Question

Let $f\left(x\right)$ be a continous and differentiable function on $[0,1]$ , such that $f\left(0\right)\ne 0$ and $f\left(1\right)=0$. We can conclude that there exists $c\in (0,1)$ such that

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

Answer 1

Answer: (D)

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

Consider $\theta \left(x\right)=x.f\left(x\right)$
Clearly Rolle's theorem is applicable to $\theta \left(x\right)$ on $xϵ[0,1]$

as $\theta \left(0\right)=0\ast f\left(0\right)=0$ and $\theta \left(1\right)=1\ast f\left(1\right)=0$
$\therefore$ There exists $cϵ(0,1)$ such that ${\theta }^{\prime }\left(c\right)=0$
$\Rightarrow c.f^{\prime }\left(c\right)+f\left(c\right)=0$
Hence, option 'D' is correct.