Question

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be twice continuously differentiable (or \(f\) " exists and is continuous) such that \(f(0)=f(1)=f^{\prime}(0)=0 .\) Then
(A) \(f^{\prime \prime}(c)=0\) for some \(c \in \mathbb{R}\)
(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 1

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

Answer 2

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