The function $f(x) = {(x – 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x – 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$  and $(4, 1)$  is

If the function $f(x) = a{x^3} + b{x^2} + 11x – 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

A value of $c$ for which the conclusion of mean value the theorem holds for the function $f(x) = log{_e}x$ on the interval $[1, 3]$ is

From mean value theorem $f(b) – f(a) = $ $(b – a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $

Suppose that $f (0) = – 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is