The number of point / points where the function $f(x) = \frac{1}{x^2 - 5|x| + 6}$ is discontinuous is......

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

Let $[x]$ represent the greatest integer not more than $x$. The discontinuous points of the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$ lie in the interval