ဒီေနရာမွာ တင္ေပးခဲ့ေသာ Section (B) ေမးခြန္းရဲ့ အေျဖျဖစ္ပါတယ္။ Section (A) ရဲ့ အေျဖကိုေတာ့ ဒီေနရာမွာ တင္ေပးခဲ့ပါတယ္။ Section (C) အေျဖကိုလည္း ဆက္လက္ တင္ေပးသြားပါ့မယ္။
Section (B)
Solution
6. (a) A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto \frac{x}{p}+q,p\ne 0$. If $ \displaystyle f(8)=1$ and$ \displaystyle {{f}^{{-1}}}(-2)=2$, show that $ \displaystyle \frac{p}{2}+{{q}^{2}}=10$.
(5 marks)
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$ \displaystyle \ \ \ \ \ f(x)=\frac{x}{p}+q,p\ne 0$
$ \displaystyle \ \ \ \ \ f(8)=1,$
$ \displaystyle \therefore \ \ \ \frac{8}{p}+q=1\,\ \ \ \ \ \ \ -----(1)$
$ \displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ {{f}^{{-1}}}(-2)=2} \\ {} \\ {\therefore \ \ \ f(2)=-2} \end{array}$
$ \displaystyle \therefore \ \ \ \frac{2}{p}+q=-2\ \ \ \ \ \ \ \ ------(2)$
$ \displaystyle \text{By Equation(1) }-\text{ Equation(2)}$
$ \displaystyle \ \ \ \ \ \frac{6}{p}=3\Rightarrow p=2$
$ \displaystyle \therefore \ \ \ \frac{8}{2}+q=1\Rightarrow q=-3$
$ \displaystyle \therefore \ \ \ \frac{p}{2}+{{q}^{2}}=\ \frac{2}{2}+{{(-3)}^{2}}=10$
$ \displaystyle \ \ \ \ \ f(8)=1,$
$ \displaystyle \therefore \ \ \ \frac{8}{p}+q=1\,\ \ \ \ \ \ \ -----(1)$
$ \displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ {{f}^{{-1}}}(-2)=2} \\ {} \\ {\therefore \ \ \ f(2)=-2} \end{array}$
$ \displaystyle \therefore \ \ \ \frac{2}{p}+q=-2\ \ \ \ \ \ \ \ ------(2)$
$ \displaystyle \text{By Equation(1) }-\text{ Equation(2)}$
$ \displaystyle \ \ \ \ \ \frac{6}{p}=3\Rightarrow p=2$
$ \displaystyle \therefore \ \ \ \frac{8}{2}+q=1\Rightarrow q=-3$
$ \displaystyle \therefore \ \ \ \frac{p}{2}+{{q}^{2}}=\ \frac{2}{2}+{{(-3)}^{2}}=10$
6. (b) Given that the $ \displaystyle (p + 1)^{\text{th}}$ term of an A.P. is twice the $ \displaystyle (q + 1)^{\text{th}}$ term. Prove that$ \displaystyle (3p + 1)^{\text{th}}$ term is twice the $ \displaystyle (p +q+ 1)^{\text{th}}$ term.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the first term be }a\ \text{and }\\\ \ \ \ \text{the common difference be}\ d.\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ {{u}_{{p+1}}}=2{{u}_{{q+1}}}\\\\\therefore \ \ a+(p+1-1)d=2\left[ {a+(q+1-1)d} \right]\\\\\ \ \ \ a+pd=2a+2qd\\\\\therefore \ \ a=pd-2qd\\\\\therefore \ \ {{u}_{{3p+1}}}=a+(3p+1-1)d\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =pd-2qd+3pd\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4pd-2qd\\\\\therefore \ \ 2{{u}_{{p+q+1}}}=2a+2(p+q+1-1)d\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2pd-4qd+2pd+2qd\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4pd-2qd\\\\\therefore \ \ {{u}_{{3p+1}}}=2{{u}_{{p+q+1}}}\end{array}$
7. (a) Find the term in $ \displaystyle x^2$ and $ \displaystyle x^3$ in the expansion of $ \displaystyle(2x+1)^5$. Hence find the term in $ \displaystyle x^3$ in the expansion of $ \displaystyle (x+3)(2x+1)^5$.
