If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is

Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then

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

Let $f$ 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

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

If $f:R \to R$  and $f(x)$ is a polynomial function of degree ten with $f(x)=0$ has all real and distinct roots. Then the equation ${\left( {f'\left( x \right)} \right)^2} - f\left( x \right)f''\left( x \right) = 0$ has

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