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)

Answer for 2019 Sample Question : Section (C)

ဒီေနရာမွာ တင္ေပးခဲ့ေသာ Section (C) ေမးခြန္းရဲ့ အေျဖျဖစ္ပါတယ္။ Section (A) ရဲ့ အေျဖကိုေတာ့ ဒီေနရာမွာ တင္ေပးခဲ့ၿပီး  Section (B) အေျဖကို ဒီေနရာမွာ တင္ေပးထားပါတယ္။

Section (C)

Solution

11.    (a) Two unequal circles are tangent externally at $ \displaystyle O$. $ \displaystyle AB$ the chord of the first circle is tangent to the second circle at $ \displaystyle C$, and $ \displaystyle AO$ meets this circle at $ \displaystyle E$. Prove that $ \displaystyle ∠BOC=∠COE$.

(5 marks)

Show/Hide Solution

    Draw a common tangent at $ \displaystyle O$ to meet $ \displaystyle BC$ at $ \displaystyle D.$

$\displaystyle \begin{array}{l}\ \ \ \text{In}\odot P,\\\\\ \ \ \alpha =\angle A\ \ \ \text{( }\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \text{in alternate segment)}\\\\\ \ \ OD=CD\ \ \text{(tangents from same exterior point)}\\\\\therefore \beta =\gamma \\\\\therefore \alpha +\beta =\angle A+\gamma \\\\\ \ \ \text{But }\alpha +\beta =\angle BOC\\\\\ \ \ \text{In}\ \vartriangle AOC,\\\text{ }\\\ \ \ \angle A+\gamma =\angle COE\\\\\therefore \angle BOC=\angle COE\end{array}$

11.    (b) In $ \displaystyle ΔABC$, $ \displaystyle D$ is a point of $ \displaystyle AC$ such that $ \displaystyle AD=2CD$. $ \displaystyle E$ is on $ \displaystyle BC$ such that $ \displaystyle DE \parallel AB$. Compare the areas of $ \displaystyle ΔCDE$ and $ \displaystyle ΔABC$. If $ \displaystyle α (ABED)=40$, what is $ \displaystyle α (ΔABC)$?

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ AD=2CD\ \ \ (\text{given)}\\\\\ \ \ \text{Since}\ DE\parallel AB,\\\\\ \ \ \vartriangle CAB\sim \vartriangle CDE\end{array}$

$ \displaystyle \therefore \frac{{\alpha (\Delta CAB)}}{{\alpha (\Delta CDE)}}=\frac{{A{{C}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(AD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(2CD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9\\\\\ \ \ \text{Let }\alpha (\Delta CDE)=x\ \text{and}\ \alpha (\Delta CAB)=9x.\\\\\therefore \alpha (ABED)=9x-x=8x\\\\\therefore 8x=40\Rightarrow x=5\\\\\therefore \alpha (\Delta CAB)=9\times 5=45\ \text{sq-units}\text{.}\end{array}$

12.    (a) The position vectors, relative to an origin $ \displaystyle O$, of three nonlinear points $ \displaystyle A, B$ and $ \displaystyle C$ are $ -2\hat{\text{i}}+3\hat{\text{j}}$, $ 3\hat{\text{i}}+2\hat{\text{j}}$ and $ -\hat{\text{i}}-5\hat{\text{j}}$. Show that $ \displaystyle ΔABC$ is isosceles.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \overrightarrow{{OA}}=-2\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OB}}=3\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OC}}=-\widehat{\text{i}}-5\widehat{\text{j}}\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =3\widehat{\text{i}}+2\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ AB=\sqrt{{{{5}^{2}}+{{{(-1)}}^{2}}}}=\sqrt{{26}}\\\\\therefore \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}-3\widehat{\text{i}}-2\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =-4\widehat{\text{i}}-7\widehat{\text{j}}\\\\\therefore \ \ BC=\sqrt{{{{{(-4)}}^{2}}+{{{(-7)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =\widehat{\text{i}}-8\widehat{\text{j}}\\\\\therefore \ \ AC=\sqrt{{{{1}^{2}}+{{{(-8)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ AC=BC\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles triangle}\text{.}\end{array}$

