WAEC SSCE - Further Mathematics - 2008 (Essay)
Tambaya 1 Rahoto
(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of employing the 4 teachers, if : (i) there is no restriction ; (ii) at least 2 of them are females.
(b) The table shows the positions awarded to 7 contestants by Judges X and Y in a competition.
| Contestant | P | Q | R | S | T | U | V |
| Judge X | 2 | 7 | 1 | 3 | 6 | 5 | 4 |
| Judge Y | 4 | 6 | 2 | 3 | 1 | 1 | 5 |
(i) Calculate, correct to one decimal place, the Spearman's rank correlation coefficient.
(ii) Interpret your answer in b(i) above.
(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.
(i) No restriction. Choose any 4 of the 12:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \](ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):
\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \](b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.
| Contestant | \(R_X\) | \(R_Y\) | \(d\) | \(d^2\) |
|---|---|---|---|---|
| P | 2 | 5 | -3 | 9 |
| Q | 7 | 7 | 0 | 0 |
| R | 1 | 3 | -2 | 4 |
| S | 3 | 4 | -1 | 1 |
| T | 6 | 1.5 | 4.5 | 20.25 |
| U | 5 | 1.5 | 3.5 | 12.25 |
| V | 4 | 6 | -2 | 4 |
| Total \(\sum d^2\) | 50.5 | |||
(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.
Bayanin Amsa
(a) Selecting 4 teachers from 5 female + 7 male = 12 teachers.
(i) No restriction. Choose any 4 of the 12:
\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \mathbf{495 \text{ ways}} \](ii) At least 2 females. The cases are exactly 2, 3 or 4 females (the rest male):
\[ \binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0} \] \[ = (10)(21) + (10)(7) + (5)(1) = 210 + 70 + 5 = \mathbf{285 \text{ ways}} \](b) Spearman's rank correlation for 7 contestants (\(n = 7\)). Judge Y has a tie (T and U both awarded position 1), so they share positions 1 and 2, each taking the average rank \(1.5\); the remaining Y values are then re-ranked \(2\to3, 3\to4, 4\to5, 5\to6, 6\to7\). Judge X already has distinct positions 1 to 7.
| Contestant | \(R_X\) | \(R_Y\) | \(d\) | \(d^2\) |
|---|---|---|---|---|
| P | 2 | 5 | -3 | 9 |
| Q | 7 | 7 | 0 | 0 |
| R | 1 | 3 | -2 | 4 |
| S | 3 | 4 | -1 | 1 |
| T | 6 | 1.5 | 4.5 | 20.25 |
| U | 5 | 1.5 | 3.5 | 12.25 |
| V | 4 | 6 | -2 | 4 |
| Total \(\sum d^2\) | 50.5 | |||
(ii) Interpretation. \(r_s \approx 0.1\) is very close to zero, so there is only a very weak positive correlation. The two judges show almost no agreement in the way they ranked the contestants.
Tambaya 2 Rahoto
The following table shows the distribution of marks obtained by some students in an examination.
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Frequency | 50 | 50 | 40 | 60 | 100 | 100 | 50 | 25 | 15 | 10 |
(a) Construct a cumulative frequency table for the distribution
(b) Draw an ogive for the distribution
(c) Use your graph in (b) to determine : (i) semi- interquartile range ; (ii) number of students who failed, if the pass mark for the examination is 37 ; (iii) probability that a student selected at random scored between 20% and 60%.
(a) Cumulative frequency table
| Marks | Frequency, \(f\) | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 0–9 | 50 | −0.5–9.5 | 50 |
| 10–19 | 50 | 9.5–19.5 | 100 |
| 20–29 | 40 | 19.5–29.5 | 140 |
| 30–39 | 60 | 29.5–39.5 | 200 |
| 40–49 | 100 | 39.5–49.5 | 300 |
| 50–59 | 100 | 49.5–59.5 | 400 |
| 60–69 | 50 | 59.5–69.5 | 450 |
| 70–79 | 25 | 69.5–79.5 | 475 |
| 80–89 | 15 | 79.5–89.5 | 490 |
| 90–99 | 10 | 89.5–99.5 | 500 |
Total number of students, \(N=500\).
(b) Ogive
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\).
(c)
(i) Semi-interquartile range
\(Q_1\) is the \(\frac{N}{4}=125\)th value and \(Q_3\) is the \(\frac{3N}{4}=375\)th value. From the ogive,
\[Q_1\approx 26.0,\qquad Q_3\approx 56.3.\]
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{56.3-26.0}{2}=\boxed{15.15\text{ marks}}.\]
(ii) Number who failed
At a mark of \(37\), the cumulative frequency read from the ogive is approximately \(190\). Hence, \(\boxed{190\text{ students}}\) failed.
