Let $f: [-1, 2] \rightarrow R$ be a differentiable function such that $0 \le f'(t) \le 1$ for $t \in [-1, 0]$ and $-1 \le f'(t) \le 0$ for $t \in [0, 2]$. Then:

Let $g: R \rightarrow R$ be a non-constant twice differentiable function 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 $f(x)=\frac{1}{2}[g(x)+g(2-x)]$,then:

Consider all functions given in List-$I$ in the interval $[1,3]$. List-$II$ has the values of '$c$' obtained by applying Lagrange's Mean Value Theorem $(LMVT)$ on the functions of List-$I$. Match the functions and values of '$c$'.

List-$I$	List-$II$
$A. |x-1|$	$I. 2 \log (e^3+e^2)$
$B. \log x$	$II. 2$
$C. x^2+x+1$	$III. \log_3 e^2$
$D. e^x$	$IV. \sqrt{2}$
	$V. \log \left(\frac{e^3-e}{2}\right)$

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

The value of $c$ in $(0, 2)$ satisfying the Mean Value Theorem for the function $f(x) = x(x - 1)^2, x \in [0, 2]$ is equal to

Let $a, b, c$ be real numbers such that $2a + 3b + 6c = 0$ and $g(x) = ax^2 + bx + c = 0$ has at least one root in the interval $(1, 2)$. If a function $f: [1, 2] \rightarrow \mathbb{R}$ for which Rolle's Theorem holds is such that $f(x)$ is a primitive of $g(x)$,then $f(x) = $