Question

Let $f$ be a function which is continuous in $[0,1]$ and differentiable in $(0,1)$ such that $f\left(1\right)=0$, then there exists some $c\in (0,1)$ such that:

• A. $c{f}^{\prime }\left(c\right)-f\left(c\right)=0$
• B. $f^{\prime }\left(c\right)+cf\left(c\right)=0$
• C. $f^{\prime }\left(c\right)-cf\left(c\right)=0$
• D. $c{f}^{\prime }\left(c\right)+f\left(c\right)=0$

Answer 1

The correct option is D $c{f}^{\prime }\left(c\right)+f\left(c\right)=0$

Let $g\left(x\right)=xf\left(x\right)$.
Now, $g\left(0\right)=g\left(1\right)=0$ & $g$ is continuous & differentiable in $[0,1]$ and $(0,1)$ respectively.

By Rolle's Theorem, there exists a $c\in (0,1)$ such that $g^{\prime }\left(c\right)=0$.
$\Rightarrow c{f}^{\prime }\left(c\right)+f\left(c\right)=0$