 | 12 | 8 | 13 | 6 | 15 | 10 | 14 | 17 |
| School Performance | 5 | 10 | 9 | 7 | 4 | 8 | 6 | 14 | 11 |
1, Calculate the spearman's rank correlation coefficient
2. What would be the researcher's from the result in a?
The data provided cover 9 students (A to I), so \(n = 9\). Rank each set with 1 for the highest score.
| Student | Entrance | Rank \(R_x\) | School | Rank \(R_y\) | \(d = R_x - R_y\) | \(d^2\) |
|---|---|---|---|---|---|---|
| A | 11 | 6 | 5 | 8 | -2 | 4 |
| B | 12 | 5 | 10 | 3 | 2 | 4 |
| C | 8 | 8 | 9 | 4 | 4 | 16 |
| D | 13 | 4 | 7 | 6 | -2 | 4 |
| E | 6 | 9 | 4 | 9 | 0 | 0 |
| F | 15 | 2 | 8 | 5 | -3 | 9 |
| G | 10 | 7 | 6 | 7 | 0 | 0 |
| H | 14 | 3 | 14 | 1 | 2 | 4 |
| I | 17 | 1 | 11 | 2 | -1 | 1 |
| Total \(\sum d^2\) | 42 | |||||
1. Spearman's rank correlation coefficient.
\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 42}{9(81 - 1)} = 1 - \frac{252}{720} = 1 - 0.35 = \mathbf{0.65} \]2. Conclusion. \(r_s = 0.65\) is a fairly strong positive correlation. Students who perform well in the entrance examination tend also to perform well subsequently in school, so the researcher would conclude that entrance-examination performance is a reasonably good indicator of later school performance.
Détails de la réponse
The data provided cover 9 students (A to I), so \(n = 9\). Rank each set with 1 for the highest score.
| Student | Entrance | Rank \(R_x\) | School | Rank \(R_y\) | \(d = R_x - R_y\) | \(d^2\) |
|---|---|---|---|---|---|---|
| A | 11 | 6 | 5 | 8 | -2 | 4 |
| B | 12 | 5 | 10 | 3 | 2 | 4 |
| C | 8 | 8 | 9 | 4 | 4 | 16 |
| D | 13 | 4 | 7 | 6 | -2 | 4 |
| E | 6 | 9 | 4 | 9 | 0 | 0 |
| F | 15 | 2 | 8 | 5 | -3 | 9 |
| G | 10 | 7 | 6 | 7 | 0 | 0 |
| H | 14 | 3 | 14 | 1 | 2 | 4 |
| I | 17 | 1 | 11 | 2 | -1 | 1 |
| Total \(\sum d^2\) | 42 | |||||
1. Spearman's rank correlation coefficient.
\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 42}{9(81 - 1)} = 1 - \frac{252}{720} = 1 - 0.35 = \mathbf{0.65} \]2. Conclusion. \(r_s = 0.65\) is a fairly strong positive correlation. Students who perform well in the entrance examination tend also to perform well subsequently in school, so the researcher would conclude that entrance-examination performance is a reasonably good indicator of later school performance.
Question 23 Rapport
(a) Find the range of value of p for which 4x\(^2\) - px + 1 = 0
(b)(i) Expand (1 + 3x)\(^6\) in ascending powers of x
(ii) Using the expression in 10
(ii) find, correct to four significant figures, the value of (1.03)\(^6\)
(a) For \(4x^2 - px + 1 = 0\) to have real roots, the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-p)^2 - 4(4)(1) \ge 0 \;\Rightarrow\; p^2 - 16 \ge 0\] \[(p-4)(p+4) \ge 0 \;\Rightarrow\; p \le -4 \ \text{ or } \ p \ge 4\](b)(i) Using \((1+3x)^6 = \displaystyle\sum_{r=0}^{6}\binom{6}{r}(3x)^r\):
\[(1+3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6\](ii) Put \(3x = 0.03\), i.e. \(x = 0.01\), so \((1+3x)^6 = (1.03)^6\):
\[(1.03)^6 \approx 1 + 18(0.01) + 135(0.01)^2 + 540(0.01)^3 + \dots\] \[= 1 + 0.18 + 0.0135 + 0.00054 + 0.0000122 + \dots = 1.19405\]To four significant figures, \((1.03)^6 \approx \mathbf{1.194}\).
