WAEC SSCE - Further Mathematics - 2012 (Essay)
Question 1 Report
Write down the first three terms of the binomial expansion \((1 + ax)^{n}\) in ascending powers of x. If the coefficients of x and x\(^{2}\) are 2 and \(\frac{3}{2}\) respectively, find the values of a and n.
The binomial expansion of \((1+ax)^n\) in ascending powers of \(x\) is:
\[(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \dots\]
So the first three terms are \(1,\; nax,\; \dfrac{n(n-1)}{2}a^2x^2\).
Forming the equations. The coefficient of \(x\) is \(2\) and the coefficient of \(x^2\) is \(\tfrac{3}{2}\):
\[na = 2 \quad\text{(1)}, \qquad \frac{n(n-1)}{2}a^2 = \frac{3}{2} \quad\text{(2)}.\]
From (1), \(a = \dfrac{2}{n}\). Substituting into (2):
\[\frac{n(n-1)}{2}\cdot\frac{4}{n^2} = \frac{3}{2} \;\Rightarrow\; \frac{2(n-1)}{n} = \frac{3}{2}.\]
Cross-multiplying: \(4(n-1) = 3n \Rightarrow 4n-4 = 3n \Rightarrow n = 4\).
Then \(a = \dfrac{2}{n} = \dfrac{2}{4} = \dfrac{1}{2}\).
Answer: \(a = \tfrac{1}{2},\; n = 4\). (Check: with these values the terms are \(1 + 2x + \tfrac{3}{2}x^2\), as required.)
Answer Details
The binomial expansion of \((1+ax)^n\) in ascending powers of \(x\) is:
\[(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \dots\]
So the first three terms are \(1,\; nax,\; \dfrac{n(n-1)}{2}a^2x^2\).
Forming the equations. The coefficient of \(x\) is \(2\) and the coefficient of \(x^2\) is \(\tfrac{3}{2}\):
\[na = 2 \quad\text{(1)}, \qquad \frac{n(n-1)}{2}a^2 = \frac{3}{2} \quad\text{(2)}.\]
From (1), \(a = \dfrac{2}{n}\). Substituting into (2):
\[\frac{n(n-1)}{2}\cdot\frac{4}{n^2} = \frac{3}{2} \;\Rightarrow\; \frac{2(n-1)}{n} = \frac{3}{2}.\]
Cross-multiplying: \(4(n-1) = 3n \Rightarrow 4n-4 = 3n \Rightarrow n = 4\).
Then \(a = \dfrac{2}{n} = \dfrac{2}{4} = \dfrac{1}{2}\).
Answer: \(a = \tfrac{1}{2},\; n = 4\). (Check: with these values the terms are \(1 + 2x + \tfrac{3}{2}x^2\), as required.)
Question 2 Report
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining :
(i) exactly three sixes ; (ii) at most three sixes.
(b) Eight percent of screws produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that :
(i) exactly two will be defective ; (ii) not more than two will be defective.
Both parts use the binomial distribution \(P(X=r) = \binom{n}{r}p^r(1-p)^{n-r}\).
(a) Die thrown \(n = 6\) times, \(p = \tfrac16\) for a six, \(q = \tfrac56\).
(i) Exactly three sixes:
\[\binom{6}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^3 = 20\cdot\frac{1}{216}\cdot\frac{125}{216} = \frac{2500}{46656} \approx 0.054.\]
(ii) At most three sixes \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.334898,\ P(1)=0.401878,\ P(2)=0.200939,\ P(3)=0.053584.\]
Sum \(= 0.991\) (to three decimal places).
(b) Screws: \(n = 10\), \(p = 0.08\) defective, \(q = 0.92\).
(i) Exactly two defective:
\[\binom{10}{2}(0.08)^2(0.92)^8 = 45(0.0064)(0.513219) \approx 0.148.\]
(ii) Not more than two \(= P(0)+P(1)+P(2)\):
\[P(0)=(0.92)^{10}=0.434390,\ P(1)=10(0.08)(0.92)^9=0.377730,\ P(2)=0.147807.\]
Sum \(= 0.960\) (to three decimal places).
Answer Details
Both parts use the binomial distribution \(P(X=r) = \binom{n}{r}p^r(1-p)^{n-r}\).
(a) Die thrown \(n = 6\) times, \(p = \tfrac16\) for a six, \(q = \tfrac56\).
(i) Exactly three sixes:
\[\binom{6}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^3 = 20\cdot\frac{1}{216}\cdot\frac{125}{216} = \frac{2500}{46656} \approx 0.054.\]
(ii) At most three sixes \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.334898,\ P(1)=0.401878,\ P(2)=0.200939,\ P(3)=0.053584.\]
Sum \(= 0.991\) (to three decimal places).
(b) Screws: \(n = 10\), \(p = 0.08\) defective, \(q = 0.92\).
(i) Exactly two defective:
\[\binom{10}{2}(0.08)^2(0.92)^8 = 45(0.0064)(0.513219) \approx 0.148.\]
(ii) Not more than two \(= P(0)+P(1)+P(2)\):
\[P(0)=(0.92)^{10}=0.434390,\ P(1)=10(0.08)(0.92)^9=0.377730,\ P(2)=0.147807.\]
Sum \(= 0.960\) (to three decimal places).
Question 3 Report
(a) Evaluate : \(\int_{1} ^{4} \frac{x(3x - 2)}{2\sqrt{x}} \mathrm {d} x\)
(b) The equation of a circle is given by \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\). Find the :
(i) coordinates of the centre ; (ii) radius of the circle .
(a) Simplify the integrand first. With \(\sqrt{x} = x^{1/2}\),
\[\frac{x(3x-2)}{2\sqrt{x}} = \frac{3x^2 - 2x}{2x^{1/2}} = \frac{3}{2}x^{3/2} - x^{1/2}.\]
Integrate:
\[\int\left(\frac{3}{2}x^{3/2} - x^{1/2}\right)dx = \frac{3}{2}\cdot\frac{x^{5/2}}{5/2} - \frac{x^{3/2}}{3/2} = \frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2}.\]
Evaluate from \(1\) to \(4\) (using \(4^{5/2}=32,\ 4^{3/2}=8\)):
\[\left[\frac{3}{5}(32) - \frac{2}{3}(8)\right] - \left[\frac{3}{5} - \frac{2}{3}\right] = \left(\frac{96}{5} - \frac{16}{3}\right) - \left(-\frac{1}{15}\right) = \frac{208}{15} + \frac{1}{15} = \frac{209}{15}.\]
So the integral is \(\dfrac{209}{15} = 13\tfrac{14}{15} \approx 13.93\).
