Question

Let $f:R\to R$ be twice continuously differentiable. Let $f\left(0\right)=f\left(1\right)=f^{\prime }\left(0\right)=0$. Then

• A. $f"\left(x\right)\ne 0$ for all $x$
• B. $f"\left(c\right)=0$ for some $cϵR$
• C. $f"\left(x\right)\ne 0$ if $x\ne 0$
• D. $f^{\prime }\left(x\right)>0$ for all $x$

Answer 1

Answer: (B)

$f"\left(c\right)=0$ for some $cϵR$

Given : $f:R\to R$ is twice continuously differentiable

Also, $f\left(0\right)=f\left(1\right)=0$

By Rolle's theorem, there exist some $x$ between $0$ and $1$ such that $f^{\prime }\left(x\right)=0$

Now, $f^{\prime }$ is also continuously differentiable and $f^{\prime }\left(0\right)=0=f^{\prime }\left(x\right)$

Again applying Rolle's theorem we get $f"\left(c\right)=0$ for some $cϵ[0,1]$.

Thus, $f"\left(c\right)=0$ for some $cϵR$