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 function $f(x) = x(x + 3) e^{-x/2} ;$ satisfies the rolle's theorem in the interval $[-3, 0],$ then find $C$

If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?

If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............

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 $(0,2)$

For a polynomial $g ( x )$ with real coefficient, let $m _{ g }$ denote the number of distinct real roots of $g ( x )$. Suppose $S$ is the set of polynomials with real coefficient defined by

$S=\left\{\left(x^2-1\right)^2\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right): a_0, a_1, a_2, a_3 \in R\right\} \text {. }$

For a polynomial $f$, let $f^{\prime}$ and $f^{\prime \prime}$ denote its first and second order derivatives, respectively. Then the minimum possible value of $\left(m_f+m_{f^{\prime}}\right)$, where $f \in S$, is. . . . . . . .