(b) Divide the circle equation by 2 to get unit leading coefficients:
\[x^2 + y^2 - 4x + \tfrac{5}{2}y - 5 = 0.\]
Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\): \(2g=-4\Rightarrow g=-2\), \(2f=\tfrac52\Rightarrow f=\tfrac54\), \(c=-5\).
(i) Centre \((-g, -f) = \left(2,\, -\tfrac{5}{4}\right)\).
(ii) Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + \tfrac{25}{16} + 5} = \sqrt{\tfrac{169}{16}} = \tfrac{13}{4} = 3.25\).
Answer Details
(a) Simplify the integrand first. With \(\sqrt{x} = x^{1/2}\),
\[\frac{x(3x-2)}{2\sqrt{x}} = \frac{3x^2 - 2x}{2x^{1/2}} = \frac{3}{2}x^{3/2} - x^{1/2}.\]
Integrate:
\[\int\left(\frac{3}{2}x^{3/2} - x^{1/2}\right)dx = \frac{3}{2}\cdot\frac{x^{5/2}}{5/2} - \frac{x^{3/2}}{3/2} = \frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2}.\]
Evaluate from \(1\) to \(4\) (using \(4^{5/2}=32,\ 4^{3/2}=8\)):
\[\left[\frac{3}{5}(32) - \frac{2}{3}(8)\right] - \left[\frac{3}{5} - \frac{2}{3}\right] = \left(\frac{96}{5} - \frac{16}{3}\right) - \left(-\frac{1}{15}\right) = \frac{208}{15} + \frac{1}{15} = \frac{209}{15}.\]
So the integral is \(\dfrac{209}{15} = 13\tfrac{14}{15} \approx 13.93\).
(b) Divide the circle equation by 2 to get unit leading coefficients:
\[x^2 + y^2 - 4x + \tfrac{5}{2}y - 5 = 0.\]
Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\): \(2g=-4\Rightarrow g=-2\), \(2f=\tfrac52\Rightarrow f=\tfrac54\), \(c=-5\).
(i) Centre \((-g, -f) = \left(2,\, -\tfrac{5}{4}\right)\).
(ii) Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + \tfrac{25}{16} + 5} = \sqrt{\tfrac{169}{16}} = \tfrac{13}{4} = 3.25\).
Question 4 Report
(a)(i) Find the sum of the series \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\).
(ii) Given that r = 8% and A = GH 40.00, find the sum of the 6th to 10th terms of the series in (i).
(b) Find the equation of the tangent to the curve \(y = \frac{1}{x}\) at the point on the curve when x = 2.
(a)(i) The series \(A(1+r) + A(1+r)^2 + \dots + A(1+r)^n\) is a geometric progression with first term \(a = A(1+r)\), common ratio \(R = (1+r)\) and \(n\) terms. Its sum is
\[S_n = \frac{a(R^n - 1)}{R - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{(1+r) - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{r}.\]
(ii) With \(r = 8\% = 0.08\) and \(A = 40\), the \(k\)-th term is \(T_k = 40(1.08)^k\). The sum of the 6th to 10th terms is
\[\sum_{k=6}^{10} 40(1.08)^k = 40\big[(1.08)^6 + (1.08)^7 + (1.08)^8 + (1.08)^9 + (1.08)^{10}\big].\]
Computing the powers: \(1.586874,\ 1.713824,\ 1.850930,\ 1.999005,\ 2.158925\); their sum is \(9.309558\). Hence
\[S = 40 \times 9.309558 \approx \text{GH}\phi\,372.38.\]
The sum of the 6th to 10th terms is about GH\(\phi\)372.38.
(b) For \(y = \dfrac{1}{x} = x^{-1}\), \(\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}\).
At \(x = 2\): gradient \(= -\dfrac{1}{4}\), and the point is \(\left(2, \tfrac12\right)\). The tangent is
\[y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2) \;\Rightarrow\; y = 1 - \tfrac{1}{4}x, \quad\text{or}\quad x + 4y = 4.\]
Answer Details
(a)(i) The series \(A(1+r) + A(1+r)^2 + \dots + A(1+r)^n\) is a geometric progression with first term \(a = A(1+r)\), common ratio \(R = (1+r)\) and \(n\) terms. Its sum is
\[S_n = \frac{a(R^n - 1)}{R - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{(1+r) - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{r}.\]
(ii) With \(r = 8\% = 0.08\) and \(A = 40\), the \(k\)-th term is \(T_k = 40(1.08)^k\). The sum of the 6th to 10th terms is
\[\sum_{k=6}^{10} 40(1.08)^k = 40\big[(1.08)^6 + (1.08)^7 + (1.08)^8 + (1.08)^9 + (1.08)^{10}\big].\]
Computing the powers: \(1.586874,\ 1.713824,\ 1.850930,\ 1.999005,\ 2.158925\); their sum is \(9.309558\). Hence
\[S = 40 \times 9.309558 \approx \text{GH}\phi\,372.38.\]
The sum of the 6th to 10th terms is about GH\(\phi\)372.38.
(b) For \(y = \dfrac{1}{x} = x^{-1}\), \(\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}\).
At \(x = 2\): gradient \(= -\dfrac{1}{4}\), and the point is \(\left(2, \tfrac12\right)\). The tangent is
\[y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2) \;\Rightarrow\; y = 1 - \tfrac{1}{4}x, \quad\text{or}\quad x + 4y = 4.\]
Question 5 Report
Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\)
= \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! (n - 3)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)}{4!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)}{4!}\)
= \(\frac{(n - 1)(n - 2) [n(n + 1) - (n - 3)(n - 4)]}{4!}\)
= \(\frac{(n - 1)(n - 2) [n^{2} + n - n^{2} + 7n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)[8n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)(2n - 3)}{6}\)
Question 6 Report
The marks scored by 35 students in a test are given in the table below.
| Marks | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 |
| Frequency | 2 | 7 | 12 | 8 | 5 | 1 |
Draw a histogram for the distribution.
All six classes have the same width. Each class spans five marks (\(1\text{-}5,\ 6\text{-}10,\ \dots,\ 26\text{-}30\)), so the class widths are equal and an ordinary frequency histogram may be drawn: the bar heights are simply the frequencies, and there is no need to adjust for frequency density. First convert the class limits to continuous class boundaries by subtracting \(0.5\) from each lower limit and adding \(0.5\) to each upper limit.
| Marks | Class boundaries | Width | Frequency |
|---|---|---|---|
| 1 - 5 | 0.5 - 5.5 | 5 | 2 |
| 6 - 10 | 5.5 - 10.5 | 5 | 7 |
| 11 - 15 | 10.5 - 15.5 | 5 | 12 |
| 16 - 20 | 15.5 - 20.5 | 5 | 8 |
| 21 - 25 | 20.5 - 25.5 | 5 | 5 |
| 26 - 30 | 25.5 - 30.5 | 5 | 1 |
The bars are drawn touching (no gaps), because the class boundaries are continuous. The frequencies total \(2+7+12+8+5+1 = 35\), matching the 35 students.
