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Question 1 Report
The position vectors of points A, B and C with respect to the origin are (8i - 2j), (2i + 6j) and (-10i + 4j) respectively. If ABCN is a parallelogram, find :
(a) the position vector of N;
(b) AN and AB ;
(c) correct to two decimal place, the acute angle between AN and AB.
Position vectors: \(A=(8,-2),\ B=(2,6),\ C=(-10,4)\).
(a) In parallelogram \(ABCN\) the diagonals \(AC\) and \(BN\) bisect each other, so they share a midpoint.
Midpoint of \(AC=\left(\tfrac{8-10}{2},\tfrac{-2+4}{2}\right)=(-1,1)\).
Set midpoint of \(BN=(-1,1)\): \(\tfrac{2+N_x}{2}=-1\Rightarrow N_x=-4\); \(\tfrac{6+N_y}{2}=1\Rightarrow N_y=-4\).
\[N=-4\mathbf{i}-4\mathbf{j}\]
(b) \(\overrightarrow{AN}=N-A=(-4-8,\,-4+2)=(-12,-2)\); \(\overrightarrow{AB}=B-A=(2-8,\,6+2)=(-6,8)\).
(c) \(\overrightarrow{AN}\cdot\overrightarrow{AB}=(-12)(-6)+(-2)(8)=72-16=56\).
\(|\overrightarrow{AN}|=\sqrt{144+4}=\sqrt{148}\approx12.17\); \(|\overrightarrow{AB}|=\sqrt{36+64}=10\).
\[\cos\theta=\frac{56}{10\sqrt{148}}=0.4603\Rightarrow \theta\approx62.59^{\circ}\]
Position vectors: \(A=(8,-2),\ B=(2,6),\ C=(-10,4)\).
(a) In parallelogram \(ABCN\) the diagonals \(AC\) and \(BN\) bisect each other, so they share a midpoint.
Midpoint of \(AC=\left(\tfrac{8-10}{2},\tfrac{-2+4}{2}\right)=(-1,1)\).
Set midpoint of \(BN=(-1,1)\): \(\tfrac{2+N_x}{2}=-1\Rightarrow N_x=-4\); \(\tfrac{6+N_y}{2}=1\Rightarrow N_y=-4\).
\[N=-4\mathbf{i}-4\mathbf{j}\]
(b) \(\overrightarrow{AN}=N-A=(-4-8,\,-4+2)=(-12,-2)\); \(\overrightarrow{AB}=B-A=(2-8,\,6+2)=(-6,8)\).
(c) \(\overrightarrow{AN}\cdot\overrightarrow{AB}=(-12)(-6)+(-2)(8)=72-16=56\).
\(|\overrightarrow{AN}|=\sqrt{144+4}=\sqrt{148}\approx12.17\); \(|\overrightarrow{AB}|=\sqrt{36+64}=10\).
\[\cos\theta=\frac{56}{10\sqrt{148}}=0.4603\Rightarrow \theta\approx62.59^{\circ}\]
Question 2 Report
The probabilities that Ali, Baba and Katty will gain admission to college are \(\frac{2}{3}, \frac{3}{4}\) and \(\frac{4}{5}\) respectively. Find the probability that:
(a) only Katty and Baba will gain admission ;
(b) none of them will gain admission ;
(c) at most two of them will gain admission.
Admission probabilities: Ali \(P(A)=\tfrac23\), Baba \(P(B)=\tfrac34\), Katty \(P(K)=\tfrac45\). Failures: \(P(A')=\tfrac13,\ P(B')=\tfrac14,\ P(K')=\tfrac15\). The three events are independent.
