WBJEE · Maths · Application of Derivatives
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
- 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 \(f^{\prime \prime}(x) < 0\)
Answer & Solution
Correct Answer
(A) \(\mathrm{f}^{\prime \prime}(\mathrm{c})=0\) for some \(\mathrm{c} \in \mathbb{R}\)
Step-by-step Solution
Detailed explanation
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…
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