Continuity and DifferentiabilityHard

Question

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

Options

A.f″(x) > 0 for all x
B.- 1 < f″(x) < 0 for all x
C.- 2 ≤ f″(x) ≤ - 1 for all x
D.f″(x) ≤ - 2 for all x

Solution

Since, f(x) is continuous and differentaible where f(0) = 1and f′(0) = - 1, f(x) > 0 for all x
Thus f(x) is decreasing for x > 0 and concave down
⇒       f″(x) > 0
Therefore,(a) is answer.

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