Let $f:[1,3] \rightarrow R$ be continuous and differentiable in $(1,3)$ such that $f^{\prime}(x)=[f(x)]^2+4$ for all $x \in (1,3)$. Then:

The value of $c$ in the Lagrange's mean value theorem for the function $f(x) = x^{3} - 4x^{2} + 8x + 11$ on the interval $x \in [0, 1]$ 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 $[0, 1]$.

Suppose $f$ is differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for all $x \in [1, 6]$,then:

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

If $f(x) = \sqrt{x}$ and $g(x) = \frac{1}{\sqrt{x}}$ for $x \in [3, 12]$,then the value of $c \in (3, 12)$ for which $\frac{f^{\prime}(c)}{g^{\prime}(c)} = \frac{f(12) - f(3)}{g(12) - g(3)}$ holds,is