Limits, Continuity and Differentiability

$\mathop {\lim }\limits_{x \to 3} \frac{x^2+(a+3) x+(a+1)}{x+3}$

It is $\frac{0}{0}$ form

$ \mathop {\lim }\limits_{x \to 3} \frac{2 x+(a+3)}{1}=-6+a+3=a-3 $

$\because f(x)$ is continuous $a-3=-\frac{5}{2}$

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

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2} \\ & =\lim _{x \rightarrow 0} \frac{2 x \tan x\left[\frac{1}{1-\tan ^2 x}-1\right]}{\frac{1-\cos 3 x}{(3 x)^2} \times(3 x)^2\left[\frac{1-\cos x}{\sin x}\right]^2} \\ & =\lim _{x \rightarrow 0} \frac{2 x \tan x\left(1-1+\tan ^2 x\right) \times \sin ^2 x}{\left(1-\tan ^2 x\right) \frac{1-\cos 3 x}{(3 x)^2}} \\ & \times 9 x^2 \frac{(1-\cos x)^2}{x^2} \times x^4 \\ & =\lim _{x \rightarrow 0} \frac{2}{9} \times \frac{\tan x}{x} \times \frac{\tan ^2 x}{\frac{1}{2}} \times \frac{\sin ^2 x}{x^4 \times\left(\frac{1}{2}\right)^2} \\ & =\lim _{x \rightarrow 0} \frac{2}{9} \times 2 \times 4=\frac{16}{9} \end{aligned} \end{aligned} $

Which is continuous and strictly increasing

(B) $f(x)=\sqrt{|x|}=\left\{\begin{aligned} \sqrt{x}, & x \geq 0 \\ \sqrt{-x}, & x<0\end{aligned}\right. f^{\prime}(x)=\frac{1}{2 \sqrt{x}}( \pm 1)$ Which is not defined at $x=0$, this is continuous but not differentiable at $x=0$

(C) $f(x)=x+[x]=\left\{\begin{array}{cc}x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ x+1, & 1 \leq x<2\end{array}\right.$

This is strictly increasing but not differentiable at integer.

$ \begin{aligned}(D) f(x) & =|x-1|+|x|+|x+1| \\ & =\left\{\begin{array}{cc} -3 x, & -1 \leq x<0 \\ 2-x, & 0 \leq x<1 \\ 3 x, & 1 \leq x<\infty \end{array}\right. \end{aligned} $

Clearly, it is not differentiable at $x=-1$, 0,1 but differentiable in $(-1,0) \cup(0,1)$

A-I, B-II, C-V, D-IV

(I) $f(x)= \begin{cases}\frac{1}{2}-x, & x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2 & x \geq \frac{1}{2}\end{cases}$

$f(x)$ is not differentiable at $x=\frac{1}{2}$

LMVT not applicable

(II) $f(x)=|3 x-1|$

$f(x)$ is not differentiable at $x=\frac{1}{3}$

$\therefore$ LMVT is not applicable.

(III) $f(x)=x|x|$ it is continuous and differentiable in $(0,1)$

= LMVT applicable.

(V) $f(x)=|x|$ it is continuous and differentiable in $(0,1)$

$\therefore$ LMVT is applicable.

$ \text { } \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{n+r}{n^2+(r)^2} \\ & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{1+\frac{r}{n}}{1+\left(\frac{r}{n}\right)^2} \\ & =\int_0^2 \frac{1+x}{1+(x)^2} d x \\ & =\int_0^2 \frac{1}{1+x^2} d x+\int_0^2 \frac{x}{1+x^2} d x \\ & =\left(\tan ^{-1} x\right)_0^2+\left(\frac{1}{2} \ln \left(1+x^2\right)\right)_0^2 \\ & =\tan ^{-1} 2+\frac{1}{2} \ln 5 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)\left[1-\cos \left(\frac{\pi}{2}-x\right)\right]}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right) \cdot 2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\left(\frac{\pi}{2}-x\right) \cdot 16\left(\frac{\pi}{4}-\frac{x}{2}\right)^2} \\ & =\frac{2}{32} \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)} \times \lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}{\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}^2=\frac{2}{32}=\frac{1}{16} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} h e^a\left(\frac{e^{h h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} e^a\left\{\frac{e^{b-a}-1}{\left(\frac{e^h-1}{h}\right)}\right\}=e^a\left(e^{b-a}-1\right)=e^b-e^a \end{aligned} $

When $x<2|x-2|=-(x-2)=2-x$.

