Answer for 2019 Sample Question : Section (B)

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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       $\displaystyle \ \ \ \ \ f(x)=\frac{x}{p}+q,p\ne 0$

       $\displaystyle \ \ \ \ \ f(8)=1,$

       $\displaystyle \therefore \ \ \ \frac{8}{p}+q=1\,\ \ \ \ \ \ \ -----(1)$

       $\displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ {{f}^{{-1}}}(-2)=2} \\ {} \\ {\therefore \ \ \ f(2)=-2} \end{array}$

       $\displaystyle \therefore \ \ \ \frac{2}{p}+q=-2\ \ \ \ \ \ \ \ ------(2)$

       $\displaystyle \text{By Equation(1) }-\text{ Equation(2)}$

       $\displaystyle \ \ \ \ \ \frac{6}{p}=3\Rightarrow p=2$

       $\displaystyle \therefore \ \ \ \frac{8}{2}+q=1\Rightarrow q=-3$

       $\displaystyle \therefore \ \ \ \frac{p}{2}+{{q}^{2}}=\ \frac{2}{2}+{{(-3)}^{2}}=10$

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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$\displaystyle \ \ \ k+4,\ k,\ 2k-15\text{ is a G}\text{.P}\text{.}$

$\displaystyle \therefore \frac{k}{{k+4}}=\frac{{2k-15}}{k}$

$\displaystyle \begin{array}{l}\therefore 2{{k}^{2}}-7k-60={{k}^{2}}\\\\\therefore {{k}^{2}}-7k-60=0\\\\\therefore (k+5)(k-12)=0\\\\\therefore k=-5\ \,\text{or}\ k=12\\\\\ \ \ \text{Since }k>0,\ k=-5\ \text{is impossible}\text{.}\\\\\therefore k=12.\\\\\therefore \text{The terms of the G}\text{.P}\text{. are 16, 12, 9}...\text{ }\text{.}\end{array}$

$\displaystyle \therefore a=16,\ r=\frac{{12}}{{16}}=\frac{3}{4}$

$\displaystyle \therefore \left| r \right|=\left| {\frac{3}{4}} \right|=\frac{3}{4}<1.$

$\displaystyle \therefore \text{The sum to infinity exists}\text{.}$

$\displaystyle \therefore S=\frac{a}{{1-r}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =\frac{{16}}{{1-\frac{3}{4}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =64$

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

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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$\displaystyle \begin{array}{l}\ \ \ f(x)=6{{x}^{3}}+3{{x}^{2}}+ax+b\\\\\ \ \ \text{When }f(x)\ \text{is divided by }x+1,\text{the remainder is }45.\\\\\therefore f(-1)=45\\\\\therefore 6{{(-1)}^{3}}+3{{(-1)}^{2}}+a(-1)+b=45\\\\\therefore -6+3-a+b=45\\\\\therefore b-a=48\ \ \ \ \ \ \ \ \ \ \ --------(1)\\\\\ \ \ \text{Since }2x+1\ \text{is a factor of }f(x),\ \end{array}$

$\displaystyle \therefore f\left( {-\frac{1}{2}} \right)=0$

$\displaystyle \therefore 6{{\left( {-\frac{1}{2}} \right)}^{3}}+3{{\left( {-\frac{1}{2}} \right)}^{2}}+a\left( {-\frac{1}{2}} \right)+b=0$

$\displaystyle \ \ \ -\frac{3}{4}+\frac{3}{4}-\frac{a}{2}+b=0$

$\displaystyle \begin{array}{l}\therefore \ 2b-a=0\ \ \ \ \ \ \ \ \ \ --------(2)\\\\\ \ \ \text{By Equation (2) }-\text{ Equation(1), }b=-48.\\\\\therefore -48-a=48\\\\\therefore a=-96\\\\\therefore f(x)=6{{x}^{3}}+3{{x}^{2}}-96x-48\end{array}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3{{x}^{2}}\ \ \ \ \ \ \ \ \ \ \ -48\\2x+1\overline{\left){\begin{array}{l}6{{x}^{3}}+3{{x}^{2}}-96x-48\\\underline{{6{{x}^{3}}+3{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ -96x-48\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{-96x-48}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$\displaystyle \begin{array}{l}\therefore f(x)=(2x+1)(3{{x}^{2}}-48)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)({{x}^{2}}-16)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =3(2x+1)(x+4)(x-4)\ \ \end{array}$

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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$\displaystyle \ \ \ \ \ \ \ \ A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)$

$\displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ \det A=-8-(-9)=1\ne 0,\\\\\therefore \ \ \ \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\end{array}$

$\displaystyle \therefore \ \ \ \ \ \ {{A}^{{-1}}}=\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)$

$\displaystyle \therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I$

$\displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)-2\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)$

$\displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 4 & {-3} \\ 3 & {-2} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 2 \end{array}} \right)$

$\displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} {-2+4-2} & {3-3} \\ {-3+3} & {4-2-2} \end{array}} \right)$

$\displaystyle \ \ \ \ =\ \ \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)$

$\displaystyle \begin{array}{l}\ \ \ \ =\ \ O\\\\\therefore \ \ \ \ \ \ A+{{A}^{{-1}}}-2I=O\end{array}$

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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$\displaystyle \ \ \ P=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right),Q=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$

$\displaystyle \ \ \ R=PQ\ \ \ \ \ \left[ {\text{given}} \right]$

$\displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$

$\displaystyle \therefore R=\left( {\begin{array}{*{20}{c}} {-2+2} & {0+1} \\ {-3+4} & {0+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 0 & 1 \\ 1 & 2 \end{array}} \right)$

$\displaystyle \begin{array}{l}\ \ \ \det P=4-3=1\ne 0,\ \text{hence }{{P}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det Q=-1-0=-1\ne 0,\ \text{hence }{{Q}^{{-1}}}\text{exists}\text{.}\\\\\ \ \ \det R=0-1=-1\ne 0,\ \text{hence }{{R}^{{-1}}}\text{exists}\text{.}\end{array}$

$\displaystyle \therefore {{P}^{{-1}}}=\frac{1}{{\det P}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\frac{1}{1}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$

$\displaystyle \ \ \ {{Q}^{{-1}}}=\frac{1}{{\det Q}}\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ {-2} & {-1} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$

$\displaystyle \ \ \ {{R}^{{-1}}}=\frac{1}{{\det R}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\frac{1}{{-1}}\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-1} & 0 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$

$\displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2 & {-1} \\ {-3} & 2 \end{array}} \right)$

$\displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-2+0} & {1+0} \\ {4-3} & {-2+2} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-2} & 1 \\ 1 & 0 \end{array}} \right)$

$\displaystyle \therefore {{Q}^{{-1}}}{{P}^{{-1}}}={{R}^{{-1}}}$

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.

(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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