$ \lim\limits_{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)= $

If a real valued function $f(x)=\left\{\begin{array}{cl}e^{\frac{\sin a(x-[x])}{x-[x]}} & , \text { if } x<1 \\ b+1 & , \text { if } x=1 \text { is } \\ \frac{\left|x^2+x-2\right|}{x-1} & , \text { if } x>1\end{array}\right.$ continuous at $x=1$, then $b \sin a=([x]$ denotes the greatest integer function)

$\mathop {\lim }\limits_{x \to {\pi \over 4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}= $

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,

$ \lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)= $

If the function $f$ defined by

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

is continuous at $x=0$, then $a=$

The domain of the derivative of the function $f(x)=\frac{x}{1+|x|}$ is

$ \mathop {\lim }\limits_{x \to 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3}-\sqrt{\sin ^2 x-2 \tan x-x+3}} $

$ \mathop {\lim }\limits_{x \to \infty } \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}}= $

If a real valued function

$ f(x)=\left\{\begin{array}{cc} \log (1+[x]), & x \geq 0 \\ \sin ^{-1}[x], & -1 \leq x<0 \\ k([x]+|x|), & x<-1 \end{array}\right. $

is continuous at $x=-1$, then $k=$

$\mathop {\lim }\limits_{n \to \infty } \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\sin \frac{3 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]= $

$[x]$ represents the greatest integer function. If $\mathop {\lim }\limits_{x \to 0 + } \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$, then $k=$

$ \mathop {\lim }\limits_{x \to 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}= $

If $f(x)=\left\{\begin{array}{cl}\frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x}, & \text { if } x>0 \\ 2, & \text { if } x=0 \\ \frac{\cos 4 x-\cos b x}{\tan ^2 x}, & \text { if } x<0\end{array}\right.$ is continuous at $x=0$, then $\sqrt{b^2-a^2}=$

$\mathop {\lim }\limits_{x \to 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}= $

If a real valued function

$ f(x)=\left\{\begin{array}{cc} (1+\sin x)^{\cos x}, & -\pi / 2 < x < 0 \\ a, & x=0 \\ \frac{e^{2 / x}+e^{3 / x}}{a e^{2 / x}+b e^{3 / x}}, & 0 < x < \pi / 2 \end{array}\right. $

is continuous at $x=0$, then $a b=$

$ \mathop {\lim }\limits_{x \to 0} \frac{(\operatorname{cosec} x-\cot x)\left(e^x-e^{-x}\right)}{\sqrt{3}-\sqrt{2+\cos x}}= $

$ \mathop {\lim }\limits_{x \to \infty } \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}}= $

If a function,

$ f(x)=\left\{\begin{array}{cc} \frac{\sqrt[3]{1+a x^2+b x^3}-\sqrt[3]{1-a x^2-b x^3}}{x^2}, & x<0 \\ 5, & x=0 \\ \frac{\tan 3 x-\sin 3 x}{b x^3}, & x>0 \end{array}\right. $

is continuous at $x=0$, then the geometric mean of $a$ and $b$ is

$[x]$ denotes the greater integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim\limits_{x \rightarrow 0}-\frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$, then $\sin \theta+\cos \theta=$

$ \mathop {\lim }\limits_{n \to \infty } \frac{1}{n^3} \sum\limits_{k=1}^n k^2 x= $

Let $f: R \rightarrow R$ be defined by

$ f(x)=\left\{\begin{array}{cc} a-\frac{\sin [x-1]}{x-1}, & \text { if } x>1 \\ 1, & \text { if } x=1 \\ b-\left[\frac{\sin [x-1]-[x-1]}{([x-1])^3},\right. & \text { if } x<1 \end{array}\right. $

where $[t]$ denotes the greatest integer less than or equal to $t$. If $f$ is continuous at $x=1$, then $a+b=$

$ \mathop {\lim }\limits_{y \to 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}= $

If $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 4 x}{1-\cos 2 x}=k$, then $\lim\limits_{x \rightarrow k} \frac{x^k-27}{x^{k+1}-81}=$

If the function $f(x)=\left\{\begin{array}{l}1+\cos x, x \leq 0 \\ a-x, 02\end{array}\right.$ everywhere, then $a^2+b^2=$

$ \mathop {\lim }\limits_{x \to - \infty } \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}= $

If $\mathop {\lim }\limits_{x \to {a^ + }} f(x)=p, \mathop {\lim }\limits_{x \to {a^ - }} f(x)=m$ and $f(a)=k$, then which one of the following is true?