(5 marks)
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$\displaystyle \begin{array}{l}\text{Binomial :}\ {{(2x+1)}^{5}}\\\\{{(r+1)}^{{\text{th}}}}\text{term}={}^{5}{{C}_{r}}{{2}^{{5-r}}}{{x}^{{5-r}}}\\\\\text{For }{{x}^{2}},5-r=2\Rightarrow r=3\\\\\therefore \text{the term in }{{x}^{2}}={}^{5}{{C}_{3}}{{2}^{2}}{{x}^{2}}=40{{x}^{2}}\\\\\text{For }{{x}^{3}},5-r=3\Rightarrow r=2\\\\\therefore \text{the term in }{{x}^{3}}={}^{5}{{C}_{2}}{{2}^{3}}{{x}^{3}}=80{{x}^{3}}\\\\\therefore \text{the term in }{{x}^{3}}\text{ in the expansion of}\\\ \ (x+3){{(2x+1)}^{5}}=x(40{{x}^{2}})+3(80{{x}^{3}})=280{{x}^{3}}\end{array}$
7. (b) Let $ \displaystyle {{\text{J}}^{+}}$ be the set of all positive integers. Is the operation $ \displaystyle \odot $ defined by $ \displaystyle x\odot y=x^2+3y$ a binary operation on $ \displaystyle {{\text{J}}^{+}}$? If it is a binary operation, solve the equation $ \displaystyle \left( {k\odot 5} \right)-\left( {3\odot k} \right)=3k+1$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ x\odot y={{x}^{2}}+3y,x,y\in {{\text{J}}^{+}}\\\\\therefore {{x}^{2}}\in {{\text{J}}^{+}}\ \text{and }3y\in {{\text{J}}^{+}}\\\\\therefore {{x}^{2}}+3y\in {{\text{J}}^{+}}\\\\\therefore x\odot y\in {{\text{J}}^{+}}\\\\\therefore \text{Closure property is satisfied}\text{.}\\\\\therefore \odot \ \text{is a binary operation}\text{.}\\\\\ \ \ \left( {k\odot 5} \right)-\left( {3\odot k} \right)=3k+1\\\\\ \ \ \left( {{{k}^{2}}+15} \right)-\left( {9+3k} \right)=3k+1\\\\\ \ \ {{k}^{2}}-3k+6=3k+1\\\\\ \ \ {{k}^{2}}-6k+5=0\\\\\ \ \ (k-1)(k-5)=0\\\\\therefore k=1\ \text{or}\ k=5\end{array}$
8. (a) If $ \displaystyle k+4, k$ and $ \displaystyle 2k-15$ where $ \displaystyle k>0$ are the first three terms of a geometric progression, find the value of $ \displaystyle k$. Hence find the first term and the common ratio and determine whether the sum to infinity exists or not. Find the sum to infinity of the progression if exists.
