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

ဒီေနရာမွာ တင္ေပးလိုက္တဲ့ ေမးခြန္းရဲ့ section (C) အေျဖျဖစ္ပါတယ္။

Section (C)

11. (a) In the diagram, two circles are tangent at $ \displaystyle A$ and have a common tangent touching them $ \displaystyle B$ and $ \displaystyle C$ respectively. If $ \displaystyle BA$ is produced to meet the second circle at $ \displaystyle D,$ show that $ \displaystyle CD$ is a diameter.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Draw a common tangent to both }\\\ \ \ \ \ \ \text{circles at }A\text{ to cut }BC\text{ at }E\text{.}\\\\\ \ \ \ \ \ \text{For smaller circle, }\\\\\ \ \ \ \ \ EA=EB\\\\\therefore \ \ \ \ \alpha =\beta \\\\\ \ \ \ \ \ \text{Similarly, for larger circle,}\\\\\ \ \ \ \ \ EA=EB\\\\\therefore \ \ \ \ \delta =\gamma \\\\\ \ \ \ \ \ \text{Since }\alpha +\beta +\delta +\gamma =180{}^\circ ,\\\\\ \ \ \ \ \ 2\alpha +2\delta =180{}^\circ \\\\\therefore \ \ \ \ \alpha +\delta =90{}^\circ \\\\\therefore \ \ \ \ \angle CAD=90{}^\circ \\\\\therefore \ \ \ \ \text{Arc }CAD\text{ is a semicircle}\text{.}\\\\\therefore \ \ \ \ CD\text{ is a diameter}\text{.}\end{array}$

     (b) $ \displaystyle ABC$ is a right triangle with $ \displaystyle A$ the right angle. $ \displaystyle E$ and $ \displaystyle D$ are points on opposite side of $ \displaystyle AC,$ with $ \displaystyle E$ on the same side of $ \displaystyle AC$ as $ \displaystyle B,$ such that $ \displaystyle ΔACD$ and $ \displaystyle ΔBCE$ are both equilateral. If $ \displaystyle α (ΔBCE) = 2 α (ΔACD),$ prove that $ \displaystyle ABC$ is an isosceles right triangle.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha \text{(}\vartriangle BCE\text{) = 2}\alpha \text{(}\vartriangle ACD\text{) }\left[ {\because \text{given}} \right]\\\\\ \ \ \ \ \displaystyle \frac{{\alpha \text{(}\vartriangle BCE\text{)}}}{{\alpha \text{(}\vartriangle ACD\text{)}}}=2\\\\\ \ \ \ \ \text{Since }\vartriangle ACD\text{ and }\vartriangle BCE\text{ are equilateral,}\\\\\ \ \ \ \ \vartriangle ACD\sim \vartriangle BCE\\\\\ \ \ \ \ \displaystyle \frac{{\alpha \text{(}\vartriangle BCE\text{)}}}{{\alpha \text{(}\vartriangle ACD\text{)}}}=\displaystyle \frac{{B{{C}^{2}}}}{{A{{C}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{B{{C}^{2}}}}{{A{{C}^{2}}}}=2\Rightarrow B{{C}^{2}}=2A{{C}^{2}}\\\\\ \ \ \ \ \text{But in rt}\text{. }\vartriangle ABC,\\\\\ \ \ \ \ B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\\\\\therefore \ \ \ A{{B}^{2}}+A{{C}^{2}}=2A{{C}^{2}}\\\\\therefore \ \ \ A{{B}^{2}}=A{{C}^{2}}\Rightarrow AB=AC\\\\\therefore \ \ \ \vartriangle ABC\ \text{is an isosceles right triangle}\\\ \ \ \ \ \end{array}$

12. (a) Two circles are drawn intersecting at $ \displaystyle A, B$ and so that the circumference of each passes through the centre of the another. Through $ \displaystyle A,$ a line is drawn meeting the circumference at $ \displaystyle C, D$ respectively. Prove that $ \displaystyle \vartriangle BCD$ is equilateral.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \text{Draw}\ OA,OB,OP,PA\ \operatorname{and}PB.\\\\\ \ \ \ \ \ \ \ \text{Since}\odot O\ \operatorname{and}\ \odot P\ \text{are congruent,}\ \ \\\ \ \ \ \ \ \ \ OA,OB,OP,PA\ \operatorname{and}PB\ \text{are equal }\\\ \ \ \ \ \ \ \ \text{radii}\ \text{of congruent circles}.\\\\\ \ \ \ \ \ \ \ \vartriangle AOP\ \operatorname{and}\ \vartriangle BOP\ \text{are equilateral }\vartriangle \text{s}.\\\\\therefore \ \ \ \ \ \ \alpha =\beta =\gamma =\delta =60{}^\circ \\\\\ \ \ \ \ \ \ \ \text{But}\ \angle C=\displaystyle \frac{1}{2}(\alpha +\beta ),\\\\\therefore \ \ \ \ \ \ \ \angle C=60{}^\circ \\\\\ \ \ \ \ \ \ \ \text{Similarly,}\ \angle D=\displaystyle \frac{1}{2}(\gamma +\delta ),\\\ \\\therefore \ \ \ \ \ \ \ \angle D=60{}^\circ \\\\\therefore \ \ \ \ \ \ \angle CBD=60{}^\circ \\\\\therefore \ \ \ \ \ \ \vartriangle BCD\ \text{is equilateral triangle}\text{.}\end{array}$

     (b) Given that $ \displaystyle \sin \alpha =\frac{3}{5}$ and $ \displaystyle \cos \beta =\frac{{12}}{{13}}$, where $ \displaystyle α$ is obtuse and $ \displaystyle β$ is acute, find the exact values of $ \displaystyle \cos (α+β)$ and $ \displaystyle \cot (α- β).$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \alpha =\displaystyle \frac{3}{5},\ 90{}^\circ <\alpha <180{}^\circ \\\\\ \ \ \ \text{Since}\ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\\\\\ \ \ \ \cos \alpha =-\sqrt{{1-{{{\sin }}^{2}}\alpha }}\ \ \ \ \left[ {\because 90{}^\circ <\alpha <180{}^\circ } \right]\\\\\therefore \ \ \cos \alpha =-\sqrt{{1-\displaystyle \frac{9}{{25}}}}=-\displaystyle \frac{4}{5}\\\\\therefore \ \ \tan \alpha =\displaystyle \frac{{\sin \alpha }}{{\cos \alpha }}=-\displaystyle \frac{3}{4}\\\\\ \ \ \ \cos \beta =\displaystyle \frac{{12}}{{13}},\ 0{}^\circ <\beta <90{}^\circ \\\\\ \ \ \ \text{Since}\ {{\sin }^{2}}\beta +{{\cos }^{2}}\beta =1\\\\\ \ \ \ \sin \beta =\sqrt{{1-{{{\cos }}^{2}}\beta }}\ \ \ \ \left[ {\because 0{}^\circ <\beta <90{}^\circ } \right]\\\\\therefore \ \ \sin \beta =\sqrt{{1-\displaystyle \frac{{144}}{{169}}}}=\displaystyle \frac{5}{{13}}\\\\\therefore \ \ \tan \beta =\displaystyle \frac{{\sin \beta }}{{\cos \beta }}=\displaystyle \frac{5}{{12}}\\\\\therefore \ \ \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\\\therefore \ \ \cos (\alpha +\beta )=\left( {-\displaystyle \frac{4}{5}} \right)\left( {\displaystyle \frac{{12}}{{13}}} \right)-\left( {\displaystyle \frac{3}{5}} \right)\left( {\displaystyle \frac{5}{{13}}} \right)\\\\\therefore \ \ \cos (\alpha +\beta )=-\displaystyle \frac{{63}}{{65}}\\\\\therefore \ \ \tan (\alpha -\beta )=\displaystyle \frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}\\\\\therefore \ \ \tan (\alpha -\beta )=\displaystyle \frac{{-\displaystyle \frac{3}{4}-\displaystyle \frac{5}{{12}}}}{{1+\left( {-\displaystyle \frac{3}{4}} \right)\left( {\displaystyle \frac{5}{{12}}} \right)}}\\\\\therefore \ \ \tan (\alpha -\beta )=\displaystyle \frac{{-\displaystyle \frac{{56}}{{48}}}}{{\displaystyle \frac{{33}}{{48}}}}=-\displaystyle \frac{{56}}{{33}}\\\\\therefore \ \ \cot (\alpha -\beta )=-\displaystyle \frac{{33}}{{56}}\\\ \ \ \ \,\end{array}$

13. (a) Solve $ \displaystyle ΔABC$ with $ \displaystyle b=12.5, c=23$ and $ \displaystyle α=38°20′.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ b=12.5,\ \ c=23,\ \ \alpha =38{}^\circ 2{0}'.\\\\\ \ \ \ \ \ \ a=?\,\ \ \ \beta =?\ \ \ \ \ \gamma =?\\\\\ \ \ \ \ \ \ \text{By the law of cosines,}\\\\\ \ \ \ \ \ \ {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos \alpha \\\\\therefore \ \ \ \ \ {{a}^{2}}={{(12.5)}^{2}}+{{23}^{2}}-2(12.5)(23)\cos 38{}^\circ 2{0}'\\\\\therefore \ \ \ \ \ {{a}^{2}}=156.25+529-575\cos 38{}^\circ 2{0}' \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline {575} & {2.7597} \\ { \cos 38 {}^\circ2{0}' } & {\overline{1}.8945}\\ \hline {451} & {2.6542} \\ \hline \end{array}\end{array} \\\\\therefore \ \ \ \ \ {{a}^{2}}=685.25-451=234.25\\\\\therefore \ \ \ \ \ a=\sqrt{{234.25}}=15.31\\\\\ \ \ \ \ \ \ \text{By the law of sines,}\\\\\ \ \ \ \ \ \ \displaystyle \frac{{\sin \beta }}{b}=\displaystyle \frac{{\sin \alpha }}{a}\\\\\therefore \ \ \ \ \ \sin \beta =\displaystyle \frac{{b\sin \alpha }}{a}\\\\\therefore \ \ \ \ \ \sin \beta =\displaystyle \frac{{12.5\sin 38{}^\circ 2{0}'}}{{15.31}}\\\\\therefore \ \ \ \ \ \sin \beta =\displaystyle \frac{{12.5\sin 38{}^\circ 2{0}'}}{{15.31}} \\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 12.5} & {1.0969} \\ { \sin 38 {}^\circ2{0}' } & {\overline{1}.7926}\\ \hline {} & {0.8895} \\ \hline {15.31} & {1.1850} \\ \hline {\sin 30 {}^\circ{25}'} & {\overline{1}.7045} \\ \hline \end{array}\end{array} \\\\\therefore \ \ \ \ \beta =30{}^\circ 2{5}'\ \\\\\therefore \ \ \ \ \gamma =180{}^\circ -(38{}^\circ 2{0}'+30{}^\circ 2{5}')=111{}^\circ 1{5}'\end{array}$

     (b) Find the stationary points on the curve $ \displaystyle y=x^4(x^2-6)$ and determine their natures.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ y={{x}^{4}}({{x}^{2}}-6)={{x}^{6}}-6{{x}^{4}}\\\\\ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=6{{x}^{5}}-24{{x}^{3}}=6{{x}^{3}}({{x}^{2}}-4)\\\\\ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=0\ \text{when}\ 6{{x}^{3}}({{x}^{2}}-4)=0\\\\\therefore \ \ \ \ x=0\ \text{(or)}\ x=\pm 2\\\\\ \ \ \ \ \ \text{When}\ x=-2,y=16(4-6)=-32\\\\\ \ \ \ \ \ \text{When}\ x=0,y=0\\\\\ \ \ \ \ \ \text{When}\ x=2,y=16(4-6)=-32\\\\\therefore \ \ \ \ \text{The stationary points are}\ (-2,-32),(0,0)\\\\\ \ \ \ \ \ \operatorname{and}\ (2,-32).\end{array}$


$ \displaystyle \begin{array}{l}\therefore \ \ \ \ (-2,-32)\ \operatorname{and}\ (2,-32)\ \text{are minimum turning} \\\ \ \ \ \ \ \text{ points and}\ (0,0)\ \text{is maximum turning point.}\end{array}$

14. (a) Show that the tangent to the curve $ \displaystyle y=e^{-2x}-3x$ at the point $ \displaystyle (a,0)$ meets the $ \displaystyle y$-axis at the point whose $ \displaystyle y$-coordinate is $ \displaystyle 2ae^{-2a} +3a.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ y={{e}^{{-2x}}}-3x\\\\\ \ \ \ \ \frac{{dy}}{{dx}}=-2{{e}^{{-2x}}}-3\\\\\ \ \ \ \ \text{At}\ (a,0),\ \frac{{dy}}{{dx}}=-2{{e}^{{-2a}}}-3\\\\\therefore \ \ \ \text{Equation of tangent at}\ (a,0)\ \text{is}\\\\\ \ \ \ \ y-0=\left( {-2{{e}^{{-2a}}}-3} \right)(x-a)\\\\\therefore \ \ \ y=\left( {-2{{e}^{{-2a}}}-3} \right)(x-a)\\\\\ \ \ \ \ \text{When the tangent meets the y-axis},\ x=0.\\\\\therefore \ \ \ y=\left( {-2{{e}^{{-2a}}}-3} \right)(0-a)=2a{{e}^{{-2a}}}+3a\\\\\therefore \ \ \ \text{The tangent to the curve}\ \text{meets the y-axis}\\\ \ \ \ \ \text{at the point whose}\ \text{y}\ \text{oordinate is}\ 2a{{e}^{{-2a}}}+3a.\\\end{array}$

     (b) Points $ \displaystyle A$ and $ \displaystyle B$ have position vectors $ \displaystyle {\vec{a}}$ and $ \displaystyle {\vec{b}}$ respectively, relative to an origin $ \displaystyle O.$ The point $ \displaystyle C$ lies on $ \displaystyle OA$ produced such that $ \displaystyle OC = 3OA,$ and $ \displaystyle D$ lies on $ \displaystyle OB$ such that $ \displaystyle OD = \frac {1}{4}OB.$ Express $ \displaystyle \overrightarrow{{AB}}$ and $ \displaystyle \overrightarrow{{CD}}$ in terms of $ \displaystyle {\vec{a}}$ and $ \displaystyle {\vec{b}}$. The line segments $ \displaystyle AB$ and $ \displaystyle CD$ intersect at $ \displaystyle P.$ If $ \displaystyle CP = hCD$ and $ \displaystyle AP = kAB,$ calculate the values of $ \displaystyle h$ and $ \displaystyle k.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=\vec{a},\ \ \overrightarrow{{OB}}=\vec{b}\\\\\ \ \ \ \ OC=\text{ }3OA\ \operatorname{and}\ O,A\ \operatorname{and}\ C\ \text{are collinear}.\\\\\therefore \ \ \ \overrightarrow{{OC}}=3\vec{a}\ \operatorname{and}\ \overrightarrow{{AC}}=2\vec{a}\\\\\ \ \ \ \ \text{Similarly,}\ OD=\text{ }\displaystyle \frac{1}{4}OB\operatorname{and}\ O,D,B\ \text{are collinear}.\\\\\therefore \ \ \ \overrightarrow{{OD}}=\displaystyle \frac{1}{4}\vec{b}\ \operatorname{and}\ \ \overrightarrow{{DB}}=\displaystyle \frac{3}{4}\vec{b}.\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}=\vec{b}-\vec{a}\\\\\therefore \ \ \ \overrightarrow{{CD}}=\overrightarrow{{OD}}-\overrightarrow{{OC}}=\displaystyle \frac{1}{4}\vec{b}-3\vec{a}\\\\\ \ \ \ \ \text{Since }CP\text{ = }hCD\ \text{and }C\text{, }P\text{ and }D\ \text{are collinear,}\\\\\ \ \ \ \ \overrightarrow{{CP}}=h\overrightarrow{{CD}}=\displaystyle \frac{h}{4}\vec{b}-3h\vec{a}\\\\\ \ \ \ \ \text{Similarly, }AP\text{ = }kAB\text{, }A\text{, }P\text{ and }P\text{ are collinear,}\\\\\therefore \ \ \ \overrightarrow{{AP}}=k\overrightarrow{{AB}}=k\vec{b}-k\vec{a}\\\\\ \ \ \ \ \text{Since}\ \overrightarrow{{AP}}=\overrightarrow{{AC}}+\overrightarrow{{CP}},\\\\\therefore \ \ \ \overrightarrow{{AC}}+\overrightarrow{{CP}}=k\vec{b}-k\vec{a}\\\\\therefore \ \ \ 2\vec{a}+\displaystyle \frac{h}{4}\vec{b}-3h\vec{a}\ =k\vec{b}-k\vec{a}\\\\\therefore \ \ \ \displaystyle \frac{h}{4}\vec{b}+(2-3h)\vec{a}\ =k\vec{b}-k\vec{a}\\\\\therefore \ \ \ \displaystyle \frac{h}{4}=k\Rightarrow h=4k\\\ \\\ \ \ \ \ 2-3h=-k\Rightarrow 2-12k=-k\\\\\therefore \ \ \ 11k=2\Rightarrow k=\displaystyle \frac{2}{{11}}\\\\\therefore \ \ \ h=\displaystyle \frac{8}{{11}}\end{array}$

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}$

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

👉 ဒီေနရာမွာ တင္ေပးလိုက္တဲ့ ေမးခြန္းရဲ့ section (A) အေျဖျဖစ္ပါတယ္။ ဒီေမးခြန္းက ထူးခၽြန္ ေက်ာင္းသားမ်ား အတြက္ ရည္ရြယ္ပါတယ္။ ေအာင္မွတ္ အတြက္သာ လုပ္ေနရတဲ့ ေက်ာင္းသား မ်ားအတြက္ အဆင္မေျပႏိုင္ပါဘူး။ ဒါ့ေၾကာင့္ သာမန္အဆင့္ ေက်ာင္းသားမ်ားကို ေလ့က်င့္ေပးရန္ မသင့္ေလ်ာ္ေၾကာင္း အႀကံျပဳ အပ္ပါတယ္။ ရည္မွန္းထားသည့္ အတိုင္း ေပါက္ေျမာက္ ေအာင္ျမင္ ႏိုင္ၾကပါေစ။...

