If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin x - \sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $f(0) = $

If the function $f(x) = \begin{cases} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{cases}$ is continuous everywhere,then $a^2+b^2=$

Let $f$ be a differentiable function on the open interval $(a, b)$. Which of the following statements must be true?
$I$. $f$ is continuous on the closed interval $[a, b]$
$II$. $f$ is bounded on the open interval $(a, b)$
$III$. If $a < a_1 < b_1 < b$,and $f(a_1) < 0 < f(b_1)$,then there is a number $c$ such that $a_1 < c < b_1$ and $f(c) = 0$

For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

If $f(x) = \begin{cases} [x] + [-x], & x \ne 2 \\ \lambda, & x = 2 \end{cases},$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is the Greatest Integer Function).

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is