2 Marks Questions

Question 12 Marks

Calculate differentiation of $e^x \sin x+x^n \cos x$ $\text {w.r.t.x.}$

Answer

Let $\quad y=e^x \sin x+x^n \cos x$
On differenting both sides $\text {w.r.t.x},$
$\therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(e^x \cdot \sin x\right)+\frac{d}{d x}\left(x^n \cdot \cos x\right)$
$\begin{array}{l}=e^x \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}\left(e^x\right)+ x^n \cdot \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}\left(x^n\right)\end{array}$
$\begin{aligned}=e^x \cos x+\sin x \cdot e^x+ & x^n(-\sin x) +\cos x \cdot n x^{n-1}\end{aligned}$
$\begin{array}{r}\therefore \quad \frac{d y}{d x}=e^x \cos x+e^x \sin x-x^n \sin x +n \cdot x^{n-1} \cos x\end{array}$

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Question 22 Marks

Evaluate $\lim _{x \rightarrow \pi} \frac{\sin ^2 2 x}{\cos ^2 \frac{1}{2} x}.$

Answer

$\begin{array}{l}\lim _{x \rightarrow \pi} \frac{\sin ^2 2 x}{\cos ^2 \frac{1}{2} x}=\lim _{x \rightarrow \pi} \frac{(2 \sin x \cos x)^2}{\cos ^2 \frac{x}{2}} \\\because \sin 2 x=2 \sin x \cos x\\ =\lim _{x \rightarrow \pi} \frac{4 \sin ^2 x \cos ^2 x}{\cos ^2 \frac{1}{2} x}=\lim _{x \rightarrow \pi} \frac{4(\sin x)^2 \cos ^2 x}{\cos ^2 \frac{x}{2}} \\ =\lim _{x \rightarrow \pi} \frac{4.4 \sin ^2 \frac{x}{2} \cdot \cos ^2 \frac{x}{2} \cdot \cos ^2 x}{\cos ^2 \frac{x}{2}} \because \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \\ =\lim _{x \rightarrow \pi}\left(16 \sin ^2 \frac{x}{2} \cdot \cos ^2 x\right) \\ =16 \sin ^2 \frac{\pi}{2} \cdot \cos ^2 \pi=16(1)^2(-1)^2 \\ =16.1 .1=16\end{array}$

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Question 32 Marks

Prove that $\lim _{x \rightarrow 1} \frac{\sqrt{4+x}-\sqrt{5}}{x-1}=\frac{\sqrt{5}}{10}.$

Answer

$\begin{array}{l}\text { L.H.S. }=\lim _{x \rightarrow 1} \frac{\sqrt{4+x}-\sqrt{5}}{x-1} \\ =\lim _{x \rightarrow 1} \frac{(\sqrt{4+x}-\sqrt{5})(\sqrt{4+x}+\sqrt{5})}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{4+x-5}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{1}{\sqrt{4+x}+\sqrt{5}} \\ =\frac{1}{\sqrt{4+1}+\sqrt{5}}=\frac{1}{\sqrt{5}+\sqrt{5}}=\frac{1}{2 \sqrt{5}} \\ \Rightarrow=\frac{\sqrt{5}}{2 \sqrt{5} \times \sqrt{5}}=\frac{\sqrt{5}}{10}=\text { R.H.S. }\end{array}$

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Question 42 Marks

Prove that $\lim _{x \rightarrow 0} \frac{\sqrt{1-x}-\sqrt{1+x}}{x}=-1.$

Answer

$\text { L.H.S. }=\lim _{x \rightarrow 0} \frac{\sqrt{1-x}-\sqrt{1+x}}{x}$
On multiplying and dividing by $(\sqrt{1-x}+\sqrt{1+x})$ in numerator and denominator,
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{(\sqrt{1-x}-\sqrt{1+x})(\sqrt{1-x}+\sqrt{1+x})}{x(\sqrt{1-x}+\sqrt{1+x})} \\=\lim _{x \rightarrow 0} \frac{1-x-(1+x)}{x(\sqrt{1-x}+\sqrt{1+x})}=\lim _{x \rightarrow 0}\frac{1-x-1-x}{x(\sqrt{1-x}+\sqrt{1+x})}\end{array}$
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{-2 x}{x(\sqrt{1-x}+\sqrt{1+x})}=\lim _{x \rightarrow 0} \frac{-2}{(\sqrt{1-x}+\sqrt{1+x})} \\ =\frac{-2}{\sqrt{1-0}+\sqrt{1+0}}=\frac{-2}{1+1}=\frac{-2}{2}=-1=\text { R.H.S. }\end{array}$