Section (A)

1.  (a) The function $ \displaystyle g : N \to N$ is defined as $ \displaystyle g : x\mapsto$ smallest prime factor of $ \displaystyle x.$ (i) Find values for $ \displaystyle g(10), g (20)$ and $ \displaystyle g (81).$ (ii) Does $ \displaystyle g$ have an inverse? Give reasons for your answer.

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ g:N\to N\\\\\ \ \ \ g(x)=\text{smallest prime factor of}\ x.\\\\\ \ \ \ 10=2\times 5\\\\\therefore \ \ g(10)=2\\\\\ \ \ \ 20=2\times 2\times 5\\\\\therefore \ \ g(20)=2\\\\\ \ \ \ 81=3\times 3\times 3\times 3\\\\\therefore \ \ g(81)=3\\\\\ \ \ \ \text{Since}\ g(10)=g(20),\\\\\ \ \ \ g\ \text{is not one to one correspondence}\text{.}\\\\\therefore \ \ {{g}^{{-1}}}\ \text{does not}\ \text{exists}\text{.}\end{array}$

(1)  (b) If $ \displaystyle 2x-1$ is a factor of $ \displaystyle 2x^3-x^2-8x+k,$ find $ \displaystyle k$ and the other factors.

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let }f(x)=2{{x}^{3}}-{{x}^{2}}-8x+k\\\\\ \ \ \ \ \ \ 2x-1\ \text{is a factor of }f(x).\\\\\therefore \ \ \ \ \ f\left( {\displaystyle \frac{1}{2}} \right)=0\\\\\therefore \ \ \ \ \ 2{{\left( {\displaystyle \frac{1}{2}} \right)}^{3}}-{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-8\left( {\displaystyle \frac{1}{2}} \right)+k=0\\\\\therefore \ \ \ \ \ \displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}-4+k=0\\\\\therefore \ \ \ \ \ k=4\\\\\therefore \ \ \ \ \ f(x)=2{{x}^{3}}-{{x}^{2}}-8x+4\\\\\ \ \ \ \ \ \ \text{Let }f(x)=(2x-1)({{x}^{2}}+ax+b)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{3}}+2a{{x}^{2}}+2bx-{{x}^{2}}-ax-b\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{3}}+(2a-1){{x}^{2}}+(2b-a)x-b\\\\\therefore \ \ \ \ \ 2a-1=-1\ \operatorname{and}\,b=-4\\\\\therefore \ \ \ \ \ a=0\ \operatorname{and}\ b=-4\\\\\therefore \ \ \ \ \ f(x)=(2x-1)({{x}^{2}}-4)=(2x-1)(x-2)(x+2)\\\\\therefore \ \ \ \ \text{The other factors are }x-2\ \text{and }x+2.\end{array}$

2.  (a) Find the term independent of $ \displaystyle x$ in the expansion of $ \displaystyle {{\left( {x-\frac{2}{{{{x}^{2}}}}} \right)}^{9}}.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of }\ {{\left( {x-\displaystyle \frac{2}{{{{x}^{2}}}}} \right)}^{9}}\\\\\ \ \ \ ={}^{9}{{C}_{r}}{{x}^{{9-r}}}{{\left( {-\displaystyle \frac{2}{{{{x}^{2}}}}} \right)}^{r}}\\\\\ \ \ \ ={}^{9}{{C}_{r}}{{(-2)}^{r}}{{x}^{{9-3r}}}\\\\\ \ \ \ \ \ \ \text{For the term independent of }x,\ 9-3r=0\\\\\therefore \ \ \ \ \ r=3\\\\\therefore \ \ \ \ \ \text{The term independent of }x={}^{9}{{C}_{3}}{{(-2)}^{3}}=\displaystyle \frac{{9\times 8\times 7}}{{1\times 2\times 3}}(-8)=-672\end{array}$

(2)  (b) If the sum of n terms of a certain sequence is $ \displaystyle 2n + 3n^2,$ find the $ \displaystyle {n}^{\text{th}}$ term.

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{S}_{n}}=2n+3{{n}^{2}}\\\\\ \ \ \ \ \ {{u}_{n}}={{S}_{n}}-{{S}_{{n-1}}}\\\\\therefore \ \ \ \ {{u}_{n}}=2n+3{{n}^{2}}-\left[ {2(n-1)+3{{{(n-1)}}^{2}}} \right]\\\\\therefore \ \ \ \ {{u}_{n}}=2n+3{{n}^{2}}-2n+2+3{{n}^{2}}+6n-3\\\\\therefore \ \ \ \ {{u}_{n}}=4n-1\end{array}$

3.  (a) If $ \displaystyle X=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)$ and $ \displaystyle X-kI$ is singular, where $ \displaystyle I$ is a unit matrix of order $ \displaystyle 2,$ find $ \displaystyle k.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)\\\\\ \ \ \ \ X-kI=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)-k\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {1-k} & 0 \\ 2 & {3-k} \end{array}} \right)\\\\\ \ \ \ \ X-kI\ \text{is singular}.\\\\\therefore \ \ \ \det (X-kI)=0\\\\\therefore \ \ \ (1-k)(3-k)=0\\\\\therefore \ \ \ k=1\ (\text{or})\ k=3\end{array}$

(3)  (b) A number $ \displaystyle x$ is chosen at random from the numbers $ \displaystyle -4, -3, -2, -1, 0, 1, 2, 3, 4.$ What is the probability that $ \displaystyle |x| \le 2?$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Set of possible outcomes}=\left\{ {-4,-3,-2,-1,0,1,2,3,4} \right\}\\\\\therefore \ \ \ \ \text{Number of possible outcomes}=9\\\\\ \ \ \ \ \ |x|\le 2\Leftrightarrow -2\le x\le 2\\\\\therefore \ \ \ \ \text{Set of favourable outcomes}=\left\{ {-2,-1,0,1,2} \right\}\\\\\therefore \ \ \ \ \text{Number of favourable outcomes}=5\\\\\therefore \ \ \ \ P\left( {|x|\le 2} \right)=\displaystyle \frac{5}{9}\end{array}$

4.  (a) $ \displaystyle TA$ is the tangent to the circle at$ \displaystyle A, AB = BC, ∠BAC = 41°$ and $ \displaystyle ∠ACT = 46°.$ Find $ \displaystyle ∠ATC.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \angle BAC=41{}^\circ ,\angle ACT=46{}^\circ (\text{given})\\\\\ \ \ \ \text{Since}\ AB=BC,\angle BCA=\angle BAC\\\\\therefore \ \ \angle BCA=41{}^\circ \\\\\therefore \ \ \angle ABC=180{}^\circ -(41{}^\circ +41{}^\circ )=98{}^\circ \\\\\ \ \ \ \text{Since}\ \angle CAT=\angle ABC,\angle CAT=98{}^\circ \\\\\ \ \ \ \text{In}\ \vartriangle CAT,\\\\\ \ \ \ \angle ATC=180{}^\circ -(\angle CAT+\angle ACT)\\\\\therefore \ \ \angle ATC=180{}^\circ -(98{}^\circ +46{}^\circ )=36{}^\circ \end{array}$

4.  (b) If $ \displaystyle 3\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}=\vec{0},$ show that the points $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{OA}}-2\overrightarrow{{OB}}+\overrightarrow{{OA}}-\overrightarrow{{OC}}=\vec{0}\\\\\therefore \ \ \ 2\left( {\overrightarrow{{OA}}-\overrightarrow{{OB}}} \right)+\left( {\overrightarrow{{OA}}-\overrightarrow{{OC}}} \right)=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}+\overrightarrow{{CA}}=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}=-\overrightarrow{{CA}}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}=\overrightarrow{{AC}}\\\\\therefore \ \ \ A,B\ \operatorname{and}\ C\ \text{are collinear}\text{.}\end{array}$

5.  (a) If $\displaystyle \tan \alpha =x+1$ and $ \displaystyle \tan \beta =x-1$, find $ \displaystyle \cot (\alpha -\beta )$ in terms of $ \displaystyle x.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \tan \ \alpha =x+1,\ \tan \beta =x-1\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{{\tan \ \alpha -\tan \beta }}{{1+\tan \ \alpha \tan \beta }}\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{{x+1-x+1}}{{1+(x+1)(x-1)}}\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{2}{{1+({{x}^{2}}-1)}}=\displaystyle \frac{2}{{{{x}^{2}}}}\\\\\therefore \ \ \ \ \cot (\alpha -\beta )=\displaystyle \frac{1}{{\tan (\alpha -\beta )}}=\displaystyle \frac{{{{x}^{2}}}}{2}\end{array}$

5.  (b) Evaluate $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(2x-3)(\sqrt{x}-1)}}{{2{{x}^{2}}+x-3}}$ and $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{3{{{\sin }}^{2}}x-2\sin {{x}^{2}}}}{{3{{x}^{2}}}}.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{2{{x}^{2}}+x-3}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{(2x+3)(x-1)}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)}}{{(2x+3)(\sqrt{x}+1)}}\\\\\ \ \ \ =\displaystyle \frac{{2-3}}{{(2+3)(1+1)}}\\\\\ \ \ \ =-\displaystyle \frac{1}{{10}}\\\\\\\ \ \ \ \ \ \ \underset{{x\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{3{{{\sin }}^{2}}x-2\sin {{x}^{2}}}}{{3{{x}^{2}}}}\\\\\ \ \ \ =\underset{{x\to 0}}{\mathop{{\lim }}}\,\left[ {\displaystyle \frac{{{{{\sin }}^{2}}x}}{{{{x}^{2}}}}-\displaystyle \frac{2}{3}\cdot \displaystyle \frac{{\sin {{x}^{2}}}}{{{{x}^{2}}}}} \right]\\\\\ \ \ \ =\underset{{x\to 0}}{\mathop{{\lim }}}\,\left[ {{{{\left( {\displaystyle \frac{{\sin x}}{x}} \right)}}^{2}}-\displaystyle \frac{2}{3}\cdot \displaystyle \frac{{\sin {{x}^{2}}}}{{{{x}^{2}}}}} \right]\\\\\ \ \ \ =1-\displaystyle \frac{2}{3}\\\\\ \ \ \ =\displaystyle \frac{1}{3}\end{array}$

က်န္ေသာ အေျဖမ်ားကို ဆက္လက္ တင္ေပးသြားပါ့မယ္။

Sample Math Paper - Set (2) - for 2019 Matriculation Examination

2019 Matriculation Examination

Sample Paper (2)

Mathematics                              Time allowed : 3 hours

WRITE YOUR ANSWER IN THE ANSWER BOOKLET.