(a) Only Katty and Baba (so Ali fails):
\[P(A')\,P(B)\,P(K)=\tfrac13\times\tfrac34\times\tfrac45=\frac{12}{60}=\frac15\]
(b) None gains admission:
\[P(A')\,P(B')\,P(K')=\tfrac13\times\tfrac14\times\tfrac15=\frac{1}{60}\]
(c) At most two = not all three. So subtract the probability that all three gain admission:
\[1-P(A)\,P(B)\,P(K)=1-\left(\tfrac23\times\tfrac34\times\tfrac45\right)=1-\frac{24}{60}=1-\frac25=\frac35\]
Admission probabilities: Ali \(P(A)=\tfrac23\), Baba \(P(B)=\tfrac34\), Katty \(P(K)=\tfrac45\). Failures: \(P(A')=\tfrac13,\ P(B')=\tfrac14,\ P(K')=\tfrac15\). The three events are independent.
(a) Only Katty and Baba (so Ali fails):
\[P(A')\,P(B)\,P(K)=\tfrac13\times\tfrac34\times\tfrac45=\frac{12}{60}=\frac15\]
(b) None gains admission:
\[P(A')\,P(B')\,P(K')=\tfrac13\times\tfrac14\times\tfrac15=\frac{1}{60}\]
(c) At most two = not all three. So subtract the probability that all three gain admission:
\[1-P(A)\,P(B)\,P(K)=1-\left(\tfrac23\times\tfrac34\times\tfrac45\right)=1-\frac{24}{60}=1-\frac25=\frac35\]
Question 3 Report
(a) Using the substitution \(u = x - 2\), write \(\frac{x^{3} + 5}{(x - 2)^{4}}\) as an expression in terms of u.
(b) Using the answer in (a), express \(\frac{x^{3} + 5}{(x - 2)^{4}}\) in partial fractions.
(a) With \(u=x-2\), we have \(x=u+2\), so
\[x^{3}+5=(u+2)^{3}+5=u^{3}+6u^{2}+12u+8+5=u^{3}+6u^{2}+12u+13\]
Therefore
\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{u^{3}+6u^{2}+12u+13}{u^{4}}=\frac{1}{u}+\frac{6}{u^{2}}+\frac{12}{u^{3}}+\frac{13}{u^{4}}\]
(b) Replace \(u\) by \(x-2\):
\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{1}{x-2}+\frac{6}{(x-2)^{2}}+\frac{12}{(x-2)^{3}}+\frac{13}{(x-2)^{4}}\]
(a) With \(u=x-2\), we have \(x=u+2\), so
\[x^{3}+5=(u+2)^{3}+5=u^{3}+6u^{2}+12u+8+5=u^{3}+6u^{2}+12u+13\]
Therefore
\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{u^{3}+6u^{2}+12u+13}{u^{4}}=\frac{1}{u}+\frac{6}{u^{2}}+\frac{12}{u^{3}}+\frac{13}{u^{4}}\]
(b) Replace \(u\) by \(x-2\):
\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{1}{x-2}+\frac{6}{(x-2)^{2}}+\frac{12}{(x-2)^{3}}+\frac{13}{(x-2)^{4}}\]
Question 4 Report
Two panel of judges, X and Y, rank 8 brands of cooking oil as follows :
| Cooking oil type | A | B | C | D | E | F | G | H |
| X | 8 | 5 | 1 | 7 | 2 | 6 | 3 | 4 |
| Y | 6 | 3 | 4 | 8 | 5 | 7 | 1 | 2 |
Calculate the Spearmann's rank correlation coefficient.
The values given by X and Y are already ranks (1 to 8), so \(n = 8\). Form \(d = X - Y\) and \(d^2\).
| Oil | X | Y | \(d = X - Y\) | \(d^2\) |
|---|---|---|---|---|
| A | 8 | 6 | 2 | 4 |
| B | 5 | 3 | 2 | 4 |
| C | 1 | 4 | -3 | 9 |
| D | 7 | 8 | -1 | 1 |
| E | 2 | 5 | -3 | 9 |
| F | 6 | 7 | -1 | 1 |
| G | 3 | 1 | 2 | 4 |
| H | 4 | 2 | 2 | 4 |
| Total \(\sum d^2\) | 36 | |||
\(r_s \approx 0.57\) indicates a moderate positive agreement between the two panels of judges in ranking the brands of cooking oil.
