Question

Let $f:[a,b]\to R$ be such that $f$ is differentiable in $(a,b),f$ is continuous at $x=a$ and $x=b$ and moreover $f\left(a\right)=0=f\left(b\right)$. Then

• A. there exists atleast one point $c$ in $(a,b)$ such that $f^{\prime }\left(c\right)=f\left(c\right)$
• B. $f^{\prime }\left(x\right)=f\left(x\right)$ does not hold at any point in $(a,b)$
• C. at every point of $(a,b),f^{\prime }\left(x\right)>f\left(x\right)$
• D. at every point of $(a,b),f^{\prime }\left(x\right)<f\left(x\right)$

Answer 1

The correct option is A there exists atleast one point $c$ in $(a,b)$ such that $f^{\prime }\left(c\right)=f\left(c\right)$

Let $h\left(x\right)=e^{-x}f\left(x\right)$
$h\left(a\right)=0,h\left(b\right)=0$
$h\left(x\right)$ is continuous and differentiable
by Rolle's theorem
$h^{\prime }\left(c\right)=0,c\in (a,b)$
$e^{-x}{f}^{\prime }\left(x\right)+(-e^{-x})f\left(x\right)=0$
$e^{-c}{f}^{\prime }\left(c\right)=e^{-c}f\left(c\right)$
$f^{\prime }\left(c\right)=f\left(c\right)$