Section (A)

Answer ALL Questions.

1.  (a) The function $ \displaystyle g : N \to N$ is defined as $ \displaystyle g : x\mapsto$ smallest prime factor of $ \displaystyle x.$ (i) Find values for $ \displaystyle g(10), g (20)$ and $ \displaystyle g (81).$ (ii) Does $ \displaystyle g$ have an inverse? Give reasons for your answer.

(3 marks)

     (b) If $ \displaystyle 2x-1$ is a factor of $ \displaystyle 2x^3-x^2-8x+k,$ find $ \displaystyle k$ and the other factors.

(3 marks)

2.  (a) Find the term independent of $ \displaystyle x$ in the expansion of $ \displaystyle {{\left( {x-\frac{2}{{{{x}^{2}}}}} \right)}^{9}}.$

(3 marks)

     (b) If the sum of n terms of a certain sequence is $ \displaystyle 2n + 3n^2,$ find the $ \displaystyle {n}^{\text{th}}$ term.

(3 marks)

3.  (a) If $ \displaystyle X=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)$ and $ \displaystyle X-kI$ is singular, where $ \displaystyle I$ is a unit matrix of order $ \displaystyle 2,$ find $ \displaystyle k.$

(3 marks)

     (b) A number $ \displaystyle x$ is chosen at random from the numbers $ \displaystyle -4, -3, -2, -1, 0, 1, 2, 3, 4.$ What is the probability that $ \displaystyle |x| \le 2?$

(3 marks)

4.  (a) $ \displaystyle TA$ is the tangent to the circle at$ \displaystyle A, AB = BC, ∠BAC = 41°$ and $ \displaystyle ∠ACT = 46°.$ Find $ \displaystyle ∠ATC.$

(3 marks)

     (b) If $ \displaystyle 3\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}=\vec{0},$ show that the points $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.

(3 marks)

5.  (a) If $\displaystyle \tan \alpha =x+1$ and $ \displaystyle \tan \beta =x-1$, find $ \displaystyle \cot (\alpha -\beta )$ in terms of $ \displaystyle x.$

(3 marks)

     (b) Evaluate $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(2x-3)(\sqrt{x}-1)}}{{2{{x}^{2}}+x-3}}$ and $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{3{{{\sin }}^{2}}x-2\sin {{x}^{2}}}}{{3{{x}^{2}}}}.$

(3 marks)

Section (B)

Answer any FOUR Questions.

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)

     (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)

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)

     (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)

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

(5 marks)

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

(5 marks)

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)

     (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)

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)

     (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)

Section (C)

Answer any THREE Questions.

11. (a) In the diagram, two circles are tangent at $ \displaystyle A$ and have a common tangent touching them $ \displaystyle B$ and $ \displaystyle C$ respectively. If $ \displaystyle BA$ is produced to meet the second circle at $ \displaystyle D,$ show that $ \displaystyle CD$ is a diameter.

(5 marks)

     (b) $ \displaystyle ABC$ is a right triangle with $ \displaystyle A$ the right angle. $ \displaystyle E$ and $ \displaystyle D$ are points on opposite side of $ \displaystyle AC,$ with $ \displaystyle E$ on the same side of $ \displaystyle AC$ as $ \displaystyle B,$ such that $ \displaystyle ΔACD$ and $ \displaystyle ΔBCE$ are both equilateral. If $ \displaystyle α (ΔBCE) = 2 α (ΔACD),$ prove that $ \displaystyle ABC$ is an isosceles right triangle.

(5 marks)

12. (a) Two circles are drawn intersecting at $ \displaystyle A, B$ and so that the circumference of each passes through the centre of the another. Through $ \displaystyle A,$ a line is drawn meeting the circumference at $ \displaystyle C, D$ respectively. Prove that $ \displaystyle \vartriangle BCD$ is equilateral.

(5 marks)

     (b) Given that $ \displaystyle \sin \alpha =\frac{3}{5}$ and $ \displaystyle \cos \beta =\frac{{12}}{{13}}$, where $ \displaystyle α$ is obtuse and $ \displaystyle β$ is acute, find the exact values of $ \displaystyle \cos (α+β)$ and $ \displaystyle \cot (α- β).$

(5 marks)

13. (a) Solve $ \displaystyle ΔABC$ with $ \displaystyle b=12.5, c=23$ and $ \displaystyle α=38°20′.$

(5 marks)

     (b) Find the stationary points on the curve $ \displaystyle y=x^4(x^2-6)$ and determine their natures.

(5 marks)

14. (a) Show that the tangent to the curve $ \displaystyle y=e^{-2x}-3x$ at the point $ \displaystyle (a,0)$ meets the $ \displaystyle y$-axis at the point whose $ \displaystyle y$-coordinate is $ \displaystyle 2ae^{-2a} +3a.$

(5 marks)

     (b) Points $ \displaystyle A$ and $ \displaystyle B$ have position vectors $ \displaystyle {\vec{a}}$ and $ \displaystyle {\vec{b}}$ respectively, relative to an origin $ \displaystyle O.$ The point $ \displaystyle C$ lies on $ \displaystyle OA$ produced such that $ \displaystyle OC = 3OA,$ and $ \displaystyle D$ lies on $ \displaystyle OB$ such that $ \displaystyle OD = \frac {1}{4}OB.$ Express $ \displaystyle \overrightarrow{{AB}}$ and $ \displaystyle \overrightarrow{{CD}}$ in terms of $ \displaystyle {\vec{a}}$ and $ \displaystyle {\vec{b}}$. The line segments $ \displaystyle AB$ and $ \displaystyle CD$ intersect at $ \displaystyle P.$ If $ \displaystyle CP = hCD$ and $ \displaystyle AP = kAB,$ calculate the values of $ \displaystyle h$ and $ \displaystyle k.$

(5 marks)

ပညာရေး ဝန်ကြီးဌာန၏ သင်္ချာ နမူနာ - မေးခွန်း ပုံစံ(၃) - အဖြေ

2019 SAMPLE QUESTION (3)

MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS                                  Time Allowed: 3 hours

WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.

SECTION (A)

(Answer ALL questions.)

1.(a)      If $ \displaystyle f: R\to R$ is defined by $ \displaystyle f(x) = x^2 + 3,$ find the function $ \displaystyle g$ such that $\displaystyle (g\circ f)(x)=2{{x}^{2}}+3.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{2}}+3\\\\\ \ \ \ (g\circ f)(x)=2{{x}^{2}}+3\\\\\ \ \ \ g\left( {f(x)} \right)=2{{x}^{2}}+3\\\\\ \ \ \ g\left( {{{x}^{2}}+3} \right)=2({{x}^{2}}+3)-3\\\\\therefore \ \ g(x)=2x-3\end{array}$

1.(b)      The expression $ \displaystyle 6x^2-2x+3$ leaves a remainder of $ \displaystyle 3$ when divided by $ \displaystyle x-p.$ Determine the values of $ \displaystyle p.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)=6{{x}^{2}}-2x+3\\\\\ \ \ \ f(x)\ \text{leaves a remainder of}\ \text{3}\ \\\\\ \ \ \ \text{when divided by}\ x-p.\\\\\therefore \ \ \ f(p)=3\\\\\therefore \ \ \ 6{{p}^{2}}-2p+3=3\\\\\therefore \ \ \ 3{{p}^{2}}-p=0\\\\\therefore \ \ \ p(3p-1)=0\\\\\therefore \ \ \ p=0\ (\text{or})\ p=\displaystyle \frac{1}{3}\\\ \ \ \end{array}$

