# 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 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.