The tallest bar, over the class \(11\text{-}15\), is the modal class. Reading the peak of the histogram gives an estimate of the mode within that interval.
Examination note: do not merge or re-band the classes. The intervals are already equal in width, so each bar height equals its frequency; changing the boundaries to unequal widths would distort the distribution and lose marks.
Answer Details
All six classes have the same width. Each class spans five marks (\(1\text{-}5,\ 6\text{-}10,\ \dots,\ 26\text{-}30\)), so the class widths are equal and an ordinary frequency histogram may be drawn: the bar heights are simply the frequencies, and there is no need to adjust for frequency density. First convert the class limits to continuous class boundaries by subtracting \(0.5\) from each lower limit and adding \(0.5\) to each upper limit.
| Marks | Class boundaries | Width | Frequency |
|---|---|---|---|
| 1 - 5 | 0.5 - 5.5 | 5 | 2 |
| 6 - 10 | 5.5 - 10.5 | 5 | 7 |
| 11 - 15 | 10.5 - 15.5 | 5 | 12 |
| 16 - 20 | 15.5 - 20.5 | 5 | 8 |
| 21 - 25 | 20.5 - 25.5 | 5 | 5 |
| 26 - 30 | 25.5 - 30.5 | 5 | 1 |
The bars are drawn touching (no gaps), because the class boundaries are continuous. The frequencies total \(2+7+12+8+5+1 = 35\), matching the 35 students.
The tallest bar, over the class \(11\text{-}15\), is the modal class. Reading the peak of the histogram gives an estimate of the mode within that interval.
Examination note: do not merge or re-band the classes. The intervals are already equal in width, so each bar height equals its frequency; changing the boundaries to unequal widths would distort the distribution and lose marks.
Question 7 Report
The gradient function of \(y = ax^{2} + bx + c\) is \(8x + 4\). If the function has a minimum value of 1, find the values of a, b and c.
The gradient function is the derivative. For \(y = ax^2 + bx + c\),
\[\frac{dy}{dx} = 2ax + b.\]
We are told this equals \(8x + 4\), so comparing coefficients:
\[2a = 8 \Rightarrow a = 4, \qquad b = 4.\]
Using the minimum value. At the minimum the gradient is zero:
\[8x + 4 = 0 \Rightarrow x = -\tfrac{1}{2}.\]
The minimum value of \(y\) there is \(1\). Substitute \(a=4,\ b=4,\ x=-\tfrac12\):
\[y = 4\left(-\tfrac12\right)^2 + 4\left(-\tfrac12\right) + c = 4\cdot\tfrac14 - 2 + c = 1 - 2 + c = c - 1.\]
Set \(c - 1 = 1 \Rightarrow c = 2\).
Answer: \(a = 4,\; b = 4,\; c = 2\), giving \(y = 4x^2 + 4x + 2\).
Answer Details
The gradient function is the derivative. For \(y = ax^2 + bx + c\),
\[\frac{dy}{dx} = 2ax + b.\]
We are told this equals \(8x + 4\), so comparing coefficients:
\[2a = 8 \Rightarrow a = 4, \qquad b = 4.\]
Using the minimum value. At the minimum the gradient is zero:
\[8x + 4 = 0 \Rightarrow x = -\tfrac{1}{2}.\]
The minimum value of \(y\) there is \(1\). Substitute \(a=4,\ b=4,\ x=-\tfrac12\):
\[y = 4\left(-\tfrac12\right)^2 + 4\left(-\tfrac12\right) + c = 4\cdot\tfrac14 - 2 + c = 1 - 2 + c = c - 1.\]
Set \(c - 1 = 1 \Rightarrow c = 2\).
Answer: \(a = 4,\; b = 4,\; c = 2\), giving \(y = 4x^2 + 4x + 2\).
Question 8 Report
(a) Given that \(p = (4i - 3j)\) and \(q = (-i + 5j)\), find r such that \(|r| = 15\) and is in the direction \((2p + 3q)\).
(b) 
Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the above diagram. Find the : (i) magnitude ; (ii) direction of the resultant force.
(a) With \(p=4i-3j\) and \(q=-i+5j\):
\[2p+3q=2(4i-3j)+3(-i+5j)=(8i-6j)+(-3i+15j)=5i+9j.\]
Its magnitude is
\[|2p+3q|=\sqrt{5^2+9^2}=\sqrt{25+81}=\sqrt{106}.\]
The unit vector in the direction of \(2p+3q\) is \(\dfrac{5i+9j}{\sqrt{106}}\). Since \(|r|=15\) and \(r\) is in this direction,
\[r=15\cdot\frac{5i+9j}{\sqrt{106}}=\frac{75}{\sqrt{106}}i+\frac{135}{\sqrt{106}}j\approx 7.28i+13.11j.\]
(b) From the diagram, taking angles anticlockwise from the positive \(x\)-axis (horizontal): the \(6\ \text{N}\) is vertical (\(90^\circ\)); the \(8\ \text{N}\) is \(30^\circ\) to the left of the \(6\ \text{N}\), so at \(120^\circ\); the \(4\ \text{N}\) is \(60^\circ\) to the right of the \(6\ \text{N}\), so at \(30^\circ\).
Resolve horizontally:
\[\sum F_x=8\cos120^\circ+6\cos90^\circ+4\cos30^\circ=-4+0+3.464=-0.536\ \text{N}.\]
Resolve vertically:
\[\sum F_y=8\sin120^\circ+6\sin90^\circ+4\sin30^\circ=6.928+6+2=14.928\ \text{N}.\]
(i) Magnitude:
\[R=\sqrt{(-0.536)^2+14.928^2}=\sqrt{0.287+222.85}=\sqrt{223.14}\approx \mathbf{14.94\ \text{N}}.\]
(ii) Direction. \(\sum F_x\) is negative and \(\sum F_y\) positive, so the resultant lies in the second quadrant (just left of the upward vertical). The acute angle to the vertical is
\[\tan^{-1}\!\left(\frac{|\sum F_x|}{\sum F_y}\right)=\tan^{-1}\!\left(\frac{0.536}{14.928}\right)\approx 2.1^\circ.\]
Measured anticlockwise from the positive \(x\)-axis, the direction is \(180^\circ-\tan^{-1}(14.928/0.536)\approx \mathbf{92.1^\circ}\); that is, the resultant of about \(14.94\ \text{N}\) acts almost vertically, inclined \(2.1^\circ\) to the left of the \(6\ \text{N}\) force.
