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Question

There exists a function f(x) satisfying f(0)=1, f′(0)=−1, f(x)>0 for all x and

A
f′′(0)<0 for all x
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B
1<f′′(x)<0 for all x
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C
2f′′(x)1 for all x
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D
f′′(x)2 for all x
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Solution

The correct option is B f′′(0)<0 for all x
Since f(x) is continuous and differentiable where f(0)=1
and f(0)=1,f(x)>0 for all x.
Thus f(x) is decreasing for x>0 and concave downward.
f′′(x)<0

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