Question
If $$f$$ and $$g$$ are differentiable functions in [0, 1] satisfying $$f\left( 0 \right) = 2{\text{ }} = g\left( 1 \right),{\text{ }}g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$
A.
$$f'\left( c \right) = g'\left( c \right)$$
B.
$$f'\left( c \right) = 2g'\left( c \right)$$
C.
$$2f'\left( c \right) = g'\left( c \right)$$
D.
$$2f'\left( c \right) = 3g'\left( c \right)$$
Answer :
$$f'\left( c \right) = 2g'\left( c \right)$$
Solution :
Since, $$f$$ and $$g$$ both are continuous functions on [0, 1] and differentiable on (0, 1) then $$\exists \,c \in \left( {0,1} \right)$$ such that
$$\eqalign{
& f'\left( c \right) = \frac{{f\left( 1 \right) - f\left( 0 \right)}}{1} = \frac{{6 - 2}}{1} = 4 \cr
& {\text{and }}g'\left( c \right) = \frac{{g\left( 1 \right) - g\left( 0 \right)}}{1} = \frac{{2 - 0}}{1} = 2 \cr
& {\text{Thus, we get }}f'\left( c \right) = 2g'\left( c \right) \cr} $$