If $f(x) = ax^3 + bx^2 + 11x - 6$ for $x \in [1, 3]$ satisfies the conditions of Rolle's theorem and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,find $a$ and $b$.

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$, then Rolle's theorem is not applicable to $f(x)$ because

Verify Rolle's Theorem for the function $f(x)=x^{2}+2x-8, x \in[-4,2]$

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

For all twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R},$ with $f(0)=f(1)=f^{\prime}(0)=0,$ which of the following is true?

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:

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

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)^{\prime \prime}$ never vanishes. Then the correct statement$(s)$ is(are):
$(A)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup (0,2)$
$(B)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$