The value of $c$ satisfied by Rolle's theorem for the function $f(x) = x^2(1 - x)^2$ on the interval $x \in [0, 1]$ is...

If $2a + 3b + 6c = 0$ and $a, b, c \in \mathbb{R}$,then the equation $ax^2 + bx + c = 0$ has at least one root between $0$ and $1$.

If the $L.M.V.T.$ holds for the function $f(x) = x + \frac{1}{x}$ on the interval $x \in [1, 3]$,then $c$ is:

Let $f: R \rightarrow R$ be a twice continuously differentiable function. Let $f(0)=f(1)=f^{\prime}(0)=0$. Then,

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in which interval?

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 $(0,2)$