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
$\left( {{7 \over 2},{1 \over 2}} \right)$
$\left( {{7 \over 2},{1 \over 4}} \right)$
$(1, 4)$
$(4, 1)$
If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?
If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f\left( 0 \right) = 2 = g\left( 1 \right)\;,\;\;g\left( 0 \right) = 0,$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$ . .
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
Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse of Roller's Theorem from these examples?
$f(x)=[x]$ for $x \in[-2,2]$
If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f ^{\prime}\left(\frac{4}{3}\right)=0,$ then ordered pair $( a , b )$ is equal to