# Answer for 2019 Sample Question : Section (A)

ဒီေနရာမွာ တင္ေပးခဲ့တဲ့ sample question Section (A) ရဲ့ အေျဖ ျဖစ္ပါတယ္။ Section (B), Section (C) တို႔ရဲ့ အေျဖကိုလဲ ဆက္လက္ တင္ေပးသြားပါ့မယ္။

Section (A)
Solution

1. (a) Let the function $\displaystyle f(x)=\frac{{1-2x}}{{1+x}},x\ne 1$. If $\displaystyle {{g}^{{-1}}}(x)={{f}^{{-1}}}(x+1)$, evaluate $\displaystyle g(2)$.

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$\displaystyle f(x)=\frac{{1-2x}}{{1+x}},x\ne -1$

$\displaystyle \text{Let}\ g(2)=x\ \text{then }{{g}^{{-1}}}(x)=2.$

$\displaystyle \therefore {{f}^{{-1}}}(x+1)=2$

$\displaystyle \therefore f(2)=x+1$

$\displaystyle \therefore \frac{{1-2(2)}}{{1+(2)}}=x+1$

$\displaystyle \therefore x+1=-1$

$\displaystyle \therefore x=-2$

$\displaystyle \therefore g(2)=-2$

1. (b) The expression $\displaystyle (x + 4)^3 + ax + b$ has a factor $\displaystyle x + 1$ but leaves a remainder of $\displaystyle 8$ when divided by $\displaystyle x + 5$. Find the values of $\displaystyle a$ and $\displaystyle b$.

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$\displaystyle \text{Let}\ f(x)={{(x+4)}^{3}}+ax+b.$

$\displaystyle x+1$ is a factor of $\displaystyle f(x)$.

$\displaystyle \therefore f(-1)=0$

$\displaystyle \therefore {{(-1+4)}^{3}}+a(-1)+b=0$

$\displaystyle \therefore a-b=27\ \ \ \ \ \ -----(1)$

When $\displaystyle f(x)$ is divided by $\displaystyle x+5$, the remainder is $\displaystyle 8$.

$\displaystyle \therefore f(-5)=8$

$\displaystyle \therefore {{(-5+4)}^{3}}+a(-5)+b=0$

$\displaystyle \therefore 5a-b=1\ \ \ \ \ \ \ -----(2)$

$\displaystyle \text{By}\ (2)-(1),$

$\displaystyle 4a=-36\Rightarrow a=-9$

$\displaystyle \therefore -9-b=27\Rightarrow b=-36$

2. (a) If the first four terms in the expansion of $\displaystyle (x^2−2)^5$ in descending powers of $\displaystyle x$ are $\displaystyle x^{10}−10x^8+40x^6+Ax^4+...$, find the value of $\displaystyle A$.

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$\displaystyle {{({{x}^{2}}-2)}^{5}}={{x}^{{10}}}-10{{x}^{8}}+40{{x}^{6}}+A{{x}^{4}}+...$

$\displaystyle \text{Using the binomial theorem,}$

$\displaystyle {{({{x}^{2}}-2)}^{5}}={{({{x}^{2}})}^{5}}+5{{({{x}^{2}})}^{4}}(-2)+10{{({{x}^{2}})}^{3}}{{(-2)}^{2}}+10{{({{x}^{2}})}^{2}}{{(-2)}^{3}}+...$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{{10}}}-10{{x}^{8}}+40{{x}^{6}}-80{{x}^{4}}+...$

$\displaystyle \text{Equating the two equations, }$

$\displaystyle A=-80$

2. (b) In a geometric progression, $\displaystyle {{u}_{1}}=\frac{1}{{81}}$ and $\displaystyle {{u}_{4}}=\frac{1}{{3}}$. Find the common ratio.

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Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$.

By the problem,

$\displaystyle {{u}_{1}}=\frac{1}{{81}}\Rightarrow a=\frac{1}{{81}}$

$\displaystyle {{u}_{4}}=\frac{1}{3}\Rightarrow a{{r}^{3}}=\frac{1}{3}$

$\displaystyle \therefore \frac{1}{{81}}{{r}^{3}}=\frac{1}{3}$

$\displaystyle \therefore {{r}^{3}}=27\Rightarrow r=3$

3. (a) Find the two matrices of the form $\displaystyle P=\left( {\begin{array}{*{20}{c}} 4 \\ {x-2} \end{array}\ \ \ \begin{array}{*{20}{c}} {{{x}^{2}}-2x} \\ {-1} \end{array}} \right)$ such that $\displaystyle P=P'$.

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$\displaystyle P=\left( {\begin{array}{*{20}{c}} 4 \\ {x-2} \end{array}\ \ \ \begin{array}{*{20}{c}} {{{x}^{2}}-2x} \\ {-1} \end{array}} \right)$

$\displaystyle {P}'=\left( {\begin{array}{*{20}{c}} 4 \\ {{{x}^{2}}-2x} \end{array}\ \ \ \begin{array}{*{20}{c}} {x-2} \\ {-1} \end{array}} \right)$

By the problem, $\displaystyle P={P}'$

$\displaystyle \therefore \left( {\begin{array}{*{20}{c}} 4 \\ {x-2} \end{array}\ \ \ \begin{array}{*{20}{c}} {{{x}^{2}}-2x} \\ {-1} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 4 \\ {{{x}^{2}}-2x} \end{array}\ \ \ \begin{array}{*{20}{c}} {x-2} \\ {-1} \end{array}} \right)$

$\displaystyle \therefore {{x}^{2}}-2x=x-2$

$\displaystyle \therefore {{x}^{2}}-3x+2=0$

$\displaystyle \therefore (x-1)(x-2)=0$

$\displaystyle \therefore x=1\ \text{or}\ x=2$

3.(b) A fair coin is tossed 5 times. What is the probability of getting at least one head?

