If f and g are differentiable functions in [0, 1] satisfying `f(0)""=""2""=g(1),g(0)""=""0` and `f(1)""=""6` , then for some `c in ]0,""1[` (1) `2f^'(c)""=g^'(c)` (2) `2f^'(c)""=""3g^'(c)` (3) `f^'(c)""=g^'(c)` (4) `f'(c)""=""2g'(c)`

Options

A. `f'(c)=g'(c)`
B. `f'(c)=2g'(c)`
C. `2f'(c)=g'(c)`
D. `2f'(c)=3g'(c)`

Answer

B

Explanation

Consider the function `h(x)=f(x)-2g(x)` for all is `x in [0,1]`. As f and g are the differentiable on (0,1). Therefore, h(x) is differentiable and continuous on [0,1].
Also, `h(0)=f(0)-2g(0)=2-2xx0=2` and `h(1)=f(1)-2g(1)=6-2xx2=2`
Thus, h(x) satisfies conditions of Rollle's theoram on [0,1]. So, there exists `c in (0,1) ` such that `h'(c )=0`
Now, `h(x)=f(x)-2g(x)`
`Rightarrow h'(x)=f'(x)-2g'(x)`
`therefore h'(c)=0 Rightarrow f'(c)=2g'(c)`
Hence, option (b) is correct.