Answer Details
(a) With \(p=4i-3j\) and \(q=-i+5j\):
\[2p+3q=2(4i-3j)+3(-i+5j)=(8i-6j)+(-3i+15j)=5i+9j.\]
Its magnitude is
\[|2p+3q|=\sqrt{5^2+9^2}=\sqrt{25+81}=\sqrt{106}.\]
The unit vector in the direction of \(2p+3q\) is \(\dfrac{5i+9j}{\sqrt{106}}\). Since \(|r|=15\) and \(r\) is in this direction,
\[r=15\cdot\frac{5i+9j}{\sqrt{106}}=\frac{75}{\sqrt{106}}i+\frac{135}{\sqrt{106}}j\approx 7.28i+13.11j.\]
(b) From the diagram, taking angles anticlockwise from the positive \(x\)-axis (horizontal): the \(6\ \text{N}\) is vertical (\(90^\circ\)); the \(8\ \text{N}\) is \(30^\circ\) to the left of the \(6\ \text{N}\), so at \(120^\circ\); the \(4\ \text{N}\) is \(60^\circ\) to the right of the \(6\ \text{N}\), so at \(30^\circ\).
Resolve horizontally:
\[\sum F_x=8\cos120^\circ+6\cos90^\circ+4\cos30^\circ=-4+0+3.464=-0.536\ \text{N}.\]
Resolve vertically:
\[\sum F_y=8\sin120^\circ+6\sin90^\circ+4\sin30^\circ=6.928+6+2=14.928\ \text{N}.\]
(i) Magnitude:
\[R=\sqrt{(-0.536)^2+14.928^2}=\sqrt{0.287+222.85}=\sqrt{223.14}\approx \mathbf{14.94\ \text{N}}.\]
(ii) Direction. \(\sum F_x\) is negative and \(\sum F_y\) positive, so the resultant lies in the second quadrant (just left of the upward vertical). The acute angle to the vertical is
\[\tan^{-1}\!\left(\frac{|\sum F_x|}{\sum F_y}\right)=\tan^{-1}\!\left(\frac{0.536}{14.928}\right)\approx 2.1^\circ.\]
Measured anticlockwise from the positive \(x\)-axis, the direction is \(180^\circ-\tan^{-1}(14.928/0.536)\approx \mathbf{92.1^\circ}\); that is, the resultant of about \(14.94\ \text{N}\) acts almost vertically, inclined \(2.1^\circ\) to the left of the \(6\ \text{N}\) force.
Question 9 Report
Express \(3x^{2} - 6x + 10\) in the form \(a(x - b)^{2} + c\), where a, b and c are integers. Hence state the minimum value of \(3x^{2} - 6x + 10\) and the value of x for which it occurs.
Express \(3x^{2}-6x+10\) as \(a(x-b)^{2}+c\).
Factor 3 from the x-terms and complete the square:
\[3x^{2}-6x+10=3(x^{2}-2x)+10=3\big[(x-1)^{2}-1\big]+10\]\[=3(x-1)^{2}-3+10=3(x-1)^{2}+7\]So \(a=3,\ b=1,\ c=7\).
Minimum value. Since \(3(x-1)^{2}\ge 0\) and is zero when \(x=1\), the least value of the expression is
\[3(0)+7=7\ \text{at } x=1\]The minimum value is \(7\), occurring at \(x=1\).
Answer Details
Express \(3x^{2}-6x+10\) as \(a(x-b)^{2}+c\).
Factor 3 from the x-terms and complete the square:
\[3x^{2}-6x+10=3(x^{2}-2x)+10=3\big[(x-1)^{2}-1\big]+10\]\[=3(x-1)^{2}-3+10=3(x-1)^{2}+7\]So \(a=3,\ b=1,\ c=7\).
Minimum value. Since \(3(x-1)^{2}\ge 0\) and is zero when \(x=1\), the least value of the expression is
\[3(0)+7=7\ \text{at } x=1\]The minimum value is \(7\), occurring at \(x=1\).
Question 10 Report
(a) Write down the matrix A of the linear transformation \(A(x, y) \to (2x -y, -5x + 3y)\).
(b) If \(B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\), find :
(i) \(A^{2} - B^{2}\) ; (ii) matrix \(C = B^{2} A\) ; (iii) the point \(M(x, y)\) whose image under the linear transformation \(C\) is \(M' (10, 18)\).
(c) What is the relationship between matrix A and matrix C?
(a) The transformation \(A(x,y)\to(2x - y,\, -5x + 3y)\) has matrix
\[A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}.\]
(b) First compute the squares.
\[A^2 = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pmatrix},\qquad B^2 = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}^2 = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}.\]
(i) \(\displaystyle A^2 - B^2 = \begin{pmatrix} 9-14 & -5-5 \\ -25-25 & 14-9 \end{pmatrix} = \begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}.\)
(ii) \(\displaystyle C = B^2 A = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}.\)
(iii) If \(C\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}10\\18\end{pmatrix}\), then \(3x + y = 10\) and \(5x + 2y = 18\).
From the first, \(y = 10 - 3x\); substituting: \(5x + 2(10-3x) = 18 \Rightarrow -x + 20 = 18 \Rightarrow x = 2\), then \(y = 4\). So \(M(2, 4)\).
(c) Note \(\det A = (2)(3)-(-1)(-5) = 1\), and \(AC = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I\). Hence \(C\) is the inverse of \(A\), that is \(C = A^{-1}\).
Answer Details
(a) The transformation \(A(x,y)\to(2x - y,\, -5x + 3y)\) has matrix
\[A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}.\]
(b) First compute the squares.
\[A^2 = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pmatrix},\qquad B^2 = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}^2 = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}.\]
(i) \(\displaystyle A^2 - B^2 = \begin{pmatrix} 9-14 & -5-5 \\ -25-25 & 14-9 \end{pmatrix} = \begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}.\)
(ii) \(\displaystyle C = B^2 A = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}.\)
(iii) If \(C\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}10\\18\end{pmatrix}\), then \(3x + y = 10\) and \(5x + 2y = 18\).
From the first, \(y = 10 - 3x\); substituting: \(5x + 2(10-3x) = 18 \Rightarrow -x + 20 = 18 \Rightarrow x = 2\), then \(y = 4\). So \(M(2, 4)\).
(c) Note \(\det A = (2)(3)-(-1)(-5) = 1\), and \(AC = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I\). Hence \(C\) is the inverse of \(A\), that is \(C = A^{-1}\).
