Question

Let f(x) be a continuous and differentiable function on [0,1], such that $f\left(0\right)\ne 0andf\left(1\right)=0.$ We can conclude that there exists $cϵ$ (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 $\varphi \left(x\right)=x.f\left(x\right)$
Clearly Rolle's theorem is applicable to $\varphi \left(x\right)$ on $xϵ[0,1]$
$\therefore$ There exists $Xϵ(0,1)suchthat{\varphi }^{\prime }\left(x\right)=0$
$\Rightarrow c.f^{\prime }\left(x\right)+f\left(x\right)=0$