Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then

If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

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

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

Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f\left( 0 \right) = 2 = g\left( 1 \right)\;,\;\;g\left( 0 \right) = 0,$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$  . .

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:

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 $(0,2)$