Question

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

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

Answer 1

The correct option is B $f^{\prime \prime }\left(c\right)=0$ for some $c\in R$

$f\left(x\right)$ is twice continuous and differentiable,
$f\left(0\right)=f\left(1\right)=f^{\prime }\left(0\right)=0$
So the function satisfies all the conditions of Rolle's theorem
Therefore,
$f^{\prime }\left(a\right)=\frac{f\left(1\right)-f\left(0\right)}{1}=0,\text{for some}a\in (0,1)$

So $f^{\prime }\left(a\right)=f^{\prime }\left(0\right)=0$
The function $f^{\prime }\left(x\right)$ satisfies all the conditions of Rolle's theorem in $x\in (0,a)$
Therefore,
$f^{\prime \prime }\left(k\right)=\frac{f^{\prime }\left(a\right)-f^{\prime }\left(0\right)}{a}=0$
For some $k\in (0,a)$