Question

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

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

Answer 1

The correct option is D $f^{\prime }\left(c\right)=2g^{\prime }\left(c\right)$

Let $h\left(f\right)=f\left(x\right)-2g\left(x\right)$
as $h\left(0\right)=h\left(1\right)=2$
Hence, using Rolle's theorem
$h^{\prime }\left(c\right)=0$
$\Rightarrow f^{\prime }\left(c\right)=2g^{\prime }\left(c\right)$