Sample Math Paper (2) - Section (B) Solution

ဒီေနရာမွာ တင္ေပးလိုက္တဲ့ ေမးခြန္းရဲ့ section (B) အေျဖျဖစ္ပါတယ္။


Section (B)

6.  (a) Given that $ \displaystyle f(x) =2x^2-1$ and $ \displaystyle g(x) = \cos x$ where $ \displaystyle x\in A=\{x|0\le x\le \frac{\pi}{2}\}.$ Solve the equation $ \displaystyle (f∘g)(x)=0,$ where $ \displaystyle x\in A.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ f(x)=2{{x}^{2}}-1,g(x)=\cos x\\\\\ \ \ \ \ \ \ (f\circ g)(x)=f\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f(\cos x)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\cos }^{2}}x-1\\\\\ \ \ \ \ \ \ \ (f\circ g)(x)=0\\\\\ \ \ \ \ \ \ \ 2{{\cos }^{2}}x-1=0\\\\\ \ \ \ \ \ \ \ \cos 2x=0\\\\\ \therefore \ \ \ \ \ 2x=\displaystyle \frac{\pi }{2}\ \ \text{(or)}\ 2x=\displaystyle \frac{{3\pi }}{2}\\\\\therefore \ \ \ \ \ \ x=\displaystyle \frac{\pi }{4}\ \text{(or)}\ x=\displaystyle \frac{{3\pi }}{4}>\displaystyle \frac{\pi }{2}\ \text{(reject)}\\\\\therefore \ \ \ \ \ \ x=\displaystyle \frac{\pi }{4}\ \end{array}$


     (b) The curve of the polynomial $ \displaystyle f(x)=-x^3+2x^2+ax-10$ cuts the $ \displaystyle x$-axis at $ \displaystyle x=p, x=2$ and $ \displaystyle x=q.$ Find the value of $ \displaystyle p$ and $ \displaystyle q.$ Hence show that $ \displaystyle a=5.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ f(x)=-{{x}^{3}}+2{{x}^{2}}+ax-10\\\\\ \ \ \ \ \ \text{Since the graph of }f(x)\text{ cuts the x-axis at }x=p\text{, }x=2\text{ and }x=q\text{,}\\\\\ \ \ \ \ \ x-p,x-2\ \operatorname{and}\ x-q\ \text{are factors of }f(x).\\\\\ \ \ \ \ \ \text{Since the coefficient of }{{x}^{3}}\ \text{is }-1,\\\\\ \ \ \ \ f(x)=-(x-p)(x-q)(x-2)\\\\\therefore \ \ \ \ f(x)=-\left( {{{x}^{2}}-(p+q)x+pq} \right)(x-2)\\\\\therefore \ \ \ \ f(x)=-\left( {{{x}^{3}}-(p+q){{x}^{2}}+pqx-2{{x}^{2}}+2(p+q)x-2pq} \right)\\\\\therefore \ \ \ \ f(x)=-{{x}^{3}}+(p+q){{x}^{2}}-pqx+2{{x}^{2}}-2(p+q)x+2pq\\\\\therefore \ \ \ \ f(x)=-{{x}^{3}}+(p+q+2){{x}^{2}}-(2p+2q+pq)x+2pq\\\\\therefore \ \ \ \ -{{x}^{3}}+2{{x}^{2}}+ax-10=-{{x}^{3}}+(p+q+2){{x}^{2}}-(2p+2q+pq)x+2pq\\\\\therefore \ \ \ \ 2pq=-10\Rightarrow pq=-5\Rightarrow q=-\displaystyle \frac{5}{p}\\\\\therefore \ \ \ \ p+q+2=2\Rightarrow p+q=0\\\\\therefore \ \ \ \ p-\displaystyle \frac{5}{p}=0\Rightarrow {{p}^{2}}=5\Rightarrow p=\sqrt{5}\\\\\therefore \ \ \ \ q=-\displaystyle \frac{5}{{\sqrt{5}}}=-\sqrt{5}\\\\\therefore \ \ \ \ a=-(2p+2q+pq)=-(2\sqrt{5}-2\sqrt{5}-5)=5\end{array}$