Question 11 Report
Two functions g and h are defined on the set R of real numbers by \(g : x \to x^{2} - 2\) and \(h : x \to \frac{1}{x + 2}\). Find :
(a) \(h^{-1}\), the inverse of h ;
(b) \(g \circ h\), when \(x = -\frac{1}{2}\).
\(g:x\to x^{2}-2\) and \(h:x\to\dfrac{1}{x+2}\).
(a) Inverse of h. Let \(y=\dfrac{1}{x+2}\). Then
\[x+2=\frac{1}{y}\;\Rightarrow\;x=\frac{1}{y}-2\]So, replacing \(y\) by \(x\),
\[h^{-1}(x)=\frac{1}{x}-2=\frac{1-2x}{x}\](b) \(g\circ h\) at \(x=-\dfrac12\). First evaluate \(h\):
\[h\left(-\tfrac12\right)=\frac{1}{-\tfrac12+2}=\frac{1}{\tfrac32}=\frac{2}{3}\]Then apply \(g\):
\[g\left(\tfrac23\right)=\left(\tfrac23\right)^{2}-2=\frac{4}{9}-2=\frac{4-18}{9}=-\frac{14}{9}\]Hence \((g\circ h)\left(-\tfrac12\right)=-\dfrac{14}{9}\).
Answer Details
\(g:x\to x^{2}-2\) and \(h:x\to\dfrac{1}{x+2}\).
(a) Inverse of h. Let \(y=\dfrac{1}{x+2}\). Then
\[x+2=\frac{1}{y}\;\Rightarrow\;x=\frac{1}{y}-2\]So, replacing \(y\) by \(x\),
\[h^{-1}(x)=\frac{1}{x}-2=\frac{1-2x}{x}\](b) \(g\circ h\) at \(x=-\dfrac12\). First evaluate \(h\):
\[h\left(-\tfrac12\right)=\frac{1}{-\tfrac12+2}=\frac{1}{\tfrac32}=\frac{2}{3}\]Then apply \(g\):
\[g\left(\tfrac23\right)=\left(\tfrac23\right)^{2}-2=\frac{4}{9}-2=\frac{4-18}{9}=-\frac{14}{9}\]Hence \((g\circ h)\left(-\tfrac12\right)=-\dfrac{14}{9}\).
Question 12 Report
(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° respectively to the downward vertical, find the tension in each of the strings. [Take \(g = 10 ms^{-2}\)].
(b) A constant force F acts on a toy car of mass 5 kg and increases its velocity from 5 ms\(^{-1}\) to 9 ms\(^{-1}\) in 2 seconds. Calculate :
(i) the magnitude of the force ; (ii) velocity of the toy car 3 seconds after attaining a velocity of 9 ms\(^{-1}\).
(a) Weight \(W = mg = 5\times10 = 50\,\text{N}\). Let \(T_1\) (string \(AP\)) and \(T_2\) (string \(BP\)) act at \(40^{o}\) and \(30^{o}\) to the downward vertical on opposite sides.
Horizontal: \(T_1\sin40^{o} = T_2\sin30^{o}\).
Vertical: \(T_1\cos40^{o} + T_2\cos30^{o} = 50\).
Using Lami's theorem is neat here. The angles at \(P\) are: between \(W\) and \(T_1 = 140^{o}\), between \(W\) and \(T_2 = 150^{o}\), between \(T_1\) and \(T_2 = 70^{o}\).
\[\frac{T_1}{\sin150^{o}} = \frac{T_2}{\sin140^{o}} = \frac{50}{\sin70^{o}}\] \[\frac{50}{\sin70^{o}} = \frac{50}{0.9397} = 53.21\] \[T_1 = 53.21\times\sin150^{o} = 53.21\times0.5 = 26.6\,\text{N}\] \[T_2 = 53.21\times\sin140^{o} = 53.21\times0.6428 = 34.2\,\text{N}\]Tension in \(AP \approx \mathbf{26.6\,\text{N}}\); tension in \(BP \approx \mathbf{34.2\,\text{N}}\).
(b)(i) \(a = \dfrac{9 - 5}{2} = 2\,\text{ms}^{-2}\), so \(F = ma = 5\times2 = \mathbf{10\,\text{N}}\).
(ii) Continuing at \(a = 2\,\text{ms}^{-2}\) for a further \(3\,\text{s}\) from \(9\,\text{ms}^{-1}\):
\[v = 9 + 2\times3 = \mathbf{15\,\text{ms}^{-1}}\]Answer Details
(a) Weight \(W = mg = 5\times10 = 50\,\text{N}\). Let \(T_1\) (string \(AP\)) and \(T_2\) (string \(BP\)) act at \(40^{o}\) and \(30^{o}\) to the downward vertical on opposite sides.
Horizontal: \(T_1\sin40^{o} = T_2\sin30^{o}\).
Vertical: \(T_1\cos40^{o} + T_2\cos30^{o} = 50\).
Using Lami's theorem is neat here. The angles at \(P\) are: between \(W\) and \(T_1 = 140^{o}\), between \(W\) and \(T_2 = 150^{o}\), between \(T_1\) and \(T_2 = 70^{o}\).
\[\frac{T_1}{\sin150^{o}} = \frac{T_2}{\sin140^{o}} = \frac{50}{\sin70^{o}}\] \[\frac{50}{\sin70^{o}} = \frac{50}{0.9397} = 53.21\] \[T_1 = 53.21\times\sin150^{o} = 53.21\times0.5 = 26.6\,\text{N}\] \[T_2 = 53.21\times\sin140^{o} = 53.21\times0.6428 = 34.2\,\text{N}\]Tension in \(AP \approx \mathbf{26.6\,\text{N}}\); tension in \(BP \approx \mathbf{34.2\,\text{N}}\).
(b)(i) \(a = \dfrac{9 - 5}{2} = 2\,\text{ms}^{-2}\), so \(F = ma = 5\times2 = \mathbf{10\,\text{N}}\).
(ii) Continuing at \(a = 2\,\text{ms}^{-2}\) for a further \(3\,\text{s}\) from \(9\,\text{ms}^{-1}\):
\[v = 9 + 2\times3 = \mathbf{15\,\text{ms}^{-1}}\]Question 13 Report
The table gives the distribution of heights in metres of 100 students.
| Height | 1.40-1.42 | 1.43-1.45 | 1.46-1.48 | 1.49-1.51 | 1.52-1.54 | 1.55-1.57 | 1.58-1.60 | 1.61-1.63 |
| Freq | 2 | 4 | 19 | 30 | 24 | 14 | 6 | 1 |
(a) Calculate the : (i) mean height ; (ii) mean deviation of the distribution.
(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?