The values given by X and Y are already ranks (1 to 8), so \(n = 8\). Form \(d = X - Y\) and \(d^2\).
| Oil | X | Y | \(d = X - Y\) | \(d^2\) |
|---|---|---|---|---|
| A | 8 | 6 | 2 | 4 |
| B | 5 | 3 | 2 | 4 |
| C | 1 | 4 | -3 | 9 |
| D | 7 | 8 | -1 | 1 |
| E | 2 | 5 | -3 | 9 |
| F | 6 | 7 | -1 | 1 |
| G | 3 | 1 | 2 | 4 |
| H | 4 | 2 | 2 | 4 |
| Total \(\sum d^2\) | 36 | |||
\(r_s \approx 0.57\) indicates a moderate positive agreement between the two panels of judges in ranking the brands of cooking oil.
Question 5 Report
Ten coins were tossed together a number of times. The distribution of the number of heads obtained is given in the following table :
| No of heads | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Frequency | 2 | 7 | 23 | 36 | 11 | 61 | 100 | 12 | 8 | 5 | 3 |
Calculate, correct to three decimal places, the :
(a) mean number of heads ;
(b) probability of getting an even head ;
(c) probability of getting an odd number.
Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(f\) | 2 | 7 | 23 | 36 | 11 | 61 | 100 | 12 | 8 | 5 | 3 | 268 |
| \(fx\) | 0 | 7 | 46 | 108 | 44 | 305 | 600 | 84 | 64 | 45 | 30 | 1333 |
(a) Mean number of heads.
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{1333}{268} = \mathbf{4.974} \](b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).
\[ P(\text{even}) = \frac{147}{268} = \mathbf{0.549} \](c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).
\[ P(\text{odd}) = \frac{121}{268} = \mathbf{0.451} \]Check: \(0.549 + 0.451 = 1.000\).
Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(f\) | 2 | 7 | 23 | 36 | 11 | 61 | 100 | 12 | 8 | 5 | 3 | 268 |
| \(fx\) | 0 | 7 | 46 | 108 | 44 | 305 | 600 | 84 | 64 | 45 | 30 | 1333 |
(a) Mean number of heads.
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{1333}{268} = \mathbf{4.974} \](b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).
\[ P(\text{even}) = \frac{147}{268} = \mathbf{0.549} \](c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).
\[ P(\text{odd}) = \frac{121}{268} = \mathbf{0.451} \]Check: \(0.549 + 0.451 = 1.000\).
Question 6 Report
(a) Solve : \(2^{3y + 2} - 7(2^{2y + 2}) - 31(2^{y}) - 8 = 0, y \in R\).
(b) Find \(\int (\sqrt{x^{2} + 1}) xdx\).
(a) Let \(u=2^{y}\). Then \(2^{3y+2}=4u^{3}\) and \(2^{2y+2}=4u^{2}\), so
\[4u^{3}-7(4u^{2})-31u-8=0\Rightarrow 4u^{3}-28u^{2}-31u-8=0\]
Testing \(u=8\): \(4(512)-28(64)-31(8)-8=2048-1792-248-8=0\). So \((u-8)\) is a factor:
\[(u-8)(4u^{2}+4u+1)=0\Rightarrow (u-8)(2u+1)^{2}=0\]
Thus \(u=8\) or \(u=-\tfrac12\). Since \(u=2^{y}>0\), reject \(u=-\tfrac12\). Then \(2^{y}=8=2^{3}\), so
\[\boxed{y=3}\]
(b) \(\displaystyle\int x\sqrt{x^{2}+1}\,dx\). Let \(w=x^{2}+1\), so \(dw=2x\,dx\Rightarrow x\,dx=\tfrac12 dw\).