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For each toss,

$\displaystyle P$ (head) = $\displaystyle \frac{1}{2}$

$\displaystyle P$ (tail) = $\displaystyle \frac{1}{2}$

For 5 tosses,

$\displaystyle P$ (getting all tail) =$\displaystyle \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{{32}}$

$\displaystyle \therefore P\text{(getting at least one head)}$ = $\displaystyle P\text{(not getting all tail)}$

$\displaystyle \therefore P\text{(getting at least one head)}$ = $\displaystyle 1-P\text{(getting all tail)}$

$\displaystyle \therefore P\text{(getting at least one head)}$ = $\displaystyle 1-\frac{1}{{32}}=\frac{{31}}{{32}}$

4. (a) In the figure, $\displaystyle ∠ABC=30°$, $\displaystyle AB=BC$ and $\displaystyle AD$ is a tangent. Find $\displaystyle ∠BDA$.

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$\displaystyle ∠ABC = 30°$ and $\displaystyle AB = BC$ (given)

$\displaystyle \therefore α = γ = \frac{1}{2}(180° – 30°) = 75°$

$\displaystyle θ = γ = 75°$ (∠ between tangent & chord = ∠ in alternate segment)

Since $\displaystyle θ = ∠ABC + ∠BDA,$

$\displaystyle 75° = 30° + ∠BDA$

$\displaystyle \therefore ∠BDA=45°$

4.(b) $\displaystyle A,B$, and $\displaystyle C$ are with position vectors $\displaystyle \hat{\text{i}}+3\hat{\text{j}}$, $\displaystyle 2\hat{\text{i}}+5\hat{\text{j}}$ and $\displaystyle k\hat{\text{i}}-4\hat{\text{j}}$  respectively. Find the value of $\displaystyle k$ if  $\displaystyle A,B$, and $\displaystyle C$ are collinear.

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$\displaystyle \overrightarrow{{OA}}=-2\widehat{\text{i}}+3\widehat{\text{j}}$

$\displaystyle \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}$

$\displaystyle \overrightarrow{{OC}}=k\widehat{\text{i}}-4\widehat{\text{j}}$

Since $\displaystyle A, B$ and $\displaystyle C$ are collinear,

Let $\displaystyle h\overrightarrow{{AB}}=\overrightarrow{{BC}}$.

$\displaystyle \therefore h\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)=\overrightarrow{{OC}}-\overrightarrow{{OB}}$

$\displaystyle \ \ h\left( {2\widehat{\text{i}}+5\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}} \right)=k\widehat{\text{i}}-4\widehat{\text{j}}-2\widehat{\text{i}}-5\widehat{\text{j}}$

$\displaystyle \ \ 4h\widehat{\text{i}}+2h\widehat{\text{j}}=(k-2)\widehat{\text{i}}-9\widehat{\text{j}}$

$\displaystyle \therefore 2h=-9\Rightarrow h=-\frac{9}{2}$
$\displaystyle \ \ \ 4h=k-2\Rightarrow k-2=4\left( {-\frac{9}{2}} \right)\Rightarrow k=-16$

5. (a) Prove that $\displaystyle {{(1-\tan x)}^{2}}+{{(1-\cot x)}^{2}}={{(\sec x-\operatorname{cosec}x)}^{2}}$.

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$\displaystyle \ \ \ \ {{(1-\tan x)}^{2}}+{{(1-\cot x)}^{2}}$

$\displaystyle =1-2\tan x+{{\tan }^{2}}x+1-2\cot x+{{\cot }^{2}}x$

$\displaystyle =(1+{{\tan }^{2}}x)+(1+{{\cot }^{2}}x)-2(\tan x+\cot x)$

$\displaystyle ={{\sec }^{2}}x+{{\operatorname{cosec}}^{2}}x-2\left( {\frac{{\sin x}}{{\cos x}}+\frac{{\cos x}}{{\sin x}}} \right)$

$\displaystyle ={{\sec }^{2}}x+{{\operatorname{cosec}}^{2}}x-2\left( {\frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{\sin x\cos x}}} \right)$

$\displaystyle ={{\sec }^{2}}x+{{\operatorname{cosec}}^{2}}x-\frac{2}{{\sin x\cos x}}$

$\displaystyle ={{\sec }^{2}}x-2\sec x\operatorname{cosec}x+{{\operatorname{cosec}}^{2}}x$

$\displaystyle ={{(\sec x-\operatorname{cosec}x)}^{2}}$

5. (b) Evaluate (i) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{x}-1}}{{x-1}}$   (ii) $\displaystyle \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{\cos \frac{x}{2}}}{{\pi -x}}$.

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(i)
$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{x}-1}}{{x-1}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{x}-1}}{{{{{\left( {\sqrt[3]{x}} \right)}}^{3}}-{{1}^{3}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{x}-1}}{{\left( {\sqrt[3]{x}-1} \right)\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}$

(ii)
$\displaystyle \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{\cos \frac{x}{2}}}{{\pi -x}}$

Let $\displaystyle \pi -x=t$, then $\displaystyle x=\pi -t$.

When $\displaystyle x\to \pi ,\ t\to$.

$\displaystyle \therefore \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{\cos \frac{x}{2}}}{{\pi -x}}=\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \frac{{\pi -t}}{2}}}{t}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \left( {\frac{\pi }{2}-\frac{t}{2}} \right)}}{t}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \frac{t}{2}}}{t}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{1}{2}\sin \frac{t}{2}}}{{\frac{t}{2}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \frac{t}{2}}}{{\frac{t}{2}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}(1)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}$