Limits, Continuity and Differentiability

$f(x) = \left\{ \matrix{ {{{{72}^x} - {9^x} - {8^x} + 1} \over {\sqrt 2 - \sqrt {1 + \cos x} }},\,x \ne 0 \hfill \cr a{\log _e}2{\log _e}3\,\,\,\,\,\,,\,\,x = 0 \hfill \cr} \right.$

$\because f(x)$ is continuous at $x=0$

$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ & \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2} \times x^2} \\ & =(\ln 9 \cdot \ln 8)(2 \sqrt{2}) \times 2 \\ & =4 \sqrt{2} \times 2 \times 3 \ln 2 \cdot \ln 3 \\ & 24 \sqrt{2} \cdot \ln 2 \cdot \ln 3 \\ & \Rightarrow \quad a=24 \sqrt{2} \\ & \quad a^2=1152 \end{aligned}$

$f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^2}, & x< \\ \alpha, & x=0 \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0 \end{array}\right.$

$f(x)$ is continuous at $x=0$

$\Rightarrow f(0)=\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)$

$\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\alpha \\ & \lim _{x \rightarrow 0^{-}}\left(\frac{1-\cos 2 x}{x^2}\right)=\alpha \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 h}{x^2}=\alpha \\ & \Rightarrow \lim _{h \rightarrow 0} \frac{2 \sin ^2}{h^2} h=\alpha \\ & \Rightarrow \alpha=2 \\ & \end{aligned}\\ &\begin{aligned} & \text { Also, } \lim _{x \rightarrow 0^{+}} f(x)=f(0) \\ & \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\beta \sqrt{1-\cos x}}{x}=2 \end{aligned} \end{aligned}$

$\Rightarrow \lim _\limits{h \rightarrow 0} \frac{\beta \sqrt{\frac{1-\cos h}{h^2}} h^2}{h}=2$

$\begin{aligned} & \Rightarrow \quad \frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \quad \beta=2 \sqrt{2} \\ & \Rightarrow \quad \alpha^2+\beta^2=4+8 \\ & \quad=12 \end{aligned}$

$\therefore$ Option (1) is correct

$\begin{aligned} & f:(0, \infty) \rightarrow R \\ & f(x)=e^{-\left|\log _e x\right|} \end{aligned}$

$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\begin{array}{l} \frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \\ \frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathrm{x} \geq 1 \end{array}\right.$

$\left\{\begin{array}{l} \frac{1}{\frac{1}{x}}=x ; 0< x<1 \\ \frac{1}{x}, x \geq 1 \end{array}\right.$

$\mathrm{m}=0$ (No point at which function is not continuous)

$\mathrm{n}=1$ (Not differentiable)

$\therefore \mathrm{m}+\mathrm{n}=1$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\ & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2} \end{aligned}$

Let $|\sin \mathrm{x}|=\mathrm{t}$

$\begin{aligned} & \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\ & =\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2 \end{aligned} $

Let $g(x)=a x+b$

Now function $\mathrm{f}(\mathrm{x})$ in continuous at $\mathrm{x}=0$

$\begin{aligned} & \therefore \lim _\limits{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \\ & \Rightarrow \lim _\limits{\mathrm{x} \rightarrow 0}\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}}=\mathrm{b} \\ & \Rightarrow 0=\mathrm{b} \\ & \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax} \end{aligned}$

Now, for $\mathrm{x}>0$

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \\ & +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \\ & \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \\ & \text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \\ & \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \\ & \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \\ & =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right) \end{aligned}$

$\begin{aligned} & f:(0,2) \rightarrow R ; f(x)=\frac{x}{2}+\frac{2}{x} \\ & f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2} \end{aligned}$

$\therefore \mathrm{f}(\mathrm{x})$ is decreasing in domain.

$g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0 < x \leq 1 \\ \frac{3}{2}+x & 1 < x < 2\end{cases}$

$\lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$

$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots .\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots .\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3}$

$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots .}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3}$

$\begin{aligned} & \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \\ & \Rightarrow \beta=-3, \alpha=1 \\ & \Rightarrow 2 \alpha-\beta=2+3=5 \end{aligned}$

$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|} \quad$ (for $f(x)$ to be cont.)