Each class has width 3; the mid-values are \(1.41,1.44,\dots,1.62\). Use a working-mean coding \(u=\dfrac{x-A}{c}\) with \(A=1.50\) and \(c=0.03\).
| Height | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 1.40-1.42 | 1.41 | 2 | -3 | -6 | 18 |
| 1.43-1.45 | 1.44 | 4 | -2 | -8 | 16 |
| 1.46-1.48 | 1.47 | 19 | -1 | -19 | 19 |
| 1.49-1.51 | 1.50 | 30 | 0 | 0 | 0 |
| 1.52-1.54 | 1.53 | 24 | 1 | 24 | 24 |
| 1.55-1.57 | 1.56 | 14 | 2 | 28 | 56 |
| 1.58-1.60 | 1.59 | 6 | 3 | 18 | 54 |
| 1.61-1.63 | 1.62 | 1 | 4 | 4 | 16 |
| Total | 100 | 41 | 203 |
(a)(i) Mean height.\[\bar{x}=A+\frac{\sum fu}{\sum f}\,c=1.50+\frac{41}{100}(0.03)=1.50+0.0123=1.5123\approx 1.51\text{ m}\]
(a)(ii) Mean deviation. \(\text{MD}=\dfrac{\sum f\,|x-\bar{x}|}{\sum f}\) with \(\bar{x}=1.5123\).
| \(x\) | \(f\) | \(|x-\bar{x}|\) | \(f|x-\bar{x}|\) |
|---|---|---|---|
| 1.41 | 2 | 0.1023 | 0.2046 |
| 1.44 | 4 | 0.0723 | 0.2892 |
| 1.47 | 19 | 0.0423 | 0.8037 |
| 1.50 | 30 | 0.0123 | 0.3690 |
| 1.53 | 24 | 0.0177 | 0.4248 |
| 1.56 | 14 | 0.0477 | 0.6678 |
| 1.59 | 6 | 0.0777 | 0.4662 |
| 1.62 | 1 | 0.1077 | 0.1077 |
| Total | 100 | 3.333 |
\[\text{MD}=\frac{3.333}{100}=0.0333\approx 0.03\text{ m}\]
(b) Probability of height greater than the mean. The mean \(1.5123\) m lies below the upper boundary \(1.515\) m of the class \(1.49-1.51\), so every student in the classes \(1.52-1.54\) and above exceeds the mean:\[24+14+6+1=45\]\[P(\text{height}>\bar{x})=\frac{45}{100}=0.45\]
Mean = 1.51 m, mean deviation \(\approx\) 0.03 m, and the required probability = 0.45.
Answer Details
Each class has width 3; the mid-values are \(1.41,1.44,\dots,1.62\). Use a working-mean coding \(u=\dfrac{x-A}{c}\) with \(A=1.50\) and \(c=0.03\).
| Height | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 1.40-1.42 | 1.41 | 2 | -3 | -6 | 18 |
| 1.43-1.45 | 1.44 | 4 | -2 | -8 | 16 |
| 1.46-1.48 | 1.47 | 19 | -1 | -19 | 19 |
| 1.49-1.51 | 1.50 | 30 | 0 | 0 | 0 |
| 1.52-1.54 | 1.53 | 24 | 1 | 24 | 24 |
| 1.55-1.57 | 1.56 | 14 | 2 | 28 | 56 |
| 1.58-1.60 | 1.59 | 6 | 3 | 18 | 54 |
| 1.61-1.63 | 1.62 | 1 | 4 | 4 | 16 |
| Total | 100 | 41 | 203 |
(a)(i) Mean height.\[\bar{x}=A+\frac{\sum fu}{\sum f}\,c=1.50+\frac{41}{100}(0.03)=1.50+0.0123=1.5123\approx 1.51\text{ m}\]
(a)(ii) Mean deviation. \(\text{MD}=\dfrac{\sum f\,|x-\bar{x}|}{\sum f}\) with \(\bar{x}=1.5123\).
| \(x\) | \(f\) | \(|x-\bar{x}|\) | \(f|x-\bar{x}|\) |
|---|---|---|---|
| 1.41 | 2 | 0.1023 | 0.2046 |
| 1.44 | 4 | 0.0723 | 0.2892 |
| 1.47 | 19 | 0.0423 | 0.8037 |
| 1.50 | 30 | 0.0123 | 0.3690 |
| 1.53 | 24 | 0.0177 | 0.4248 |
| 1.56 | 14 | 0.0477 | 0.6678 |
| 1.59 | 6 | 0.0777 | 0.4662 |
| 1.62 | 1 | 0.1077 | 0.1077 |
| Total | 100 | 3.333 |
\[\text{MD}=\frac{3.333}{100}=0.0333\approx 0.03\text{ m}\]
(b) Probability of height greater than the mean. The mean \(1.5123\) m lies below the upper boundary \(1.515\) m of the class \(1.49-1.51\), so every student in the classes \(1.52-1.54\) and above exceeds the mean:\[24+14+6+1=45\]\[P(\text{height}>\bar{x})=\frac{45}{100}=0.45\]
Mean = 1.51 m, mean deviation \(\approx\) 0.03 m, and the required probability = 0.45.
Question 14 Report
(a) The polynomial \(f(x) = x^{3} + px^{2} - 10x + q\) is exactly divisible by \(x^{2} + x - 6\). Find the :
(i) values of p and q ; (ii) third factor.
(b) The volume of a cube is increasing at the rate of \(2\frac{1}{2} cm^{3} s^{-1}\). Find the rate of change of the side of the base when its length is 2cm.
(a) Since \(x^2 + x - 6 = (x+3)(x-2)\), the polynomial \(f(x) = x^3 + px^2 - 10x + q\) is zero at \(x = -3\) and \(x = 2\).
(i) Using \(f(2) = 0\): \(8 + 4p - 20 + q = 0 \Rightarrow 4p + q = 12\).
Using \(f(-3) = 0\): \(-27 + 9p + 30 + q = 0 \Rightarrow 9p + q = -3\).
Subtracting: \(5p = -15 \Rightarrow p = -3\), and \(q = 12 - 4(-3) = 24\).
So \(p = -3,\; q = 24\).
(ii) With \(f(x) = x^3 - 3x^2 - 10x + 24\), divide by \(x^2 + x - 6\):
\[x^3 - 3x^2 - 10x + 24 = (x^2 + x - 6)(x - 4).\]
The third factor is \((x - 4)\).
(b) For a cube of side \(s\), volume \(V = s^3\), so \(\dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt}\).
