Let $f(x)=2+\cos x$ for all real $x$.

$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because

$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.

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 $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is....................

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

Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $ [-1, 1] $ because