If a function $f$ defined by

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

is continuous at $x=0$, then $a=$

$ \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt{2})-\sqrt{1+\cos x}}{\sqrt{15+\cos 2 x-4}}= $

If a real valued function

$ f(x)=\left\{\begin{array}{cl} \frac{x^2+(a+3) x+(a+1)}{x+3} & , \text { when } x \neq-3 \\ -\frac{5}{2} & , \text { when } x=-3 \end{array}\right. $

is continuous at $x=-3$, then $\lim _{x \rightarrow a}\left(x^2+x+1\right)=$

$ \mathop {\lim }\limits_{x \to 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}= $

Match the functions in Column I with their properties in Column II. In the following [ $x$ ] denotes the greatest integer less than or equal to $x$.

Consider the following functions

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

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

III. $f(x)=x|x|$

IV. $f(x)=|x|$

Then, on $[0,1]$ Lagrange's mean value theorem is applicable to the functions

$ \mathop {\lim }\limits_{x \to \infty }\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{n+2 n}{n^2+4 n^2}\right]= $

Let $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \\ a x, & 1 < x \leq 2\end{array}\right.$.If $\lim _{x \rightarrow 1} f(x)$ exists, then the sum of the cubes of the possible values of $a$ is

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$, then the number of integral values of ' $a$ ' such that $\lim \limits_{x \rightarrow a}\left(\left[x^2\right]-[x]^2\right)$ does not exist is

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

If a real valued function $f(x)=\left\{\begin{array}{cl}\frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}, & \text { for } x \neq 3 \\ 1, & \text { for } x=3\end{array}\right.$ is continuous at $x=3$ and $l$ is a finite value, then $l-k$ is equal to

$\mathop {\lim }\limits_{x \to o} \left[\frac{1}{x}-\frac{1}{e^x-1}\right]= $

Let $f(x)=\left\{\begin{array}{cl}0, & x=0 \\ 2-x, & \text { for } 0 < x < 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 < x < 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.$

then which of the following is true

The values of $a$ and $b$ for which the function

$ f(x)=\left\{\begin{array}{cl}1+|\sin x|^{\frac{a}{\sin x \mid}} & \frac{-\pi}{6} < x < 0 \\ b, & x=0 \quad \text { is continuous at } x=0 \\ e^{\frac{\tan 2 x}{\tan 3 x},} & 0 < x < \frac{\pi}{6}\end{array}\right. $

are

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

is differentiable, $\forall x \in R$, then $f^{\prime}(2)=$

If a function $f(x)=\left\{\begin{array}{cl}\frac{\tan (\alpha+1) x+\tan 2 x}{x} & \text { if } x>0 \\ \beta & \text { at } x=0 \text { is } \\ \frac{\sin 3 x-\tan 3 x}{x^3} & \text { if } x<0\end{array}\right.$

continuous at $x=0$, then $|\alpha|+|\beta|=$

If $f(x)=\left\{\begin{array}{ll}3 a x-2 b, & x>1 \\ a x+b+1, & x<1\end{array}\right.$ and

$\lim \limits_{x \rightarrow 1} f(x)$ exists, then the relation between $a$ and $b$ is

If $f(x)=\left\{\begin{array}{cl}x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$

which of the following is true?

$f(x)=\left\{\begin{array}{cl}\frac{\left(2 x^2-a x+1\right)-\left(a x^2+3 b x+2\right)}{x+1}, & \text { if } x \neq-1 \\ k_k, & \text { if } x=-1\end{array}\right.$

is a real valued function. If $a, b, k \in R$ and $f$ is continuous on $R$, then $k=$

$\lim _\limits{x \rightarrow-\infty} \log _e(\cosh x)+x=$

If $a, b$ and $c$ are three distinct real numbers and $\lim _\limits{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{2}$, then $a+2 c=$

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

If $[\cdot]$ denotes greatest integer function, then $\lim _\limits{x \rightarrow \frac{-3}{5}} \frac{1}{\dot{x}}\left[\frac{-1}{x}\right]=$

If $l, m(l< m)$ are roots of $a x^2+b x+c=0$, then $\lim _\limits{x \rightarrow \alpha} \frac{\left|a x^2+b x+c\right|}{a x^2+b x+c}=$