(iii) Probability of scoring between \(20\%\) and \(60\%\)
From the cumulative-frequency curve, the number scoring below \(20\%\) is \(100\), while the number scoring below \(60\%\) is \(400\).
\[\text{Number scoring between }20\%\text{ and }60\%=400-100=300.\]
\[P(20\%<X<60\%)=\frac{300}{500}=\boxed{\frac35}.\]
Bayanin Amsa
(a) Cumulative frequency table
| Marks | Frequency, \(f\) | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 0–9 | 50 | −0.5–9.5 | 50 |
| 10–19 | 50 | 9.5–19.5 | 100 |
| 20–29 | 40 | 19.5–29.5 | 140 |
| 30–39 | 60 | 29.5–39.5 | 200 |
| 40–49 | 100 | 39.5–49.5 | 300 |
| 50–59 | 100 | 49.5–59.5 | 400 |
| 60–69 | 50 | 59.5–69.5 | 450 |
| 70–79 | 25 | 69.5–79.5 | 475 |
| 80–89 | 15 | 79.5–89.5 | 490 |
| 90–99 | 10 | 89.5–99.5 | 500 |
Total number of students, \(N=500\).
(b) Ogive
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\).
(c)
(i) Semi-interquartile range
\(Q_1\) is the \(\frac{N}{4}=125\)th value and \(Q_3\) is the \(\frac{3N}{4}=375\)th value. From the ogive,
\[Q_1\approx 26.0,\qquad Q_3\approx 56.3.\]
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{56.3-26.0}{2}=\boxed{15.15\text{ marks}}.\]
(ii) Number who failed
At a mark of \(37\), the cumulative frequency read from the ogive is approximately \(190\). Hence, \(\boxed{190\text{ students}}\) failed.
(iii) Probability of scoring between \(20\%\) and \(60\%\)
From the cumulative-frequency curve, the number scoring below \(20\%\) is \(100\), while the number scoring below \(60\%\) is \(400\).
\[\text{Number scoring between }20\%\text{ and }60\%=400-100=300.\]
\[P(20\%<X<60\%)=\frac{300}{500}=\boxed{\frac35}.\]
Tambaya 3 Rahoto
(a) Forces \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N\) and \(F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\) act on a body. Find the :
(i) resultant of these forces ; (ii) fifth force that will keep the body in equilibrium.
(b) A body moving at 20 ms\(^{-1}\) accelerates uniformly at 2.5 ms\(^{-2}\) for 4 seconds. It continues the journey at the speed for 8 seconds, before coming to rest in t seconds with a uniform retardation. If the ratio of the acceleration to the retardation is 3 : 4,
(i) sketch the velocity- time graph of the journey ; (ii) find t ; (iii) find the total distance of the journey.
(a)
(i) Resultant force
Writing each force as a column vector,
\[\begin{aligned}\mathbf R&=\begin{pmatrix}-5\\4\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}+\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}3\\-5\end{pmatrix}\\&=\begin{pmatrix}-5+2+2+3\\4+5-1-5\end{pmatrix}\\&=\begin{pmatrix}2\\3\end{pmatrix}\text{ N}.\end{aligned}\]
Hence, \(\boxed{\mathbf R=2\mathbf i+3\mathbf j\text{ N}}\).
(ii) Fifth force for equilibrium
For equilibrium, the fifth force \(\mathbf F_5\) must be the negative of the resultant:
\[\mathbf F_5=-\mathbf R=-\begin{pmatrix}2\\3\end{pmatrix}=\boxed{\begin{pmatrix}-2\\-3\end{pmatrix}\text{ N}}.\]
(b)
During the first 4 s,
\[v=20+(2.5)(4)=30\text{ m s}^{-1}.\]
Let the magnitude of the retardation be \(r\text{ m s}^{-2}\). Since
\[2.5:r=3:4,\qquad r=\frac{4}{3}(2.5)=\frac{10}{3}\text{ m s}^{-2}.\]
(i) Velocity-time graph
The graph consists of straight-line sections joining \((0,20)\), \((4,30)\), \((12,30)\), and \((21,0)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(ii) Time taken to come to rest
For the retardation stage, \(u=30\text{ m s}^{-1}\), \(v=0\), and acceleration \(=-\frac{10}{3}\text{ m s}^{-2}\).
\[0=30-\frac{10}{3}t\]
\[\frac{10}{3}t=30\]
\[\boxed{t=9\text{ s}}.\]
(iii) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned}A_1&=\frac12(20+30)\times4=100\text{ m},\\A_2&=30\times8=240\text{ m},\\A_3&=\frac12\times30\times9=135\text{ m}.\end{aligned}\]
\[\text{Total distance}=100+240+135=\boxed{475\text{ m}}.\]
Bayanin Amsa
(a)
(i) Resultant force
Writing each force as a column vector,
\[\begin{aligned}\mathbf R&=\begin{pmatrix}-5\\4\end{pmatrix}+\begin{pmatrix}2\\5\end{pmatrix}+\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}3\\-5\end{pmatrix}\\&=\begin{pmatrix}-5+2+2+3\\4+5-1-5\end{pmatrix}\\&=\begin{pmatrix}2\\3\end{pmatrix}\text{ N}.\end{aligned}\]
Hence, \(\boxed{\mathbf R=2\mathbf i+3\mathbf j\text{ N}}\).
(ii) Fifth force for equilibrium
For equilibrium, the fifth force \(\mathbf F_5\) must be the negative of the resultant:
\[\mathbf F_5=-\mathbf R=-\begin{pmatrix}2\\3\end{pmatrix}=\boxed{\begin{pmatrix}-2\\-3\end{pmatrix}\text{ N}}.\]
(b)
During the first 4 s,
\[v=20+(2.5)(4)=30\text{ m s}^{-1}.\]
Let the magnitude of the retardation be \(r\text{ m s}^{-2}\). Since
\[2.5:r=3:4,\qquad r=\frac{4}{3}(2.5)=\frac{10}{3}\text{ m s}^{-2}.\]
(i) Velocity-time graph
The graph consists of straight-line sections joining \((0,20)\), \((4,30)\), \((12,30)\), and \((21,0)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(ii) Time taken to come to rest
For the retardation stage, \(u=30\text{ m s}^{-1}\), \(v=0\), and acceleration \(=-\frac{10}{3}\text{ m s}^{-2}\).
\[0=30-\frac{10}{3}t\]
\[\frac{10}{3}t=30\]
\[\boxed{t=9\text{ s}}.\]
(iii) Total distance travelled
The total distance is the area under the velocity-time graph:
\[\begin{aligned}A_1&=\frac12(20+30)\times4=100\text{ m},\\A_2&=30\times8=240\text{ m},\\A_3&=\frac12\times30\times9=135\text{ m}.\end{aligned}\]
\[\text{Total distance}=100+240+135=\boxed{475\text{ m}}.\]
Tambaya 4 Rahoto
(a) If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x - 6 = 0\), find the equation whose roots are \((\alpha - 2)\) and \((\beta - 2)\).
(b) Given that \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = 4\), find the values of k.
(a) New equation whose roots are \((\alpha-2)\) and \((\beta-2).\)
For \(2x^2+5x-6=0,\) sum and product of roots:
\[\alpha+\beta=-\frac{5}{2},\qquad \alpha\beta=\frac{-6}{2}=-3.\]
Sum of new roots:
\[(\alpha-2)+(\beta-2)=(\alpha+\beta)-4=-\frac{5}{2}-4=-\frac{13}{2}.\]
Product of new roots:
\[(\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=-3-2\!\left(-\frac{5}{2}\right)+4=-3+5+4=6.\]
Required equation \(x^2-(\text{sum})x+(\text{product})=0:\)
\[x^2+\frac{13}{2}x+6=0\ \Rightarrow\ 2x^2+13x+12=0.\]
(b) Solve \(\displaystyle\int_0^k(x^2-2x)\,dx=4.\)
\[\int_0^k(x^2-2x)\,dx=\left[\frac{x^3}{3}-x^2\right]_0^k=\frac{k^3}{3}-k^2.\]
Set equal to 4 and multiply through by 3:
\[\frac{k^3}{3}-k^2=4\ \Rightarrow\ k^3-3k^2-12=0.\]
This cubic has a single real root. Testing shows \(f(3)=27-27-12=-12<0\) and \(f(4)=64-48-12=4>0,\) so the root lies between 3 and 4. Solving numerically,
\[k\approx3.8.\]
Bayanin Amsa
(a) New equation whose roots are \((\alpha-2)\) and \((\beta-2).\)
For \(2x^2+5x-6=0,\) sum and product of roots:
\[\alpha+\beta=-\frac{5}{2},\qquad \alpha\beta=\frac{-6}{2}=-3.\]
Sum of new roots:
\[(\alpha-2)+(\beta-2)=(\alpha+\beta)-4=-\frac{5}{2}-4=-\frac{13}{2}.\]
Product of new roots:
\[(\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=-3-2\!\left(-\frac{5}{2}\right)+4=-3+5+4=6.\]
Required equation \(x^2-(\text{sum})x+(\text{product})=0:\)
\[x^2+\frac{13}{2}x+6=0\ \Rightarrow\ 2x^2+13x+12=0.\]
(b) Solve \(\displaystyle\int_0^k(x^2-2x)\,dx=4.\)
\[\int_0^k(x^2-2x)\,dx=\left[\frac{x^3}{3}-x^2\right]_0^k=\frac{k^3}{3}-k^2.\]
Set equal to 4 and multiply through by 3:
\[\frac{k^3}{3}-k^2=4\ \Rightarrow\ k^3-3k^2-12=0.\]
This cubic has a single real root. Testing shows \(f(3)=27-27-12=-12<0\) and \(f(4)=64-48-12=4>0,\) so the root lies between 3 and 4. Solving numerically,
\[k\approx3.8.\]
Tambaya 5 Rahoto
A survey indicated that 65% of the families in an area have cars. Find, correct to three decimal places, the probability that among 7 families selected at random in the area
(a) exactly 5 ;
(b) 3 or 4 ;
(c) at most 2 of them have cars.
