Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.

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

Consider a 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 $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1].$

Let $f$ be any continuously differentiable function on $[a, b]$ and twice differentiable on $(a, b)$ such that $f(a)=f^{\prime}(a)=0$ and $f(b)=0$. Then:

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

$A$ 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