(5 marks)
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$ \displaystyle \ \ \ k+4,\ k,\ 2k-15\text{ is a G}\text{.P}\text{.}$
$ \displaystyle \therefore \frac{k}{{k+4}}=\frac{{2k-15}}{k}$
$ \displaystyle \begin{array}{l}\therefore 2{{k}^{2}}-7k-60={{k}^{2}}\\\\\therefore {{k}^{2}}-7k-60=0\\\\\therefore (k+5)(k-12)=0\\\\\therefore k=-5\ \,\text{or}\ k=12\\\\\ \ \ \text{Since }k>0,\ k=-5\ \text{is impossible}\text{.}\\\\\therefore k=12.\\\\\therefore \text{The terms of the G}\text{.P}\text{. are 16, 12, 9}...\text{ }\text{.}\end{array}$
$ \displaystyle \therefore a=16,\ r=\frac{{12}}{{16}}=\frac{3}{4}$
$ \displaystyle \therefore \left| r \right|=\left| {\frac{3}{4}} \right|=\frac{3}{4}<1.$
$ \displaystyle \therefore \text{The sum to infinity exists}\text{.}$
$ \displaystyle \therefore S=\frac{a}{{1-r}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{16}}{{1-\frac{3}{4}}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =64$
$ \displaystyle \therefore \frac{k}{{k+4}}=\frac{{2k-15}}{k}$
$ \displaystyle \begin{array}{l}\therefore 2{{k}^{2}}-7k-60={{k}^{2}}\\\\\therefore {{k}^{2}}-7k-60=0\\\\\therefore (k+5)(k-12)=0\\\\\therefore k=-5\ \,\text{or}\ k=12\\\\\ \ \ \text{Since }k>0,\ k=-5\ \text{is impossible}\text{.}\\\\\therefore k=12.\\\\\therefore \text{The terms of the G}\text{.P}\text{. are 16, 12, 9}...\text{ }\text{.}\end{array}$
$ \displaystyle \therefore a=16,\ r=\frac{{12}}{{16}}=\frac{3}{4}$
$ \displaystyle \therefore \left| r \right|=\left| {\frac{3}{4}} \right|=\frac{3}{4}<1.$
$ \displaystyle \therefore \text{The sum to infinity exists}\text{.}$
$ \displaystyle \therefore S=\frac{a}{{1-r}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{16}}{{1-\frac{3}{4}}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =64$
8. (b) Find the solution set in $\displaystyle \text{R}$ of the in equations $\displaystyle (x+2)^2>2x+7$ and illustrate it on the number line.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ {{(x+2)}^{2}}>2x+7\\\\\ \ \ \ {{x}^{2}}+4x+4>2x+7\\\\\ \ \ \ {{x}^{2}}+2x-3>0\\\\\ \ \ \ \text{Let }y={{x}^{2}}+2x-3\ \\\\\ \ \ \ \text{When }y=0,\ {{x}^{2}}+2x-3=0.\\\\\ \ \ \ (x+3)(x-1)=0\\\\\therefore \ \ x=-3\ \text{or}\ x=1\\\\\therefore \ \ \text{The graph cuts the }x\ \text{axis at (}-3,0)\ \text{and}\ \text{(}1,0).\\\\\ \ \ \ \text{When }x=0,\ y=-3.\\\\\therefore \ \ \text{The graph cuts the }y\ \text{axis at (}0,-3).\end{array}$
$ \displaystyle \therefore \ \ \text{Solution set}=\{x|x<-3\ \text{or}\ x>1\}.$
Number Line
$ \displaystyle \therefore \ \ \text{Solution set}=\{x|x<-3\ \text{or}\ x>1\}.$
Number Line
9. (a) Given that when $ \displaystyle f(x)=6x^3+3x^2+ax+b$, where $ \displaystyle a$ and $ \displaystyle b$ are constants, is divided by $ \displaystyle (x+1)$ the remainder is $ \displaystyle 45$, show that $ \displaystyle b - a=48$. Given also that $ \displaystyle (2x+1)$ is a factor of $ \displaystyle f(x)$, find the value of $ \displaystyle a$ and the of $ \displaystyle b$. Hence factorise $ \displaystyle f(x)$ completely.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ f(x)=6{{x}^{3}}+3{{x}^{2}}+ax+b\\\\\ \ \ \text{When }f(x)\ \text{is divided by }x+1,\text{the remainder is }45.\\\\\therefore f(-1)=45\\\\\therefore 6{{(-1)}^{3}}+3{{(-1)}^{2}}+a(-1)+b=45\\\\\therefore -6+3-a+b=45\\\\\therefore b-a=48\ \ \ \ \ \ \ \ \ \ \ --------(1)\\\\\ \ \ \text{Since }2x+1\ \text{is a factor of }f(x),\ \end{array}$
$ \displaystyle \therefore f\left( {-\frac{1}{2}} \right)=0$
$ \displaystyle \therefore 6{{\left( {-\frac{1}{2}} \right)}^{3}}+3{{\left( {-\frac{1}{2}} \right)}^{2}}+a\left( {-\frac{1}{2}} \right)+b=0$
$ \displaystyle \ \ \ -\frac{3}{4}+\frac{3}{4}-\frac{a}{2}+b=0$