This is a binomial situation with \(n=7\) families, probability of owning a car \(p=0.65,\) and \(q=1-p=0.35.\) The probability of exactly \(r\) successes is \(P(r)=\binom{7}{r}p^r q^{7-r}.\)
(a) Exactly 5 have cars.
\[P(5)=\binom{7}{5}(0.65)^5(0.35)^2=21\times0.116029\times0.1225\approx0.298.\]
(b) 3 or 4 have cars.
\[P(3)=\binom{7}{3}(0.65)^3(0.35)^4=35\times0.274625\times0.015006\approx0.144.\]
\[P(4)=\binom{7}{4}(0.65)^4(0.35)^3=35\times0.178506\times0.042875\approx0.268.\]
\[P(3\text{ or }4)=0.144+0.268\approx0.412.\]
(c) At most 2 have cars \((r=0,1,2).\)
\[P(0)=(0.35)^7\approx0.000643,\]
\[P(1)=\binom{7}{1}(0.65)(0.35)^6\approx0.008364,\]
\[P(2)=\binom{7}{2}(0.65)^2(0.35)^5\approx0.046590.\]
\[P(\text{at most }2)=0.000643+0.008364+0.046590\approx0.056.\]
Bayanin Amsa
This is a binomial situation with \(n=7\) families, probability of owning a car \(p=0.65,\) and \(q=1-p=0.35.\) The probability of exactly \(r\) successes is \(P(r)=\binom{7}{r}p^r q^{7-r}.\)
(a) Exactly 5 have cars.
\[P(5)=\binom{7}{5}(0.65)^5(0.35)^2=21\times0.116029\times0.1225\approx0.298.\]
(b) 3 or 4 have cars.
\[P(3)=\binom{7}{3}(0.65)^3(0.35)^4=35\times0.274625\times0.015006\approx0.144.\]
\[P(4)=\binom{7}{4}(0.65)^4(0.35)^3=35\times0.178506\times0.042875\approx0.268.\]
\[P(3\text{ or }4)=0.144+0.268\approx0.412.\]
(c) At most 2 have cars \((r=0,1,2).\)
\[P(0)=(0.35)^7\approx0.000643,\]
\[P(1)=\binom{7}{1}(0.65)(0.35)^6\approx0.008364,\]
\[P(2)=\binom{7}{2}(0.65)^2(0.35)^5\approx0.046590.\]
\[P(\text{at most }2)=0.000643+0.008364+0.046590\approx0.056.\]
Tambaya 6 Rahoto
(a) The sum of the first n terms of a sequence is given by \(S_{n} = \frac{5n^{2}}{2} + \frac{5n}{2}\). Write down the first four terms of the sequence and an expression for the nth term.
(b) The equation of a circle is given by \(x^{2} + y^{2} - 10x - 8y + 25 = 0\).
(i) Show that the circle touches the x- axis ; (ii) Find the coordinates of the point of contact.
(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)
Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:
\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)
Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)
First four terms: 5, 10, 15, 20.
General term:
\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]
So \(T_n=5n.\)
(b) Circle \(x^2+y^2-10x-8y+25=0.\)
Complete the square:
\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]
\[(x-5)^2+(y-4)^2=16.\]
Centre \((5,4),\) radius \(r=4.\)
(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.
(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)
Bayanin Amsa
(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)
Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:
\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)
Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)
First four terms: 5, 10, 15, 20.