$\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x}<3 \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}$

Hence $f\left(3^{-}\right)=\frac{-a}{b}$

Then $f\left(3^{+}\right)=2^{\lim _\limits{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$ and $f(3)=b$.

Hence $\mathrm{f}(3)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3^{-}\right)$

$\begin{aligned} & \Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}} \\ & \mathrm{b}=2, \mathrm{a}=-4 \end{aligned}$

Hence only 1 ordered pair $(-4,2)$.

$\begin{aligned} a= & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)} \end{aligned}$

Applying limit $\mathrm{a}=\frac{1}{4 \sqrt{2}}$

$\begin{aligned} & b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\ & b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) \end{aligned}$

Applying limits $b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}$

Now, $a b^3=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^3=32$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1)=6 \\ & \lim _{x \rightarrow 0} \frac{\sin (\sin k x) \cdot \sin k x}{(\sin k x) k x} \cdot k+1-\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} \cdot x=3 \\ & k+1=3 \Rightarrow k=2 \end{aligned}$

Option-A : $\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}$

$\mathrm{f}(\mathrm{x})=0 \Rightarrow \sin \frac{\pi}{\mathrm{x}^2}=0 \Rightarrow \frac{\pi}{\mathrm{x}^2}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{N}$

$\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}$

$\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}$, finite number of solutions

Option-B : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n}<\pi^2$, Number of solutions is 9

Option-C : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}$, Infinite number of solutions

Option-D : $\frac{1}{\pi^2}<\frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)$, Definitely more than 25 solutions

$f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; \quad x=0 \end{array} \quad g(x)=\left\{\begin{array}{cc} 1-2 x & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.\right.$

$g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{cc} 2 x & ; \quad 0 \leq \frac{1}{2}-x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}=\left\{\begin{array}{cc} 2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right\}\right.$

$g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{ll} 1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 & ; \text { otherwise } \end{array}\right\}$

(P) $\quad$ Now $a=0, b=1, c=0, d=0$

$\because h(x)=g(x)+g\left(\frac{1}{2}-x\right)= \begin{cases}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{cases}$

Hence Range of h(x) is {0, 1}

(Q) $\mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=0$

$h(x)=f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; x=0 \end{array}\right.$

$\begin{aligned} & R H D=\lim _{x \rightarrow 0} \frac{x^2 \sin \frac{1}{x}-0}{x}=0 \\ & \text { LHD }=\lim _{x \rightarrow 0} \frac{-x^2 \sin \frac{1}{x}-0}{x}=0 \end{aligned}$

Hence $h(x)$ is differentiable on $R$

(R) $\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=1, \mathrm{~d}=0$

$h(x)=x-g(x)=\left\{\begin{array}{cl} 3 x-1 & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

$\therefore \mathrm{h}(\mathrm{x}) \text { is ONTO }$

(S) $a=0, b=0, c=0, d=1$

$h(x)=g(x)=\left\{\begin{array}{cl} 1-2 x & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

Range of h(x) is [0, 1]

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sin x^2 \cdot\left(\log _e x\right)^\alpha \cdot \sin \frac{1}{x^2}}{x^{\alpha \beta} \cdot\left(\log _c(1+x)\right)^\beta}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^\alpha}{\left(\log _e(x+1)\right)^\beta \cdot x^{\alpha \beta+2}}=0 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{\log _c x}{\log _e(x+1)}\right)^\beta \cdot \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t \\ & \lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0 \end{aligned}$

As we know $\lim _\limits{x \rightarrow \infty} \frac{x}{e^x}=0$

$\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

$ \lim \limits_{x \rightarrow \frac{3}{2}} \frac{2 x(2 x-3)\left(4 x^{2}+6 x+9\right)}{(2 x)^{1 / 3}-3^{1 / 3}} $

$ \lim \limits_{x \rightarrow \frac{3}{2}} 2 x\left|\frac{8 x^{3}-27}{(2 x)^{1 / 3}-3^{1 / 3}}\right| $

$ 3 \lim \limits_{x \rightarrow \frac{3}{2}} \frac{24 x^{2}}{\frac{1}{3}(2 x)^{-\frac{2}{3}} \cdot 2}=\frac{72 \times 3\left(\frac{3}{2}\right)^{2}}{3^{-\frac{2}{3}}} $

$ =72 \times 3 \times \frac{9}{4} \times 3^{\frac{2}{3}} \times \frac{1}{2} $

$ =\frac{1}{2} \times 18 \times 27 \times 3^{2 / 3}=9 \times 27 \times 3^{2 / 3} $