Given \(\dfrac{dV}{dt} = 2\tfrac12 = 2.5\ \text{cm}^3\text{s}^{-1}\) and \(s = 2\ \text{cm}\):
\[2.5 = 3(2)^2\frac{ds}{dt} = 12\frac{ds}{dt} \Rightarrow \frac{ds}{dt} = \frac{2.5}{12} = \frac{5}{24} \approx 0.21\ \text{cm s}^{-1}.\]
The side is increasing at \(\dfrac{5}{24}\ \text{cm s}^{-1}\).
Answer Details
(a) Since \(x^2 + x - 6 = (x+3)(x-2)\), the polynomial \(f(x) = x^3 + px^2 - 10x + q\) is zero at \(x = -3\) and \(x = 2\).
(i) Using \(f(2) = 0\): \(8 + 4p - 20 + q = 0 \Rightarrow 4p + q = 12\).
Using \(f(-3) = 0\): \(-27 + 9p + 30 + q = 0 \Rightarrow 9p + q = -3\).
Subtracting: \(5p = -15 \Rightarrow p = -3\), and \(q = 12 - 4(-3) = 24\).
So \(p = -3,\; q = 24\).
(ii) With \(f(x) = x^3 - 3x^2 - 10x + 24\), divide by \(x^2 + x - 6\):
\[x^3 - 3x^2 - 10x + 24 = (x^2 + x - 6)(x - 4).\]
The third factor is \((x - 4)\).
(b) For a cube of side \(s\), volume \(V = s^3\), so \(\dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt}\).
Given \(\dfrac{dV}{dt} = 2\tfrac12 = 2.5\ \text{cm}^3\text{s}^{-1}\) and \(s = 2\ \text{cm}\):
\[2.5 = 3(2)^2\frac{ds}{dt} = 12\frac{ds}{dt} \Rightarrow \frac{ds}{dt} = \frac{2.5}{12} = \frac{5}{24} \approx 0.21\ \text{cm s}^{-1}.\]
The side is increasing at \(\dfrac{5}{24}\ \text{cm s}^{-1}\).
Question 15 Report
(a) Find the angle between the vectors \(a = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\) and \(b = \begin{pmatrix} -8 \\ -15 \end{pmatrix}\).
(b) Given that \(a = (4N, 060°)\) and \(b = (3N, 120°)\), find, in component form, the unit vector along \(a - b\).
(a) With \(a = \begin{pmatrix}-3\\4\end{pmatrix},\ b = \begin{pmatrix}-8\\-15\end{pmatrix}\):
\[a\cdot b = (-3)(-8) + (4)(-15) = 24 - 60 = -36\] \[|a| = \sqrt{9+16} = 5,\qquad |b| = \sqrt{64+225} = \sqrt{289} = 17\] \[\cos\theta = \frac{a\cdot b}{|a||b|} = \frac{-36}{85} = -0.4235\] \[\theta = 115.1^{o}\ (\text{to 1 d.p.})\](b) \(a = (4\cos60^{o},\ 4\sin60^{o}) = (2,\ 2\sqrt3)\); \(b = (3\cos120^{o},\ 3\sin120^{o}) = (-1.5,\ 1.5\sqrt3)\).
\[a - b = (2 - (-1.5),\ 2\sqrt3 - 1.5\sqrt3) = (3.5,\ 0.5\sqrt3)\] \[|a - b| = \sqrt{3.5^2 + (0.5\sqrt3)^2} = \sqrt{12.25 + 0.75} = \sqrt{13}\]Unit vector along \(a - b\):
\[\hat u = \frac{1}{\sqrt{13}}\begin{pmatrix} 3.5 \\ 0.5\sqrt3 \end{pmatrix} \approx \begin{pmatrix} 0.971 \\ 0.240 \end{pmatrix}\]Answer Details
(a) With \(a = \begin{pmatrix}-3\\4\end{pmatrix},\ b = \begin{pmatrix}-8\\-15\end{pmatrix}\):
\[a\cdot b = (-3)(-8) + (4)(-15) = 24 - 60 = -36\] \[|a| = \sqrt{9+16} = 5,\qquad |b| = \sqrt{64+225} = \sqrt{289} = 17\] \[\cos\theta = \frac{a\cdot b}{|a||b|} = \frac{-36}{85} = -0.4235\] \[\theta = 115.1^{o}\ (\text{to 1 d.p.})\](b) \(a = (4\cos60^{o},\ 4\sin60^{o}) = (2,\ 2\sqrt3)\); \(b = (3\cos120^{o},\ 3\sin120^{o}) = (-1.5,\ 1.5\sqrt3)\).
\[a - b = (2 - (-1.5),\ 2\sqrt3 - 1.5\sqrt3) = (3.5,\ 0.5\sqrt3)\] \[|a - b| = \sqrt{3.5^2 + (0.5\sqrt3)^2} = \sqrt{12.25 + 0.75} = \sqrt{13}\]Unit vector along \(a - b\):
\[\hat u = \frac{1}{\sqrt{13}}\begin{pmatrix} 3.5 \\ 0.5\sqrt3 \end{pmatrix} \approx \begin{pmatrix} 0.971 \\ 0.240 \end{pmatrix}\]Question 16 Report
Three forces \(-63j , 32.14i + 38.3j\) and \(14i - 24.25j\) act on a body of mass 5kg. Find, correct to one decimal place, the :
(a) magnitude of the resultant force ;
(b) acceleration of the body.
Add the forces component by component. The three forces are
\[F_1 = -63\mathbf{j},\quad F_2 = 32.14\mathbf{i} + 38.3\mathbf{j},\quad F_3 = 14\mathbf{i} - 24.25\mathbf{j}.\]
Resultant components.
\[\sum F_x = 0 + 32.14 + 14 = 46.14,\qquad \sum F_y = -63 + 38.3 - 24.25 = -48.95.\]
So \(\mathbf{R} = 46.14\mathbf{i} - 48.95\mathbf{j}\).
(a) Magnitude of the resultant force.
\[|\mathbf{R}| = \sqrt{46.14^2 + 48.95^2} = \sqrt{2128.90 + 2396.10} = \sqrt{4525.00} \approx 67.3\ \text{N}.\]
(b) Acceleration. By Newton's second law \(\mathbf{R} = m\mathbf{a}\), with \(m = 5\ \text{kg}\):
\[a = \frac{|\mathbf{R}|}{m} = \frac{67.3}{5} \approx 13.5\ \text{m s}^{-2}.\]
The acceleration is directed along the resultant force, with magnitude \(13.5\ \text{m s}^{-2}\).