\[\int\sqrt{w}\cdot\tfrac12\,dw=\tfrac12\cdot\tfrac{2}{3}w^{3/2}+C=\tfrac13\left(x^{2}+1\right)^{3/2}+C\]
(a) Let \(u=2^{y}\). Then \(2^{3y+2}=4u^{3}\) and \(2^{2y+2}=4u^{2}\), so
\[4u^{3}-7(4u^{2})-31u-8=0\Rightarrow 4u^{3}-28u^{2}-31u-8=0\]
Testing \(u=8\): \(4(512)-28(64)-31(8)-8=2048-1792-248-8=0\). So \((u-8)\) is a factor:
\[(u-8)(4u^{2}+4u+1)=0\Rightarrow (u-8)(2u+1)^{2}=0\]
Thus \(u=8\) or \(u=-\tfrac12\). Since \(u=2^{y}>0\), reject \(u=-\tfrac12\). Then \(2^{y}=8=2^{3}\), so
\[\boxed{y=3}\]
(b) \(\displaystyle\int x\sqrt{x^{2}+1}\,dx\). Let \(w=x^{2}+1\), so \(dw=2x\,dx\Rightarrow x\,dx=\tfrac12 dw\).
\[\int\sqrt{w}\cdot\tfrac12\,dw=\tfrac12\cdot\tfrac{2}{3}w^{3/2}+C=\tfrac13\left(x^{2}+1\right)^{3/2}+C\]
Question 7 Report
The sum of the first twelve terms of an Arithmetic Progression is 168. If the third term is 7, find the values of the common difference and the first term.
Let the first term be \(a\) and common difference \(d\).
Sum of first 12 terms is 168:
\[S_{12}=\frac{12}{2}\big(2a+11d\big)=168\Rightarrow6(2a+11d)=168\Rightarrow2a+11d=28\quad(1)\]
Third term is 7:
\[a+2d=7\quad(2)\]
From (2), \(a=7-2d\). Substitute into (1):
\[2(7-2d)+11d=28\Rightarrow14+7d=28\Rightarrow7d=14\Rightarrow d=2\]
Then \(a=7-2(2)=3\).
\[a=3,\qquad d=2\]
Let the first term be \(a\) and common difference \(d\).
Sum of first 12 terms is 168:
\[S_{12}=\frac{12}{2}\big(2a+11d\big)=168\Rightarrow6(2a+11d)=168\Rightarrow2a+11d=28\quad(1)\]
Third term is 7:
\[a+2d=7\quad(2)\]
From (2), \(a=7-2d\). Substitute into (1):
\[2(7-2d)+11d=28\Rightarrow14+7d=28\Rightarrow7d=14\Rightarrow d=2\]
Then \(a=7-2(2)=3\).
\[a=3,\qquad d=2\]
Question 8 Report
Given that \(m = 3i - 2j ; n = 2i - 3j\) and \(p = -i + 6j\), find \(4m + 2n - 3p\).
Given \(m=3i-2j,\ n=2i-3j,\ p=-i+6j\).
\[4m=12i-8j,\qquad 2n=4i-6j,\qquad 3p=-3i+18j\]
\[4m+2n-3p=(12i-8j)+(4i-6j)-(-3i+18j)\]
Collect \(i\) terms: \(12+4+3=19\). Collect \(j\) terms: \(-8-6-18=-32\).
\[4m+2n-3p=19i-32j\]
Given \(m=3i-2j,\ n=2i-3j,\ p=-i+6j\).
\[4m=12i-8j,\qquad 2n=4i-6j,\qquad 3p=-3i+18j\]
\[4m+2n-3p=(12i-8j)+(4i-6j)-(-3i+18j)\]
Collect \(i\) terms: \(12+4+3=19\). Collect \(j\) terms: \(-8-6-18=-32\).
\[4m+2n-3p=19i-32j\]
Question 9 Report
(a) The probability that Kunle solves a particular question is \(\frac{1}{3}\) while that of Tayo is \(\frac{1}{5}\). If both of them attempt the question, find the probability that only one of them will solve the question.
