Question

If $f\left(x\right)$ is a continuous function on $[0,1]$, differentiable in $(0,1)$ such that $f\left(1\right)=0$, then there exists some $cϵ(0,1)$ 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

Answer: (D)

$c{f}^{\prime }\left(c\right)+f\left(c\right)=0$

Let $g\left(x\right)=xf\left(x\right)$.
Then g(x) is continuous in [0, 1] and differentiable in (0, 1).
Also, $g\left(0\right)=g\left(1\right)=0$.
So, by Rolle's theorem, $g^{\prime }\left(c\right)=0$ for some $c$ in $(0,1)$,
$\Rightarrow c{f}^{\prime }\left(c\right)+f\left(c\right)=0$
Hence, option D is the correct answer.