Question

For all twice differentiable functions $f:R\to R$, with $f\left(0\right)=f\left(1\right)=f\text{'}\left(0\right)=0$,

• A. $f\text{'}\text{'}\left(x\right)=0$, at every point $x\in (0,1)$
• B. $f\text{'}\text{'}\left(x\right)\ne 0$, at every point $x\in (0,1)$
• C. $f\text{'}\text{'}\left(x\right)=0$ for some $x\in (0,1)$
• D. $f\text{'}\text{'}\left(0\right)=0$

Answer 1

The correct option is C

$f\text{'}\text{'}\left(x\right)=0$ for some $x\in (0,1)$

The explanation of the correct option :

Given a twice differentiable functions $f:R\to R$, with $f\left(0\right)=f\left(1\right)=f\text{'}\left(0\right)=0$

Apply Rolle's theorem on $y=f\left(x\right)$ in $x\in [0,1]$

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

$\Rightarrow f\text{'}\left(\alpha \right)=0$ where $\alpha \in (0,1)$

Now, again apply Rolles theorem on $y=f\text{'}\left(x\right)$ in $x\in [0,\alpha ]$

$f\text{'}\left(0\right)=f\text{'}\left(\alpha \right)=0$ and $f\text{'}\left(x\right)$is continuous and differentiable.

$\Rightarrow f\text{'}\text{'}\left(\beta \right)=0$ for some ,$\beta \in (0,\alpha )\in (0,1)$

$\Rightarrow f\text{'}\text{'}\left(x\right)=0$ for some $x\in (0,1)$.

Hence the correct option is C.