If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

If $f$ is a differentiable function such that $f(2x + 1) = f(1 - 2x)$ for all $x \in R$,then the minimum number of roots of the equation $f'(x) = 0$ in $x \in (-5, 10)$,given that $f(2) = f(5) = f(10)$,is:

Let $f(x)$ satisfy the requirements of Lagrange's Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \leqslant \frac{1}{2}$ for all $x \in [0, 2]$,then-

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then $a-b$ is equal to