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)

Given, $f(0)=2=g(1), g(0)=0$ and $f(1)=6$
$f$ and $g$ are differentiable in $(0,1)$.
Let $h(x) = f(x) - 2g(x) ....$ (i)
$h(0) = f(0) - 2g(0) = 2 - 0 = 0$
and $h(1) = f(1) - 2g(1) = 6-2(2) = 2$
$h(0) = h(1)= 2$
Hence, using Rolle's theorem.
$h^{\prime}(c)=0$, such that $c \in(0,1)$
Differentiating Eq. (i) at $c$, we get
$\Rightarrow f^{\prime}(c)-2 g^{\prime}(c) =0$
$\Rightarrow f^{\prime}(c) =2 g^{\prime}(c)$