Question

Statement $1:$ If $f\left(x\right)$ is differentiable in $[0,1]$ such that $f\left(0\right)$ $=f\left(1\right)=0$, then for any $\lambda \in R$, there exists $c$ such that $f^{\prime }\left(c\right)=\lambda f\left(c\right),0<c<1$
Statement $2:$ If $g\left(x\right)$ is differentiable in $[0,1]$, where $g\left(0\right)$ $=g\left(1\right)$, then there exists $c$ such that $g^{\prime }\left(c\right)=0,0<c<1$

• A. Only statement $1$ is true
• B. Only statement $2$ is true
• C. Both are true
• D. Both are false

Answer 1

The correct option is C Both are true

Statement $1$ :
Consider, $F\left(x\right)=e^{-\lambda x}f\left(x\right),\lambda \in R$ $F\left(0\right)=f\left(0\right)=0F\left(1\right)=e^{-1}f\left(1\right)=0$
$\therefore$ By Rolle's theorem, $F^{\prime }\left(c\right)=0$ $F^{\prime }\left(x\right)=e^{-\lambda x}\left(f^{\prime }\right(x)-\lambda f(x\left)\right)F^{\prime }\left(c\right)=0\Rightarrow e^{-\lambda c}\left(f^{\prime }\right(c)-\lambda f(c\left)\right)=0f^{\prime }\left(c\right)=\lambda f\left(c\right),0<c<1$

Statement $2$ :
By rolle's theorem, second statement is true.