Question

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 $c \in ] 0, 1 [$

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

Answer 1

Answer: (D)

$f ' (c) = 2 g ' (c)$

Answer 2

Using, mean value theorem
$f'(c)=\frac{f(1)-f(0)}{1-0}=4$
$g'(c)=\frac{g(1)-g(0)}{1-0}=2$
so, $f'(c)=2 g'(c)$