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Question

# Let f : R→R be a twice continuously differentiable function such that f(0)=f(1)=f′(0)=0. Then

A
f′′(0)=0
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B
f′′(c)=0 for some cR
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C
if c0, then f′′(c)0
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D
f(x)>0 for all x0
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Solution

## The correct option is B f′′(c)=0 for some c∈Rf(x) is twice continuous and differentiable, f(0)=f(1)=f′(0)=0 So the function satisfies all the conditions of Rolle's theorem Therefore, f′(a)=f(1)−f(0)1=0, for some a∈(0,1) So f′(a)=f′(0)=0 The function f′(x) satisfies all the conditions of Rolle's theorem in x∈(0,a) Therefore, f′′(k)=f′(a)−f′(0)a=0 For some k∈(0,a)

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