Q89 / 155 · WBJEE 2020 · Mathematics

Letf $: \mathbb{R} \rightarrow \mathbb{R}$, be twice continuously differentiable (or $f$ " exists and is continuous) such that $f(0)=f(1)=f^{\prime}(0)=0$. Then

  • Correct answer
    A
    $\mathrm{f}^{\prime \prime}(\mathrm{c})=0$ for some $\mathrm{c} \in \mathbb{R}$
  • B
    there is no point for which $f^{\prime \prime}(x)=0$
  • C
    at all points $f^{\prime \prime}(x)>0$
  • D
    at all points $f^{\prime \prime}(x) < 0$

Solution

Hint: $f(0)=f(1)=0$ By Rolle's theorem $\mathrm{f}^{\prime}(\mathrm{c})=0$ for some $\mathrm{c} \in(0,1)$ Now, $\mathrm{f}^{\prime}(0)=\mathrm{f}^{\prime}(\mathrm{c})=0$. Again by Rolle's theorem $\mathrm{f}^{\prime \prime}\left(\mathrm{c}_{1}\right)=0 \quad \mathrm{c}_{1} \in(0, \mathrm{c})$

Hence, the correct answer is option: 'A'.

Relevant Questions