General term:
\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]
So \(T_n=5n.\)
(b) Circle \(x^2+y^2-10x-8y+25=0.\)
Complete the square:
\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]
\[(x-5)^2+(y-4)^2=16.\]
Centre \((5,4),\) radius \(r=4.\)
(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.
(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)
Tambaya 7 Rahoto
The position vectors of points P, Q and R with respect to the origin are \((4i - 5j), (i + 3j)\) and \((-5i + 2j)\) respectively. If PQRM is a parallelogram, find:
(a) the position vector of M ;
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|\) ;
(c) the acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}\), correct to 1 decimal place ;
(d) the area of PQRM.
Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)
(a) Position vector of M.
\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)
\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)
\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)
(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)
\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]
(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):
\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]
Bayanin Amsa
Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)
(a) Position vector of M.
\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)
\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)
\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)
(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)
\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]
(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):
\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]
Tambaya 8 Rahoto
The magnitude of a force \(xi + 15j\) is 17N. Find the :
(a) possible values of x ;
(b) directions of the forces, correct to the nearest degree.
The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.
(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)
\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]
So \(x=8\) or \(x=-8.\)
(b) Directions (to the nearest degree), measured from the positive x-axis.
When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.
\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]
When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.
\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]
So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.
Bayanin Amsa
The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.
(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)
\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]
So \(x=8\) or \(x=-8.\)
(b) Directions (to the nearest degree), measured from the positive x-axis.
When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.
\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]
When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.
\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]
So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.
Tambaya 9 Rahoto
The table shows the distribution of marks obtained by some candidates in a test.
| Marks | 10-14 | 15-24 | 25-29 | 30-39 | 40-44 | 45-49 |
| No of candidates | 14 | 30 | 22 | 18 | 12 | 4 |
Draw a histogram for the distribution.
Since the class intervals have unequal widths, the height of each rectangle is the frequency density:
\[\text{Frequency density}=\frac{\text{Frequency}}{\text{Class width}}\]
| Marks | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10-14 | 9.5-14.5 | 5 | 14 | 2.8 |
| 15-24 | 14.5-24.5 | 10 | 30 | 3.0 |
| 25-29 | 24.5-29.5 | 5 | 22 | 4.4 |
| 30-39 | 29.5-39.5 | 10 | 18 | 1.8 |
| 40-44 | 39.5-44.5 | 5 | 12 | 2.4 |
| 45-49 | 44.5-49.5 | 5 | 4 | 0.8 |
The required histogram is:
The rectangles are contiguous, with their bases at the stated class boundaries. Their areas are proportional to the corresponding frequencies.
Bayanin Amsa
Since the class intervals have unequal widths, the height of each rectangle is the frequency density:
\[\text{Frequency density}=\frac{\text{Frequency}}{\text{Class width}}\]
| Marks | Class boundaries | Class width | Frequency | Frequency density |
|---|---|---|---|---|
| 10-14 | 9.5-14.5 | 5 | 14 | 2.8 |
| 15-24 | 14.5-24.5 | 10 | 30 | 3.0 |
| 25-29 | 24.5-29.5 | 5 | 22 | 4.4 |
| 30-39 | 29.5-39.5 | 10 | 18 | 1.8 |
| 40-44 | 39.5-44.5 | 5 | 12 | 2.4 |
| 45-49 | 44.5-49.5 | 5 | 4 | 0.8 |
The required histogram is:
The rectangles are contiguous, with their bases at the stated class boundaries. Their areas are proportional to the corresponding frequencies.
Tambaya 10 Rahoto
(a) Express \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\) in partial fractions.
(b) If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\), find the value of x.
(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)
Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)
Write
\[\frac{2x^2-5x+1}{x(x-1)(x-3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-3}.\]
Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)
\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)
\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)
\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)
\[\frac{2x^2-5x+1}{x^3-4x^2+3x}=\frac{1}{3x}+\frac{1}{x-1}+\frac{2}{3(x-3)}.\]
(b) Solve \(\begin{vmatrix}x-3&-4&3\\5&2&2\\2&-4&6-x\end{vmatrix}=-24.\)
Expand along the first row:
\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]
\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]
Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is
\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]
Set equal to \(-24:\)
\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2.\]
Bayanin Amsa
(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)
Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)
Write
\[\frac{2x^2-5x+1}{x(x-1)(x-3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-3}.\]
Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)
\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)
\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)
\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)
\[\frac{2x^2-5x+1}{x^3-4x^2+3x}=\frac{1}{3x}+\frac{1}{x-1}+\frac{2}{3(x-3)}.\]
(b) Solve \(\begin{vmatrix}x-3&-4&3\\5&2&2\\2&-4&6-x\end{vmatrix}=-24.\)
Expand along the first row:
\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]
\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]
Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is
\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]
Set equal to \(-24:\)
\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2.\]
Tambaya 11 Rahoto
A function F is defined on the set R, of real numbers by \(f : x \to px^{2} + qx + 2\), where p and q are constants. If \(f(-2) = 0\) and \(f(1) = 3\), find \(f(-4)\).
