Q42 / 90 · JEE Main 2014 (06 Apr) · Mathematics

If $f & g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0 & f (1) = 6,$ then for some $c \in]0, 1[$

Solution

Let

h (x) = f (x) - 2 g (x)

clearly

h (0) = h (1)

Also

h (x)

is continuous and differentiable.

Hence

h (x)

satisfy all conditions of Rolle's theorem in

(0, 1)

. According to Rolle's theorem there exist atleast one

c \in (0, 1)

such that

h^{'} (x) = 0 \Rightarrow f^{'} (x) - 2 g^{'} (x) = 0

\Rightarrow f^{'} (c) = 2 g^{'} (c)

Hence, the correct answer is option: 'B'.