(b) A committee of 8 is to be chosen from 10 persons. In how many ways can this be done if there is no restriction?
(a) \(P(K)=\tfrac{1}{3}\Rightarrow P(K')=\tfrac{2}{3}\); \(P(T)=\tfrac{1}{5}\Rightarrow P(T')=\tfrac{4}{5}\). The events are independent.
Only one solves = (Kunle solves, Tayo fails) OR (Kunle fails, Tayo solves):
\[P=\left(\tfrac{1}{3}\times\tfrac{4}{5}\right)+\left(\tfrac{2}{3}\times\tfrac{1}{5}\right)=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\]
(b) Choosing 8 from 10 with no restriction:
\[\binom{10}{8}=\binom{10}{2}=\frac{10\times9}{2}=45\ \text{ways}\]
(a) \(P(K)=\tfrac{1}{3}\Rightarrow P(K')=\tfrac{2}{3}\); \(P(T)=\tfrac{1}{5}\Rightarrow P(T')=\tfrac{4}{5}\). The events are independent.
Only one solves = (Kunle solves, Tayo fails) OR (Kunle fails, Tayo solves):
\[P=\left(\tfrac{1}{3}\times\tfrac{4}{5}\right)+\left(\tfrac{2}{3}\times\tfrac{1}{5}\right)=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\]
(b) Choosing 8 from 10 with no restriction:
\[\binom{10}{8}=\binom{10}{2}=\frac{10\times9}{2}=45\ \text{ways}\]
Question 10 Report
A uniform beam, XY, 4m long and weighing 350N rests on two pivots P and Q. It is kept in equilibrium by weights of 80N attached at X and 1000N attached at a point between P and Q such that it is 0.6m from Q. If XP = 0.8m and PQ = 2.2m.
(a) calculate the reactions at P and Q ;
(b) if the 1000N weight is replaced with a 1200N weight, at what point from Q should it be placed in order to maintain the equilibrium.
Measure positions from \(X\): \(X=0,\ P=0.8\text{m},\ Q=0.8+2.2=3.0\text{m},\ Y=4.0\text{m}\). The uniform weight \(350\text{N}\) acts at the centre, \(2.0\text{m}\). The \(80\text{N}\) acts at \(X=0\); the \(1000\text{N}\) is \(0.6\text{m}\) from \(Q\), i.e. at \(2.4\text{m}\).
(a) Vertical equilibrium: \(R_P+R_Q=80+350+1000=1430\).
Take moments about \(P\) (anticlockwise positive):
\[80(0.8)-350(1.2)-1000(1.6)+R_Q(2.2)=0\]
\[64-420-1600+2.2R_Q=0\Rightarrow R_Q=\frac{1956}{2.2}=889.09\text{N}\]
Then \(R_P=1430-889.09=540.91\text{N}\).
(b) Replace with \(1200\text{N}\) at distance \(s\) from \(Q\) (position \(3.0-s\)). Keeping the reaction at \(P\) unchanged at \(540.91\text{N}\), take moments about \(Q\):
\[R_P(2.2)=80(3.0)+350(1.0)+1200\,s\]
\[540.91(2.2)=240+350+1200s\Rightarrow 1190=590+1200s\Rightarrow s=0.5\text{m}\]
So the \(1200\text{N}\) weight should be placed \(0.5\text{m}\) from \(Q\) (between \(P\) and \(Q\)).
Measure positions from \(X\): \(X=0,\ P=0.8\text{m},\ Q=0.8+2.2=3.0\text{m},\ Y=4.0\text{m}\). The uniform weight \(350\text{N}\) acts at the centre, \(2.0\text{m}\). The \(80\text{N}\) acts at \(X=0\); the \(1000\text{N}\) is \(0.6\text{m}\) from \(Q\), i.e. at \(2.4\text{m}\).