Given \(f(x)=px^2+qx+2.\)
From \(f(-2)=0:\ p(-2)^2+q(-2)+2=0\Rightarrow 4p-2q+2=0\Rightarrow 2p-q=-1.\)
From \(f(1)=3:\ p+q+2=3\Rightarrow p+q=1.\)
Add the two results: \((2p-q)+(p+q)=-1+1\Rightarrow 3p=0\Rightarrow p=0.\) Then \(q=1.\)
So \(f(x)=x+2.\) Therefore
\[f(-4)=-4+2=-2.\]
Check: \(f(-2)=-2+2=0\) and \(f(1)=1+2=3,\) as required.
Bayanin Amsa
Given \(f(x)=px^2+qx+2.\)
From \(f(-2)=0:\ p(-2)^2+q(-2)+2=0\Rightarrow 4p-2q+2=0\Rightarrow 2p-q=-1.\)
From \(f(1)=3:\ p+q+2=3\Rightarrow p+q=1.\)
Add the two results: \((2p-q)+(p+q)=-1+1\Rightarrow 3p=0\Rightarrow p=0.\) Then \(q=1.\)
So \(f(x)=x+2.\) Therefore
\[f(-4)=-4+2=-2.\]
Check: \(f(-2)=-2+2=0\) and \(f(1)=1+2=3,\) as required.
Tambaya 12 Rahoto
A straight line passes through the point P(-1, 3). Another line which passes through Q(-4, 4) intersects the first line at the point R(k, 5), where k is a constant. If \(<PRQ = 90°\), find the values of k.
The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.
\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]
\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]
Set the dot product to zero:
\[(-1-k)(-4-k)+(-2)(-1)=0.\]
Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so
\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]
Factorise:
\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]
Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)
Bayanin Amsa
The point of intersection is \(R(k,5),\) with \(P(-1,3)\) and \(Q(-4,4).\) The condition \(\angle PRQ=90^\circ\) means the segments \(RP\) and \(RQ\) are perpendicular at R, so the vectors \(\overrightarrow{RP}\) and \(\overrightarrow{RQ}\) have zero dot product.
\[\overrightarrow{RP}=P-R=(-1-k,\,3-5)=(-1-k,\,-2),\]
\[\overrightarrow{RQ}=Q-R=(-4-k,\,4-5)=(-4-k,\,-1).\]
Set the dot product to zero:
\[(-1-k)(-4-k)+(-2)(-1)=0.\]
Expand \((-1-k)(-4-k)=(1+k)(4+k)=k^2+5k+4,\) so
\[k^2+5k+4+2=0\Rightarrow k^2+5k+6=0.\]
Factorise:
\[(k+2)(k+3)=0\Rightarrow k=-2\ \text{or}\ k=-3.\]
Both values give a right angle at R, so the required values are \(k=-2\) and \(k=-3.\)
Tambaya 13 Rahoto
Differentiate, with respect to x, \(x^{3} + 2x\) from the first principle.
Differentiate \(f(x)=x^3+2x\) from first principles.
By definition,
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.\]
Compute \(f(x+h):\)
\[f(x+h)=(x+h)^3+2(x+h)=x^3+3x^2h+3xh^2+h^3+2x+2h.\]
Then
\[f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h.\]
Divide by \(h\) (for \(h\neq0\)):
\[\frac{f(x+h)-f(x)}{h}=3x^2+3xh+h^2+2.\]
Now let \(h\to0:\)
\[f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2+2\right)=3x^2+2.\]
Bayanin Amsa
Differentiate \(f(x)=x^3+2x\) from first principles.
By definition,
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.\]
Compute \(f(x+h):\)
\[f(x+h)=(x+h)^3+2(x+h)=x^3+3x^2h+3xh^2+h^3+2x+2h.\]
Then
\[f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h.\]
Divide by \(h\) (for \(h\neq0\)):
\[\frac{f(x+h)-f(x)}{h}=3x^2+3xh+h^2+2.\]
Now let \(h\to0:\)
\[f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2+2\right)=3x^2+2.\]
Tambaya 14 Rahoto
A car travelling at a velocity of 50kmh\(^{-1}\), covers a distance of 20km. If it was accelerating at 6kmh\(^{-1}\), calculate, correct to one decimal place, the time the car took to cover the distance.
