Limits, Continuity and Differentiability

Given, $f(x)=2 x-[2 x]$

$[x] \rightarrow$ Greatest Integer Function

Now,

$ \begin{aligned} & l_1=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h) \\ & =\lim _{h \rightarrow 0} 2(2-h)=\lim _{h \rightarrow 0} 4-2 h-3=1 \end{aligned} $

Now, $l_2=\mathop {\lim }\limits_{x \to {2^ + }} f(x)=\mathop {\lim }\limits_{h \to {0 }} f(2+h)$

$ \begin{aligned} & =\lim _{h \rightarrow 0} 2(2+h)-[4+2 h] \\ & =\lim _{h \rightarrow 0}(4+2 h-4)=0 \\ \therefore l_1+l_2 & =0+1=1 \end{aligned} $

$\mathop {\lim }\limits_{x \to 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}\left[\frac{0}{0}\right.$ form $]$

$ =\frac{1}{2} \frac{\mathop {\lim }\limits_{x \to 0} \frac{2^x-1}{x}}{\mathop {\lim }\limits_{x \to 0} \frac{\log (1+2 x)}{2 x}} \cdot e^{\mathop {\lim }\limits_{x \to 0} (1+\sin x-1) \times \frac{2}{\sin x}} $

Using $L^{\prime}$ hospital rule,

$ \begin{aligned} & \frac{1}{2} \frac{\mathop {\lim }\limits_{x \to 0} 2^x \log 2}{\mathop {\lim }\limits_{x \to 0} \frac{1}{1+2 x}}, e^{\mathop {\lim }\limits_{x \to 0} \sin x \times \frac{2}{\sin x}} \\ & =\frac{1}{2} \frac{\log 2}{1} e^2=e^2\left(\frac{1}{2} \log 2\right)=e^2 \log \sqrt{2} \end{aligned} $

Given $f(x)$ is a differentiable function, and

$ f(0)=0, f^{\prime}(0)=20, x \in\left(0, \frac{\pi}{2}\right] $

$ \begin{aligned} &\begin{aligned}& A(x)=2 f(x) \operatorname{cosec}(4 x) +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x &\\ &\lim _{x \rightarrow 0} A(x) =\lim _{x \rightarrow 0} \frac{2 f(x)}{\sin 4 x} +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x \end{aligned}\\ &\text { Using L'Hospital rule }\\ &\begin{aligned} & \lim _{x \rightarrow 0} A(x)=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)}{4 \cos 4 x} +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x \\ & =\frac{2 f^{\prime}(0)}{4 \cos 0}+4 f(0)\left(\cos ^2 0+1\right)-4 \cos ^2 0 \\ & =\frac{2 \times 20}{4 \times 1}+0-4 \times 1=10-4=6 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } x=\log _e\left(\cot \left(\frac{\pi}{4}+\theta\right)\right) \\ & =\lim _{\theta \rightarrow 0} \frac{\theta}{\sinh x \cosh x} \\ & \begin{array}{l} \lim _{\theta \rightarrow 0} \frac{\theta}{\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^x+e^{-x}}{2}\right)} \\ =\lim _{\theta \rightarrow 0} \frac{4 \theta}{e^{2 x}-e^{-2 x}} \\ =\lim _{\theta \rightarrow 0} \frac{4 \theta}{\cot ^2\left(\frac{\pi}{4}+\theta\right)-\tan ^2\left(\frac{\pi}{4}+\theta\right)} \end{array} \end{aligned} $

$ \begin{aligned} &\begin{array}{r} =\lim _{\theta \rightarrow 0} \frac{4}{-2 \cot \left(\frac{\pi}{4}+\theta\right) \operatorname{cosec}^2\left(\frac{\pi}{4}+\theta\right)} \\ -2 \tan \left(\frac{\pi}{4}+\theta\right) \sec ^2\left(\frac{\pi}{4}+\theta\right) \end{array}\\ &\text { [ ∵ Apply L' Hospital rule] } \end{aligned} $

$ \begin{aligned} & =\frac{4}{-2(1)(2)-2(1)(2)}=\frac{4}{-4-4} \\ & =-\frac{4}{8}=-\frac{1}{2} \end{aligned} $