12.    (b) The tangent at the point $ \displaystyle C$ on the circle meets the diameter $ \displaystyle AB$ produced at $ \displaystyle T$, If $ \displaystyle ∠BCT=27°$, calculate $ \displaystyle ∠CTA$. If $ \displaystyle CT=t$ and $ \displaystyle BT=x$, prove that the radius of the circle is $ \large {\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}}$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \angle BAC=\angle BCT=\text{ }27{}^\circ \ \ \ (\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=}\ \angle \ \text{in alternate segment)}\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ \ \ \ \ \ (\angle \text{ in semicircle)}\\\\\ \ \ \ \text{In}\ \vartriangle \text{ACT,}\\\\\ \ \ \ \angle BAC+\angle ACT+\angle CTA=180{}^\circ \\\\\therefore \ \ 27{}^\circ +27{}^\circ +90{}^\circ +\angle CTA=180{}^\circ \\\\\therefore \ \ \angle CTA=36{}^\circ \\\\\ \ \ \ \text{Let}\ AB=2r,\text{where }r\ \text{is the radius of the circle}\text{.}\\\\\ \ \ \ \text{Since}\ AT\cdot BT=C{{T}^{2}},\\\\\ \ \ \ (2r+x)\cdot x={{t}^{2}}\\\\\therefore \ \ 2xr+{{x}^{2}}={{t}^{2}}\end{array}$

$ \displaystyle \therefore \ \ r=\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}$

13.    (a) Find the angles of a triangle, given that $ \displaystyle \angle A$ is obtuse and $ \sec (B+C)=\operatorname{cosec}(B-C)=2$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\angle A\ \text{is obtuse and}\\\\\ \ \ \ \sec (B+C)=\operatorname{cosec}(B-C)=2\end{array}$

$ \displaystyle \therefore \ \ \cos (B+C)=\sin (B-C)=\frac{1}{2}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ \angle A\ \text{is obtuse,}\\\\\ \ \ \angle B+\angle C=60{}^\circ \\\\\ \ \ \angle B-\angle C=30{}^\circ \\\\\therefore \ \ 2\angle B=90{}^\circ \Rightarrow \angle B=45{}^\circ \\\\\therefore \ \ 45{}^\circ +\angle C=60{}^\circ =\angle C=15{}^\circ \\\\\ \ \ \ \text{Since}\ \angle A+\angle B+\angle C=180{}^\circ ,\\\\\ \ \ \ \angle A+45{}^\circ +15{}^\circ =180{}^\circ \Rightarrow \angle A=120{}^\circ \end{array}$

13.    (b) Differentiate $ \displaystyle \frac{1}{\sqrt{x}}$ with respect to $ \displaystyle x$ from the first principles.

(5 marks)

Show/Hide Solution

$ \displaystyle \ \ \ \ \text{Let}\ f(x)=\frac{1}{{\sqrt{x}}}.$

$ \displaystyle \therefore \ \ f(x+\delta x)=\frac{1}{{\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{1}{{\sqrt{{x+\delta x}}}}-\frac{1}{{\sqrt{x}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}\times \frac{{\sqrt{x}+\sqrt{{x+\delta x}}}}{{\sqrt{x}+\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{x-x-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore \ \ \frac{{f(x+\delta x)-f(x)}}{{\delta x}}=\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+0}}\left[ {\sqrt{x}+\sqrt{{x+0}}} \right]}}=-\frac{1}{{2x\sqrt{x}}}$

14.    (a) A cruise ship travels at a bearing of $ \displaystyle 45°$ at $ \displaystyle 15$ mph for $ \displaystyle 3$ hours, and changes course to a bearing of $ \displaystyle 120°$. It then travels $ \displaystyle 10$ mph for $ \displaystyle 2$ hours. Find the distance of the ship from its original position and also its bearing from the original position.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}AB=3\ \text{hours}\ \times 15\ \text{mph}=45\ \text{mi}\\\\BC=2\ \text{hours}\ \times 10\ \text{mph}=20\ \text{mi}\\\\{{\beta }_{2}}=45{}^\circ ,{{\beta }_{1}}=180{}^\circ -120{}^\circ =60{}^\circ \\\\\beta =45{}^\circ +60{}^\circ =105{}^\circ \\\\\text{By the law of cosines,}\\\\A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2(AB)(BC)\cos \beta \\\\\ \ \ \ \ \ \ \,\ =\text{ }{{45}^{2}}+{{20}^{2}}-2\left( {45} \right)\left( {20} \right)\cos 105{}^\circ \\\\\ \ \ \ \ \ \ \ \ =\text{ }2025+400+1800\cos 75{}^\circ ~\\\\\ \ \ \ \ \ \ \ \ =\text{ }2425+465.9\\\\\ \ \ \ \ \ \ \ \ =\text{ }2890.9\\\\AC~~~~=\text{ }53.77\text{ mi}\end{array}$