$ =3^{2+3+\frac{2}{3}}=3^{\frac{17}{3}}=\sqrt[3]{3^{17}} $

To evaluate the limit of the function $ f(x) $ as $ x $ approaches 0, we start with the given expression and simplify step-by-step:

$ \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{\left( 4^x - 1 \right)^4 \cot(x \ln 4)}{\sin(x \ln 4) \ln\left(1 + x^2 \ln 4\right)} $

First, rewrite the terms using a convenient limit form:

$ = \lim\limits_{x \to 0} \frac{\left(\frac{4^x - 1}{x}\right)^4 x^4 \cos(x \ln 4)}{\left(\frac{\sin(x \ln 4)}{x \ln 4}\right)^2 x^2 (\ln 4)^2} \times \frac{\ln\left(1 + x^2 \ln 4\right)}{x^2 \ln 4} \times x^2 \ln 4 $

Here, consider the limit:

$\lim\limits_{x \to 0} \frac{4^x - 1}{x} = \ln 4$

$\lim\limits_{x \to 0} \frac{\sin(x \ln 4)}{x \ln 4} = 1$

$\lim\limits_{x \to 0} \frac{\ln(1 + x^2 \ln 4)}{x^2 \ln 4} = 1$

Substituting these values into the expression:

$ = \lim\limits_{x \to 0} \frac{(\ln 4)^4}{(\ln 4)^3} = \ln 4 = k $

Thus, the value of $ e^k $ is:

$ e^k = e^{\ln 4} = 4 $

If $ x \in\left[0, \frac{\pi}{3}\right) \cos x > \frac{1}{2} $

then, $ \lim _{x \rightarrow a}=\frac{(2 \cos x-1)}{2 \cos x-1}=1 $

The function $ f(x) = \frac{|x-a|}{x-a} $ can be analyzed by considering different cases for the value of $ x $ relative to $ a $.

For $ x > a $, the expression $ |x-a| $ simplifies to $ x-a $. Thus, the function becomes:

$ f(x) = \frac{x-a}{x-a} = 1 $

For $ x < a $, the expression $ |x-a| $ simplifies to $-(x-a)$, which means the function is:

$ f(x) = \frac{-(x-a)}{x-a} = -1 $

Thus, the function $ f(x) $ can be rewritten as a piecewise function:

$ f(x) = \begin{cases} 1, & \text{if } x > a \\ -1, & \text{if } x < a \end{cases} $

Based on this function:

$ f(x) $ is a constant function (equal to 1) when $ x > a $.

It is discontinuous at $ x = a $ because the limits from the left and the right of $ a $ do not match. Specifically, the left-hand limit approaches $-1$ while the right-hand limit approaches $1$, and thus the function does not have a limit at this point.

In summary, $ f(x) $ is not continuous at $ x = a $ and is constant when $ x > a $.

Given the function:

$ f(x) = 3x^{15} - 5x^{10} + 7x^{5} + 50 \cos(x-1) $

We need to evaluate the limit:

$ \lim\limits_{h \rightarrow 0} \frac{f(1-h) - f(1)}{h^3 + 3h} $

This expression is initially an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's rule.

Using L'Hôpital's rule, the limit becomes:

$ \lim\limits_{h \rightarrow 0} \frac{-f'(1-h)}{3h^2 + 3} $

We then evaluate this limit by substituting $h = 0$:

$ \frac{-f'(1)}{3} $

To find $f'(x)$, we differentiate $f(x)$:

$ f'(x) = 45x^{14} - 50x^{9} + 35x^{4} - 50 \sin(x-1) $

Substitute $x = 1$ into $f'(x)$:

$ f'(1) = 45(1)^{14} - 50(1)^{9} + 35(1)^{4} - 50 \sin(0) $

$ f'(1) = 45 - 50 + 35 + 0 $

$ f'(1) = 30 $

Substitute back into the limit expression:

$ \frac{-30}{3} = -10 $

Thus, the value of the limit is $-10$.