$ \displaystyle \begin{array}{l}\therefore \ 2b-a=0\ \ \ \ \ \ \ \ \ \ --------(2)\\\\\ \ \ \text{By Equation (2) }-\text{ Equation(1), }b=-48.\\\\\therefore -48-a=48\\\\\therefore a=-96\\\\\therefore f(x)=6{{x}^{3}}+3{{x}^{2}}-96x-48\end{array}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3{{x}^{2}}\ \ \ \ \ \ \ \ \ \ \ -48\\2x+1\overline{\left){\begin{array}{l}6{{x}^{3}}+3{{x}^{2}}-96x-48\\\underline{{6{{x}^{3}}+3{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ -96x-48\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{-96x-48}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$
$ \displaystyle \begin{array}{l}\therefore f(x)=(2x+1)(3{{x}^{2}}-48)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)({{x}^{2}}-16)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)(x+4)(x-4)\ \ \end{array}$
$ \displaystyle \therefore f\left( {-\frac{1}{2}} \right)=0$
$ \displaystyle \therefore 6{{\left( {-\frac{1}{2}} \right)}^{3}}+3{{\left( {-\frac{1}{2}} \right)}^{2}}+a\left( {-\frac{1}{2}} \right)+b=0$
$ \displaystyle \ \ \ -\frac{3}{4}+\frac{3}{4}-\frac{a}{2}+b=0$
$ \displaystyle \begin{array}{l}\therefore \ 2b-a=0\ \ \ \ \ \ \ \ \ \ --------(2)\\\\\ \ \ \text{By Equation (2) }-\text{ Equation(1), }b=-48.\\\\\therefore -48-a=48\\\\\therefore a=-96\\\\\therefore f(x)=6{{x}^{3}}+3{{x}^{2}}-96x-48\end{array}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3{{x}^{2}}\ \ \ \ \ \ \ \ \ \ \ -48\\2x+1\overline{\left){\begin{array}{l}6{{x}^{3}}+3{{x}^{2}}-96x-48\\\underline{{6{{x}^{3}}+3{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ -96x-48\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{-96x-48}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$
$ \displaystyle \begin{array}{l}\therefore f(x)=(2x+1)(3{{x}^{2}}-48)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)({{x}^{2}}-16)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)(x+4)(x-4)\ \ \end{array}$
9. (b) If $ \displaystyle A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)$ show that $ \displaystyle A+{{A}^{{-1}}}-2I=O$ where $\displaystyle I$ is a unit matrix of order 2.
(5 marks)
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$ \displaystyle \ \ \ \ \ \ \ \ A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)$
$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ \det A=-8-(-9)=1\ne 0,\\\\\therefore \ \ \ \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\end{array}$
$ \displaystyle \therefore \ \ \ \ \ \ {{A}^{{-1}}}=\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)$
$ \displaystyle \therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)-2\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 2 \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2+4-2} & {3-3} \\ {-3+3} & {4-2-2} \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)$
$ \displaystyle \begin{array}{l}\ \ \ \ =\ \ O\\\\\therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I=O\end{array}$
$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ \det A=-8-(-9)=1\ne 0,\\\\\therefore \ \ \ \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\end{array}$
$ \displaystyle \therefore \ \ \ \ \ \ {{A}^{{-1}}}=\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)$
$ \displaystyle \therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)-2\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 2 \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2+4-2} & {3-3} \\ {-3+3} & {4-2-2} \end{array}} \right)$
$ \displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)$
$ \displaystyle \begin{array}{l}\ \ \ \ =\ \ O\\\\\therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I=O\end{array}$
10. (a) The matrices $ \displaystyle P,Q$ and $ \displaystyle R$ such that $ \displaystyle P=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right)$, $ \displaystyle Q=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$ and $ \displaystyle R=PQ$. Verify that $ \displaystyle {{Q}^{{-1}}}{{P}^{{-1}}}={{R}^{{-1}}}$.