(a) Vertical equilibrium: \(R_P+R_Q=80+350+1000=1430\).
Take moments about \(P\) (anticlockwise positive):
\[80(0.8)-350(1.2)-1000(1.6)+R_Q(2.2)=0\]
\[64-420-1600+2.2R_Q=0\Rightarrow R_Q=\frac{1956}{2.2}=889.09\text{N}\]
Then \(R_P=1430-889.09=540.91\text{N}\).
(b) Replace with \(1200\text{N}\) at distance \(s\) from \(Q\) (position \(3.0-s\)). Keeping the reaction at \(P\) unchanged at \(540.91\text{N}\), take moments about \(Q\):
\[R_P(2.2)=80(3.0)+350(1.0)+1200\,s\]
\[540.91(2.2)=240+350+1200s\Rightarrow 1190=590+1200s\Rightarrow s=0.5\text{m}\]
So the \(1200\text{N}\) weight should be placed \(0.5\text{m}\) from \(Q\) (between \(P\) and \(Q\)).
Question 11 Report
A circle is drawn through the points (3, 2), (-1, -2) and (5, -4). Find the :
(a) coordinates of the centre of the circle ;
(b) radius of the circle ;
(c) equation of the circle.
Use the general circle \(x^2+y^2+2gx+2fy+c=0\), with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\). Substitute the three points.
\((3,2)\): \(6g+4f+c=-13\) (i)
\((-1,-2)\): \(-2g-4f+c=-5\) (ii)
\((5,-4)\): \(10g-8f+c=-41\) (iii)
(i)\(-\)(ii): \(8g+8f=-8\Rightarrow g+f=-1\). (iii)\(-\)(i): \(4g-12f=-28\Rightarrow g-3f=-7\).
Solving: \(f=\tfrac32,\ g=-\tfrac52\). From (ii): \(5-6+c=-5\Rightarrow c=-4\).
(a) Centre \((-g,-f)=\left(\tfrac52,\,-\tfrac32\right)\).
(b) Radius \(=\sqrt{g^2+f^2-c}=\sqrt{\tfrac{25}{4}+\tfrac94+4}=\sqrt{\tfrac{50}{4}}=\dfrac{5\sqrt2}{2}\approx3.54\).
(c) Equation:
\[x^2+y^2-5x+3y-4=0\]
(Check with \((3,2)\): \(9+4-15+6-4=0\).)
Use the general circle \(x^2+y^2+2gx+2fy+c=0\), with centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\). Substitute the three points.
\((3,2)\): \(6g+4f+c=-13\) (i)
\((-1,-2)\): \(-2g-4f+c=-5\) (ii)
\((5,-4)\): \(10g-8f+c=-41\) (iii)
(i)\(-\)(ii): \(8g+8f=-8\Rightarrow g+f=-1\). (iii)\(-\)(i): \(4g-12f=-28\Rightarrow g-3f=-7\).
Solving: \(f=\tfrac32,\ g=-\tfrac52\). From (ii): \(5-6+c=-5\Rightarrow c=-4\).
(a) Centre \((-g,-f)=\left(\tfrac52,\,-\tfrac32\right)\).
(b) Radius \(=\sqrt{g^2+f^2-c}=\sqrt{\tfrac{25}{4}+\tfrac94+4}=\sqrt{\tfrac{50}{4}}=\dfrac{5\sqrt2}{2}\approx3.54\).
(c) Equation:
\[x^2+y^2-5x+3y-4=0\]
(Check with \((3,2)\): \(9+4-15+6-4=0\).)
Question 12 Report
Given that \(\log_{3} x - 3\log_{x} 3 + 2 = 0\), find the values of x.
