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