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 $h\left(x\right)=f\left(x\right)-2g\left(x\right)...\left(1\right)$
$\therefore h\left(0\right)=f\left(0\right)-2g\left(0\right)=2-0=2$
and $h\left(1\right)=f\left(1\right)-2g\left(1\right)=6-2\left(2\right)=2$
Thus, $h\left(0\right)=h\left(1\right)=2$
Now apply Rolle's theorem on equation (1),
$h^{\prime }\left(c\right)=0$ where $c\in (0,1)$
Differentiating equation (1) w.r.t $x$
$\therefore h^{\prime }\left(x\right)=f^{\prime }\left(x\right)-2g^{\prime }\left(x\right)$
At $x=c,h^{\prime }\left(c\right)=f^{\prime }\left(c\right)-2g^{\prime }\left(c\right)$
Hence, $0=f^{\prime }\left(c\right)-2g^{\prime }\left(c\right)$
$\therefore f^{\prime }\left(c\right)=2g^{\prime }\left(c\right)$