In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$  is

The value of $\left[ {\frac{{\log \left( {\frac{x}{e}} \right)}}{{x - \,e}}} \right]\,\forall x\, > \,e$ is equal to (where [.] denotes greatest integer function)

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 $[0, 1 ].$

Mean value theorem $f(b) -f(a) = (b -a) f '(x_1);$ from $a < x_1 < b,$ if $f(x) = 1/x$ then $x_1 = ?$

Let $f$ be any function defined on $R$ and let it satisfy the condition

$|f( x )-f( y )| \leq\left|( x - y )^{2}\right|, \forall( x , y ) \in R$ If $f(0)=1,$ then

Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is