If $f: R \rightarrow R$ is a twice differentiable function such that $f^{\prime \prime}(x) > 0$ for all $x \in R$,and $f(\frac{1}{2}) = \frac{1}{2}$,$f(1) = 1$,then

Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$,$I_2=[0,1]$,$I_3=[1,2]$,$I_4=[-2,-1]$. Then,

If $f(x) = x^\alpha \log x$ and $f(0) = 0$,then the value of $\alpha$ for which Rolle's theorem can be applied in $[0, 1]$ is

The value of $c$ for which Rolle's theorem holds for the function $f(x)=x^3-3x^2+2x$ in the interval $[0,2]$ is:

The value of $c$ for the function $f(x) = \log x$ on $[1, e]$ if Lagrange's Mean Value Theorem $(LMVT)$ is applied,is

If $f(x) = x^{3}$ and $g(x) = x^{3} - 4x$ in the interval $[-2, 2]$,consider the following statements:
$(a)$ $f(x)$ and $g(x)$ satisfy the Mean Value Theorem.
$(b)$ $f(x)$ and $g(x)$ both satisfy Rolle's Theorem.
$(c)$ Only $g(x)$ satisfies Rolle's Theorem.
Which of these statements is correct?