Note that \(\log_{x}3=\dfrac{1}{\log_{3}x}\). Let \(y=\log_{3}x\). The equation becomes
\[y-\frac{3}{y}+2=0\]
Multiply through by \(y\):
\[y^{2}+2y-3=0\Rightarrow(y+3)(y-1)=0\Rightarrow y=-3\ \text{or}\ y=1\]
Convert back with \(y=\log_{3}x\):
\[y=1\Rightarrow x=3^{1}=3,\qquad y=-3\Rightarrow x=3^{-3}=\frac{1}{27}\]
\[x=3\quad\text{or}\quad x=\frac{1}{27}\]
Note that \(\log_{x}3=\dfrac{1}{\log_{3}x}\). Let \(y=\log_{3}x\). The equation becomes
\[y-\frac{3}{y}+2=0\]
Multiply through by \(y\):
\[y^{2}+2y-3=0\Rightarrow(y+3)(y-1)=0\Rightarrow y=-3\ \text{or}\ y=1\]
Convert back with \(y=\log_{3}x\):
\[y=1\Rightarrow x=3^{1}=3,\qquad y=-3\Rightarrow x=3^{-3}=\frac{1}{27}\]
\[x=3\quad\text{or}\quad x=\frac{1}{27}\]
Question 13 Report
A body of mass 20kg moving with a velocity of 80ms\(^{-1}\) collides with another body of mass 30kg moving with a velocity of 50ms\(^{-1}\). If they both moved in the same direction after collision, find their common velocity if they moved in the :
(a) same direction before collision ; (b) opposite direction before collision.
By conservation of linear momentum, total momentum before = total momentum after. After collision they move together with common velocity \(v\); total mass \(=20+30=50\,\text{kg}\).
(a) Same direction before collision. Take both velocities as positive:
\[20(80)+30(50)=50v\]
\[1600+1500=50v\Rightarrow3100=50v\Rightarrow v=62\ \text{m/s}\]
(b) Opposite directions before collision. Take the second body's velocity as negative:
\[20(80)+30(-50)=50v\]
\[1600-1500=50v\Rightarrow100=50v\Rightarrow v=2\ \text{m/s}\]
The positive result shows the combined body moves in the direction of the \(20\,\text{kg}\) body.
By conservation of linear momentum, total momentum before = total momentum after. After collision they move together with common velocity \(v\); total mass \(=20+30=50\,\text{kg}\).
(a) Same direction before collision. Take both velocities as positive:
\[20(80)+30(50)=50v\]
\[1600+1500=50v\Rightarrow3100=50v\Rightarrow v=62\ \text{m/s}\]
(b) Opposite directions before collision. Take the second body's velocity as negative:
\[20(80)+30(-50)=50v\]
\[1600-1500=50v\Rightarrow100=50v\Rightarrow v=2\ \text{m/s}\]
The positive result shows the combined body moves in the direction of the \(20\,\text{kg}\) body.
Question 14 Report
If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24 \), find the values of x.
Expand the determinant along the first row.
\[\Delta=(x-3)\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}-(-4)\begin{vmatrix}5&2\\2&6-x\end{vmatrix}+3\begin{vmatrix}5&2\\2&-4\end{vmatrix}\]
The \(2\times2\) minors:
\(\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}=2(6-x)+8=20-2x\).
\(\begin{vmatrix}5&2\\2&6-x\end{vmatrix}=5(6-x)-4=26-5x\).
\(\begin{vmatrix}5&2\\2&-4\end{vmatrix}=-20-4=-24\).
So
\[\Delta=(x-3)(20-2x)+4(26-5x)+3(-24)\]
\[=(-2x^{2}+26x-60)+(104-20x)-72=-2x^{2}+6x-28\]
Set \(\Delta=-24\):
\[-2x^{2}+6x-28=-24\Rightarrow-2x^{2}+6x-4=0\Rightarrow x^{2}-3x+2=0\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2\]
Expand the determinant along the first row.
\[\Delta=(x-3)\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}-(-4)\begin{vmatrix}5&2\\2&6-x\end{vmatrix}+3\begin{vmatrix}5&2\\2&-4\end{vmatrix}\]
The \(2\times2\) minors:
\(\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}=2(6-x)+8=20-2x\).
