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 $t$.

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

Let $a > 0$ and $f$ be continuous in $[- a, a]$. Suppose that $f ' (x) $ exists and $f ' (x) \le 1$ for all $x \in (- a, a)$. If $f (a) = a$ and $f (- a) = - a$ then $f (0)$

Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$

$h (x) = \{x\}$   $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable to

Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,