Then for $x<2$ :

$ \begin{aligned} g(x) & =\frac{x^2-4}{x^2-2(2-x)-4}=\frac{x^2-4}{x^2+2 x-8} \\ & =\frac{(x-2)(x+2)}{(x-2)(x+4)}=\frac{x+2}{x+4} \end{aligned} $

The limit from the left is

$ \mathop {\lim }\limits_{x \to {2^ - }} g(x)=\mathop {\lim }\limits_{x \to {2^ - }} \frac{x+2}{x+4}=\frac{2+2}{2+4}=\frac{4}{6}=\frac{2}{3} $

When $x>2|x-2|=x-2$.

Then, for $x>2$ :

$ \begin{aligned} g(x) & =\frac{x^2-4}{x^2-2(x-2)-4}=\frac{x^2-4}{x^2-2 x} \\ & =\frac{(x-2)(x+2)}{x(x-2)}=\frac{x+2}{x} \end{aligned} $

The limit from the right is:

$ \mathop {\lim }\limits_{x \to {2^ + }} g(x)=\mathop {\lim }\limits_{x \to {2^ + }} \frac{x+2}{x}=\frac{2+2}{2}=\frac{4}{2}=2 $

Since, $\mathop {\lim }\limits_{x \to {2^ - }} g(x)=\frac{2}{3}$ and $\mathop {\lim }\limits_{x \to {2^ + }} g(x)=2$, the limit $\mathop {\lim }\limits_{x \to {2}} g(x)$ does not exist because the left and right limits are not equal.

Since, the limit does not exists at $x=2, g(x)$ is not continuous at $x=2$

We have, $\mathop {\lim }\limits_{x \to \infty }\left\{\frac{x^2-1}{x+1}-a x-b\right\}=2$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{x^2(1-a)-x(a+b)-1-b}{x+1}=2 $

Since, the limit of the given expression is a finite non-zero number, therefore numerator and denominator are of the same degree.

$ \begin{aligned} \therefore & & 1-a & =0 \\ \Rightarrow & & a & =1 \end{aligned} $

$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$

Using expansion

$\begin{aligned} & =\lim _\limits{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\ & =\lim _\limits{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e \end{aligned}$

$f(x)=\left\{\begin{array}{cc} \frac{\tan ((a+1) x)+b \tan x}{x}, & x<0 \\ 3 & x=0 \\ \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x>0 \end{array}\right.$

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

$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\ & \lim _{x \rightarrow 0^{-}} f(x)=3 \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\ & \Rightarrow a+1+b=3 \\ & \Rightarrow a+b=2 \quad \text{..... (1)} \end{aligned}$

also, $\lim _\limits{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3$

$\begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\ & =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\ & \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\ & \Rightarrow \frac{b}{2 a}=3 \\ & \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)} \end{aligned}$

$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$

$\begin{aligned} & \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\ & =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\ & =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n \end{aligned}$

We will take term with the greatest power of $n$

$=\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4$

$\begin{aligned} & \text { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \\ & =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right) \end{aligned}$

Greatest power of $n$ is $\frac{n^4}{4}$

$\lim _\limits{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}$

$\begin{aligned} & f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\ & f(-1)=2+1+0=3 \\ & f\left(-1^{+}\right)=2+0+0=2 \\ & f\left(0^{-}\right)=0+1=1 \\ & f\left(0^{+}\right)=0+0+0=0 \\ & f\left(1^{+}\right)=2+1+0=3 \end{aligned}$

$\begin{aligned} & f\left(1^{-}\right)=2+0+1=3 \\ & f\left(2^{-}\right)=8+3+1=12 \\ & f\left(2^{+}\right)=8+4+0=12 \end{aligned}$

$\therefore$ discontinuous at $x=0, \sqrt{2}, \sqrt{3},-1$

$\begin{aligned} & \lim _\limits{x \rightarrow 0} f(x)=f(0) \quad \text { (continuous at } x=0) \\ & \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3} \end{aligned}$

For limit to exist $\beta=0$

$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3} \end{aligned}$

For limit to exist $\alpha+3=0 \Rightarrow \alpha=-3$

$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)$

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

$\because f(x)$ 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} $