Continuity practice · Mixed difficulty

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

If $f(x)$ is a differentiable function such that $f: R \rightarrow r$ and $f\left(\frac{1}{n}\right)=0 \forall n \geq 1, n \in l$ then

If $f\left( x \right) = {\left| x \right|^3}$, then $f'\left( 0 \right)$ equals

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

If $f:R \to R$ given by
$f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {2\cos x,\mathrm{if}\,\,x \le - \frac{\pi }{2}}\\ {a + \sin x + b,\mathrm{if}\,\, - \frac{\pi }{2} < x < \frac{\pi }{2}}\\ {1 + {{\cos }^2}x,\mathrm{if}\,\,x \ge \frac{\pi }{2}} \end{array}} \right.$ is a continuous Function on $R$, then $\left( {a,b} \right)$ is equal to

At the point $x = 1$, the function $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^3} - 1,1 < x < \infty }\\ {x - 1, - \infty < x \le 1} \end{array}} \right.$

Let $f(x)=\lim\limits _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^n-(2 \cos x)^{2 n}}, n \in$ Z. Then,

Let $f(x)=\left\{\begin{array}{cc}x^{2} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$ Then at $x=0$

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 $f\left( x \right)$ defined by $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\left| {{x^2} - x} \right|}}{{{x^2} - x}},x \ne 0,1}\\ {1,x = 0}\\ { - 1,x = 1} \end{array}} \right.$ then $f\left( x \right)$ is continuous for all

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 the function

$f(x)=\left\{\begin{array}{cc} (1+|\cos x|) \frac{\lambda}{|\cos x|} , & 0 < x < \frac{\pi}{2} \\ \mu, & x=\frac{\pi}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{\pi}{2} < x < \pi \end{array}\right.$

Is continuous at $x=\frac{\pi}{2}$, then $9 \lambda+6 \log _{e} \mu+\mu^{6}-e^{6 \lambda}$ is equal to

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1\,\,\mathrm{for}{\mkern 1mu} {\mkern 1mu} x < 0}\\ {1 + \sin x{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathrm{for}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0 \le x \le \pi /2} \end{array}} \right.$, then at $x = 0$, the derivative $f'\left( x \right)$ is

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

Let $f(x)=\left[x^2-x\right]+|-x+[x]|$, where $x \in \mathbb{R}$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ 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 $f(x)=\int\limits_{{-1}}^x|t| d t, x \geq-1$, then

Let $\left[ t \right]$ denote the greatest integer $\le t$ and $\lim\limits_{x \to 0} x\left[ {\frac{4}{x}} \right] = A.$ Then the function, $f(x) = \left[ {{x^2}} \right]\sin (\pi x)$ is discontinuous, when x is equal to

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