$ \begin{aligned} \\ \begin{array}{r} \text { } \mathop {\lim }\limits_{x \to 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)\right. \left.+\frac{x^2-4}{x^3-2 x-4}\right] \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right. \left.+\frac{(x-2)(x+2)}{x^3-2 x^2+2 x^2-4 x+2 x-4}\right] \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right] +\frac{(x-2)(x+2)}{(x-2)\left(x^2+2 x+2\right)} \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right] +\mathop {\lim }\limits_{x \to 2} \frac{(x+2)}{\left(x^2+2 x+2\right)} \,...(i)\end{array} \end{aligned} $

$ \cos \left(\frac{2}{x-2}\right) \in[-1,1] $

So, $\mathop {\lim }\limits_{x \to 2}(x-2)^2 \cos \left(\frac{2}{x-2}\right)=0$

Now, substitute $x=2$ in Eq. (i),

$ 0+\frac{4}{10}=\frac{2}{5} $

$ \text { } \begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan 2 x-2 \tan x}{(1-\cos x)\left(2^x-1\right)} \\ = & \lim _{x \rightarrow 0}\left[\frac{2 \tan x}{1-\tan ^2 x}-2 \tan x\right] \times \frac{1}{(1-\cos x)\left(2^x-1\right)} \\ = & \lim _{x \rightarrow 0} 2 \tan x\left[\frac{1-1+\tan ^2 x}{1-\tan ^2 x}\right] \times \frac{1}{2 \sin ^2 \frac{x}{2} \times\left(2^x-1\right)} \end{aligned} $

$ \begin{aligned} & =4 \lim _{x \rightarrow 0} \frac{1}{1-\tan ^2 x} \times \lim _{x \rightarrow 0} \frac{\tan ^3 x}{x^3} \times \lim _{x \rightarrow 0} \frac{x^2 / 4}{\frac{\sin ^2 x}{4}} \times \frac{1}{(-1)} \lim _{x \rightarrow 0} \frac{x}{2^x-1} \\ & =4\left[1 \times 1 \times 1 \times \frac{1}{\log 2}\right]=\frac{4}{\log 2} \end{aligned} $

$ \text { } \begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi\left(1+\tan ^2 x\right)^2\right]}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi+\pi \tan ^4 x+2 \pi \tan ^2 x\right)}{\pi^2 x^4} \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \tan ^4 x+2 \pi \tan ^2 x\right)}{\pi^2 x^4} \\ & {[\because \tan (\pi+\theta)=\tan \theta] } \\ &= \lim _{x \rightarrow 0} \frac{\tan ^2\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]}{\pi^2 x^4} \times \frac{\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]^2}{\left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]^2} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\pi^2 \tan ^4 x\left(\tan ^2 x+2\right)^2}{\pi^2 x^4}\times\left\{\frac{\tan \left[\pi \tan ^2 x\left(\tan ^2 x+2\right)\right]}{\pi \tan ^2 x\left(\tan ^2 x+2\right)}\right\}^2 \\ & \Rightarrow(1)^4 \times(0+2)^2 \times(1)^2=4 \\ & \quad\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right] \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{r} \lim _{x \rightarrow 0}\left(\frac { 4 ! } { x ^ { 8 } } \left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}\right.\right. \left.\left.+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right) \\ =\lim _{x \rightarrow 0}\left(\frac { 4 ! } { x ^ { 8 } } \left[1\left(1-\cos \frac{x^2}{3}\right)\right.\right. \left.\left.-\cos \frac{x^2}{4}\left(1-\cos \frac{x^2}{3}\right)\right]\right) \end{array} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0}\left[\frac{4!}{x^8}\left(1-\cos \frac{x^2}{3}\right)\left(1-\cos \frac{x^2}{4}\right)\right] \\ & =4!\lim _{x \rightarrow 0}\left[2 \sin ^2 \frac{x^2}{6} \times 2 \sin ^2 \frac{x^2}{8}\right] \times \frac{1}{x^8} \\ & =\frac{4 \times 4!}{64 \times 36} \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{x^2}{6}}{\frac{x^4}{36}} \times \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{x^2}{8}}{\frac{x^4}{64}} \\ & =\frac{4 \times 24}{64 \times 36}=\frac{1}{24} \end{aligned} $