7.  (a) If $ \displaystyle f(x+y,x-y)=xy$ where $ \displaystyle x,y\in R$, show that $ \displaystyle f(x,y)+f(y,x) =0$.
(5 marks)

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$ \displaystyle \begin{array}{*{20}{l}} \begin{array}{l}\ \ \ \ \ \ \ \ f(x+y,x-y)=xy\\\end{array} \\\\ \begin{array}{l}\ \ \ \ \ \ \ \ \text{Let }x+y=a\ \ \operatorname{and}x-y=b\\\end{array} \\\\ \begin{array}{l}\therefore \ \ \ \ \ \ \ 2x=a+b\Rightarrow x=\displaystyle \frac{{a+b}}{2}\\\end{array} \\\\ \begin{array}{l}\ \ \ \ \ \ \ \ \ 2y=a-b\Rightarrow y=\displaystyle \frac{{a-b}}{2}\\\end{array} \\\\ \begin{array}{l}\therefore \ \ \ \ \ \ f(a,b)=\displaystyle \frac{{a+b}}{2}\times \displaystyle \frac{{a-b}}{2}=\displaystyle \frac{{{{a}^{2}}-{{b}^{2}}}}{4}\\\end{array} \\\\ \begin{array}{l}\therefore \ \ \ \ \ \ f(x,y)=\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{4}\\\end{array} \\\\ \begin{array}{l}\therefore \ \ \ \ \ \ f(y,x)=\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{4}\\\end{array} \\\\ {\therefore \ \ \ \ \ \ f(x,y)+f(y,x)=\displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{4}+\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{4}=0} \end{array}$


     (b) If the coefficients of $ \displaystyle (2p + 4)^{\text{th}}$ and $ \displaystyle (p - 2)^{\text{th}}$ terms in the expansion of $ \displaystyle (1 + x)^{18}$ are equal, find the value of $ \displaystyle p.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}\ {{(1+x)}^{{18}}}\\\\\ \ \ \ \ \ ={}^{{18}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ \ \ {{(2p+4)}^{{\text{th}}}}\ \text{term}={{(2p+3+1)}^{{\text{th}}}}\ \text{term}={}^{{18}}{{C}_{{2p+3}}}{{x}^{{2p+3}}}\\\\\ \ \ \ \ \ {{(p-2)}^{{\text{th}}}}\ \text{term}={{(p-3+1)}^{{\text{th}}}}\ \text{term}={}^{{18}}{{C}_{{p-3}}}{{x}^{{r-3}}}\\\\\ \ \ \ \ \ \text{By the problem, }\\\\\ \ \ \ \ \ \text{coefficient of}\ {{(2p+4)}^{{\text{th}}}}\ \text{term}=\text{coefficient of}\ {{(r-2)}^{{\text{th}}}}\ \text{term}\\\\\therefore \ \ \ \ {}^{{18}}{{C}_{{2p+3}}}={}^{{18}}{{C}_{{p-3}}}\ \text{(or)}{}^{{18}}{{C}_{{2p+3}}}={}^{{18}}{{C}_{{18-(p-3)}}}\ \left[ {\because {}^{n}{{C}_{r}}={}^{n}{{C}_{{n-r}}}} \right]\\\\\therefore \ \ \ \ 2p+3=p-3\ \text{(or)}\ 2p+3=18-(p-3)\\\\\therefore \ \ \ \ p=-6\ \ \text{(or)}\ p=6\\\\\ \ \ \ \ \ \text{Since}\ p>0,p=6\end{array}$