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Question 52 Marks

Evaluate $\lim _{n \rightarrow \infty}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots . .+\frac{1}{3^n}\right].$

Answer

$\lim _{n \rightarrow \infty}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots . .+\frac{1}{3^n}\right]$
We know that sum of geometric series $=\frac{a\left(1-r^n\right)}{1-r}$
while $r<1$. So sum of given series is :
$\begin{array}{l}\frac{\frac{1}{3}\left[1-\left(\frac{1}{3}\right)^n\right]}{1-\frac{1}{3}}=\frac{1-\left(\frac{1}{3}\right)^n}{2} \\\therefore \lim _{n \rightarrow \infty}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots ..+\frac{1}{3^n}\right]=\lim _{n \rightarrow \infty} \frac{\left(1-\left(\frac{1}{3}\right)^n\right)}{2} \\\frac{\left(1-\frac{1}{3^{\infty}}\right)}{2}=\frac{\left(1-\frac{1}{\infty}\right)}{2}=\frac{1-0}{2}=\frac{1}{2}\end{array}$

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Question 62 Marks

Evaluate $\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots\ldots+n^2}{n^3}.$

Answer

We know that $1^2+2^2+3^2+\ldots \ldots \ldots+n^2=\Sigma n^2 =\frac{1}{6} n(n+1)(2 n+1)$
$\therefore \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots \ldots+n^2}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{6} n(n+1)(2 n+1)}{n^3}=\frac{1}{6} \times \lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right)\left(\frac{2 n+1}{n}\right)$
$=\frac{1}{6} \times \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)=\frac{1}{6}\left(1+\frac{1}{\infty}\right)\left(2+\frac{1}{\infty}\right)$
$=\frac{1}{6} \times(1+0)(2+0)=\frac{1}{6} \times 2=\frac{1}{3}$

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Question 72 Marks

Evaluate $\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}.$

Answer

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}$
Let $\frac{\pi}{2}+h=x$
$\begin{array}{l}\because \text { If } x=\frac{\pi}{2}, \text { then } h \rightarrow 0 \\=\lim _{h \rightarrow 0} \frac{1+\cos \left\{2\left(\frac{1}{2} \pi+h\right)\right\}}{\left\{\pi-2\left(\frac{1}{2}\pi+h\right)\right\}^2}\left[\because \lim _{x \rightarrow a} f(x)=\lim _{h \rightarrow 0} f(a+h)\right] \\=\lim _{h \rightarrow 0} \frac{1+\cos (\pi+2 h)}{4 h^2}=\lim _{h \rightarrow 0} \frac{1-\cos 2 h}{4 h^2}\end{array}$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{2 \sin ^2 h}{4 h^2}=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2 \\ =\frac{1}{2} \cdot(1)^2=\frac{1}{2}\end{array}$

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Question 82 Marks

Evaluate $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}.$

Answer

$\begin{array}{l}\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\ =\lim _{x \rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})} \\ =\lim _{x \rightarrow a} \frac{a+2 x-3 x}{3 a+x-4 x} \cdot \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \\ =\lim _{x \rightarrow a} \frac{a-x}{3(a-x)} \cdot \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \\ =\lim _{x \rightarrow a} \frac{\sqrt{3 a+x}+2 \sqrt{x}}{3(\sqrt{a+2 x}+\sqrt{3 x})} \\ =\frac{\sqrt{3 a+a}+2 \sqrt{a}}{3(\sqrt{a+2 a}+\sqrt{3 a})}=\frac{1}{3} \cdot \frac{4 \sqrt{a}}{2 \sqrt{3 a}} \\ =\frac{2}{3 \sqrt{3}}\end{array}$

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Question 92 Marks

Evaluate $\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}.$

Answer

$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1} & \\ & =\lim _{x \rightarrow 1} \frac{x^m-1}{x-1} \cdot \frac{x-1}{x^n-1} \\ & =\lim _{x \rightarrow 1} \frac{x^m-1}{x-1} \cdot \frac{1}{\frac{x^n-1}{x-1}} \\ & =\lim _{x \rightarrow 1} \frac{x^m-1}{x-1} \times \lim _{x \rightarrow 1} \frac{1}{\frac{x^n-1}{x-1}}=\frac{m(1)^{m-1}}{m(1)^{n-1}} \\ & =\frac{m}{n} \quad\left[\because \lim _{x \rightarrow a} \frac{x^m-a^m}{x-a}=m a^{m-1}\right]\end{aligned}$

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Question 102 Marks

Calculate the value of $\lim _{x \rightarrow a} \frac{a \sin x-x \sin a}{a x^2-a^2 x}.$

