Limits, Continuity and Differentiability · Mathematics · TS EAMCET

$$ \mathop {\lim }\limits_{x \to 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin ^2 x}= $$

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=\left[x^2-3\right]$ where $[$. denotes greatest integer function, then the number of points of discontinuity for the function $f$ in $(-1,2)$ is

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

The set of all values of $x$ for which $f(x)=\| x|-1|$ is differentiable is

If $\mathop {\lim }\limits_{x \to 0} \frac{3^{x^3}-\left(1-x^3\right)^{\frac{2}{3}}}{x^2 \sin x}=p+\log q$, then $p q=$

If $[x]$ is the greatest integer function and

$$ f(x)=\left\{\begin{array}{cc} 2[x]-\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0 \end{array}\right. $$

is a real valued function, then $f$ is

If $[t]$ represents the greatest integer $\leq t$, then the value of $\lim\limits_{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$ is

If the real valued function

$$ f(x)=\left\{\begin{array}{ccc} \frac{\cos 3 x-\cos x}{x \sin x}, & \text { if } & x<0 \\ p, & \text { if } & x=0 \\ \frac{\log (1+q \sin x)}{x}, & \text { if } & x>0 \end{array}\right. $$

is continuous at $x=0$, then $p+q=$

If $\{x\}=x-[x]$, where $[x]$ is the greatest integer $\leq x$ and $\mathop {\lim }\limits_{x \to {0^ - }} \frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^4}=\theta$, then $\tan \theta$

For $a \neq 0$ and $b \neq 0$, if the real valued function $f(x)=\frac{\sqrt[5]{a(625+x)}-5}{\sqrt[4]{625+b x}-5}$ is continuous at $x=0$, then $f(0)=$

The value of $x$ at which the real valued function $f(x)=7|2 x+1|-19|3 x-5|$ is not differentiable is

If $f(x)=\frac{x\left(a^x-1\right)}{1-\cos x}$ and $g(x)=\frac{x\left(1-a^x\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)}$, then $\lim _{x \rightarrow 0}(f(x)-g(x))=$

If $f(x)=\left\{\begin{array}{cc}\frac{a \sin x-b x+c x^2+x^3}{2 \log (1+x)-2 x^3+x^4} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$

is continuous at $x=0$, then

If the function $g(x)=\left\{\begin{array}{cl}K \sqrt{x+1} & , 0 \leq x \leq 3 \\ m x+2 & , 3 < x \leq 5\end{array}\right.$ is differentiable, then $K+m=$

If $[x]$ is the greatest integer function, then

$$ \mathop {\lim }\limits_{x \to 3} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} $$

Let ' $a$ ' be a positive real number. If a real valued function

$f(x)=\left\{\begin{array}{cl}\frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text { if } x \neq 0 \\ \log 3 \log 4 & \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then $a=$

Define $ f: R \rightarrow R $ by $ f(x)=\left\{\begin{array}{cl}\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. $

Then, the value of $ a $ so that $ f $ is continuous at $ x=0 $ is

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

If the function

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

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

If $ 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. $ is a real valued function, then at $ x=1, f $ is

$$ \begin{array}{r} \lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}}= \\ -\sqrt{3 \tan ^2 x+\sin x+1} \end{array} $$

If a function $f$ is defined by $f(x)=\frac{\cot ^3 x-\tan x}{\cos (x+\pi / 4)},(x \neq \pi / 4)$, then $\lim _{x \rightarrow \pi / 4} f(x)=$

$$ \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)= $$

The quadratic equation whose roots are

$$ l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\theta}\right) \text { and } m=\lim _{\theta \rightarrow 0}\left(\frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}\right) \text { is } $$

$$ \mathop {\lim }\limits_{x \to 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}= $$

$\mathop {\lim }\limits_{x \to 0} \frac{\tan ^4 x-\sin ^4 x}{x^6}=$

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

$$ \lim _{x \rightarrow 0} \frac{\left(3^{2 x}-\sqrt{x+1}\right) \sin 5 x}{1-\cos 4 x}= $$

$$ \lim\limits_{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right)= $$

If $f(9)=9$ and $f^{\prime}(9)=4$, then $\lim\limits_{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$

$$ \lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}= $$

$$ \lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]}= $$

Let [ $x$ ] denote the greatest integer less than or equal to $x$ and $f(x)=2 x-[2 x]$. If $\mathop {\lim }\limits_{x \to {2^ - }} f(x)=l_1$ and $\mathop {\lim }\limits_{x \to {2^ + }} f(x)=l_2$, then $l_1+l_2=$