Let $f(x)=\left\{\begin{array}{cl}\frac{1}{|x|}, & \text { for }|x|>1 \\ a x^2+b, & \text { for }|x| \leq 1\end{array}\right.$. If $\lim _\limits{x \rightarrow 1^{+}} f(x)$ and $\lim _\limits{x \rightarrow 1^{-}} f(x)$ exist, then the possible values for $a$ and $b$ are

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

If $f(x)=\left\{\begin{array}{cc}\frac{x^2 \log (\cos x)}{\log (1+x)} & , \quad x \neq 0 \\ 0 & , x=0\end{array}\right.$, then at $x=0, f(x)$ is

Let $f: R^{+} \longrightarrow R^{+}$ be a function satisfying $f(x)-x=\lambda$ (constant), $\forall x \in R^{+}$ and $f(x f(y))=f(x y)+x, \forall x, y, \in R^{+}$. Then, $\lim _\limits{x \rightarrow 0} \frac{(f(x))^{1 / 3}-1}{(f(x))^{1 / 2}-1}=$

$\begin{aligned} & \text { If } \lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}}=k \\ & \lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=l \text {. Then, } k+l= \end{aligned}$

If $\lim _\limits{n \rightarrow \infty} x^n \log _e x=0$, then $\log _x 12=$

If $f(x)=\operatorname{Max}\{3-x, 3+x, 6\}$ is not differentiable at $x=a$, and $x=b$, then $|a|+|b|=$

$\lim _\limits{n \rightarrow \infty}\left(\frac{1}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)=$

If $\lim _\limits{x \rightarrow 0}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)=\frac{a}{b}$, then the value of $a+b$ equals

$\lim _\limits{x \rightarrow 1} \frac{(1-x)\left(1-x^2\right) \ldots\left(1-x^{2 n}\right)}{\left\{(1-x)\left(1-x^2\right) \ldots \ldots\left(1-x^n\right)\right\}^2}= $ _____________, $\forall n \in N$

If $f(x)=\frac{\log _e\left(1+x^2(\tan x)\right)}{\sin x^3}, x \neq 0$ is to be continuous at $x=0$, then $f(0)$ must be equal to

$\mathop {\lim }\limits_{n \to \infty } {{n{{(2n + 1)}^2}} \over {(n + 2)({n^2} + 3n - 1)}}$ is equal to

If the function $f(x)$, defined below, is continuous on the interval $[0,8]$, then $f(x)=\left\{\begin{array}{cc}x^2+a x+b & , \quad 0 \leq x < 2 \\ 3 x+2, & 2 \leq x \leq 4 \\ 2 a x+5 b & , 4 < x \leq 8\end{array}\right.$

If $f(x)$, defined below, is continuous at $x=4$, then

$f(x) = \left\{ {\matrix{ {{{x - 4} \over {|x - 4|}} + a} & , & {x < 4} \cr {a + b} & , & {x = 4} \cr {{{x - 4} \over {|x - 4|}} + b} & , & {x > 4} \cr } } \right.$

If $f(x)=\left\{\begin{array}{cc}\frac{e^{\alpha x}-e^x-x}{x^2}, & x \neq 0 \\ \frac{3}{2}, & x=0\end{array}\right.$

Find the value of $\alpha$ for which the function $f$ is continuous

The value of $k(k > 0)$, for which the function $f(x)=\frac{\left(e^x-1\right)^4}{\sin \left(\frac{x^2}{k^2}\right) \log \left(1+\frac{x^2}{2}\right)}$, where $x \neq 0$ and $f(0)=8$

If $f^{\prime \prime}(x)$ is continuous at $x=0$ and $f^{\prime \prime}(0)=4$, then find the following value. $\lim _\limits{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^2}$ is equal to

$\lim _\limits{z \rightarrow 1} \frac{z^{(1 / 3)}-1}{z^{(1 / 6)}-1}$ is equal to

$f(x)=\left\{\begin{array}{cc} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ K \log 2 \log 3, & x=0 \end{array}\right.$

Find the value of $k$ for which the function $f$ is continuous.

If the function $f(x)$, defined below is continuous in the interval $[0, \pi]$, then $f(x)=\left\{\begin{array}{cc}x+a \sqrt{2}(\sin x) & , \quad 0 \leq x < \frac{\pi}{4} \\ 2 x(\cot x)+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a(\cos 2 x)-b(\sin x), & \frac{\pi}{2} < x \leq \pi\end{array}\right.$