According to the question,

$ \begin{aligned} & m=a_{11}+a_{22}+\ldots .+a_{n n} \\ \Rightarrow \quad & m=k+k^2+\ldots .+k^n \end{aligned} $

Again, $\lim _{k \rightarrow 1} \frac{n-m}{1-k}=171$

$ \Rightarrow \lim _{k \rightarrow 1} \frac{m-n}{k-1}=171 $

$ \begin{gathered} \Rightarrow \lim _{k \rightarrow 1} \frac{\left(k+k^2+k^3+\ldots+k^n\right)-(1+1+\ldots \text { to } n \text { times })}{(k-1)}=171 \\ \Rightarrow \lim _{k \rightarrow 1}\left[\frac{(k-1)}{(k-1)}+\frac{\left(k^2-1\right)}{(k-1)}+\frac{\left(k^3-1\right)}{(k-1)}\right. \\ \left.+\ldots+\frac{\left(k^n-1\right)}{(k-1)}\right]=171 \end{gathered} $

$ \begin{aligned} & \Rightarrow 1+2+3+\ldots+n=171 \\ & \quad\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] \\ & \Rightarrow \frac{n(n+1)}{2}=171 \\ & \Rightarrow \quad n=18 \end{aligned} $

$ \begin{aligned} & \text { 66. (c) } \lim _{x \rightarrow \infty} x^3\left[\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2} x\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\begin{array}{l} \left(\sqrt{x^2+\sqrt{x^4+1}}\right. \\ \frac{-\sqrt{2} x)\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)} \end{array}\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\frac{x^2+\sqrt{x^4+1}-2 x^2}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)}\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\frac{\left(\sqrt{x^4+1}-x^2\right)\left(\sqrt{x^4+1}+x^2\right)}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)\left(\sqrt{x^4+1}+x^2\right)}\right] \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \infty} x^3\left[\frac{x^4+1-x^4}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right)\left(\sqrt{x^4+1}+x^2\right)}\right] \\ & =\lim _{x \rightarrow \infty} x^3\left[\frac{1}{x\left(\sqrt{1+\sqrt{1+\frac{1}{x^4}}}+\sqrt{2}\right)x^2\left(\sqrt{1+\frac{1}{x^4}}+1\right)} .\right] \\ & =\frac{1}{(\sqrt{1+\sqrt{1+0}})+\sqrt{2})(\sqrt{1+0}+1)} \\ & =\frac{1}{2 \sqrt{2} \cdot 2}=\frac{1}{4 \sqrt{2}} \end{aligned} $

At $x=-3$

$ \begin{aligned} \mathbf{L H L} & =\lim _{h \rightarrow 0} f(-3-h) \\ & =\lim _{h \rightarrow 0}(3-(-3-h))=6 \end{aligned} $

RHL $\lim _{h \rightarrow 0} f(-3+h)=6 \Rightarrow f(-3)=6$

$\Rightarrow f(x)$ is continuous at $x=-3$

At $x=3$

$ \begin{aligned} \mathbf{L H L} & =\lim _{h \rightarrow 0} f(3-h)=6 \\ \mathbf{R H L} & =\lim _{h \rightarrow 0} f(3+h) \\ & =\lim _{h \rightarrow 0}(3+(3+h))=6 \\ f(3) & =6 \end{aligned} $

$\Rightarrow f(x)$ is continuous at $x=3$

$\therefore f(x)$ is continuous $\forall x \in R$

$ \therefore \alpha=0 $

Again, $f(x)=\left\{\begin{array}{cc}3-x, & x<-3 \\ 6, & -3 \leq x \leq 3 \\ 3+x, & x>3\end{array}\right.$

$ f^{\prime}(x)=\left\{\begin{array}{cc} -1, & x<-3 \\ 0, & -33 \end{array}\right. $

$ \begin{aligned} \because \quad f^{\prime}\left(-3^{-}\right) & =-1 \text { and } f^{\prime}\left(-3^{+}\right)=0 \\ f^{\prime}\left(-3^{-}\right) & \neq f^{\prime}\left(-3^{+}\right) \end{aligned} $

$\Rightarrow f(x)$ is not differentiable at $x=-3$ and $f^{\prime}\left(3^{-}\right)=0$ and $f^{\prime}\left(3^{+}\right)=1 \neq f^{\prime}\left(3^{+}\right)$

$\Rightarrow f(x)$ is not differentiable at $x=3$

$ \therefore \beta=2 $

Hence, $\alpha+\beta=0+2=2$.

