If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)

Let $m$ and $n$ be the number of points at which the function $f(x) = \max \{x, x^3, x^5, \dots, x^{21}\}$,$x \in R$,is not differentiable and not continuous,respectively. Then $m + n$ is equal to . . . . . . .

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

If $f: [0, 2) \to R$ is defined by $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$ where $k > 0$ and $f$ is such that $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$,then the value of $k^2$ is:

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$