Continuity practice · Easy

If the tangent to the curve $x^3-y^2=0$ at $\left(m^2,-m^3\right)$ is parallel to $y=-\frac{1}{m} x-2 m^3$, then the value of $m^2$ is

The function $f(x)=(x)$, where $(x)$ denotes the smallest integer $\geq x$, is

Let $f$ be a continuous function satisfying $f(x+y)=$ $f(x)+f(y)$ for all $x, y \in \mathbf{R}$ and $f(1)=5$ then $\lim \limits_{{x \rightarrow 4}}$ $f(x)$ is equal to

If $f(x)=\left\{\begin{array}{c}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1, \text { when } x=a\end{array}\right.$, then

If $f(x)=\left\{\begin{array}{cl}\frac{((a-n) n x-\tan x) \sin n x}{x^2} & \text { at } x \neq 0, \\ 0 & \text { at } x=0\end{array}\right.$ where $n$ is a non-zero real number, and $f$ is continuous at $x=0$, then a is equal to

If for $p \neq q \neq 0$, the function $f(x)=\frac{\sqrt[7]{p(729+x)-3}}{\sqrt[3]{729+q x-9}}$ is continuous at $x=0$, then

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {mx + 1,x \le \frac{\pi }{2}}\\ {\sin x + n,x > \frac{\pi }{2}} \end{array}} \right.$ is continuous at $x = \frac{\pi }{2}$, then

Let $f:[1,3] \rightarrow R$ be a function satisfying $\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}$, for all $x \neq 2$ and $f(2)=1$, where $R$ is the set of all real numbers and $[x]$ denotes the largest integer less than or equal to $x$.

Statement 1: $\lim \limits_{x \rightarrow 2^{-}} f(x)$ exists.
Statement 2: $f$ is continuous at $x=2$.

If $f(x)=\left\{\begin{array}{cc}\frac{x}{e^{1 / x}+1}, & \text { when } x \neq 0 \\ 0 & \text {, when } x=0\end{array}\right.$, then

Let $f(x)=\left\{\begin{array}{cc}3 \sin x+a^2-10 a+30, & x \notin Q \\ 4 \cos x & , x \in Q\end{array}\right.$.
Which one of the following statements is correct?

Let $f\left( x \right) = \left| x \right| + \left| {x - 1} \right|$, then

A function $f$ on $R$ into itself is continuous at a point a in $R$, iff for each $\in > 0$, there exists, $\delta > 0$ such that

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

The value of $a$ so that $f$ is continuous is

Let $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\tan x - \cot x}}{{x - \frac{\pi }{4}}},x \ne \frac{\pi }{4}}\\ {a,x = \frac{\pi }{4}} \end{array}} \right.$ the value of a so that $f\left( x \right)$ is continuous at $x = \frac{\pi }{4}$ is

Let
$f(x)= \begin{cases}\frac{\sin (a+1) x+\sin x}{x} & \text { for } x < 0 \\ c & \text { for } x=0 \\ \frac{\left(x+b x^2\right)^{1 / 2}-x^{1 / 2}}{b x^{3 / 2}} & \text { for } x > 0\end{cases}$
If $f$ is continuous at $x=0$, then

If $f(x)=\left\{\begin{array}{l}1+x^2, \text { when } 0 \leq x \leq 1 \\ 1-x, \text { when } x>1\end{array}\right.$, then

The question given below consist of an assertion $(A)$ and the reason $(R)$ use the following key to choose the appropriate answer.
Consider the function $f(x)=x^2-\left|x^2-1\right|+2|| x|-1|+2|x|-7$
A: $f$ is not differentiable at $x=1,-1$ and 0
$\mathrm{R}:|x|$ is not differentiable at $x=0$ and $\left|x^2-1\right|$ is not differentiable at $x=1$ and -1

The set of points of discontinuity of the function, $f(x)=\frac{1}{\log |x|}$, is

Let $f(x)=\frac{\sin x}{x}, x \neq 0$. Then, $f(x)$ can be continuous at $x=0$, if

If $f\left( x \right) = {x^2} + \frac{{{x^2}}}{{1 + {x^2}}} + \frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} + \ldots + \frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^n}}} + \ldots ,$ then at $x = 0,f\left( x \right)$

If $f(x)= \begin{cases}x^m \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{cases}$ is continuous at $x=0$, then

$f(x)=\left\{\begin{array}{cc}L t \frac{x^n-\sin x^n}{x^n+\sin x^n} & \text { if } x>0, x \neq 1 \\ 1 & \text { if } x=1\end{array}\right.$ Then at $x=1$

$f\left( x \right) = x + \left| x \right|$ is continuous for

If $f\left( x \right) = x\sin \left( {\frac{1}{x}} \right),x \ne 0$, then the value of function at $x = 0$, so that the function is continuous at $x = 0$ is

If $f(x) = \left\{ {\begin{array}{llllllllllllllllllll} { - x - \frac{\pi }{2},}&{x \le - \frac{\pi }{2}}\\ { - \cos x,}&{ - \frac{\pi }{2} < x \le 0}\\ {x - 1,}&{0 < x \le 1}\\ {\ln x}&{,x > 1} \end{array}} \right.$

then which one of the following is not correct?

Let $y=\frac{1}{\mu^2+\mu-2}$ where $\mu=\frac{1}{x-1}$ Then ${y}$ is discontinuous only at $x=$

If $f(x)=\left\{\begin{array}{l}\frac{5}{2}-x, \text { when } x<2 \\ 1, \text { when } x=2 \\ x-\frac{3}{2}, \text { when } x>2\end{array}\right.$, then

If $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\int\limits_0^x {(5 + |1 - t|)dt} }&,&{x > 2}\\ {5x + 1}&,&{x < 2} \end{array}} \right.{\mkern 1mu} {\mkern 1mu} \mathrm{then}$

If $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {a{x^2} - b,}&{a \le x < 1}\\ {2,}&{x = 1}\\ {x + 1,}&{1 < x \le 2} \end{array}} \right.$ ，

then the value of the pair $(a, b)$ for which $f(x)$ cannot be continuous at $x=1$, is

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {a{x^2} + b,b \ne 0,x \le 1}\\ {{x^2}b + ax + c,x > 1} \end{array}} \right.$, then, $f\left( x \right)$ is continuous and differentiable at $x = 1$, if