$ \begin{aligned} & \text { Given, } \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3} \\ & \lim _{x \rightarrow 3^{-}} \frac{(x-2)\left(x^2-x-6\right)}{(x-1)\left(2 x^2-5 x-3\right)} \\ & =\lim _{x \rightarrow 3^{-}} \frac{(x-2)(x-3)(x+2)}{(x-1)(2 x+1)(x-3)}=\frac{1 \times 5}{2 \times 7}=\frac{5}{14} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\left(2^x\right)^2-2 \cdot 2^x+1+1-\cos 2 x}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2+2 \sin ^2 x}{x^2} \\ & =\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^2+2\left(\frac{\sin x}{x}\right)^2=(\ln 2)^2+2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &f(x)=\left\{\begin{array}{ll} \frac{x^2-16}{x-4} ; & x>4 \\ 2 x ; & x \leq 4 \end{array}= \begin{cases}x+4 ; & x>4 \\ 2 x ; & x \leq 4\end{cases}\right. \end{aligned} $

$ \begin{aligned} f^{\prime}\left(4^{-}\right) & =\lim _{h \rightarrow 0} \frac{f(4-h)-f(4)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{2(4-h)-8}{-h}=+2 \\ f^{\prime}\left(4^{+}\right) & =\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(4+h+4)-8}{h}=1 \\ & f^{\prime}\left(4^{-}\right)+f^{\prime}\left(4^{+}\right)=3 \end{aligned} $

$\begin{aligned} & \text { Here, } \lim _\limits{x \rightarrow-\infty} \log _e(\cosh x)+x \\ & =\lim _{x \rightarrow-\infty} \log \left(e^x+e^{-x}\right)+\ln 2 \\ & \Rightarrow \quad \lim _{x \rightarrow-\infty} \log \left(1+e^{2 x}\right)+\ln 2 \\ & =-\ln 2 \\ \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{a} \\ & =\lim _{x \rightarrow \infty} \frac{x^2}{x^2}\left[\frac{(b-c)+(c-a) \frac{1}{x}+\frac{(a-b)}{x^2}}{(a-b)+(b-c) \frac{1}{x}+\frac{(c-a)}{x^2}}\right]=\frac{1}{2} \end{aligned}$

$\begin{aligned} & =\frac{b-c}{a-b}=\frac{1}{2} \\ \Rightarrow & 2 b-2 c=a-b \\ \Rightarrow & 3 b=a+2 c \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{\sin ^3 x} \\ & \lim _{x \rightarrow-\infty} \frac{-3 x-x}{-x-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{x^3 \frac{\left(\sin x^3 x\right)}{x^3}} \\ \Rightarrow & \lim _{x \rightarrow-\infty} \frac{-4 x}{-3 x}-\lim _{x \rightarrow 0} \frac{1}{\left(\frac{\sin ^3 x}{x^3}\right)} \\ & \lim _{x \rightarrow-\infty} \frac{4}{3}-1=\frac{1}{3} \end{aligned}$

$\begin{aligned} & \lim _\limits{x \rightarrow \frac{-3}{5}} \frac{1}{x}\left[-\frac{1}{x}\right] \\ & =\frac{1}{\frac{-3}{5}}\left[-\frac{1}{\left(-\frac{3}{5}\right)}\right]=\frac{-5}{3}[1.66] \\ & =\frac{-5}{3} \cdot(1)=\frac{-5}{3} \end{aligned}$

If $ l $ and $ m $ (where $ l < m $) are the roots of the quadratic equation $ ax^2 + bx + c = 0 $, then the following limit expression is considered:

$ \lim\limits_{x \to \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} $

Given that $ \alpha \in (l, m) $, by applying L'Hôpital's rule, we analyze the limit:

$ \lim\limits_{x \to \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} $

This results in:

$ \frac{|a|}{a}, \text{ when } \alpha \in (l, m) $

Hence, the above contributes to the solution for the considered limit.

