Q47 / 155 · WBJEE 2018 · Mathematics

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

  • A
    $f^{\prime \prime}(0)=0$
  • Correct answer
    B
    $f^{\prime \prime}=0$ for some $C \in A$
  • C
    if c $\neq 0,$ then $f^{\prime \prime}\neq 0$
  • D
    $f^{\prime}(x)>0$ for all $x \neq 0$

Solution

We have. $$f: \mathbb{R} \rightarrow \mathbb{R} $$ be a twice continuousty differentiable function such that $f(0)=f(1)=f'(0)=0$ Now, for atleast one value of $\epsilon_{1} \in(0,1)$ $f'\left(c_{1}\right)=0 \quad$ (by Rolle's theorem) Again. $\quad f'(0)=0=f(c_{1})$ $f'(c)=$ ofor some $c \in\left(0, c_{1}\right)$ (by Rolle's theorem)

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

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