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Question:

If f:R→R is a twice differentiable function such that f''(x) > 0 for all x∈R, and f(12)=12, f(1)=1, then

f'(1)≤0

0<f'(1)≤12

12<f'(1)≤1

f'(1)>1

Solution:

The correct option is f'(1)>1
Since f''(x)>0 → f'(x) is increasing.
Using LMVT, There exists c such that 1<c<12 and f'(c) = f(1)-f(12)/1-12 = 11/11 = 1
It is given that f''(x)>0, hence f'(x) is an increasing function.
Since f'(x) is increasing and c<1 then f'(1)>f'(c)
Therefore f'(1)>1
Hence, option A is correct.