If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?

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

Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that  $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be

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

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

Let $f (1) = - 2$ and $f ' (x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f (6)$, is