To ensure the function $ f(x) $ is differentiable at $ x = 0 $, it must also be continuous at this point. Therefore, we need:

$ \lim\limits_{x \to 0} \frac{(e^{kx} - 1) \sin kx}{4 \tan x} = P $

Let's simplify the limit step-by-step:

Convert the expression using known limits:

$ \lim\limits_{x \to 0} \frac{kx}{4} \left[\frac{e^{kx} - 1}{kx}\right] \left[\frac{\sin kx}{kx}\right] \cdot \frac{kx}{x \left(\frac{\tan x}{x}\right)^2} $

Recognize standard limits for small $ x $:

$\lim\limits_{x \to 0} \frac{e^{kx} - 1}{kx} = 1$

$\lim\limits_{x \to 0} \frac{\sin kx}{kx} = 1$

$\lim\limits_{x \to 0} \frac{\tan x}{x} = 1$

Plug these into the equation:

$ \lim\limits_{x \to 0} \frac{k^2 x^2}{4x} = \lim\limits_{x \to 0} \frac{k^2 x}{4} = P $

As $ x \to 0 $, the expression $ \frac{k^2 x}{4} \to 0 $, so $ P = 0 $.

Since $ P = 0 $, let's find the derivative at $ x = 0 $:

$ f'(x) = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{\frac{(e^{kh} - 1) \sin kh}{4 \tan h}}{h} $

Applying the same logic and limits as before, the derivative becomes:

$ f'(0) = \frac{k^2}{4} $

Therefore, $ P = 0 $ and $ f'(0) = \frac{k^2}{4} $.

To apply Rolle's Theorem, we need the function to satisfy $ f(a) = f(b) $ while $ a \neq b $. For this function, consider :

$ \lim\limits_{x \rightarrow 0} x^{p} \log x = 0 $

Expanding the logarithm using Taylor series, we get :

$ \lim\limits_{x \rightarrow 0} x^{p} \left[x - 1 - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3} + \ldots\right] = 0 $

If $ p = 0 $, the limit does not approach zero.

Therefore, the minimum value of $ p $ that satisfies the condition for the limit to be zero is $ p = 1 $.

Given, $ \lim \limits_{x \rightarrow 4} \frac{2 x^{2}+(3+2 a) x+3 a}{x^{3}-2 x^{2}-23 x+60}=\frac{11}{9} $

Put $ x=4 $

$ =\frac{4}{32+12+8 a+3 a} \frac{64-32-92+60}{64+11 a} $

$ \because $ Numerator must be zero

$ \Rightarrow \quad a=\frac{-44}{11}=-4 $

Then, $ \lim \limits_{x \rightarrow-4} \frac{x^{2}+9 x+20}{x^{2}-x-20} $

$ \lim \limits_{x \rightarrow-4} \frac{(x+5)(x+4)}{(x-5)(x+4)} \Rightarrow \frac{-4+5}{-4-5}=\frac{-1}{9} $

$ \because f(x) $ is continuous at $ x=4 $.

$ \begin{aligned} f(4) & =\frac{64-125}{16-25}=\frac{61}{9} \\\\ f\left(4^{+}\right) & =\frac{b^{4}-1}{4}=\frac{61}{9} \\\\ b^{4} & =\frac{244+9}{9}=\frac{253}{9} \Rightarrow 9 b^{4}=253 \end{aligned} $

$ \because f(x) $ is continuous at $ x=1 $

So, $ f\left(1^{+}\right)=f(1)=\frac{1-125}{1-25}=\frac{31}{6} \Rightarrow f\left(1^{-}\right)=a $

$ \Rightarrow 6 a+9 b^{4} \Rightarrow 31+253 \Rightarrow 284 $

$ \begin{aligned} & \lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} \\\\ & =\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \sin ^4 \theta+4 \sin ^2 \theta \cos ^2 \theta+5 \cos ^4 \theta}{(3 \cos \theta-2 \sin \theta)^4} \\\\ & =\frac{8(1)^4+4(1)^2(0)^2+5(0)^4}{(3(0)-2(1))^4}=\frac{8}{16}=\frac{1}{2} \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{cc} \frac{1-\cos 4 x}{x^2} & , \quad x<0 \\\\ a & , x=0 \\\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & , x>0 \end{array}\right. $

$ \text { is continuous at } x=0 $

$ \begin{array}{lr} \therefore & \lim _{x \rightarrow 0^{-}} f(x)=f(0) \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{1-\left(1-2 \sin ^2 2 x\right)}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} 8\left(\frac{\sin 2 x}{2 x}\right)^2=a \\\\ \Rightarrow & 8(1)^2=a \Rightarrow a=8 \end{array} $