2.(a)      Find the coefficient of $ \displaystyle {x}^{-10}$ in the expansion of $ \displaystyle {{\left( {2-\frac{1}{{{{x}^{2}}}}} \right)}^{8}}$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {2-\displaystyle \frac{1}{{{{x}^{2}}}}} \right)}^{8}}\\\\\,\ \ \ ={}^{8}{{C}_{r}}{{2}^{{8-r}}}{{(-1)}^{r}}{{x}^{{-2r}}}\\\\\ \ \ \ \ \ \text{For}\ {{x}^{{-10}}},\ -2r=-10\Rightarrow r=5\\\\\therefore \ \ \ \ \text{Coefficient of}\ {{x}^{{-10}}}={}^{8}{{C}_{5}}{{2}^{3}}{{(-1)}^{5}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={}^{8}{{C}_{3}}{{2}^{3}}{{(-1)}^{5}}\ \ \left[ {\because {}^{n}{{C}_{r}}={}^{n}{{C}_{{n-r}}}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{8\times 7\times 6}}{{1\times 2\times 3}}(8)(-1)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-448\end{array}$

2.(b)      Which term of the A.P. $ \displaystyle 6, 13, 20, 27, … is 111?$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ 6,13,20,27,...,111\ \text{is an A}\text{.P}\text{.}\\\\\therefore \ \ a=6,\ d=13-6=7,{{u}_{n}}=111\\\\\ \ \ \ {{u}_{n}}=a+(n-1)d\\\\\therefore \ \ 6+(n-1)7=111\\\\\,\ \ \ 7(n-1)=105\\\\\ \ \ \ n-1=15\\\\\therefore \ \ n=16\ \ \end{array}$

3.(a)      If $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right),B=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)$ find the value of $ \displaystyle k$ such that $ \displaystyle AB = BA.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ A=\displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right),B=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ \\\\\ \ \ \ AB=BA\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ =\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right)\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {2+0} & {0+0} \\ {1+10} & {0+5k} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {2+0} & {0+0} \\ {4+k} & {0+5k} \end{array}} \right)\ \\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ {11} & {5k} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ {4+k} & {5k} \end{array}} \right)\ \\\\\therefore \ \ 4+k=11\Rightarrow k=7\end{array}$

3.(b)      If a die is rolled $ \displaystyle 60$ times, what is the expected frequency of a number divisible by $ \displaystyle 3$ turns up?

(3 marks)

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$ \displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ \text{When a die is rolled}\ \text{once,}} \\ {} \\ {\ \ \ \ \ \text{Set of all possible outcomes = }\left\{ {\text{1, 2, 3, 4, 5, 6}} \right\}\text{ }} \\ {} \\ {\ \ \ \ \ \text{Number of possible outcomes = 6}} \\ {} \\ {\ \ \ \ \ \text{Set of favourable outcomes for a number divisible by 3 = }\left\{ {\text{3, 6 }} \right\}} \\ {} \\ {\ \ \ \ \ \text{Number of favourable outcomes = 2}} \\ {} \\ {\ \ \ \ \ P\text{(a number divisible by 3)}=\displaystyle \frac{2}{6}=\displaystyle \frac{1}{3}} \\ {} \\ {\ \ \ \ \ \text{When a die is rolled}\ \text{60 times,}} \\ {} \\ {\ \ \ \ \ \text{Expected frequency for a number divisible by 3 turns up}} \\ {} \\ {\ \ \ \ \ \ =\text{probability}\ \times \text{number of trials}} \\ {} \\ {\ \ \ \ \ \ =\displaystyle \frac{1}{3}\times 60} \\ {} \\ {\ \ \ \ \ \ =20} \end{array}$

4.(a)      In the given figure, $ \displaystyle AB = BC.$ Find $ \displaystyle x$ and $ \displaystyle y.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ ABPC\ \text{is a cyclic quadrilateral}\text{.}\\\\\therefore \ \ \ \ y+36{}^\circ =180{}^\circ \\\\\therefore \ \ \ \ y=144{}^\circ \\\\\ \ \ \ \ \ \text{Since}\ AB=BC,\\\\\ \ \ \ \ \ \angle ACB=36{}^\circ \\\\\therefore \ \ \ \ \angle ABC=180{}^\circ -(36{}^\circ +36{}^\circ )=108{}^\circ \\\\\ \ \ \ \ \ \text{Since}\ ABCD\ \text{is a cyclic quadrilateral,}\\\\\ \ \ \ \ \ x+108{}^\circ =180{}^\circ \\\\\therefore \ \ \ \ x=72{}^\circ \end{array}$

4.(b)      It is given that $ \displaystyle {\vec{a}}$ and $ \displaystyle {\vec{b}}$ are non-zero and non-parallel vectors. If $ \displaystyle 3\vec{a}\ +x\left( {\vec{b}-\vec{a}} \right)=y\left( {\vec{a}+2\vec{b}} \right)$ find the values of $ \displaystyle x$ and $ \displaystyle y.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \vec{a}\ \text{and}\ \vec{b}\ \text{are non-zero and non-parallel vectors}\text{.}\\\\\ \ \ \ 3\vec{a}\ +x\left( {\vec{b}-\vec{a}} \right)=y\left( {\vec{a}+2\vec{b}} \right)\\\\\therefore \ \ \ 3\vec{a}\ +x\vec{b}-x\vec{a}=y\vec{a}+2y\vec{b}\\\\\therefore \ \ \ (3-x)\vec{a}\ +x\vec{b}=y\vec{a}+2y\vec{b}\\\\\therefore \ \ \ 3-x=y\ \text{and}\ x=2y\\\\\therefore \ \ \ 3-2y=y\Rightarrow 3y=3\Rightarrow y=1\\\\\therefore \ \ \ x=2(1)=2\end{array}$

5.(a)      Prove that $ \displaystyle \operatorname{cosec}\theta =\frac{{\cos 2\theta }}{{\sin \theta }}+\frac{{\sin 2\theta }}{{\cos \theta }}.$

(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \text{LHS }=\operatorname{cosec}\theta \\\\\ \ \ \text{RHS }=\displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos (2\theta -\theta )}}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sin \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\operatorname{cosec}\theta \\\\\therefore \ \ \ \operatorname{cosec}\theta =\displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\end{array}$

5.(b)      Evaluate $ \displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-8}}{{{{x}^{2}}+3x-10}}\ \operatorname{and}\ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{{2x}}-\sqrt{a}}}{{\sqrt{{2x}}+\sqrt{a}}}.$

(3 marks)

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$ \displaystyle \ \ \ \ \ \ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{{{x}^{3}}-8}}{{{{x}^{2}}+3x-10}}$

$ \displaystyle \ \ \ =\ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{(x-2)({{x}^{2}}+2x+4)}}{{(x-2)(x+5)}}$

$ \displaystyle \ \ \ =\ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{{{x}^{2}}+2x+4}}{{x+5}}$

$ \displaystyle \ \ \ =\ \ \displaystyle \frac{{{{2}^{2}}+2(2)+4}}{{2+5}}$

$ \displaystyle \ \ \ =\ \ \displaystyle \frac{{12}}{7}$


$ \displaystyle \ \ \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{{2x}}-\sqrt{a}}}{{\sqrt{{2x}}+\sqrt{a}}}$

$ \displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{x}\left( {\sqrt{2}-\displaystyle \frac{{\sqrt{a}}}{{\sqrt{x}}}} \right)}}{{\sqrt{x}\left( {\sqrt{2}+\displaystyle \frac{{\sqrt{a}}}{{\sqrt{x}}}} \right)}}$

$ \displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{2}-\sqrt{{\displaystyle \frac{a}{x}}}}}{{\sqrt{2}+\sqrt{{\displaystyle \frac{a}{x}}}}}$

$ \displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{2}-\sqrt{0}}}{{\sqrt{2}+\sqrt{0}}}$

$ \displaystyle \ \ \ =\ \ 1$

SECTION (B)

(Answer any FOUR questions.)

6.(a)      A function f is defined by $ \displaystyle f(x) = 3x - 1.$ Determine whether $ \displaystyle {{(f\circ f)}^{{-1}}}(x)$ is the same as $ \displaystyle ({{f}^{{-1}}}\circ {{f}^{{-1}}})(x).$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)=3x-1\\\\\ \ \ \ \ \text{Let}\ {{f}^{{-1}}}(x)=y,\text{then}\ f(y)=x\\\\\therefore \ \ \ 3y-1=x\\\\\therefore \ \ \ y= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ {{f}^{{-1}}}(x)= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ \left( {{{f}^{{-1}}}\circ {{f}^{{-1}}}} \right)(x)={{f}^{{-1}}}\left( {{{f}^{{-1}}}(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( { \displaystyle \frac{{x+1}}{3}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{ \displaystyle \frac{{x+1}}{3}+1}}{3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{x+4}}{9}\\\\\ \ \ \ \ \text{Let}\ {{(f\circ f)}^{{-1}}}(x)=z\ \text{then}\ (f\circ f)(z)=x\\\\\therefore \ \ \ f\left( {f(z)} \right)=x\\\\\therefore \ \ \ f(3z-1)=x\\\\\therefore \ \ \ 3(3z-1)-1=x\\\\\therefore \ \ \ 3z-1= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ 3z= \displaystyle \frac{{x+4}}{3}\\\\\therefore \ \ \ z= \displaystyle \frac{{x+4}}{9}\\\\\therefore \ \ \ {{(f\circ f)}^{{-1}}}(x)= \displaystyle \frac{{x+4}}{9}\\\\\therefore \ \ \ {{(f\circ f)}^{{-1}}}(x)=\ \left( {{{f}^{{-1}}}\circ {{f}^{{-1}}}} \right)(x)\end{array}$

6.(b)      Given that $ \displaystyle f(x) = x^3 + px^2 - 2x + 4\sqrt{3}$ has a factor $ \displaystyle x + \sqrt{2},$ find the value of $ \displaystyle p.$ Show that $ \displaystyle x - 2\sqrt{3}$ is also a factor and solve the equation f(x) = 0.

