If from mean value theorem, $f'({x_1}) = {{f(b) - f(a)} \over {b - a}}$, then

Let  $f(x)$  satisfy the requirement of lagranges mean value theorem in $[0,2]$ . If $f(x)=0$ ; $\left| {f'\left( x \right)} \right| \leqslant \frac{1}{2}$ for all $x \in \left[ {0,2} \right]$, then-

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

If $(1 -x + 2x^2)^n$ = $a_0 + a_1x + a_2x^2+..... a_{2n}x^{2n}$ , $n \in N$ , $x \in R$ and $a_0$ , $a_2$ and $a_1$ are in $A$ . $P$ .,then there exists 

The abscissa of the points of the curve $y = {x^3}$ in the interval $ [-2, 2]$, where the slope of the tangents can be obtained by mean value theorem for the interval  $[-2, 2], $ are

Let $f$ be any function continuous on $[\mathrm{a}, \mathrm{b}]$ and twice differentiable on $(a, b) .$ If for all $x \in(a, b)$ $f^{\prime}(\mathrm{x})>0$ and $f^{\prime \prime}(\mathrm{x})<0,$ then for any $\mathrm{c} \in(\mathrm{a}, \mathrm{b})$ $\frac{f(\mathrm{c})-f(\mathrm{a})}{f(\mathrm{b})-f(\mathrm{c})}$ is greater than