Number of points of discontinuity of $f\left( x \right) = \left[ {2{x^3} - 5} \right]$ in $\left[ {1,2} \right)$ is where $\left[ x \right]$ denotes greatest integer function are

$f(x)=\displaystyle\lim _{n\rightarrow \infty}\dfrac{(x-1)^{2n}-1}{(x-1)^{2n}+1}$ is discontinuous at

If $f\left( x \right) ={ \left( \tan { \left( \dfrac { \pi  }{ 4 } +\ell nx \right)  }  \right)  }^{ \log _{ x }{ e }  }$ is to be made continuous at $X=1$, then $f(1)$ should be equal to

The function $f\left( x \right)=\left[ x \right] \cos { \left( \pi \left( \dfrac { 2x-1 }{ 2 }  \right)  \right)  } $. (where [.] denotes the greatest integer function ) is discontinuous.  

If $f(x)=\dfrac {1}{x^{2}-17x+66}$ then $f\left(\dfrac {2}{x-2}\right)$ is discontinuous at $x=$

The sum of all values of $x$ for which $f(x)=[3\sin x]$ is discontinous in $[0,\ 2\pi]$ is (where [.] represents greatest integers function)

Consider the function defined on $[0,\ 1]\rightarrow R,\ f(x)=\dfrac {\sin x-x\cos x}{x^{2}}$ if $x\neq 0$ and $f(0)=0$ then the function of $f(x)$. 

The function $f(x)={ sin }^{ -1 }(cosx)$ is :

If $f\left( x \right) =\begin{cases} -1,if\ x<0\ \ 0,if\ x=0\ \ 1,if\ x>0\ \end{cases}$ and $g\left(x\right)=\sin x +\cos x$, then point discontinuity of $(fog)(x)$ in $(0,2\pi)$ are 

$f(x)=\min { \left{ x,{ x }^{ 2 } \right} ,\forall x\epsilon R } $ then $f(x)$ is 

If $f(x)=\dfrac{1}{1-x}$, the number of points of discontinuity of $f\left{f[f(x)]\right}$ is:

Consider  $f ( x ) = \sin x \forall x \in \left[ 0 , \dfrac { \pi } { 2 } \right] ; f ( x ) + f ( \pi - x ) =2 \forall x \in \left( \dfrac { \pi } { 2 } , \pi \right) \text { and } f ( x ) = f ( 2 \pi - x ) \forall x \in ( \pi , 2 \pi ) . \text { If } n , m$ denotes number of points where  $f(x)$  is discontinuous and non derivable respectively in  $[ 0,2 \pi ]$  then value of  $n \div  m$  is

f(x) = $\dfrac{\sin2x + 1}{\sin x - \cos x}$ is discontinuous at $x =$ ____________.

The function $\displaystyle f\left ( x \right )=\frac{\log \left ( 1+ax \right )-\log \left ( 1-bx \right )}{x}$ is not defined at $ x = 0$. The value which should be assigned to $f$ at $x =0$ so that it is continuous there, is

The function 
  $\displaystyle f\left ( x \right )=\frac{\cos x-\sin x}{\cos 2x}$  is not defined at $\displaystyle x=\frac{\pi }{4}$. The value of $\displaystyle  f\left ( \frac{\pi }{4}\right )$ so that $ f\left ( x \right)$  is continuous everywhere, is

$\displaystyle g(x)= \begin{cases}1 & \, x\leq -2 \ \displaystyle \frac{1}{2} x& \, -2< x< 4 \ \sqrt{x} & , x\geq 4 \end{cases}$.then

Given $\displaystyle f(x) = \begin{cases} 3-\left [ \cot ^{-1}\left ( \frac{2x^{3}-3}{x^{2}} \right ) \right ] & \mbox{for } x> 0 \ \left { x^{2} \right }\cos \left ( e^{1/x} \right ) & \mbox{for } x< 0 \end{cases}$ where { } & [ ] denotes the fractional part and the integral part functions respectively, then which of the following statement does not hold good -

Consider $\displaystyle  f(x) = \begin{cases} x\left [ x \right ]^{2}\log _{(1+x)}2& \mbox{ for } -1< x< 0 \ \dfrac{\ln e^{x^{2}}+2\sqrt{\left { x \right }}}{\tan \sqrt{x}} & \mbox{ for } 0< x< 1  \end{cases}$ where [] & {} are the greatest integer function & fractional part function respectively, then -

Let $f(x)=\begin{cases} \dfrac{1-\cos 2x}{2x^2}&:& x\ne 0\k &:& x=0 \end{cases}$.

Let $f\left( x \right) =\dfrac { \log { \left( 1+x+{ x }^{ 2 } \right)  } +\log { \left( 1-x+{ x }^{ 2 } \right)  }  }{ \sec { x } -\cos { x }  } ,x\neq 0$ The value of $f\left (0\right)$ so that $f$ is continuous at $x=0$ is 

Given $f(x)=\dfrac{\left[ \left{ \left| x \right|  \right}  \right] { e }^{ { x }^{ 2 } }\left{ \left[ \left| x+\left{ x \right}  \right|  \right]  \right} }{\left( { e }^{ 1/{ x }^{ 2 } }-1 \right) sgn\left( \sin { x }  \right) }$ for $x\neq 0$
$=0, for\  x=0$
Where $\left{ x \right} $ is the fractional part function; $[x]$ is the step up function and $sgn{(x)}$ is the signum function of $x$ then, $f(x)$

Given that $\displaystyle \prod _{n=1}^n cos \dfrac{x}{2^n}= \dfrac{\sin  x}{2^n  \sin \left ( \dfrac{x}{2^n} \right )}$ and $\displaystyle f(x) = \left{\begin{matrix}\lim _{n \rightarrow \infty}\sum _{n = 1}^n \dfrac{1}{2^n} \tan \left (\dfrac{x}{2^n} \right ), & x \in (0, \pi) - \left {\dfrac{\pi}{2} \right }\ \dfrac{2}{\pi} & x = \dfrac{\pi}{2}\end{matrix}\right.$
Then which one of the following is true?

The value of f(0) so that the function
$f(x)=\displaystyle \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$
becomes continuous, is equal to

Let $\displaystyle f(x)=\left ( 2-\dfrac{x}{a} \right )^{\tan \left ( \dfrac{\pi :x }{2:a} \right )}, x\neq a$. The value which should be assigned to $f$ at $x=a$ so that it is continuous everywhere is

If $f(x)=\left{\begin{matrix} |x|-3, & x < 1\ |x-2|+a, & x\geq 1\end{matrix}\right.$ and $g(x)=\left{\begin{matrix} 2-|x|, & x < 2 \ sgn(x)-b, & x\geq 2\end{matrix}\right.$ and $h(x)=f(x)+g(x)$ is discontinuous at exactly one point, then which of the following values of a and b are possible.