Let $f(x) = \begin{cases} x^3 - x^2 + 10x - 5, & x \le 1 \\ -2x + \log_2(b^2 - 2), & x > 1 \end{cases}$. The set of values of $b$ for which $f(x)$ has the greatest value at $x = 1$ is given by:

If $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2} & \text{if } x < 0 \\ a & \text{if } x = 0 \\ \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{x}} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$,then find the value of $a$.

If $f: R \rightarrow R$ defined as $f(x) = \frac{x^3+2x^2+x+2}{x^2+x-2}$ (when $x \neq -2$) is continuous at $x = -2$, then $f(-2)$ is equal to

If $f(x) = \begin{cases} \frac{\sin [x]}{[x] + 1}, & \text{for } x > 0 \\ \frac{\cos \frac{\pi }{2}[x]}{[x]}, & \text{for } x < 0 \\ k, & \text{at } x = 0 \end{cases}$; where $[x]$ denotes the greatest integer less than or equal to $x$,then in order that $f$ be continuous at $x = 0$,the value of $k$ is

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is:

If $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then