Answer

$\begin{array}{l}\lim _{x \rightarrow a} \frac{a \sin x-x \sin a}{a x^2-a^2 x} \\=\lim _{x \rightarrow a} \frac{a \sin x-x \sin x+x \sin x-x \sin a}{a x^2-a^2 x}\end{array}$
on adding and subtracting $x \sin x$ in numerator
$\begin{array}{l}=\lim _{x \rightarrow a} \frac{\sin x \cdot(a-x)}{a x(x-a)}+\lim _{x \rightarrow a}\frac{x(\sin x-\sin a)}{a x(x-a)} \\=-\lim _{x \rightarrow a} \frac{\sin x \cdot(x-a)}{a x(x-a)}+\lim _{x\rightarrow a} \frac{2 \cos \frac{x+a}{2} \sin \frac{x-a}{2}}{2 a. \frac{x-a}{2}} \\=-\frac{\sin a}{a^2}+\frac{1}{a} . 1 . \cos a \\=-\frac{\sin a}{a^2}+\frac{\cos a}{a}\left[\because \lim _{\theta\rightarrow 0} \frac{\sin \theta}{\theta}=1\right]\end{array}$

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Question 112 Marks

If $f(x)=\left\{\begin{array}{ll}2 x+3, & x \leq 2 \\ 3 x+K, & x>2\end{array}\right.$ and $\lim _{x \rightarrow 2} f(x)$ are exist, then calculate the value of $K.$

Answer

$\begin{array}{l}\text { L.H.L. }=\lim _{h \rightarrow 0} 2(2-h)+3=7 \\\text { R.H.L. }=\lim _{h \rightarrow 0} 3(2+h)+K=6+K\end{array}$
As $\lim _{x \rightarrow 2} f(x)$ exists,
$\therefore \quad$ $\text {L.H.L. = R.H.L.}$
$\therefore \quad 7=6+K$
$\quad \quad \text {K} =1$

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Question 122 Marks

If $\lim _{x \rightarrow 0} k x \operatorname{cosec} x=\lim _{x \rightarrow 0} x . \operatorname{cosec} k x$, then calculate all possible values of $k.$

Answer

According to question,
$\lim _{x \rightarrow 0} k x \operatorname{cosec} x=\lim _{x \rightarrow 0} x.\operatorname{cosec} k x$
$\begin{array}{l}\Rightarrow \quad \lim _{x \rightarrow 0} \frac{k x}{\sin x}=\lim _{x \rightarrow 0} \frac{x}{\sin k \cdot x} \\ \Rightarrow \quad k \cdot \frac{1}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}=\frac{1}{\lim _{x \rightarrow 0} k \cdot \frac{\sin k x}{k x}} \\ \Rightarrow \quad k \cdot \frac{1}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}=\frac{1}{k} \cdot \frac{1}{\lim _{x \rightarrow 0} \frac{\sin k x}{k x}}\end{array}$
$\begin{array}{lc}\Rightarrow & k \times \frac{1}{1}=\frac{1}{k} \cdot \frac{1}{1} \\ \Rightarrow & k^2=1 \quad \therefore k= \pm 1\end{array}$

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Question 132 Marks

If $f(x)=\frac{|x-1|}{x-1}$, then show that $\lim _{x \rightarrow 1} f(x)$ does not exist.

Answer

When $\quad x>1,|x-1|=x-1$
$\begin{aligned}\therefore \quad \text { R.H.L. } & =\lim _{h \rightarrow 0} f(1+h) \\& =\lim _{h \rightarrow 0} \frac{1+h-1}{1+h-1}=1\end{aligned}$
When $x<1,|x-1|=-(x-1)=1-x$
$\begin{aligned}\therefore \quad \text { L.H.L. } & =\lim _{h \rightarrow 0} \frac{-(1-h)+1}{1-h-1} \\& =\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \\\text { R.H.L. } & \neq \text { L.H.L. }\end{aligned}$
So $\lim _{x \rightarrow 1} f(x)$ does not exist.

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Question 142 Marks

$\lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}$

Answer

$\begin{array}{l}\lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1} \\ =\lim _{x \rightarrow 1} \frac{(x-1)\left(x^3-2 x^2-2 x-2\right)}{(x-1)\left(x^2-4 x-1\right)} \\ =\lim _{x \rightarrow 1} \frac{x^3-2 x^2-2 x-2}{x^2-4 x-1} \\ =\frac{(1)^3-2(1)^2-2(1)-2}{(1)^2-4(1)-1}=\frac{1-2-2-2}{1-4-1}=\frac{5}{4}\end{array}$

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