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

Let $f(x)$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(0)=20$. For $x \in\left(0, \frac{\pi}{2}\right]$, if

$A(x)=2 f(x) \operatorname{cosec} 4 x+4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x$, then $\mathop {\lim }\limits_{x \to 0} A(x)=$

If $x=\log _e\left(\cot \left(\frac{\pi}{4}+\theta\right)\right)$, then $\lim _{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)}=$

$$ \mathop {\lim }\limits_{x \to 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]= $$

$$ \lim _{x \rightarrow 0} \frac{\tan 2 x-2 \tan x}{(1-\cos x)\left(2^x-1\right)}= $$

$$ \mathop {\lim }\limits_{x \to 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4}= $$

$$\mathop {\lim }\limits_{x \to 0}\left(\frac{4!}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)= $$

Let $A=\left(a_{i j}\right)$ be an $n \times n$ matrix defined by $a_{i j}=\left\{\begin{array}{cc}k^i, & \forall i=j \\ 0, & \text { otherwise }\end{array}\right.$. If $m=$ trace of $A$ and $\lim _{k \rightarrow 1} \frac{n-m}{1-k}=171$, then the value of $n$ is

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

Let $f(x)=\left\{\begin{array}{ccc}3-x & \text { if } & x<-3 \\ 6 & \text { if } & -3 \leq x \leq 3 . \text { Let } \alpha \text { be the number } \\ 3+x & \text { if } & x>3\end{array}\right.$ of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then, $\alpha+\beta=$

$$ \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}= $$

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

If $f(x)=\left\{\begin{array}{l}\frac{x^2-16}{x-4} \text { if } x>4 \\ 2 x \quad \text { if } x \leq 4\end{array}\right.$ then $f^{\prime}\left(4^{-}\right)+f^{\prime}\left(4^{+}\right)=$

Let $[x]$ denote the greatest integer less than or equal to $x$ and $k \geq 2$ be an integer. Then

$$ \mathop {Lt}\limits_{x \to k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k}= $$

Define $f(x)=\left\{\begin{array}{ll}1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3\end{array}\right.$.

If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0,3]$ and $a

If $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=\left\{\begin{array}{cc}x+2, & x>0 \\ 2-x, & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}x^2-2 x-2, & 1 \leq x<2 \\ x-7 & x \geq 2 \\ x+5, & x<1\end{array}\right.$ then $\lim _{x \rightarrow 0} g \circ f(x)$

Define $f: R \rightarrow R$ by $f(x)= \begin{cases}(x-a) \frac{e^{\frac{1}{(x-a)}}-1}{\frac{1}{(x-a)}}+1 & \text { for } x \neq a \\ 0, \quad \text { at } x=a\end{cases}$

Then which one of the following is true?

Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be functions defined by

$$ f(x)=\left\{\begin{array}{cc} x \sin \left(\frac{1}{x}\right), & \text { for } x \neq 0 \\ 0, & \text { for } x=0 \end{array}\right. $$

and $g(x)=x f(x)$

Consider the following statements

(i) $f(x)$ is continuous at $x=0$ but not differentiable at $x=0$

(ii) $g(x)$ is differentiable at $x=0$, but $g^1(x)$ is not continuous at $x=0$

Then, which one of the following is true?

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

At $x=0, f(x)=\left\{\begin{array}{l}\frac{x}{|x|+2 x^2}, x \neq 0 \\ k, \quad x=0\end{array}\right.$ is

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

Assertion (A) $f(x)=|x-a|+|x-b|$, is continuous on $\mathbf{R}$

Reason (R) $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in \mathbf{R}-\{\alpha\}$.

The correct option among the following is

A function $y=f(x)$ with $f(-1)=-249$ has no maximum and has only one minimum at $x=5$ with $f(5)=75$. Which one of the following is true?

$$ \mathop {\lim }\limits_{x \to 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2}= $$

The value of ' $a$ ' for which the function

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

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \ldots \infty$ and $\mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$, then $12 k=$

If $f(x)=\left\{\begin{array}{ll}k, & \text { for } x=1 \\ \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5}, & \text { for } x \neq 1\end{array}\right.$ is continuous on $[0, \infty)$, then $k=$

In each of the choices given below, a function and an interval are given. The correct choice having a function and the associated interval for which the Lagrange's mean value theorem is not valid is