Détails de la réponse
(a) For \(4x^2 - px + 1 = 0\) to have real roots, the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-p)^2 - 4(4)(1) \ge 0 \;\Rightarrow\; p^2 - 16 \ge 0\] \[(p-4)(p+4) \ge 0 \;\Rightarrow\; p \le -4 \ \text{ or } \ p \ge 4\](b)(i) Using \((1+3x)^6 = \displaystyle\sum_{r=0}^{6}\binom{6}{r}(3x)^r\):
\[(1+3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6\](ii) Put \(3x = 0.03\), i.e. \(x = 0.01\), so \((1+3x)^6 = (1.03)^6\):
\[(1.03)^6 \approx 1 + 18(0.01) + 135(0.01)^2 + 540(0.01)^3 + \dots\] \[= 1 + 0.18 + 0.0135 + 0.00054 + 0.0000122 + \dots = 1.19405\]To four significant figures, \((1.03)^6 \approx \mathbf{1.194}\).
Question 24 Rapport
Differentiate from first principles, with respect to x, (3x\(^2\) + 2x - 1)
Let \(f(x)=3x^2+2x-1\). By first principles \(f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).
\[f(x+h)=3(x+h)^2+2(x+h)-1=3x^2+6xh+3h^2+2x+2h-1.\] \[f(x+h)-f(x)=6xh+3h^2+2h.\] \[\frac{f(x+h)-f(x)}{h}=6x+3h+2.\] \[f'(x)=\lim_{h\to 0}(6x+3h+2)=6x+2.\]So \(\dfrac{d}{dx}(3x^2+2x-1)=6x+2\).
Détails de la réponse
Let \(f(x)=3x^2+2x-1\). By first principles \(f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).
\[f(x+h)=3(x+h)^2+2(x+h)-1=3x^2+6xh+3h^2+2x+2h-1.\] \[f(x+h)-f(x)=6xh+3h^2+2h.\] \[\frac{f(x+h)-f(x)}{h}=6x+3h+2.\] \[f'(x)=\lim_{h\to 0}(6x+3h+2)=6x+2.\]So \(\dfrac{d}{dx}(3x^2+2x-1)=6x+2\).
Question 25 Rapport
(a) P(-1, 4), Q(2, 3), R(x, y) and S(-2, 3) are the verticles of a parallelogram. Find the value of x and y.
(b) A particle starts from rest and moves in a straight line. It attains a velocity of 20ms\(^{-1}\) after travelling a distance of 8 metres. Calculate;
(ii) Iis acceleration
(ii) the time taken to travel 40 metres
(a) For parallelogram \(PQRS\) the diagonals bisect each other, so midpoint of \(PR\) = midpoint of \(QS\).
\[\left(\frac{-1+x}{2},\frac{4+y}{2}\right)=\left(\frac{2+(-2)}{2},\frac{3+3}{2}\right)=(0,3).\] \[\frac{-1+x}{2}=0\Rightarrow x=1,\qquad \frac{4+y}{2}=3\Rightarrow y=2.\]So \(R(1,2)\), i.e. \(x=1,\ y=2\). (Check: \(\vec{PQ}=(3,-1)=\vec{SR}\).)
(b) From rest, \(u=0\), reaches \(v=20\ \text{m/s}\) after \(s=8\ \text{m}\).
(i) Acceleration: \(v^2=u^2+2as\Rightarrow 20^2=2a(8)\Rightarrow a=\dfrac{400}{16}=25\ \text{m/s}^2\).
(ii) Time to travel 40 m: \(s=ut+\tfrac12 at^2\Rightarrow 40=\tfrac12(25)t^2=12.5t^2\).
\[t^2=3.2\Rightarrow t=\sqrt{3.2}\approx 1.79\ \text{s}.\]Détails de la réponse
(a) For parallelogram \(PQRS\) the diagonals bisect each other, so midpoint of \(PR\) = midpoint of \(QS\).
\[\left(\frac{-1+x}{2},\frac{4+y}{2}\right)=\left(\frac{2+(-2)}{2},\frac{3+3}{2}\right)=(0,3).\] \[\frac{-1+x}{2}=0\Rightarrow x=1,\qquad \frac{4+y}{2}=3\Rightarrow y=2.\]So \(R(1,2)\), i.e. \(x=1,\ y=2\). (Check: \(\vec{PQ}=(3,-1)=\vec{SR}\).)
(b) From rest, \(u=0\), reaches \(v=20\ \text{m/s}\) after \(s=8\ \text{m}\).
(i) Acceleration: \(v^2=u^2+2as\Rightarrow 20^2=2a(8)\Rightarrow a=\dfrac{400}{16}=25\ \text{m/s}^2\).
(ii) Time to travel 40 m: \(s=ut+\tfrac12 at^2\Rightarrow 40=\tfrac12(25)t^2=12.5t^2\).
\[t^2=3.2\Rightarrow t=\sqrt{3.2}\approx 1.79\ \text{s}.\]Question 26 Rapport
The table shows the age distribution in years of a group of people
| Age(in years) | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 | 26 - 30 |
| Number of people | 18 | 12 | 25 | 15 | 20 | 10 |
Using an assume mean of 13 years, find the mean age of the people.