$\begin{aligned} \text { Let } f(x) & =\left\{\begin{array}{cc} \frac{1}{|x|}, & |x|>1 \\ a x^2+b, & |x| \leq 1 \end{array}\right. \\ & =\left\{\begin{array}{cc} \frac{1}{x}, & x>1 \\ a x^2+b & x \leq 1 \end{array}\right\} \end{aligned}$

Given, that $\lim _\limits{x \rightarrow 1^{+}} f(x)$ and $\lim _\limits{x \rightarrow 1^{+}} f(x)$ exist.

Thus, $\lim _\limits{x \rightarrow 1^{+}}(x)=\lim _\limits{x \rightarrow 1^{+}} \frac{1}{x}=1$

$\lim _\limits{x \rightarrow 1^{-}} f(x)=\lim _\limits{x \rightarrow 1^{-}}\left(a x^2+b\right)=a+b$

According to question, $a+b=1$ if take $a=\frac{3}{2}$ and $b=-\frac{1}{2}$.

Then, option (c) is only true all other option are incorrect.

$\begin{aligned} & \frac{d}{d x}\left(\lim _{y \rightarrow 2} \frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right)\right) \\ & =\frac{d}{d x} \lim _{y \rightarrow 2}\left(\frac{\left(\frac{1}{x}-\frac{1}{x+y-2}\right)}{(y-2)}\right) \end{aligned}$

Applying L Hospital's rule,

$\begin{aligned} & \frac{d}{d x} \lim _{y \rightarrow 2} \frac{0+\frac{1}{(x+y-2)^2}}{1} \\ & =\frac{d}{d x}\left(\frac{1}{x^2}\right)=-\frac{2}{x^3} \end{aligned}$

Given, $\left\{\begin{array}{cc}\frac{x^2 \log (\cos x)}{\log (1+x)} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$

Now, $\lim _\limits{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{x \log (1+x)}=\lim _{x \rightarrow 0} \frac{x \cdot \log (\cos x)}{\log (1+x)} \\ & =\lim _{x \rightarrow 0} \frac{x \log [1-(1-\cos x)]}{1-\cos x} \cdot \frac{1-\cos x}{\log (1+x)} \\ & =\lim _{x \rightarrow 0} \frac{\log [1-(1-\cos x)]}{1-\cos x} \cdot \frac{2 \sin ^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}, x^2 \frac{x}{\log (1+x)}=0 \end{aligned}$

So, $f$ is differentiable hence it is also continuous.

$\begin{aligned} f(x \cdot f(y)) & =f(x y)+x \quad \text{... (i)}\\ x & \leftrightarrow y \end{aligned}$

$ \begin{aligned} & \Rightarrow \quad f(y \cdot f(x))=f(x y)+y \quad \text{... (ii)}\\ & x \leftrightarrow f(x) \text { in Eq. (i), } \\ \end{aligned}$

$ \begin{aligned} & f(f(x) \cdot f(y))=f(y \cdot f(x))+f(x) \quad \text{... (iii)}\\ \Rightarrow \quad & f(f x) \cdot f(y))=f(x y)+y+f(x) \\ & \text { [from Eq. (ii)]} \quad \text{.... (iv)} \end{aligned}$

$\begin{aligned} & \text { Again } x \leftrightarrow y \text { in Eq. (iv), } \\ & f(f(y) \cdot f(x))=f(y x)+x+f(y) \quad \text{... (v)}\\ & \text { Eq. (iv) = Eq. (v), } \\ & f(x y)+y+f(x)=f(x y)+x+f(y) \end{aligned}$

$\begin{aligned} & \Rightarrow f(x)-x=f(y)-y \quad \text{.... (vi)}\\ & \Rightarrow f(x)-x=\lambda=f(y)-y \end{aligned}$

