If the function $f(x) = ax^3 - bx^2 - 8x - 4$ satisfies Rolle's theorem in the interval $[1, 3]$, and $f'(2) = 0$, then the value of $a - b$ is:

Let $f :[0,1] \rightarrow R$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$,then the least number of points $x \in(0,1)$,at which $f^{\prime\prime}(x)=0$,is $......$

Consider the quadratic equation $ax^2 + bx + c = 0$ where $2a + 3b + 6c = 0$ and let $g(x) = a \frac{x^3}{3} + b \frac{x^2}{2} + cx$.
Statement-$1$: The quadratic equation has at least one root in the interval $(0, 1)$.
Statement-$2$: Rolle's theorem can be applied to the function $g(x)$ in the interval $[0, 1]$.

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

Let $f$ be differentiable for all $x$. If $f(1) = -2$ and $f'(x) \ge 2$ for $x \in [1, 6]$,then

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: