Courses
Courses for Kids
Free study material
Free LIVE classes
More
LIVE
Join Vedantu’s FREE Mastercalss

# If $f$ and $g$ are differentiable functions in $\left[ {0,1} \right]$ 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]$${\text{A}}{\text{. 2}}f'\left( c \right) = g'\left( c \right) \\ {\text{B}}{\text{. }}2f'\left( c \right) = 3g'\left( c \right) \\ {\text{C}}{\text{. }}f'\left( c \right) = g'\left( c \right) \\ {\text{D}}{\text{. }}f'\left( c \right) = 2g'\left( c \right) \\ Answer Verified 361.2k+ views Hint: - Here two differentiable functions are given and their differentiability range is also given so use Lagrange’s mean value theorem (LMVT) for further solution. And proceed further from what is given in question. We have given: f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0 and f\left( 1 \right) = 6 As given in question both f and g are differentiable in \left[ {0,1} \right] From LMVT f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} here \left( {b = 1,a = 0} \right)$$$$So, f'\left( c \right) = \left( {\dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}}} \right)$$ = \dfrac{{6 - 2}}{1}$
$\therefore f'\left( c \right) = 4$.$\ldots \ldots \left( 1 \right)$
Now apply LMVT on function $g$.
Now $g'\left( c \right) = \dfrac{{g\left( b \right) - g\left( a \right)}}{{b - a}}\left( {b = 1,a = 0} \right)$
$g'\left( c \right) = \dfrac{{g\left( 1 \right) - g\left( 0 \right)}}{{1 - 0}} = \dfrac{{2 - 0}}{1} = 2$$\ldots \ldots \left( 2 \right)$
From equation $\left( 1 \right)$and $\left( 2 \right)$
$f'\left( c \right) = 2g'\left( c \right)$
Hence the option ${\text{D}}$ is the correct option.

Note: -Whenever you get this type of question the key concept of solving is you have to understand from the question that by using LMVT you can proceed further. You have to consider the range in which you have to apply LMVT.
Last updated date: 19th Sep 2023
Total views: 361.2k
Views today: 4.61k