If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \geq 0 \\ \sin^{-1}[x], & -1 \leq x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$,then $k =$

The values of $p$ and $q$ such that the function $f(x) = \begin{cases} (1+|\sin x|)^{\frac{p}{|\sin x|}}, & \frac{-\pi}{6} < x < 0 \\ q, & x = 0 \\ e^{\frac{\sin 2x}{\sin 3x}}, & 0 < x < \frac{\pi}{6} \end{cases}$ is continuous at $x=0$ are:

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

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

Prove that the identity function on real numbers given by $f(x) = x$ is continuous at every real number.

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