Question

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

• A. $2f\text{'}\left(c\right)=g\text{'}\left(c\right)$
• B. $2f\text{'}\left(c\right)=3g\text{'}\left(c\right)$
• C. $f\text{'}\left(c\right)=g\text{'}\left(c\right)$
• D. $f\text{'}\left(c\right)=2g\text{'}\left(c\right)$

Answer 1

The correct option is D

$f\text{'}\left(c\right)=2g\text{'}\left(c\right)$

Finding the value:

Given,

$f\left(0\right)=2=g\left(1\right),g\left(0\right)=0$ and $f\left(1\right)=6$

$f$ and $g$ are differentiable in $\left(0,1\right)$

Let $h\left(x\right)=f\left(x\right)-2g\left(x\right)$

$$\begin{gathered} h\left(0\right)=f\left(0\right)-2g\left(0\right) \\ =2-0 \\ =2 \end{gathered}$$

$$\begin{gathered} h\left(1\right)=f\left(1\right)-2g\left(1\right) \\ =6-2\left(2\right) \\ =2 \end{gathered}$$

$h\left(0\right)=h\left(1\right)=2$

Using Rolle's theorem,

$h\text{'}\left(c\right)=0,$ such that $c\in \left(0,1\right)$

$\Rightarrow f\text{'}\left(c\right)-2g\text{'}\left(c\right)=0$

$\Rightarrow f\text{'}\left(c\right)=2g\text{'}\left(c\right)$

Hence , option D is correct.