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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) \\
 $

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Last updated date: 16th May 2024
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Answer
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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.