8.  (a) Find the solution set of the in equation $ \displaystyle 3(x-\frac{3}{2})^2>2x^2-4x+\frac{3}{4}$ and illustrate it on the number line.
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ 3{{\left( {x-\displaystyle \frac{3}{2}} \right)}^{2}}>2{{x}^{2}}-4x+\displaystyle \frac{3}{4}\\\\\ \ \ \ \ \ \ 3\left( {{{x}^{2}}-3x+\displaystyle \frac{9}{4}} \right)>2{{x}^{2}}-4x+\displaystyle \frac{3}{4}\\\\\ \ \ \ \ \ \ 3{{x}^{2}}-9x+\displaystyle \frac{{27}}{4}>2{{x}^{2}}-4x+\displaystyle \frac{3}{4}\\\\\ \ \ \ \ \ \ 3{{x}^{2}}-9x+\displaystyle \frac{{27}}{4}-2{{x}^{2}}+4x-\displaystyle \frac{3}{4}>0\\\\\ \ \ \ \ \ \ {{x}^{2}}-5x+6>0\\\\\ \ \ \ \ \ \ \text{Let}\ y={{x}^{2}}-5x+6.\\\\\ \ \ \ \ \ \ \text{When}\ y=0,{{x}^{2}}-5x+6=0\ \\\\\therefore \ \ \ \ \ (x-2)(x-3)=0\\\\\therefore \ \ \ \ \ x=2\ (\text{or})\ x=3\\\\\therefore \ \ \ \ \ \text{The graph cuts the x-axis at (2,0) and (3,0)}\text{.}\\\\\ \ \ \ \ \ \ \text{When}\ x=0,y=6\ \\\\\therefore \ \ \ \ \ \text{The graph cuts the y-axis at (0,6)}\text{.}\end{array}$


$ \displaystyle \begin{array}{*{20}{l}} {\therefore \ \ \ \ \ \text{Solution set}=\left\{ {x\ |\ x<2\ \text{(or)}\ x>3} \right\}} \\ \begin{array}{l}\\\ \ \ \ \ \ \text{Number line}\end{array} \end{array}$




     (b) Find three numbers in A.P. whose sum is $ \displaystyle 21$ and whose product is $ \displaystyle 315.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let the three numbers in A}\text{.P}\text{. be}\\\ \ \ \ \ \ \ a,a+d\ \operatorname{and}\ a+2d.\\\\\ \ \ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ \ \ a+a+d\ +\ a+2d=21\\\\\therefore \ \ \ \ \ a+d=7\ \ \ \ \ \ \ \ \ -------(1)\\\\\ \ \ \ \ \ \ a(a+d)(a+2d)=315\\\\\ \ \ \ \ \ \ 7a(7+d)=315\ \ \ \ \left[ {\because a+d=7} \right]\\\\\therefore \ \ \ \ \ a=\displaystyle \frac{{45}}{{7+d}}\ \ \ \ \ \ \ \ -------(2)\\\\\therefore \ \ \ \ \ \displaystyle \frac{{45}}{{7+d}}+d=7\\\ \\\therefore \ \ \ \ \ {{d}^{2}}=4\Rightarrow d=\pm 2\\\\\ \ \ \ \ \ \text{When}\ d=-2,a=9\\\\\ \ \ \ \ \ \text{When}\ d=2,a=5\\\\\therefore \ \ \ \ \text{The numbers are}\ 5,7\ \operatorname{and}\ 9.\end{array}$