We have,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x} \,\,\,\,\,\,\,\,\,\,\,\,\left(\frac{0}{0}\right. form ) \\\\ = & \lim _{x \rightarrow 0} \frac{\ln 3\left(3^{\sin x} \cdot \cos x\right)-\ln 2\left(2^{\operatorname{an} x} \cdot \sec ^2 x\right)}{\cos x} \\\\ = & \frac{\ln 3 \cdot 1-\ln 2 \cdot 1}{1}=\ln 3-\ln 2=\ln \left(\frac{3}{2}\right) \end{aligned} $

We have,

$ f(x)=\left\{\begin{array}{cc} \frac{\cos a x-\cos 9 x}{x^2} & , \text { if } x \neq 0 \\\\ 16 & \text {, if } x=0 \end{array}\right. $

Now, $\lim _{x \rightarrow 0} f(x)=16 \quad \ldots \text { (ii) }$

$[\because f(x)$ is continuous at $x=0$ ]

$ \lim _{x \rightarrow 0} \frac{\cos a x-\cos 9 x}{x^2}=16 $

$ \begin{aligned} &\text { Take LHS }\\\\ &\begin{aligned} & \text { HS } \\\\ & \begin{array}{l} \lim _{x \rightarrow 0} \frac{\cos a x-\cos 9 x}{x^2} \\\\ =\lim _{x \rightarrow 0} \frac{-a \sin a x+9 \sin 9 x}{2 x} \\\\ \text { [using L.' Hospital's rule] } \\\\ =\lim _{x \rightarrow 0} \frac{-a^2 \cos a x+81 \cos 9 x}{2} \\\\ =\frac{81-a^2}{2}=16 \quad \text { [from, Eq. (i)] } \\\\ 81-a^2=32 \Rightarrow a^2=49 \Rightarrow a= \pm 7 \end{array} \end{aligned} \end{aligned} $

We have,

$ f(x)=\left\{\begin{array}{ll} \frac{8}{x^{3}}-6 x, & \text { if } 0 < x \leq 1 \\\\ \frac{x-1}{\sqrt{x}-1}, & \text { if } x > 1 \end{array}\right. $

$ \begin{array}{l} f(1)=2 \text { and } \\\\ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(\frac{8}{x^{3}}-6 x\right)=2=\text { LHL } \end{array} $

$ \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{x-1}{\sqrt{x}-1} $

$ =\lim _{x \rightarrow 1^{+}} \frac{1-0}{\frac{1}{2 \sqrt{x}}-0}=\frac{21}{1}=2 $

$ \because \quad f(\mathrm{l})=\mathrm{LHL}=\text { RHL } $

$ \Rightarrow \quad f(x) $ is continuous at $ x=1 $

And for differentiability at $ x=1 $,

$ \begin{array}{l} \text { LHD }=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\\\ =\lim _{h \rightarrow 0} \frac{\frac{8}{(1-h)^{3}}-6(1-h)-f(2)}{-h} \\\\ =\lim _{h \rightarrow 0} \frac{8-6(1-h)^{4}-2(1-h)^{3}}{-h(1-h)^{3}} \end{array} $

$ \begin{array}{l}=\lim _{h \rightarrow 0} \frac{0+24(1-h)^{3}+6(1-h)^{2}}{-\left[(1-h)^{3}-3 h(1-h)^{2}\right]} \\\\ =-\lim _{h \rightarrow 0} \frac{24(1-h)+6}{1-h-3 h}=-30 \\\\ \text { RHD }=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \\\\ =\lim _{h \rightarrow 0} \frac{\frac{1+h-1}{\sqrt{1+h}-1}-2}{h} \\\\ =\lim _{h \rightarrow 0} \frac{h-2 \sqrt{1+h}+2}{(\sqrt{1+h}-1) h}\end{array} $

$ \begin{array}{l} =\lim _{h \rightarrow 0} \frac{1-\frac{1}{\sqrt{1+h}}}{\frac{1}{2 \sqrt{1+h}} \cdot h+\sqrt{1+h}-1} \\\\ =2 \lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h+2(1+h)-2 \sqrt{1+h}} \\\\ =2 \lim _{h \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+h}}}{1+2-\frac{1}{\sqrt{1+h}}}=2 \times \frac{\frac{1}{2}}{3-1}=\frac{1}{2} \end{array} $

$ \because \mathrm{LHD} \neq \mathrm{RHD} $

Hence, at $ x=1, f(x) $ is continuous but not differentiable.