Method. Assumed mean \(A = 13\), class width \(c = 5\), \(u = \dfrac{x - 13}{5}\) where \(x\) is the class midpoint.
| Age (years) | Midpoint \(x\) | \(u\) | \(f\) | \(fu\) |
|---|---|---|---|---|
| 1 - 5 | 3 | -2 | 18 | -36 |
| 6 - 10 | 8 | -1 | 12 | -12 |
| 11 - 15 | 13 | 0 | 25 | 0 |
| 16 - 20 | 18 | 1 | 15 | 15 |
| 21 - 25 | 23 | 2 | 20 | 40 |
| 26 - 30 | 28 | 3 | 10 | 30 |
| Total | 100 | 37 |
Mean age.
\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 13 + \frac{37}{100}\times 5 = 13 + 1.85 = \mathbf{14.85 \text{ years}} \]Détails de la réponse
Method. Assumed mean \(A = 13\), class width \(c = 5\), \(u = \dfrac{x - 13}{5}\) where \(x\) is the class midpoint.
| Age (years) | Midpoint \(x\) | \(u\) | \(f\) | \(fu\) |
|---|---|---|---|---|
| 1 - 5 | 3 | -2 | 18 | -36 |
| 6 - 10 | 8 | -1 | 12 | -12 |
| 11 - 15 | 13 | 0 | 25 | 0 |
| 16 - 20 | 18 | 1 | 15 | 15 |
| 21 - 25 | 23 | 2 | 20 | 40 |
| 26 - 30 | 28 | 3 | 10 | 30 |
| Total | 100 | 37 |
Mean age.
\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 13 + \frac{37}{100}\times 5 = 13 + 1.85 = \mathbf{14.85 \text{ years}} \]Question 27 Rapport
If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1 - tan\(^2\)x
Given \(\sin x = \dfrac{P-Q}{P+Q}\) with \(0^{o}\le x\le 90^{o}\).
\[\cos^2 x = 1 - \sin^2 x = 1 - \frac{(P-Q)^2}{(P+Q)^2} = \frac{(P+Q)^2-(P-Q)^2}{(P+Q)^2} = \frac{4PQ}{(P+Q)^2}\] \[\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{(P-Q)^2/(P+Q)^2}{4PQ/(P+Q)^2} = \frac{(P-Q)^2}{4PQ}\]Hence
\[1 - \tan^2 x = 1 - \frac{(P-Q)^2}{4PQ} = \frac{4PQ - (P^2 - 2PQ + Q^2)}{4PQ} = \frac{6PQ - P^2 - Q^2}{4PQ}\]\(\displaystyle 1 - \tan^2 x = \frac{6PQ - P^2 - Q^2}{4PQ}\).
Détails de la réponse
Given \(\sin x = \dfrac{P-Q}{P+Q}\) with \(0^{o}\le x\le 90^{o}\).
\[\cos^2 x = 1 - \sin^2 x = 1 - \frac{(P-Q)^2}{(P+Q)^2} = \frac{(P+Q)^2-(P-Q)^2}{(P+Q)^2} = \frac{4PQ}{(P+Q)^2}\] \[\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{(P-Q)^2/(P+Q)^2}{4PQ/(P+Q)^2} = \frac{(P-Q)^2}{4PQ}\]Hence
\[1 - \tan^2 x = 1 - \frac{(P-Q)^2}{4PQ} = \frac{4PQ - (P^2 - 2PQ + Q^2)}{4PQ} = \frac{6PQ - P^2 - Q^2}{4PQ}\]\(\displaystyle 1 - \tan^2 x = \frac{6PQ - P^2 - Q^2}{4PQ}\).
Question 28 Rapport
Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0
First find the y-intercept of \(3x-2y+6=0\): put \(x=0\Rightarrow -2y+6=0\Rightarrow y=3\). The point is \((0,3)\).
The radius is the distance from the centre \((2,3)\) to \((0,3)\):
\[r=\sqrt{(2-0)^2+(3-3)^2}=\sqrt{4}=2.\]Equation of the circle:
\[(x-2)^2+(y-3)^2=2^2\Rightarrow (x-2)^2+(y-3)^2=4,\]or \(x^2+y^2-4x-6y+9=0\).
Détails de la réponse
First find the y-intercept of \(3x-2y+6=0\): put \(x=0\Rightarrow -2y+6=0\Rightarrow y=3\). The point is \((0,3)\).