9.  (a) If $ \displaystyle S_1, S_2,$ and $ \displaystyle S_3$ are the sums of $ \displaystyle n, 2n$ and $ \displaystyle 3n$ terms of a G.P., show that $ \displaystyle S_1(S_3- S_2) = (S_2-S_1)^2.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let the first term and the common ratio }\\\ \ \ \ \ \ \ \text{of given G}\text{.P}\text{. be }a\text{ and }r\text{ respectively}\text{.}\\\\\ \ \ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ \ {{S}_{1}}={{S}_{n}}=\displaystyle \frac{{a(1-{{r}^{n}})}}{{1-r}}\\\\\ \ \ \ \ \ {{S}_{2}}={{S}_{{2n}}}=\displaystyle \frac{{a(1-{{r}^{{2n}}})}}{{1-r}}\\\\\ \ \ \ \ \ {{S}_{3}}={{S}_{{3n}}}=\displaystyle \frac{{a(1-{{r}^{{3n}}})}}{{1-r}}\\\\\therefore \ \ \ \ {{S}_{3}}-{{S}_{2}}=\displaystyle \frac{a}{{1-r}}\left( {1-{{r}^{{3n}}}-1+{{r}^{{2n}}}} \right)\\\\\therefore \ \ \ \ {{S}_{3}}-{{S}_{2}}=\displaystyle \frac{a}{{1-r}}\left( {{{r}^{{2n}}}-{{r}^{{3n}}}} \right)\\\\\therefore \ \ \ \ {{S}_{1}}\left( {{{S}_{3}}-{{S}_{2}}} \right)=\displaystyle \frac{{a(1-{{r}^{n}})}}{{1-r}}\times \displaystyle \frac{a}{{1-r}}\left( {{{r}^{{2n}}}-{{r}^{{3n}}}} \right)\\\\\therefore \ \ \ \ {{S}_{1}}\left( {{{S}_{3}}-{{S}_{2}}} \right)=\displaystyle \frac{{{{a}^{2}}}}{{{{{\left( {1-r} \right)}}^{2}}}}\left( {{{r}^{{2n}}}-{{r}^{{3n}}}-{{r}^{{3n}}}+{{r}^{{4n}}}} \right)\\\\\therefore \ \ \ \ {{S}_{1}}\left( {{{S}_{3}}-{{S}_{2}}} \right)=\displaystyle \frac{{{{a}^{2}}}}{{{{{\left( {1-r} \right)}}^{2}}}}\left( {{{r}^{{2n}}}-2{{r}^{{3n}}}+{{r}^{{4n}}}} \right)---(1)\\\\\ \ \ \ \ \ {{S}_{2}}-{{S}_{1}}=\displaystyle \frac{a}{{1-r}}\left( {1-{{r}^{{2n}}}-1+{{r}^{n}}} \right)\\\\\therefore \ \ \ \ {{S}_{2}}-{{S}_{1}}=\displaystyle \frac{a}{{1-r}}\left( {{{r}^{n}}-{{r}^{{2n}}}} \right)\\\\\therefore \ \ \ \ {{\left( {{{S}_{2}}-{{S}_{1}}} \right)}^{2}}=\displaystyle \frac{{{{a}^{2}}}}{{{{{\left( {1-r} \right)}}^{2}}}}{{\left( {{{r}^{n}}-{{r}^{{2n}}}} \right)}^{2}}\\\\\therefore \ \ \ \ {{\left( {{{S}_{2}}-{{S}_{1}}} \right)}^{2}}=\displaystyle \frac{{{{a}^{2}}}}{{{{{\left( {1-r} \right)}}^{2}}}}\left( {{{r}^{{2n}}}-2{{r}^{{3n}}}+{{r}^{{4n}}}} \right)---(2)\\\\\therefore \ \ \ \ \text{By}\ (1)\ \operatorname{and}\ (2),\\\\\ \ \ \ \ \ {{S}_{1}}\left( {{{S}_{3}}-{{S}_{2}}} \right)={{\left( {{{S}_{2}}-{{S}_{1}}} \right)}^{2}}\end{array}$


     (b) Given that $ \displaystyle A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right).$ If $ \displaystyle A + A' = I$ where $ \displaystyle I$ is a unit matrix of order $ \displaystyle 2,$ find the value of $ \displaystyle \theta$ for $ \displaystyle 0°<\theta< 90°.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ A=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)\\\\\therefore \ \ \ {A}'=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\ {-\sin \theta } & {\cos \theta } \end{array}} \right)\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ A+{A}'=I\\\\\therefore \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)+\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\ {-\sin \theta } & {\cos \theta } \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\therefore \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {2\cos \theta } & 0 \\ 0 & {2\cos \theta } \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\therefore \ \ \ 2\cos \theta =1\Rightarrow \cos \theta =\displaystyle \frac{1}{2}\\\\\therefore \ \ \ \theta =60{}^\circ \end{array}$