$ \begin{aligned} & \text L=\lim _{n \rightarrow \infty}\left[\begin{array}{c} \left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right) \\\\ \left(1+\frac{9}{n^2}\right) \ldots(2) \end{array}\right]^{1 / n} \\\\ & =\lim _{n \rightarrow \infty}\left(\prod_{K=1}^n\left(1+\frac{K^2}{n^2}\right)\right)^{1 / n} \\\\ & L=\lim _{n \rightarrow \infty} e^{\left(\frac{1}{n} \sum_{K=1}^n \ln \left(1+\frac{K^2}{n^2}\right)\right)} \quad \ldots \text { (ii) } \end{aligned} $

Sum can be approximated by an integral in the limit as $n \rightarrow \infty$

$ \Rightarrow \frac{1}{n} \sum_{K=1}^n \ln \left(1+\frac{K^2}{n^2}\right) \approx \int_0^1 \ln \left(1+x^2\right) d x $

Using integration by parts

$ \begin{aligned} \int_0^1 \ln \left(1+x^2\right) d x=\left[\ln \left(1+x^2\right)\right. & \cdot x]_0^1 \\\\ & -\int \frac{2 x^2+2-2}{1+x^2} d x \end{aligned} $

$ =\ln 2-\int_0^1 2 d x+2 \int_0^1 \frac{1}{1+x^2} d x=\ln 2-2+\frac{\pi}{2} $

$ \begin{aligned} & \text { So, from Eq. (i), we get } \\\\ & L=e^{\ln 2-2+\frac{\pi}{2}}=2^{\frac{\pi}{2}-2}=2^{\left(\frac{\pi-4}{2}\right)} \end{aligned} $

$ \begin{aligned} & \text { } \because \lim _{x \rightarrow 1} f(x) \text { exists } \\ & \therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned} $

$ \begin{aligned} &Now, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ & =\lim _{h \rightarrow 0}\left(1+\frac{2(1-h)}{a}\right)=1+\frac{2}{a} \end{aligned} $

$ \begin{array}{ll} \text { and } & \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ & =\lim _{h \rightarrow 0} a(1+h)=a \\ \therefore & 1+\frac{2}{a}=a \\ \Rightarrow & a+2=a^2 \\ \Rightarrow & a^2-a-2=0 \\ \Rightarrow & a^2-2 a+a-2=0 \\ \Rightarrow & a(a-2)+1(a-2)=0 \\ \Rightarrow & (a-2)(a+1)=0 \\ \Rightarrow & a=-1,2 \end{array} $

$\therefore$ Sum of the cubes of the possible values of $a$ is $-1+8=7$.

At $x=0$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}}\left(\left[x^2\right]-[x]^2\right)=0-1=-1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 0^{+}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 \end{aligned} $

$\therefore$ LHL $\neq$ RHL

Limit does not exists at $x=0$.

At $x=1$

$ \mathrm{LHL}=\lim _{x \rightarrow 1^{-}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 $

$\mathrm{RHL}=\lim _{x \rightarrow 1^{+}}\left(\left[x^2\right]-[x]^2\right)=1-1=0$

As LHL $=$ RHL, so limit exits at $x=1$

At $x=2$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 2^{-}}\left(\left[x^2\right]-[x]^2\right)=3-1=2 \\ & \mathrm{RHL}=\lim _{x \rightarrow 2^{+}}\left(\left[x^2\right]-[x]^2\right)=4-4=0 \end{aligned} $

As LHL $\neq$ RHL, so limit does not exists. Hence, there are two integral points where the limits does not exists.

