Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:

	$x=-1$	$x=0$	$x=2$
$f(x)$	$3$	$6$	$0$
$g(x)$	$0$	$1$	$-1$

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)

$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$

$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$

$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$

$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

For the function$x + {1 \over x},x \in [1,\,3]$, the value of $ c$  for the mean value theorem is

Let $f$ be any function defined on $R$ and let it satisfy the condition

$|f( x )-f( y )| \leq\left|( x - y )^{2}\right|, \forall( x , y ) \in R$ If $f(0)=1,$ then

Suppose that $f$ is differentiable for all $x$ and that $f '(x) \le 2$ for all x. If $f (1) = 2$ and $f (4) = 8$ then $f (2)$ has the value equal to

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f\left( x \right) = \log x$ on the interval $[1,3]$ is