(5 marks)

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$ \displaystyle \begin{array}{*{20}{l}} {\ \ \ \ f(x)={{x}^{3}}+p{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\ \ \ \ x+\sqrt{2}\ \text{is a factor of }f(x).} \\ {} \\ {\therefore \ \ f(-\sqrt{2})=0} \\ {} \\ {\ \ \ \ {{{(-\sqrt{2})}}^{3}}+p{{{(-\sqrt{2})}}^{2}}-2(-\sqrt{2})+4\sqrt{3}=0} \\ {} \\ {\ \ \ \ -2\sqrt{2}+2p+2\sqrt{2}+4\sqrt{3}=0} \\ {} \\ {\therefore \ \ p=-2\sqrt{3}} \\ {} \\ {\therefore \ \ f(x)={{x}^{3}}-2\sqrt{3}{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\therefore \ \ f(2\sqrt{3})={{{(2\sqrt{3})}}^{3}}-2\sqrt{3}{{{(2\sqrt{3})}}^{2}}-2(2\sqrt{3})+4\sqrt{3}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{{(2\sqrt{3})}}^{3}}-{{{(2\sqrt{3})}}^{3}}-4\sqrt{3}+4\sqrt{3}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0} \\ {} \\ {\therefore \ \ \ x-2\sqrt{3}\ \text{is also a factor of }f(x).} \\ {} \\ {\ \ \ \ \text{Let}\ f(x)=(x+\sqrt{2})(x-2\sqrt{3})(x-k)} \\ {} \\ {\therefore \ \ \ (x+\sqrt{2})(x-2\sqrt{3})(x-k)={{x}^{3}}-2\sqrt{3}{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\therefore \ \ \ \sqrt{2}(-2\sqrt{3})(-k)=4\sqrt{3}} \\ {} \\ {\therefore \ \ \ k=\sqrt{2}} \\ {} \\ \begin{array}{l}\therefore \ \ f(x)=(x+\sqrt{2})(x-2\sqrt{3})(x-\sqrt{2})\\\\\ \ \ f(x)=0\Rightarrow (x+\sqrt{2})(x-2\sqrt{3})(x-\sqrt{2})=0\\\\\therefore \ \ x=-\sqrt{2}\ (\text{or})\ x=2\sqrt{3}\ (\text{or})\ x=\sqrt{2}\end{array} \end{array}$

7.(a)      Let $ \displaystyle J^{+}$ be the set of all positive integers. Is the function $ \displaystyle \odot $ defined by $ \displaystyle x\odot y = x + 3y$ a binary operation on $ \displaystyle J^{+}?$ Why? If it is a binary operation, find $ \displaystyle (2\odot 3) \odot 4.$ Solve the equation $ \displaystyle (k\odot 5) - (3\odot k) = 2k + 8.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ {{J}^{+}}=\text{the set of all positive integers}\text{.}\\\\\ \ \ x\odot y=x+3y\ \ \\\\\ \ \ \text{Since}\ x,y\in {{J}^{+}},3y\in \ \ {{J}^{+}}.\\\\\therefore \ x+3y\in {{J}^{+}}\\\\\therefore \ x\odot y\in {{J}^{+}}\\\\\therefore \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \odot \ \text{is a binary operation}.\\\\\therefore \ 2\odot 3=2+3(3)=11\\\\\therefore \ \left( {2\odot 3} \right)\odot 4=11\odot 4\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =11+3(4)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =23\\\\\ \ \ k\odot 5=k+3(5)=15+k\\\\\ \ \ 3\odot k=3+3(k)=3+3k\\\\\ \ \ (k\odot 5)-(3\odot k)=2k+8\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ (15+k)-(3+3k)=2k+8\\\\\therefore \ 12-2k=2k+8\\\\\therefore \ 4k=4\Rightarrow k=1\end{array}$

7.(b)      The first three terms in the expansion of $ \displaystyle (a + b)^n,$ in ascending powers of $ \displaystyle b,$ are denoted by $ \displaystyle p, q$ and $ \displaystyle r$ respectively. Show that $ \displaystyle \frac{{{{q}^{2}}}}{{pr}}=\frac{{2n}}{{n-1}}.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ {{(a+b)}^{n}}=p+q+r+...\\\\\ \ \ \ \ {}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{{n-1}}}b+{}^{n}{{C}_{2}}{{a}^{{n-2}}}{{b}^{2}}+...=p+q+r+...\\\\\ \ \ \ \ {{a}^{n}}+n{{a}^{{n-1}}}b+ \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}+...=p+q+r+...\\\\\therefore \ \ \ p={{a}^{n}},\ \ q=n{{a}^{{n-1}}}b,\ \ r= \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{{{{\left( {n{{a}^{{n-1}}}b} \right)}}^{2}}}}{{{{a}^{n}}\cdot \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{{{n}^{2}}{{a}^{{2n-2}}}{{b}^{2}}}}{{ \displaystyle \frac{{n(n-1)}}{2}{{a}^{{2n-2}}}{{b}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}={{n}^{2}}\cdot \displaystyle \frac{2}{{n(n-1)}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{2n}}{{n-1}}\end{array}$

8.(a)      Find the solution set of the inequation $ \displaystyle 2 + 3x > 5x^2$ and illustrate it on the number line.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ 2+3x>5{{x}^{2}}\\\\\therefore \ \ 5{{x}^{2}}-3x-2<0\\\\\ \ \ \ \text{Let}\ y=5{{x}^{2}}-3x-2\\\\\ \ \ \ \text{When}\ y=0,\\\\\ \ \ \ 5{{x}^{2}}-3x-2=0\\\\\ \ \ \ (5x+2)(x-1)=0\\\\\therefore \ \ x=-\displaystyle \frac{2}{5}\ (\text{or})\ x=1\\\\\therefore \ \ \text{The graph cuts the x-axis at (}-\displaystyle \frac{2}{5},0)\ \text{and}\ \text{(1,0)}\text{.}\\\\\ \ \ \ \text{When}\ x=0,y=-2\\\\\therefore \ \ \text{The graph cuts the y-axis at (}0,-2)\text{.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ \text{Solution set}=\{x|-\displaystyle \frac{2}{5}<x<0\}\\\\\ \ \ \ \text{Number Line:}\end{array}$

8.(b)      The sum of four consecutive numbers in an A.P. is $ \displaystyle 28.$ The product of the second and third numbers exceeds that of the first and last by $ \displaystyle 18.$ Find the numbers.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the four consecutive numbers in A}\text{.P}\text{. be}\\\\\ \ \ \ \ a,a+d,a+2d\ \operatorname{and}\ a+3d.\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ a+a+d+a+2d+a+3d=28\\\\\ \ \ \ \ 4a+6d=28\\\\\therefore \ \ \ 2a+3d=14\ \ \ \ -----\ (1)\\\\\ \ \ \ \ (a+d)(a+2d)-a(a+3d)=18\\\\\therefore \ \ \ {{a}^{2}}+3ad+2{{d}^{2}}-{{a}^{2}}-3ad=18\\\\\therefore \ \ \ {{d}^{2}}=9\Rightarrow d=\pm 3\ -----\ (2)\\\\\ \ \ \ \ \text{When }d=-3,\\\\\ \ \ \ \ \ 2a-9=14\ \Rightarrow a= \displaystyle \frac{{23}}{2}\\\\\therefore \ \ \ \ \ {{\text{1}}^{{\text{st}}}}\text{ number}\ \text{=} \displaystyle \frac{{23}}{2}\\\\\ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\text{ number}\ \text{=} \displaystyle \frac{{17}}{2}\\\\\ \ \ \ \ \ {{\text{3}}^{{\text{rd}}}}\text{ number}\ \text{=} \displaystyle \frac{{11}}{2}\\\\\ \ \ \ \ \ {{\text{4}}^{{\text{th}}}}\text{ number}\ \text{=} \displaystyle \frac{5}{2}\\\\\ \ \ \ \ \text{When }d=3,\\\\\ \ \ \ \ \ 2a+9=14\ \Rightarrow a= \displaystyle \frac{5}{2}\\\\\therefore \ \ \ \ {{\text{1}}^{{\text{st}}}}\text{ number}\ \text{=} \displaystyle \frac{5}{2}\\\\\ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\text{ number}\ \text{=} \displaystyle \frac{{11}}{2}\\\\\ \ \ \ \ \ {{\text{3}}^{{\text{rd}}}}\text{ number}\ \text{=} \displaystyle \frac{{17}}{2}\\\\\ \ \ \ \ \ {{\text{4}}^{{\text{th}}}}\text{ number}\ \text{=} \displaystyle \frac{{23}}{2}\end{array}$