Answer Details
Add the forces component by component. The three forces are
\[F_1 = -63\mathbf{j},\quad F_2 = 32.14\mathbf{i} + 38.3\mathbf{j},\quad F_3 = 14\mathbf{i} - 24.25\mathbf{j}.\]
Resultant components.
\[\sum F_x = 0 + 32.14 + 14 = 46.14,\qquad \sum F_y = -63 + 38.3 - 24.25 = -48.95.\]
So \(\mathbf{R} = 46.14\mathbf{i} - 48.95\mathbf{j}\).
(a) Magnitude of the resultant force.
\[|\mathbf{R}| = \sqrt{46.14^2 + 48.95^2} = \sqrt{2128.90 + 2396.10} = \sqrt{4525.00} \approx 67.3\ \text{N}.\]
(b) Acceleration. By Newton's second law \(\mathbf{R} = m\mathbf{a}\), with \(m = 5\ \text{kg}\):
\[a = \frac{|\mathbf{R}|}{m} = \frac{67.3}{5} \approx 13.5\ \text{m s}^{-2}.\]
The acceleration is directed along the resultant force, with magnitude \(13.5\ \text{m s}^{-2}\).
Question 17 Report
The twenty-first term of an Arithmetic Progression is \(5\frac{1}{2}\) and the sum of the first twenty-one terms is \(94\frac{1}{2}\). Find the :
(a) first term ; (b) common difference ; (c) sum of the first thirty terms.
A.P.: \(T_{21}=5\tfrac12\) and \(S_{21}=94\tfrac12\).
Write the two facts algebraically:
\[T_{21}=a+20d=\frac{11}{2}\qquad(1)\]\[S_{21}=\frac{21}{2}\big(2a+20d\big)=\frac{189}{2}\;\Rightarrow\;2a+20d=9\;\Rightarrow\;a+10d=\frac{9}{2}\qquad(2)\]Subtract (2) from (1):
\[10d=\frac{11}{2}-\frac{9}{2}=1\;\Rightarrow\;d=\frac{1}{10}\]From (2), \(a=\dfrac92-10\left(\tfrac{1}{10}\right)=\dfrac92-1=\dfrac72\).
(a) First term \(a=\dfrac72=3\tfrac12\).
(b) Common difference \(d=\dfrac{1}{10}=0.1\).
(c) Sum of the first 30 terms:
\[S_{30}=\frac{30}{2}\big(2a+29d\big)=15\left(7+29\times 0.1\right)=15(7+2.9)=15(9.9)=148.5\]So \(S_{30}=148\tfrac12\).
Answer Details
A.P.: \(T_{21}=5\tfrac12\) and \(S_{21}=94\tfrac12\).
Write the two facts algebraically:
\[T_{21}=a+20d=\frac{11}{2}\qquad(1)\]\[S_{21}=\frac{21}{2}\big(2a+20d\big)=\frac{189}{2}\;\Rightarrow\;2a+20d=9\;\Rightarrow\;a+10d=\frac{9}{2}\qquad(2)\]Subtract (2) from (1):
\[10d=\frac{11}{2}-\frac{9}{2}=1\;\Rightarrow\;d=\frac{1}{10}\]From (2), \(a=\dfrac92-10\left(\tfrac{1}{10}\right)=\dfrac92-1=\dfrac72\).
(a) First term \(a=\dfrac72=3\tfrac12\).
(b) Common difference \(d=\dfrac{1}{10}=0.1\).
(c) Sum of the first 30 terms:
\[S_{30}=\frac{30}{2}\big(2a+29d\big)=15\left(7+29\times 0.1\right)=15(7+2.9)=15(9.9)=148.5\]So \(S_{30}=148\tfrac12\).
Question 18 Report
(a) Two items are selected at random from four items labelled (p, q, r, s).
(i) List the sample space if sampling is done (1) with replacement ; (2) without replacement.
(ii) Find the probability that r is at least one of the two objects selected : (1) in a(i)1 ; (2) in a(i)2.
(b) How many whole numbers from 100 to 999 are divisible by (i) 4 ; (ii) both 3 and 4?
(a)(i)(1) With replacement
Since an item is replaced after the first selection, there are \(4 \times 4 = 16\) possible ordered outcomes:
| p | q | r | s | |
|---|---|---|---|---|
| p | pp | pq | pr | ps |
| q | qp | qr | qs | |
| r | rp | rq | rr | rs |
| s | sp | sq | sr | ss |
\(S=\{pp,pq,pr,ps,qp,qq,qr,qs,rp,rq,rr,rs,sp,sq,sr,ss\}\).
(2) Without replacement
There are \(4 \times 3=12\) possible ordered outcomes:
\(S'=\{pq,pr,ps,qp,qr,qs,rp,rq,rs,sp,sq,sr\}\).
(a)(ii)(1) With replacement
The outcomes containing at least one \(r\) are:
\(\{pr,qr,rp,rq,rr,rs,sr\}\).
Therefore,
\[ P(\text{at least one }r)=\frac{7}{16}. \](2) Without replacement
The outcomes containing \(r\) are:
\(\{pr,qr,rp,rq,rs,sr\}\).
Therefore,
\[ P(\text{at least one }r)=\frac{6}{12}=\frac{1}{2}. \](b)(i) Numbers from 100 to 999 divisible by 4
The multiples of 4 are:
\(100,104,108,\ldots,996\).
\[ 996=100+4(n-1) \] \[ 996=100+4n-4 \] \[ 996=96+4n
Answer Details
(a)(i)(1) With replacement
Since an item is replaced after the first selection, there are \(4 \times 4 = 16\) possible ordered outcomes:
| p | q | r | s | |
|---|---|---|---|---|
| p | pp | pq | pr | ps |
| q | qp | qr | qs | |
| r | rp | rq | rr | rs |
| s | sp | sq | sr | ss |
\(S=\{pp,pq,pr,ps,qp,qq,qr,qs,rp,rq,rr,rs,sp,sq,sr,ss\}\).
(2) Without replacement
There are \(4 \times 3=12\) possible ordered outcomes:
\(S'=\{pq,pr,ps,qp,qr,qs,rp,rq,rs,sp,sq,sr\}\).
(a)(ii)(1) With replacement
The outcomes containing at least one \(r\) are:
\(\{pr,qr,rp,rq,rr,rs,sr\}\).
Therefore,
\[ P(\text{at least one }r)=\frac{7}{16}. \](2) Without replacement
The outcomes containing \(r\) are:
\(\{pr,qr,rp,rq,rs,sr\}\).
Therefore,
\[ P(\text{at least one }r)=\frac{6}{12}=\frac{1}{2}. \](b)(i) Numbers from 100 to 999 divisible by 4
The multiples of 4 are:
\(100,104,108,\ldots,996\).
\[ 996=100+4(n-1) \] \[ 996=100+4n-4 \] \[ 996=96+4n