\(\begin{vmatrix}5&2\\2&6-x\end{vmatrix}=5(6-x)-4=26-5x\).
\(\begin{vmatrix}5&2\\2&-4\end{vmatrix}=-20-4=-24\).
So
\[\Delta=(x-3)(20-2x)+4(26-5x)+3(-24)\]
\[=(-2x^{2}+26x-60)+(104-20x)-72=-2x^{2}+6x-28\]
Set \(\Delta=-24\):
\[-2x^{2}+6x-28=-24\Rightarrow-2x^{2}+6x-4=0\Rightarrow x^{2}-3x+2=0\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2\]
Question 15 Report
(a)(i) Write down the binomial expansion of \((2 - \frac{1}{2}x)^{5}\) in ascending powers of x.
(ii) Using the expansion in (a)(i), find, correct to two decimal places, the value of \((1.99)^{5}\).
(b) The polynomial \(x^{3} + qx^{2} + rx + 9\), where q and r are constants, has (x + 1) as a factor and has a remainder -17 when divided by (x + 2). Find the values of q and r.
(a)(i) Expand \(\left(2-\tfrac12 x\right)^{5}\) using \(\binom{5}{k}2^{5-k}\left(-\tfrac12 x\right)^{k}\):
\[32-40x+20x^{2}-5x^{3}+\tfrac58 x^{4}-\tfrac{1}{32}x^{5}\]
(ii) Put \(2-\tfrac12 x=1.99\Rightarrow \tfrac12 x=0.01\Rightarrow x=0.02\). Substitute:
\(32-40(0.02)+20(0.02)^{2}-5(0.02)^{3}+\cdots\)
\(=32-0.8+0.008-0.00004+\cdots\approx31.20796\).
\[(1.99)^{5}\approx31.21\ \text{(2 d.p.)}\]
(b) Let \(P(x)=x^{3}+qx^{2}+rx+9\).
\((x+1)\) is a factor \(\Rightarrow P(-1)=0\): \(-1+q-r+9=0\Rightarrow q-r=-8\).
Remainder \(-17\) on division by \((x+2)\Rightarrow P(-2)=-17\): \(-8+4q-2r+9=-17\Rightarrow 4q-2r=-18\Rightarrow 2q-r=-9\).
Subtracting: \((2q-r)-(q-r)=-9-(-8)\Rightarrow q=-1\), then \(r=q+8=7\).
\[\boxed{q=-1,\ r=7}\]
(a)(i) Expand \(\left(2-\tfrac12 x\right)^{5}\) using \(\binom{5}{k}2^{5-k}\left(-\tfrac12 x\right)^{k}\):
\[32-40x+20x^{2}-5x^{3}+\tfrac58 x^{4}-\tfrac{1}{32}x^{5}\]
(ii) Put \(2-\tfrac12 x=1.99\Rightarrow \tfrac12 x=0.01\Rightarrow x=0.02\). Substitute:
\(32-40(0.02)+20(0.02)^{2}-5(0.02)^{3}+\cdots\)
\(=32-0.8+0.008-0.00004+\cdots\approx31.20796\).
\[(1.99)^{5}\approx31.21\ \text{(2 d.p.)}\]
(b) Let \(P(x)=x^{3}+qx^{2}+rx+9\).
\((x+1)\) is a factor \(\Rightarrow P(-1)=0\): \(-1+q-r+9=0\Rightarrow q-r=-8\).
Remainder \(-17\) on division by \((x+2)\Rightarrow P(-2)=-17\): \(-8+4q-2r+9=-17\Rightarrow 4q-2r=-18\Rightarrow 2q-r=-9\).
Subtracting: \((2q-r)-(q-r)=-9-(-8)\Rightarrow q=-1\), then \(r=q+8=7\).
\[\boxed{q=-1,\ r=7}\]
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