$ \begin{aligned} &\text { As } f(x) \text { is continuous at } x=0\\ &\begin{aligned} & f(0)=\lim _{x \rightarrow 0} f(x) \\ & K=\lim _{x \rightarrow 0} \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \Rightarrow K=\lim _{x \rightarrow 0} \\ & \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \quad \quad \times \frac{\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}}{\left.\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}\right)} \\ & \quad \quad \times \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-2 a x(\sqrt{a+x}+\sqrt{a-x)}}{2 x\left(\sqrt{x^2-a x+a^2}+\sqrt{x^2+a x+a^2}\right)} \\ & =\frac{-a(2 \sqrt{a})}{2 a}=-\sqrt{a} \\ & \therefore \quad K=-\sqrt{a} \end{aligned} $

$f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \\ b x^3+1, & x > 2\end{array}\right.$

$ f^{\prime}(x)=\left\{\begin{array}{cc} 2 a x+b, & x \leq 1 \\ 3, & 1 < x \leq 2 \\ 3 b x^2, & x > 2 \end{array}\right. $

As, the function is differentiable at $x \in R$,

$ \begin{aligned} & \lim _{x \rightarrow 1} 2 a x+b=3 \text { and } \lim _{x \rightarrow 2} 3 b x^2=3 \\ & \Rightarrow 2 a+b=3 \text { and } 12 b=3 \Rightarrow b=\frac{1}{4} \\ & \quad 2 a=3-\frac{1}{4} \Rightarrow 2 a=\frac{11}{4} \Rightarrow a=\frac{11}{8} \\ & \therefore \quad a-b=\frac{11}{8}-\frac{1}{4}=\frac{9}{8} \end{aligned} $

For option (a), $f(x)=|x|=x$ as $1 \leq x \leq 5$

As $f(x)$ is a polynomial function in its domain, so it is continuous and differentiable.

$ \begin{aligned} & f^{\prime}(x)=1, f(1)=1 \text { and } f(3)=5 \\ & f^{\prime}(x)=\frac{f(5)-f(1)}{5-1}=1 \in[1,5] \end{aligned} $

$\therefore$ Lagrange's mean value theorem is applicable.

For option (b)

$ f(x)=[x]=1 \text { for } x \in[\sqrt{2}, \sqrt{3}] $

As $f(x)$ is a constant function, so Lagranye's mean value theorem is applicable.

For option (c),

The function $f(x)=x^2-1$ is continuous if $x^2-1>0$ i.e. either $x<-1$ or $x>1$.

For $x=\frac{1}{e}, x^2-1>0$, but $\frac{1}{e^2}-1<0$

So, the value is not within the domain Thus, lagrange mean value theorem is not applicable.

For option (d), $f(x)=e^x$ is continuous and differentiable over $[-e, e]$

$ f^{\prime}(x)=e^x \text { and } f(-e)=\frac{1}{e^i} f(l)=e^e{} $

$ \begin{aligned} f^{\prime}(e) & =\frac{f(b)-f(a)}{b-a}=\frac{e^e-\frac{1}{e^e}}{e+e} \\ & =\frac{\left(e^e\right)^2-1}{e^e \times 2 e} \\ & =\frac{e^{e-1}}{2}-\frac{1}{2 e^{e+1}} \in[-e, e] \end{aligned} $

$\therefore$ Lagrange mean value theorem is applicable.

$ \begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0 \end{array}\right. \\ & = \begin{cases}2, & x<0 \\ 0, & x>0 \\ 2, & x=0\end{cases} \\ & \quad\left[\because|x|=\left\{\begin{array}{c} x, \text { if } x>0 \\ -x, \text { if } x<1 \end{array}\right.\right. \end{aligned} $

$\therefore f(x)$ has maximum value 2 and minimum value 0 .

$\mathop {\lim }\limits_{x \to \infty } \frac{[2 x-3]}{x} \text { } $

$ \begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{2 x}{x}-\frac{3}{x}\right) \Rightarrow 2-\lim _{x \rightarrow \infty} \frac{3}{x} \\ & \Rightarrow \quad 2-0 \Rightarrow 2 \end{aligned} $

$\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 3 x}{\cos 4 x-\cos 5 x} \quad\left(\frac{0}{0}\right.$ form $)$

By L' Hospital rule,

$ =\mathop {\lim }\limits_{x \to 0} \frac{-2 \sin 2 x+3 \sin 3 x}{-4 \sin 4 x+5 \sin 5 x} \quad\left(\frac{0}{0} \text { form }\right) $

Again by L' Hospital rule

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-4 \cos 2 x+9 \cos 3 x}{-16 \cos 4 x+25 \cos 5 x} \\ & =\frac{-4 \times 1+9 \times 1}{-16 \times 1+25 \times 1}=\frac{5}{9} \end{aligned} $