9.(a)      The product of the first $ \displaystyle 3$ terms of a G.P. is $ \displaystyle 1$ and the product of the third, fourth and fifth terms is $ \displaystyle 11\frac{25}{64}.$ Find the fifth term of the G.P.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let}\ {{u}_{1}},{{u}_{2}},{{u}_{3}},...\ \text{be the given G}\text{.P}\text{.}\\\\\ \ \ \ \ \text{Let}\ a\text{ be the first term and }r\ \text{be }\\\ \ \ \ \ \text{the}\ \text{common ratio of given G}\text{.P}\text{.}\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ {{u}_{1}}\times {{u}_{2}}\times {{u}_{3}}=1\\\\\therefore \ \ \ a\times ar\times a{{r}^{2}}=1\\\\\ \ \ \ \ {{(ar)}^{3}}=1\Rightarrow a= \displaystyle \frac{1}{r}\\\\\ \ \ \ \ {{u}_{3}}\times {{u}_{4}}\times {{u}_{5}}=11 \displaystyle \frac{{25}}{{64}}\\\\\ \ \ \ \ a{{r}^{2}}\times a{{r}^{3}}\times a{{r}^{4}}=11 \displaystyle \frac{{25}}{{64}}\\\\\therefore \ \ \ {{a}^{3}}{{r}^{9}}= \displaystyle \frac{{729}}{{64}}\\\\\ \ \ \ \ \displaystyle \frac{1}{{{{r}^{3}}}}\times {{r}^{9}}= \displaystyle \frac{{729}}{{64}}\\\\\ \ \ \ \ {{r}^{6}}= \displaystyle \frac{{729}}{{64}}\Rightarrow r=\pm \displaystyle \frac{3}{2}\\\\\therefore \ \ \ {{u}_{5}}=a{{r}^{4}}= \displaystyle \frac{1}{r}\times {{r}^{4}}={{r}^{3}}=\pm \displaystyle \frac{{27}}{8}\end{array}$

9.(b)      Show that the matrix $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)$ satisfies the equation $ \displaystyle A^2 - 4A - 5I = O,$ where $ \displaystyle I$ is the unit matrix of order $ \displaystyle 2.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ A= \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right),I= \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ {{A}^{2}}-4A-5I\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)-4 \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)-5 \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} {4+9} & {6+6} \\ {6+6} & {9+4} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-8} & {-12} \\ {-12} & {-8} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & 0 \\ 0 & {-5} \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} {4+9-8-5} & {6+6-12+0} \\ {6+6-12+0} & {9+4-8-5} \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\ \ =\ O\\\\\therefore \ \ {{A}^{2}}-4A-5I=O\end{array}$

10.(a)    Find the inverse of the matrix $ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)$ and use it to solve the system of equations $ \displaystyle 7x + 8y = 10, 5x + 6y = 7.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let}\ A= \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)\\\\\therefore \ \ \ \det A=42-40=2\ne 0\\\\\therefore \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\\\\\therefore \ \ {{A}^{{-1}}}=\ \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-8} \\ {-5} & 7 \end{array}} \right)\\\\\ \ \ \ 7x+8y=10\\\\\ \ \ \ 5x+6y=7\\\\\ \ \ \ \text{Transforming into matrix form,}\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ {{ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)}^{{-1}}} \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)={{ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)}^{{-1}}} \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-8} \\ {-5} & 7 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} {60-56} \\ {-50+49} \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 4 \\ {-1} \end{array}} \right)\ \\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {- \displaystyle \frac{1}{2}} \end{array}} \right)\ \\\\\therefore \ \ x=2,y=- \displaystyle \frac{1}{2}\end{array}$

10.(b)    A coin is tossed three times. Head or tail is recorded each time. Drawing a tree diagram, find the probabilities of getting exactly one head, and getting no head.

(5 marks)

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{Number of possible outcomes}=8\\\\\ \ \ \ \ \ \text{Set of favourable outcomes for getting }\\\ \ \ \ \ \ \text{exactly one head}=\ \{HTT,THT,TTH\}\\\\\therefore \ \ \ \ \text{Number}\ \text{favourable outcomes for getting }\\\ \ \ \ \ \ \text{exactly one head}=3\\\\\therefore \ \ \ \ P(\text{getting exactly one head)}= \displaystyle \frac{3}{8}\\\\\ \ \ \ \ \ \text{Set of favourable outcomes for getting }\\\ \ \ \ \ \ \text{no head}=\ \{TTT\}\\\\\therefore \ \ \ \ \text{Number}\ \text{favourable outcomes for getting }\\\ \ \ \ \ \ \text{no head}=1\\\\\therefore \ \ \ \ P(\text{getting no head)}= \displaystyle \frac{1}{8}\ \ \ \ \ \ \end{array}$

SECTION (C)

(Answer any THREE questions.)

11.(a)    Two circles intersect at $ \displaystyle A, B.$ At $ \displaystyle A$ a tangent is drawn to each circle meeting the circles again at $ \displaystyle P$ and $ \displaystyle Q$ respectively. Prove that $ \displaystyle ∠ ABP = ∠ ABQ.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{In}\odot ABP,\\\\\ \ \ \ \angle ABP=\alpha \ \ \ \ \ \ \text{(}\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment)}\\\\\ \ \ \ \text{Similarly,}\ \text{in}\odot ABQ,\\\\\ \ \ \ \angle ABQ=\beta \ \ \ \ \ \ \text{(}\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment)}\\\\\ \ \ \ \text{But}\ \alpha =\beta \ \ \ \ \ \ \ (\text{vertically opposite }\ \angle \text{s})\\\\\therefore \ \ \angle ABP=\angle ABQ\end{array}$

11.(b)    In $ \displaystyle ∆ABC, AD$ and $ \displaystyle BE$ are altitudes. If $ \displaystyle ∠ACB = 45°,$ prove that $ \displaystyle α(∆DEC) = α(ABDE).$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Since }\angle AEB\text{ }=\angle ADB=\text{90}{}^\circ \text{,}\\\\\ \ \ \ \ \ \ ABDE\text{ is cyclic}\text{.}\\\\\ \ \ \ \ \ \ \angle ABC\text{ }=\angle DEC\\\\\ \ \ \ \ \ \ \text{In }\vartriangle DEC\text{ and }\vartriangle ABC\text{,}\\\text{ }\\\ \ \ \ \ \ \ \angle ABC\text{ }=\angle DEC\ \ \text{(proved)}\\\\\ \ \ \ \ \ \ \angle C=\angle C\ \text{(common }\angle \text{)}\\\\\therefore \ \ \ \ \ \vartriangle DEC\sim \vartriangle ABC\ \ \ \text{(AA corollary)}\\\\\ \ \ \ \ \ \ \displaystyle \frac{{\alpha (\vartriangle DEC)}}{{\alpha (\vartriangle ABC)}}= \displaystyle \frac{{D{{C}^{2}}}}{{A{{C}^{2}}}}\\\\\ \ \ \ \ \ \ \ \text{In right }\vartriangle ACD,\\\\\ \ \ \ \ \ \ \ \displaystyle \frac{{DC}}{{AC}}=\cos (\angle ACB)=\cos 45{}^\circ = \displaystyle \frac{1}{{\sqrt{2}}}\\\\\therefore \ \ \ \ \ \ \displaystyle \frac{{D{{C}^{2}}}}{{A{{C}^{2}}}}= \displaystyle \frac{1}{2}\Rightarrow \displaystyle \frac{{\alpha (\vartriangle DEC)}}{{\alpha (\vartriangle ABC)}}= \displaystyle \frac{1}{2}\\\\\therefore \ \ \ \ \ \ 2\alpha (\vartriangle DEC)=\alpha (\vartriangle ABC)\\\\\therefore \ \ \ \ \ \ 2\alpha (\vartriangle DEC)=\alpha (\vartriangle DEC)+\alpha (ABDE)\\\\\therefore \ \ \ \ \ \ \alpha (\vartriangle DEC)=\alpha (ABDE)\end{array}$

12.(a)    In $ \displaystyle ∆ABC, AB = AC.\ P$ is a point on $ \displaystyle BC,$ and $ \displaystyle Y$ is a point on $ \displaystyle AP.$ The circles $ \displaystyle BPY$ and $ \displaystyle CPY$ cut $ \displaystyle AB$ and $ \displaystyle AC$ respectively at $ \displaystyle X$ and $ \displaystyle Z.$ Prove $ \displaystyle XZ ∥ BC.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{In}\odot BPY\text{,}\\\\\ \ \ \ \ \ \ AX\cdot AB=AY\cdot AP\\\\\ \ \ \ \ \ \ \text{Similarly, in}\odot CPY\text{,}\\\\\ \ \ \ \ \ \ AZ\cdot AC=AY\cdot AP\\\\\therefore \ \ \ \ \ AX\cdot AB=AZ\cdot AC\\\\\ \ \ \ \ \ \ BCZX\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \ \ \theta =\gamma \ \ \ \text{(exterior }\angle \ \text{of cyclic quadrilateral}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\text{interior opposite }\angle \text{)}\\\\\ \ \ \ \ \ \ \text{Since}\ AB=AC,\beta =\gamma .\\\\\therefore \ \ \ \ \ \theta =\beta \\\\\ \ \ \ \ \ \ \text{Since }\theta \text{ and }\beta \text{ are corresponding }\\\ \ \ \ \ \ \ \text{angles, we can say}\ XZ\parallel BC.\end{array}$

