Continuity practice · Medium

If $f(x)=\int\limits_{{-1}}^x|t| d t, x \geq-1$, then

Consider the function :
$f(x)=[x]+|1-x|,-1 \leq x \leq 3$ where $[x]$ is the greatest integer function.
Statement 1 : $f$ is not continuous at $x=0,1,2$ and 3 .
Statement $2: f(x)=\left(\begin{array}{cc}-x, & -1 \leq x<0 \\ 1-x, & 0 \leq x<1 \\ 1+x, & 1 \leq x<2 \\ 2+x, & 2 \leq x \leq 3\end{array}\right.$

$f(x)=\frac{\left(4^x-1\right)^3}{\sin \left(\frac{x}{p}\right) \log \left(1+\frac{x^2}{3}\right)}$ is continuous at ${x}=0$ and $f(0)=12(\log 4)^3$ Then ${p}$

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

Number of points at which $f\left( x \right) = \frac{{2{x^2} + 7}}{{{x^3} + 3{x^2} - x - 3}}$ is discontinuous is

If function $f(x)$ given by $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{(\sin x)}^{1/(\pi - 2x)}}}&{,x \ne \pi /2}\\ \lambda &{,x = \pi /2} \end{array}} \right.$ is continuous at $x=\frac{\pi}{2}$, then $\lambda=$

If $f(x)=\left\{\begin{array}{c}1 \text { if } x \text { is rational } \\ -1 \text { if } x \text { is irrational }\end{array}\right.$ is continuous on

The function $f\left( x \right) = x - \left| {x - {x^2}} \right|, - 1 \le x \le 1$ is continuous on the interval

Let $f(x)=1 /(x-1)$ and $g(x)=1 /\left(x^2+x-2\right)$. Then the set of points where $(g \circ f)(x)$ is discontinuous, is

For the function $f(x)=\left\{\begin{array}{c}\frac{\sin ^2 a x}{x^2}, \text { when } x \neq 0 \\ 1 \text {, when } x=0\end{array}\right.$ which one is a true statement

$f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {2x,x < 0}\\ {2x + 1,x \ge 0} \end{array}} \right.$ Then

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left[ {\cos \pi x} \right],x < 1}\\ {\left| {x - 2} \right|,2 > x \ge 1} \end{array}} \right.$, then $f\left( x \right)$ is

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left( {\cos x} \right)}^{1/\sin x}}}&{{\text{for}}}&{x \ne 0} \\ k&{{\text{for}}}&{x = 0} \end{array}} \right.$
The value of $k$, so that $f$ is continuous at $x=0$ is

If $f\left( x \right) = \left( {x - {x_0}} \right)\phi \left( x \right)$ and $\phi \left( x \right)$ is continuous at $x = {x_0}$, then $f'\left( {{x_0}} \right)$ is equal to

Let $f(x)=\left\{\begin{array}{cc}-1 & x < 0 \\ 0 & x=0 \\ 1, & x > 0\end{array}\right.$ and $g(x)=\sin x+\cos$ $x$, then points of discontinuity of $f\{g(x)\}$ in $(0,2 \pi)$ is :

If $f(x)$ is continuous and $f\left(\frac{9}{2}\right)=\frac{2}{9}$, then $\lim _{x \rightarrow 0} f\left(\frac{1-\cos 3 x}{x^{2}}\right)$ is equal to:

The function $f\left( x \right)$ is defined as $f\left( x \right) = \frac{{2x - {{\sin }^{ - 1}}x}}{{2x + {{\tan }^{ - 1}}x}}$, if $x \ne 0$. The value of $f$ to be assigned at $x = 0$ so that the function is continuous there, is

Let $f$ be a composite function of $x$ defined by $f(u)=\frac{1}{u^{2}+u-2}, u(x)=\frac{1}{x-1}$.
Then the number of points $x$ where $f$ is discontinuous is :

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\frac{x}{a}} \right)\log \left( {1 + \frac{{{x^2}}}{3}} \right)}},x \ne 0}\\ {9{{\left( {\log 4} \right)}^3},x = 0} \end{array}} \right.$ is continuous function at $x = 0$, then the value of $a$ is equal to

Let $f(x)= \begin{cases}x\left(1+\frac{1}{3} \sin \left(\log x^2\right)\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}$
Then

Suppose $f$ is continuous on the closed interval $[0,2]$ and $f(0)=f(2)$. Then

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function. Then,

Suppose $f$ and $g$ are continuous functions on the closed interval $[a, b]$ such that $f(a) \geq g(a)$ and $f(b) \leq$ $g(b)$. Then

Assertion : If a function ${f}$ is discontinuous at ${c}$, then ${c}$ is called a point of discontinuity.

Reason : A function is continuous at ${x}={c}$, if the function is defined at ${x}={c}$ and the value of the function at ${x}={c}$ equals the limit of the function at ${x}={c}$.

A function $f\left( x \right)$ is defined as fallows for real $x$,$f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1 - {x^2},\mathrm{for}\,\,x < 1}\\ {0,\mathrm{for}\,\,x = 1}\\ {1 + {x^2},\mathrm{for}\,\,x > 1} \end{array}} \right.$ Then,

On the interval $I = \left[ { - 2,2} \right]$, the function $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left( {x + 1} \right){e^{ - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}}}\\ {0,x = 0} \end{array}} \right.,x \ne 0$

If the derivative of the function $f\left( x \right)$ is everywhere continuous and is given by $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {b{x^2} + ax + 4;x \ge - 1}\\ {a{x^2} + b;x < - 1} \end{array}} \right.$, then

If $f(x)=x(\sqrt{x}+\sqrt{x+1})$, then

The function $f\left( x \right) = \frac{{2{x^2} + 7}}{{{x^3} + 3{x^2} - x - 3}}$ is discontinuous for

$f(x)=\lim \limits_{{n \rightarrow \infty}} \cos ^{2 n}\left(\pi x^2\right)+[x]($ where, [.]) denote the greatest integer function and $n \in N$ is