ဒီေနရာမွာ တင္ေပးခဲ့တဲ့ sample question Section (A) ရဲ့ အေျဖ ျဖစ္ပါတယ္။ Section (B), Section (C) တို႔ရဲ့ အေျဖကိုလဲ ဆက္လက္ တင္ေပးသြားပါ့မယ္။
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)$.
$ \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$.
$ \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$.
$ \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.
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'$.
$ \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?
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$.
$ \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.
$ \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}}$.
$ \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}}$.
(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}$
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)$.
Show/Hide Solution
$ \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$.
Show/Hide Solution
$ \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$.
Show/Hide Solution
$ \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.
Show/Hide Solution
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'$.
Show/Hide Solution
$ \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?
Show/Hide Solution
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$.
Show/Hide Solution
$ \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.
Show/Hide Solution
$ \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}}$.
Show/Hide Solution
$ \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}}$.
Show/Hide Solution
(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}$
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