12.(b)    Given that $ \displaystyle \sin α = \frac{15}{17}$ and that $ \displaystyle \cos β = -\frac{3}{5}$ and that $ \displaystyle α$ and $ \displaystyle β$ are in the same quadrant, find without using tables, the values of $ \displaystyle \sin 2α , \cos \frac{α}{2}$ and $ \displaystyle \cos 2β.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \alpha = \displaystyle \frac{{15}}{{17}},\ \ \ \cos \beta =- \displaystyle \frac{3}{5}\\\ \ \ \ \text{and }\alpha \ \text{and}\ \beta \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \text{Since}\sin \alpha \ \text{is positive and}\cos \beta \ \text{is negative,}\\\ \ \ \ \alpha \ \text{and}\ \beta \ \text{will be in }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ \sin \alpha =\displaystyle \frac{{15}}{{17}},\ \cos \alpha =-\displaystyle \frac{8}{{17}}\\\\\therefore \ \ \sin 2\alpha =2\sin \alpha \ \cos \alpha \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ \left( {\displaystyle \frac{{15}}{{17}}} \right)\left( {-\displaystyle \frac{8}{{17}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{240}}{{289}}\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\pm \sqrt{{\displaystyle \frac{{1+\cos \alpha }}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \sqrt{{\displaystyle \frac{{1-\displaystyle \frac{8}{{17}}}}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{3}{{\sqrt{{34}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \text{Since}\ {{90}^{{}^\circ }}<\alpha <{{180}^{{}^\circ }},{{45}^{{}^\circ }}<\displaystyle \frac{\alpha }{2}<{{90}^{{}^\circ }},\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \ \cos \beta =-\displaystyle \frac{3}{5}\ \ \ \ \ \left[ {\text{given}} \right]\\\\\ \ \ \ \cos 2\beta =2{{\cos }^{2}}\beta -1\\\\\therefore \ \ \cos 2\beta =2{{\left( {-\displaystyle \frac{3}{5}} \right)}^{2}}-1\\\\\therefore \ \ \cos 2\beta =-\displaystyle \frac{7}{{25}}\end{array}$

13.(a)    In A town $ \displaystyle A$ is $ \displaystyle 50$ miles away from a town $ \displaystyle B$ in the direction $ \displaystyle N 35° E$ and a town $ \displaystyle C$ is $ \displaystyle 68$ miles from $ \displaystyle B$ in the direction $ \displaystyle N 42° 1{2}' W.$ Calculate the distance and bearing of $ \displaystyle A$ from $ \displaystyle C.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ PQ=50\ \text{mi,}\ QR=68\ \text{mi}\\\\\ \ \ \ \ \angle PQR=35{}^\circ +42{}^\circ 1{2}'=77{}^\circ 1{2}'\\\\\ \ \ \ \ \text{By the law of cosines,}\\\\\,\ \ \ \ P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}-2\cdot PQ\cdot QR\cos Q\\\\\,\ \ \ \ P{{R}^{2}}={{50}^{2}}+{{68}^{2}}-2(50)(68)\cos 77{}^\circ 1{2}'\\\\\,\ \ \ \ P{{R}^{2}}=2500+4624-6800\cos 77{}^\circ 1{2}'\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 6800} & {3.8325} \\ { \cos 77 {}^\circ1{2}' } & {\overline{1}.3455}\\ \hline {1507} & {3.1780} \\ \hline \end{array}\end{array}\\\\\ \ \ \ \ P{{R}^{2}}=7124-1507\\\\\ \ \ \ \ P{{R}^{2}}=5617\\\\\ \ \ \ \ PR=74.95\ \text{mi}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{P{{R}^{2}}+Q{{R}^{2}}-P{{Q}^{2}}}}{{2(PR)(QR)}}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{5617+4624-2500}}{{2(74.95)(68)}}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{7741}}{{74.95\times 136}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 7741} & {3.8888} \\ \hline { 74.95 } & {1.8748}\\ {136} & {2.1335} \\ \hline {} & {4.0083} \\ \hline {\cos 40 {}^\circ3{5}'} & {\overline{1}.8805} \\ \hline \end{array}\end{array} \\\\\ \ \ \ \ \cos \gamma =\cos 40{}^\circ 3{5}'\\\\\ \ \ \ \ \gamma =40{}^\circ 3{5}'\\\\\therefore \ \ \ \theta =42{}^\circ 1{2}'+40{}^\circ 3{5}'=82{}^\circ 4{7}'\\\\\therefore \ \ \ P\ \text{is 74}\text{.95 mi away from}\,R.\\\ \ \ \ \ \text{It is in the direction }S\text{ }82{}^\circ 4{7}'\text{ }E\text{ from }R\text{.}\\\ \ \ \ \ \end{array}$

13.(b)    If $ \displaystyle y=x^2+2x+3$ show that $ \displaystyle {{\left( {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right)}^{3}}+{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=4y.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ y={{x}^{2}}+2x+3\\\\\ \ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2x+2\\\\\ \ \ \ \ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=2\\\\\therefore \ \ \ \ \ {{\left( {\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right)}^{3}}+{{\left( {\displaystyle \frac{{dy}}{{dx}}} \right)}^{2}}\\\\\ \ \ =\ \ {{2}^{3}}+{{(2x+2)}^{2}}\\\\\ \ \ =\ \ 8+4{{x}^{2}}+8x+4\\\\\ \ \ =\ \ 4{{x}^{2}}+8x+12\\\\\ \ \ =\ \ 4({{x}^{2}}+x+3)\\\\\ \ \ =\ \ 4y\end{array}$

14.(a)    The vector $ \displaystyle \overrightarrow{{OP}}$ has a magnitude of $ \displaystyle 26$ units and has the same direction as $ \displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right).$ The vector $ \displaystyle \overrightarrow{{OQ}}$ has a magnitude of $ \displaystyle 20$ units and has the same direction as $ \displaystyle \left( {\begin{array}{*{20}{c}} {3} \\ {4} \end{array}} \right).$ Find the magnitude of $ \displaystyle PQ.$

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let}\ \vec{a}=\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right).\\\\\therefore \ \ \ \ \ \displaystyle \left| {\vec{a}} \right|=\sqrt{{{{{(-5)}}^{2}}+{{{12}}^{2}}}}=\sqrt{{169}}=13\\\\\therefore \ \ \ \ \ \hat{a}\ =\displaystyle \frac{{\vec{a}}}{{\displaystyle \left| {\vec{a}} \right|}}=\displaystyle \frac{1}{{13}}\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right)\\\\\ \ \ \ \ \ \overrightarrow{{OP}}\ \text{has}\ \text{magnitude 26 units and the same direction as}\ \vec{a}.\\\\\therefore \ \ \ \ \ \overrightarrow{{OP}}=\text{26}\hat{a}=26\times \displaystyle \frac{1}{{13}}\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {-10} \\ {24} \end{array}} \right)\\\\\ \ \ \ \ \ \ \text{Let}\ \vec{b}=\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right).\\\\\therefore \ \ \ \ \ \displaystyle \left| {\vec{b}} \right|=\sqrt{{{{3}^{2}}+{{4}^{2}}}}=\sqrt{{25}}=5\\\\\therefore \ \ \ \ \ \hat{b}\ =\displaystyle \frac{{\vec{b}}}{{\displaystyle \left| {\vec{b}} \right|}}=\displaystyle \frac{1}{5}\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right)\\\\\ \ \ \ \ \ \overrightarrow{{OQ}}\ \text{has}\ \text{magnitude 20 units and the same direction as}\ \vec{b}.\\\\\therefore \ \ \ \ \ \overrightarrow{{OQ}}=\text{20}\hat{b}=20\times \displaystyle \frac{1}{5}\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {12} \\ {16} \end{array}} \right)\\\\\therefore \ \ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}=\displaystyle \left( {\begin{array}{*{20}{c}} {12} \\ {16} \end{array}} \right)-\displaystyle \left( {\begin{array}{*{20}{c}} {-10} \\ {24} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {22} \\ {-8} \end{array}} \right)\\\\\therefore \ \ \ \ \ \displaystyle \left| {\overrightarrow{{PQ}}} \right|=\sqrt{{{{{22}}^{2}}+{{{(-8)}}^{2}}}}=\sqrt{{548}}=2\sqrt{{137}}\ \text{units}.\end{array}$

14.(b)    Find the stationary points of the curve $ \displaystyle y = x^2(3 - x)$ and determine their natures.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ y={{x}^{2}}(3-x)\\\\\therefore \ \ \ y=3{{x}^{2}}-{{x}^{3}}\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=6x-3{{x}^{2}}=3x(2-x)\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(2-x)=0\\\\\therefore \ \ \ \ x=0\ (\text{or)}\ x=2\\\\\ \ \ \ \ \text{When}\ x=0,y=0.\\\\\ \ \ \ \ \text{When}\ x=2,y={{2}^{2}}(3-2)=4.\\\\\therefore \ \ \ \text{The stationary points are (0,0) and (2,4)}\text{.}\\\\\ \ \ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6-6x=6(1-x)\\\\\ \ \ \ \ \text{When}\ x=0,\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(1-0)=6>0\\\\\therefore \ \ \ \text{(0,0) is a minimum turning point}\text{.}\\\\\ \ \ \ \ \text{When}\ x=2,\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(1-2)=-6<0\\\\\therefore \ \ \ \text{(2,4) is a maximum turning point}\text{.}\end{array}$