If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $ad-bc = $

Let $f(x) = \begin{cases} 3x, & x < 0 \\ \min \{1+x+[x], x+2[x]\}, & 0 \leq x \leq 2 \\ 5, & x > 2 \end{cases}$ where $[.]$ denotes the greatest integer function. If $\alpha$ and $\beta$ are the number of points where $f$ is not continuous and is not differentiable,respectively,then $\alpha + \beta$ equals.......

If $f(x) = \begin{cases} ax^2 - bx + 2, & x < 3 \\ bx^2 - 3, & x \geq 3 \end{cases}$ is differentiable at every $x \in R$,then the area (in sq units) of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

Match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+|x-1|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

Let $f: R \rightarrow (0, \infty)$ and $g: R \rightarrow R$ be twice differentiable functions such that $f^{\prime \prime}$ and $g^{\prime \prime}$ are continuous functions on $R$. Suppose $f^{\prime}(2) = g(2) = 0$,$f^{\prime \prime}(2) \neq 0$ and $g^{\prime}(2) \neq 0$. If $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$,then:

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to