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$.
(5 marks)
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$\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)$?
(5 marks)
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$ \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)
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$ \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}}}$.
(5 marks)
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$ \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}}$
$ \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$.
(5 marks)
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$ \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}$
$ \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)
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$ \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}}}$
$ \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)
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$ \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)
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$ \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.$
$ \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.$
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