Substituting $f(x)=\lambda+x$ in Eq. (i), we have

$\begin{aligned} & x \cdot f(y)+\lambda=(x y+\lambda)+x \\ & \Rightarrow \quad x f(y)=x y+x \quad [\because f(y)=\lambda+y]\\ & \therefore x(y+\lambda)=x y+x \\ & \Rightarrow \lambda x=x \\ & \Rightarrow \lambda=1 \quad [\because x>0]\\ & \therefore \quad f(x)=x+\lambda=x+1 \end{aligned}$

$\text { Now, } \lim _\limits{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}}-1}{(f(x))^{\frac{1}{2}}-1}=\lim _\limits{x \rightarrow 0} \frac{(1+x)^{\frac{1}{3}}-1}{(1+x)^{\frac{1}{2}}-1}$

$=\lim _\limits{x \rightarrow 0}\left(\frac{(1+x)^{\frac{1}{3}}-1}{1+x-1}\right)\left(\frac{1+x-1}{(1+x)^{\frac{1}{2}}-1}\right)$

$=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\begin{aligned} &\begin{aligned} k & =\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}} \\ & =\frac{|0|}{\sqrt{5}}=0 \\ l & =\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=(0)^4 \sin \left(\frac{1}{3 \sqrt{0}}\right) \\ \end{aligned}\\ &\begin{aligned} & =0 \times(\text { a value which belongs to }[-1,1]) \\ &=0\\ \therefore k+l & =0+0=0 \end{aligned} \end{aligned}$

$\lim _\limits{n \rightarrow \infty} x^n \log _e x=0$

It is possible only when $x \in(0,1)$ i.e. $x>0$ and $x<1$.

In this case,

$\begin{gathered} x^n \text { or } x^{\infty}=0 \\ \Rightarrow \quad x^{\infty} \log _c x=0 \end{gathered}$

Now, $\log _x 12$ :

We know, for $\log _a b$ : If $a \in(0,1)$, then $\log _a b=$ Negative

$\therefore \log _x 12=$ Negative \quad $[\because x \in(0,1)]$

$f(x)=\operatorname{Max}\{3-x, 3+x, 6\}$

Plotting the graph :

We know that a function is not differentiable at the points where it has sharp corner point.

In the above graph, we can see that only two sharp corner points are present.

Therefore, the above function is not differentiable at two points which are

$\begin{array}{r} x=-3 \text { and } x=3 \\ a=-3 \text { and } b=3 \\ \therefore|a|+|b|=3+3=6 \end{array}$

$\begin{aligned} & \lim _\limits{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots .\right. \\ & \left.=\quad+\frac{n^4}{n^5+n^5}\right) \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^4}{r^5+n^5} \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{(r / n)^4}{(r / n)^5+1} \cdot\left(\frac{1}{n}\right) \end{aligned}$

On putting $\frac{r}{n} \rightarrow x, \frac{1}{n} \rightarrow d x$ and $\Sigma \rightarrow \int$

$\begin{aligned} & =\int_\limits0^1 \frac{x^4}{1+x^5} \cdot d x=\frac{1}{5} \int_\limits0^1 \frac{5 x^4}{1+x^5} \cdot d x \\ & =\frac{1}{5}\left[\log \left(1+x^5\right)\right]_0^1=\frac{1}{5}[\log 2-\log 1] \\ & =\frac{1}{5}[\log 2-0]=\frac{1}{5} \log 2 \\ & =\log \sqrt[5]{2} \end{aligned}$

f(0) = 0, f(1) = 1 and f(2) = 2

Let h(x) = f(x) $-$ x

Clearly h(x) is continuous and twice differentiable on (0, 2)

Also, h(0) = h(1) = h(2) = 0

$\therefore$ h(x) satisfies all the condition of Rolle's theorem.

$\therefore$ there exist C₁ $\in$(0, 1) such that h'(c₁) = 0

$\Rightarrow$ f'(₁) $-$ 1 = 0 $\Rightarrow$ f'(c₁) = 1

also there exist c₂ $\in$(1, 2) such that h'(c₂) = 0

$\Rightarrow$ f'(c₂) = 1

Now, using Rolle's theorem on [c₁, c₂] for f'(x)

We have f''(c) = 0, c$\in$(c₁, c₂)

Hence, f''(x) = 0 for some x$\in$(0, 2).