If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1 \leq x < 0 \\ 2x^2+3x-2, & 0 \leq x \leq 1 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

The value of $k$ which makes $f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$ is

If $f(x) = |x|/x$ for $x \neq 0$ and $1$ for $x = 0$,then the function is

If $f(x) = \max(\sin x, \sin^{-1}(\cos x))$,then

If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then the value of $f(\pi)$ is