No	Log
1800 cos 75°	3.2553 $ \displaystyle \overline{1}.4130$
465.9	2.6683

$ \displaystyle \ \ \ \ \text{By the law of sines,}$

$ \displaystyle \ \ \ \ \frac{{\sin \alpha }}{{BC}}=\frac{{\sin \beta }}{{AC}}$

$ \displaystyle \ \ \ \ \sin \alpha ~~=~\frac{{BC}}{{AC}}\times \sin \beta $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 105{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 75{}^\circ \ (\sin 105{}^\circ =\sin 75{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\sin 21{}^\circ 3'$

$ \displaystyle \therefore \ \ \alpha =21{}^\circ 3'$

$ \displaystyle \therefore \ \ \theta =45{}^\circ +21{}^\circ 3'=66{}^\circ 3'$

No	Log
20 sin 75°	1.3010 $ \displaystyle \overline{1}.9849$
53.77	1.02859 1.7305
sin21°3′	$ \displaystyle \overline{1}.5554$

Hence, the ship is $ \displaystyle 53.77$ mi from the original position.

It is in the direction $ \displaystyle N 66° 3' E.$

  14.    (b) Find the two positive numbers $ \displaystyle x$ and $ \displaystyle y$ such that their sum is $ \displaystyle 60$ and $ \displaystyle xy^3$ is maximum.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ x+y=60\\\\\therefore \ \ \ \ y=60-x\\\\\ \ \ \ \ \ \text{Let}\ z=x{{y}^{3}}\\\\\therefore \ \ \ \ z=x{{\left( {60-x} \right)}^{3}}\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}={{\left( {60-x} \right)}^{3}}-3x{{\left( {60-x} \right)}^{2}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-x-3x} \right)\\\\\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}=0,\text{when}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ {{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)=0\\\\\ \ \ \ \ x=60\ \ \text{or}\,\ x=15\\\\\ \ \ \ \ \text{When}\ x=60,\ y=0\ \text{which is impossible as }y>0.\\\\\ \ \ \ \ \text{When}\ x=15,\ y=45\ \text{which is possible}\text{.}\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-x} \right)}^{2}}-2\left( {60-x} \right)\left( {60-4x} \right)$

$ \displaystyle \ \ \ \ \ \text{When}\ x=15,$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}-2\left( {60-15} \right)\left( {60-60} \right)$

$ \displaystyle \therefore \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}<0$

$ \displaystyle \therefore \ \ \ \text{z is maximum}\ \text{when}\ x=15\ \text{and}\ y=45.$

Problem Study: The Remainder Theorem and Division Algorithm

Given that x⁵ + ax³ + bx² − 3 = (x² − 1) Q(x) − x − 2 where Q(x) is a polynomial. State the degree of Q(x) and find the value of a and b. Find also the remainder when Q(x) is divided by x + 2. 

Solution

Since $ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3$ is divided by (x² − 1), Q(x) is a polynomial of degree 3.  

$ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$

$ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=(x-1)(x+1)Q(x)-x-2$

 When x = − 1, $ \displaystyle -1-a+b-3=(-1-1)(-1+1)Q(x)-(-1)-2$

$ \displaystyle \therefore -a+b=3$      --------------(1)

When x = 1, $ \displaystyle 1+a+b-3=(1-1)(1+1)Q(x)-1-2$

$ \displaystyle \therefore a+b=-1$    --------------(2)

$ \displaystyle (2)+(1)\Rightarrow 2b=2\Rightarrow b=1$

$ \displaystyle (2)-(1)\Rightarrow 2a=-4\Rightarrow a=-2$ 

$ \displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$

$ \displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)$ 

$ \displaystyle \therefore Q(x)=\frac{{{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1}}{{{{x}^{2}}-1}}$

When Q(x) is divided by x + 2, 

the remainder $ \displaystyle =Q(-2)$ 

                       $ \displaystyle =\frac{{{{{(-2)}}^{5}}-2{{{(-2)}}^{3}}+{{{(-2)}}^{2}}+(-2)-1}}{{{{{(-2)}}^{2}}-1}}$

                       $ \displaystyle =\frac{{-32+16+4-2-1}}{3}$

                       $ \displaystyle =-\frac{{15}}{3}$

                       $ \displaystyle =-5$