10. (a) Given that $ \displaystyle A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right).$ Determine whether $\displaystyle {{A}^{{-1}}}$ exists or not, if exists find $\displaystyle {{A}^{{-1}}}.$ Hence solve the system of equations $\displaystyle x\cos \theta -y\sin \theta =2$ and $\displaystyle x\sin \theta +y\cos \theta =2\sqrt{3}$ when $\displaystyle \theta=30°.$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ A=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)\\\\\therefore \ \ \ \det A={{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\ne 0\\\\\therefore \ \ \ {{A}^{{-1}}}\ \text{exists}.\\\\\therefore \ \ \ {{A}^{{-1}}}=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\ {-\sin \theta } & {\cos \theta } \end{array}} \right)\\\\\ \ \ \ \ \displaystyle \left. \begin{array}{l}x\cos \theta -y\sin \theta =2\\x\sin \theta +y\cos \theta =2\sqrt{3}\end{array} \right\}-------(1)\\\\\ \ \ \ \ \text{Transforming into matrix form},\\\\\ \ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {2\sqrt{3}} \end{array}} \right)---(2)\\\\\ \ \ \ \ \text{Let}\ X=\displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\operatorname{and}\ B=\displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {2\sqrt{3}} \end{array}} \right).\\\\\ \ \ \ \ \text{Equation (2) becomes,}\ AX=B.\\\\\therefore \ \ \ {{A}^{{-1}}}AX={{A}^{{-1}}}B\\\\\therefore \ \ \ IX={{A}^{{-1}}}B\\\\\therefore \ \ \ X={{A}^{{-1}}}B\\\\\therefore \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\ {-\sin \theta } & {\cos \theta } \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {2\sqrt{3}} \end{array}} \right)\\\\\ \ \ \ \ \text{When}\ \theta =30{}^\circ ,\\\\\ \ \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos 30{}^\circ } & {\sin 30{}^\circ } \\ {-\sin 30{}^\circ } & {\cos 30{}^\circ } \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {2\sqrt{3}} \end{array}} \right)\\\\\ \ \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} {\displaystyle \frac{{\sqrt{3}}}{2}} & {\displaystyle \frac{1}{2}} \\ {-\displaystyle \frac{1}{2}} & {\displaystyle \frac{{\sqrt{3}}}{2}} \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {2\sqrt{3}} \end{array}} \right)\\\\\therefore \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} {\displaystyle \frac{{2\sqrt{3}}}{2}+\displaystyle \frac{{2\sqrt{3}}}{2}} \\ {-1+3} \end{array}} \right)\\\\\therefore \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {2\sqrt{3}} \\ 2 \end{array}} \right)\Rightarrow x=2\sqrt{3},y=2.\end{array}$


     (b) A set of cards bearing the number from $ \displaystyle 200$ to $ \displaystyle 299$ is used in a game. If a card is drawn at random, what is the probability that it is divisible by $ \displaystyle 3?$
(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Set of possible outcomes}=\left\{ {200,201,202,...,299} \right\}\\\\\therefore \ \ \ \ \text{Number of possible outcomes}=100\\\\\ \ \ \ \ \ \text{The numbers between 200 and 299 }\\\ \ \ \ \ \ \text{which are divisible by 3 are}\ \\\\\ \ \ \ \ \ 201,204,205,...,297.\\\\\ \ \ \ \ \ \text{It is an A}\text{.P with }a=201\ \operatorname{and}\ d=3.\\\\\therefore \ \ \ \ {{u}_{n}}=99\\\\\therefore \ \ \ \ a+(n-1)d=297\\\\\therefore \ \ \ \ 201+(n-1)3=297\\\\\therefore \ \ \ \ n=33\\\\\therefore \ \ \ \ \text{Number of favourable outcomes}=33\\\\\therefore \ \ \ \ P(\text{a number divisible by3)}=\displaystyle \frac{{33}}{{100}}.\end{array}$