Take consistent units of kilometres and hours. Initial velocity \(u=50\text{ km h}^{-1},\) distance \(s=20\text{ km},\) acceleration \(a=6\text{ km h}^{-2}.\)
Using \(s=ut+\tfrac12 at^2:\)
\[20=50t+\tfrac12(6)t^2\Rightarrow 20=50t+3t^2.\]
Rearrange into a standard quadratic in \(t:\)
\[3t^2+50t-20=0.\]
Apply the quadratic formula \(t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) with \(a=3,\ b=50,\ c=-20:\)
\[t=\frac{-50\pm\sqrt{2500+240}}{6}=\frac{-50\pm\sqrt{2740}}{6}.\]
Since \(\sqrt{2740}\approx52.35,\) the positive (physical) root is
\[t=\frac{-50+52.35}{6}=\frac{2.35}{6}\approx0.39\text{ h}.\]
\[t\approx0.4\text{ hours (to one decimal place)}.\]
(The negative root is rejected as time cannot be negative.)
Bayanin Amsa
Take consistent units of kilometres and hours. Initial velocity \(u=50\text{ km h}^{-1},\) distance \(s=20\text{ km},\) acceleration \(a=6\text{ km h}^{-2}.\)
Using \(s=ut+\tfrac12 at^2:\)
\[20=50t+\tfrac12(6)t^2\Rightarrow 20=50t+3t^2.\]
Rearrange into a standard quadratic in \(t:\)
\[3t^2+50t-20=0.\]
Apply the quadratic formula \(t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) with \(a=3,\ b=50,\ c=-20:\)
\[t=\frac{-50\pm\sqrt{2500+240}}{6}=\frac{-50\pm\sqrt{2740}}{6}.\]
Since \(\sqrt{2740}\approx52.35,\) the positive (physical) root is
\[t=\frac{-50+52.35}{6}=\frac{2.35}{6}\approx0.39\text{ h}.\]
\[t\approx0.4\text{ hours (to one decimal place)}.\]
(The negative root is rejected as time cannot be negative.)
Tambaya 15 Rahoto
If \(\sin A = \frac{3}{5}\) and \(\cos B = \frac{15}{17}\), where A is an obtuse angle and B is acute, find the value of \(\cos (A + B)\).
Given \(\sin A=\dfrac{3}{5}\) with A obtuse, and \(\cos B=\dfrac{15}{17}\) with B acute.
Find cos A. A is obtuse (second quadrant), so cosine is negative. Using \(\cos^2A=1-\sin^2A:\)
\[\cos A=-\sqrt{1-\tfrac{9}{25}}=-\sqrt{\tfrac{16}{25}}=-\frac{4}{5}.\]
Find sin B. B is acute, so sine is positive:
\[\sin B=\sqrt{1-\tfrac{225}{289}}=\sqrt{\tfrac{64}{289}}=\frac{8}{17}.\]
Apply the compound-angle formula.
\[\cos(A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\!\left(\frac{15}{17}\right)-\left(\frac{3}{5}\right)\!\left(\frac{8}{17}\right).\]
\[=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}.\]
Bayanin Amsa
Given \(\sin A=\dfrac{3}{5}\) with A obtuse, and \(\cos B=\dfrac{15}{17}\) with B acute.
Find cos A. A is obtuse (second quadrant), so cosine is negative. Using \(\cos^2A=1-\sin^2A:\)
\[\cos A=-\sqrt{1-\tfrac{9}{25}}=-\sqrt{\tfrac{16}{25}}=-\frac{4}{5}.\]
Find sin B. B is acute, so sine is positive:
\[\sin B=\sqrt{1-\tfrac{225}{289}}=\sqrt{\tfrac{64}{289}}=\frac{8}{17}.\]
Apply the compound-angle formula.
\[\cos(A+B)=\cos A\cos B-\sin A\sin B=\left(-\frac{4}{5}\right)\!\left(\frac{15}{17}\right)-\left(\frac{3}{5}\right)\!\left(\frac{8}{17}\right).\]
\[=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}.\]
Tambaya 16 Rahoto
(a) The point P(3, -5) is rotated through an angle 60° anticlockwise about the origin. (i) Obtain the matrix for the rotation ; (ii) Find the image P' of the point P under the rotation.
(b) A linear transformation is given by \(N : (x, y) \to (2x + 3y, 3x - y)\).