The radius is the distance from the centre \((2,3)\) to \((0,3)\):
\[r=\sqrt{(2-0)^2+(3-3)^2}=\sqrt{4}=2.\]Equation of the circle:
\[(x-2)^2+(y-3)^2=2^2\Rightarrow (x-2)^2+(y-3)^2=4,\]or \(x^2+y^2-4x-6y+9=0\).
Question 29 Rapport
The distribution of the masses of a group of persons is shown in the following table
| Mass/kg | 10.5 - 14.4 | 14.5 - 24.4 | 24.5 - 44.4 | 44.5 - 47.4 | 47.5 - 49.4 |
| Number of Persons | 2 | 6 | 18 | 2 | 1 |
Draw a histogram for the distribution
Histogram of Mass Distribution
Since the class intervals are unequal, the vertical axis is frequency density:
\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}\]
| Mass (kg) | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10.5–14.4 | 10.45–14.45 | 4 | 2 | 0.50 |
| 14.5–24.4 | 14.45–24.45 | 10 | 6 | 0.60 |
| 24.5–44.4 | 24.45–44.45 | 20 | 18 | 0.90 |
| 44.5–47.4 | 44.45–47.45 | 3 | 2 | 0.67 |
| 47.5–49.4 | 47.45–49.45 | 2 | 1 | 0.50 |
The required histogram is:
For example, the area of the bar from \(24.45\) kg to \(44.45\) kg is \(20\times0.90=18\), equal to its frequency.
Détails de la réponse
Histogram of Mass Distribution
Since the class intervals are unequal, the vertical axis is frequency density:
\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}\]
| Mass (kg) | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10.5–14.4 | 10.45–14.45 | 4 | 2 | 0.50 |
| 14.5–24.4 | 14.45–24.45 | 10 | 6 | 0.60 |
| 24.5–44.4 | 24.45–44.45 | 20 | 18 | 0.90 |
| 44.5–47.4 | 44.45–47.45 | 3 | 2 | 0.67 |
| 47.5–49.4 | 47.45–49.45 | 2 | 1 | 0.50 |
The required histogram is:
For example, the area of the bar from \(24.45\) kg to \(44.45\) kg is \(20\times0.90=18\), equal to its frequency.
Question 30 Rapport
(a) In a bakery, 30% of loaves of bread produced are of bad quality. If twelve loaves are selected at random from the bakery, calculate, correct to four decimal places. the probabshty of getting
(i) exactly 6 bad ones:
(ii) at least 4 bad ones;
(ii) no bad one.
(b) A group consists of 8 boys and 5 girls. A committee of 7 members is chosen from the group. Find the probability that the committee is made up of 4 boys and 3 girls.
(a) \(P(\text{bad})=0.3,\ P(\text{good})=0.7,\ n=12\), \(P(X=r)=\binom{12}{r}(0.3)^r(0.7)^{12-r}\) (\(X\)=number bad).
(i) exactly 6 bad: \(\binom{12}{6}(0.3)^6(0.7)^6=924(0.000729)(0.117649)=0.0793\).
(ii) at least 4 bad \(=1-\big[P(0)+P(1)+P(2)+P(3)\big]\):
\[P(0)=0.0138,\ P(1)=0.0712,\ P(2)=0.1678,\ P(3)=0.2397;\ \text{sum}=0.4925.\] \[P(X\ge 4)=1-0.4925=0.5075.\](iii) no bad one: \(P(0)=(0.7)^{12}=0.0138\).
(b) 8 boys, 5 girls; committee of 7 with 4 boys and 3 girls:
\[P=\frac{\binom{8}{4}\binom{5}{3}}{\binom{13}{7}}=\frac{70\times 10}{1716}=\frac{700}{1716}=\frac{175}{429}\approx 0.4079.\]Détails de la réponse
(a) \(P(\text{bad})=0.3,\ P(\text{good})=0.7,\ n=12\), \(P(X=r)=\binom{12}{r}(0.3)^r(0.7)^{12-r}\) (\(X\)=number bad).
(i) exactly 6 bad: \(\binom{12}{6}(0.3)^6(0.7)^6=924(0.000729)(0.117649)=0.0793\).
(ii) at least 4 bad \(=1-\big[P(0)+P(1)+P(2)+P(3)\big]\):
\[P(0)=0.0138,\ P(1)=0.0712,\ P(2)=0.1678,\ P(3)=0.2397;\ \text{sum}=0.4925.\] \[P(X\ge 4)=1-0.4925=0.5075.\](iii) no bad one: \(P(0)=(0.7)^{12}=0.0138\).
(b) 8 boys, 5 girls; committee of 7 with 4 boys and 3 girls:
\[P=\frac{\binom{8}{4}\binom{5}{3}}{\binom{13}{7}}=\frac{70\times 10}{1716}=\frac{700}{1716}=\frac{175}{429}\approx 0.4079.\]