(5 marks)
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$ \displaystyle \ \ \ P=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right),Q=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$
$ \displaystyle \ \ \ R=PQ\ \ \ \ \ \left[ {\text{given}} \right]$
$ \displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$
$ \displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} {-2+2} & {0+1} \\ {-3+4} & {0+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 0 & 1 \\ 1 & 2 \end{array}} \right)$
$ \displaystyle \begin{array}{l}\ \ \ \det P=4-3=1\ne 0,\ \text{hence }{{P}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det Q=-1-0=-1\ne 0,\ \text{hence }{{Q}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det R=0-1=-1\ne 0,\ \text{hence }{{R}^{{-1}}}\text{exists}\text{.}\end{array}$
$ \displaystyle \therefore {{P}^{{-1}}}=\frac{1}{{\det P}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\frac{1}{1}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$
$ \displaystyle \ \ \ {{Q}^{{-1}}}=\frac{1}{{\det Q}}\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ {-2} & {-1} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$
$ \displaystyle \ \ \ {{R}^{{-1}}}=\frac{1}{{\det R}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-2+0} & {1+0} \\ {4-3} & {-2+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}={{R}^{{-1}}}$
$ \displaystyle \ \ \ R=PQ\ \ \ \ \ \left[ {\text{given}} \right]$
$ \displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$
$ \displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} {-2+2} & {0+1} \\ {-3+4} & {0+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 0 & 1 \\ 1 & 2 \end{array}} \right)$
$ \displaystyle \begin{array}{l}\ \ \ \det P=4-3=1\ne 0,\ \text{hence }{{P}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det Q=-1-0=-1\ne 0,\ \text{hence }{{Q}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det R=0-1=-1\ne 0,\ \text{hence }{{R}^{{-1}}}\text{exists}\text{.}\end{array}$
$ \displaystyle \therefore {{P}^{{-1}}}=\frac{1}{{\det P}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\frac{1}{1}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$
$ \displaystyle \ \ \ {{Q}^{{-1}}}=\frac{1}{{\det Q}}\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ {-2} & {-1} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$
$ \displaystyle \ \ \ {{R}^{{-1}}}=\frac{1}{{\det R}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-2+0} & {1+0} \\ {4-3} & {-2+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$
$ \displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}={{R}^{{-1}}}$
10. (b) The probability that a student will receive an $ \displaystyle A, B, C$ or $ \displaystyle D$ grade are 0.3, 0.38, 0.22 and 0.1 respectively. What is the probability that student will receive
(i) at least $ \displaystyle B$ grade.
(ii) at most $ \displaystyle C$ grade.
(iii) not an $ \displaystyle A$ grade.
(iv) $ \displaystyle B$ or $ \displaystyle C$ grade.
(i) at least $ \displaystyle B$ grade.
(ii) at most $ \displaystyle C$ grade.
(iii) not an $ \displaystyle A$ grade.
(iv) $ \displaystyle B$ or $ \displaystyle C$ grade.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ P(A\ \text{grade})=0.3\\\\\ \ \ \ \ \ P(B\ \text{grade})=0.38\\\\\ \ \ \ \ \ P(C\ \text{grade})=0.22\\\\\ \ \ \ \ \ P(D\ \text{grade})=0.1\\\\\text{(i)}\ \ \ P(\text{at least}\ B\ \text{grade})=P(A\ \text{grade or}\ B\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =P(A\ \text{grade) +}\ P(B\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.3+0.38\\\ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.68\\\\\text{(ii)}\ \ P(\text{at most}\ C\ \text{grade})=P(C\ \text{grade or}\ D\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =P(C\ \text{grade) +}\ P(D\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.22+0.1\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.32\\\\\text{(iii)}\ P(\text{not an }A\ \text{grade})=1-P(A\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-0.3\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.7\\\\\text{(iv)}\ P(B\text{ or}\ C\ \text{grade})=P(B\ \text{grade) +}\ P(C\ \text{grade})\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.38+0.22\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.6\end{array}$
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