If the function $f$ is continuous at $x = \pi$, where $f(x) = \frac{1 - \cos[7(x - \pi)]}{5(x - \pi)^2}$ for $x \neq \pi$, then $f(\pi) =$

Find all the points of discontinuity of the function $f$ defined by
$f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ 0, & \text{if } x = 1 \\ x - 2, & \text{if } x > 1 \end{cases}$

If $f(x) = \frac{x^2 - bx + 25}{x^2 - 7x + 10}$ for $x \neq 5$ and $f$ is continuous at $x = 5$,then the value of $f(5)$ is:

For every pair of continuous functions $f, g: [0, 1] \rightarrow \mathbb{R}$ such that $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$,the correct statement$(s)$ is (are):

Define $f: R \rightarrow R$ by $f(x) = [x] + \sqrt{x - [x]}$ for $x \in R$,where $[x]$ denotes the greatest integer function. Then the set of points at which $f$ is continuous is

Let $[t]$ represent the greatest integer not exceeding $t$ and $C=1-2e^2$. If the function $f(x)=\begin{cases} [e^x], & x < 0 \\ ae^x+[x-2], & 0 \leq x < 2 \\ [e^{-x}]-C, & x \geq 2 \end{cases}$ is continuous at $x=2$,then $f(x)$ is discontinuous at