(i) Write down the matrix N of the transformation ; (ii) If \(N^{2} + aN + bI = 0\), where a, b \(\in\) R, \(I\) is the \(2 \times 2\) matrix and \(0\) is the \(2 \times 2\) null matrix, find the values of a and b.
(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.
\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]
(ii) Image of \(P(3,-5).\)
\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]
Numerically \(P'\approx(5.83,\ 0.10).\)
(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)
(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)
(ii) Find a and b in \(N^2+aN+bI=0.\)
\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]
Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)
Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)
Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)
Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.
\[a=-1,\qquad b=-11.\]
(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))
Bayanin Amsa
(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.
\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]
(ii) Image of \(P(3,-5).\)
\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]
Numerically \(P'\approx(5.83,\ 0.10).\)
(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)
(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)
(ii) Find a and b in \(N^2+aN+bI=0.\)
\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]
Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)
Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)
Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)
Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.
\[a=-1,\qquad b=-11.\]
(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))
Tambaya 17 Rahoto
In a hotel, the breakfast is a choice between yam (Y) or plantain (P) or both. The Venn diagram shows the choices made by 25 guests of the hotel.
(a) Find the value of x;
(b) What is the probability that a guest chosen at random chose only one of the two?
From the Venn diagram the three disjoint regions are:
- Plantain only (\(P\) region outside the overlap): \(2x+1\)
- Both yam and plantain (the overlap): \(x\)
- Yam only (\(Y\) region outside the overlap): \((x-2)^2\)
Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.
(a) Finding the value of \(x\)
\[ (2x+1) + x + (x-2)^2 = 25 \]Expand \((x-2)^2 = x^2 - 4x + 4\):
\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]Factorising:
\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]A number of guests cannot be negative, so we reject \(x = -4\).
\[ \boxed{x = 5} \](b) Probability that a guest chose only one of the two
Substitute \(x = 5\) into each region:
\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.
Guests who chose only one of the two are those in "plantain only" or "yam only":
\[ 11 + 9 = 20 \]Therefore the required probability is:
\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]Bayanin Amsa
From the Venn diagram the three disjoint regions are:
- Plantain only (\(P\) region outside the overlap): \(2x+1\)
- Both yam and plantain (the overlap): \(x\)
- Yam only (\(Y\) region outside the overlap): \((x-2)^2\)
Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.
(a) Finding the value of \(x\)
\[ (2x+1) + x + (x-2)^2 = 25 \]Expand \((x-2)^2 = x^2 - 4x + 4\):
\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]Factorising:
\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]A number of guests cannot be negative, so we reject \(x = -4\).
\[ \boxed{x = 5} \](b) Probability that a guest chose only one of the two
Substitute \(x = 5\) into each region:
\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.
Guests who chose only one of the two are those in "plantain only" or "yam only":
\[ 11 + 9 = 20 \]Therefore the required probability is:
\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]Tambaya 18 Rahoto
(a) An object is thrown up a smooth plane inclined at an angle of 30° to the horizontal. If the plane is 15m long and the object comes to rest at the top, find the :
(i) initial speed of the object ; (ii) time taken to reach the top.
(b) 
Force of magnitudes \(5 N, 5\sqrt{3} N, 10 N, 5\sqrt{3} N\) and \(5 N\) act on a body P, of mass 5 kg as shown in the diagram. Find the :
(i) magnitude of the resultant force ; (ii) acceleration of the body.
(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation
\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]
(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):
\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]
(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):
\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]
(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):
- \(5\ \text{N}\) at \(60^\circ\)
- \(5\sqrt3\ \text{N}\) at \(30^\circ\)
- \(10\ \text{N}\) at \(0^\circ\)
- \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
- \(5\ \text{N}\) at \(-60^\circ\)
(i) Magnitude of the resultant. Resolve horizontally:
\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]
\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]
Vertically the pairs cancel by symmetry:
\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]
Hence the resultant is
\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]
(ii) Acceleration of the body (mass \(5\ \text{kg}\)):
\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]
Bayanin Amsa
(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation
\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]
(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):
\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]
(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):
\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]
(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):
- \(5\ \text{N}\) at \(60^\circ\)
- \(5\sqrt3\ \text{N}\) at \(30^\circ\)
- \(10\ \text{N}\) at \(0^\circ\)
- \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
- \(5\ \text{N}\) at \(-60^\circ\)
(i) Magnitude of the resultant. Resolve horizontally:
\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]
\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]
Vertically the pairs cancel by symmetry:
\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]
Hence the resultant is
\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]
(ii) Acceleration of the body (mass \(5\ \text{kg}\)):
\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]
