JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 1)
If f : R $$ \to $$ R is a twice differentiable function such that f"(x) > 0 for all x$$ \in $$R, and $$f\left( {{1 \over 2}} \right) = {1 \over 2}$$, f(1) = 1, then
f'(1) $$ \le $$ 0
f'(1) > 1
0 < f'(1) $$ \le $$ $${1 \over 2}$$
$${1 \over 2}$$ < f'(1) $$ \le $$ 1
Explanation
f'(x) is increasing
For some x in $$\left( {{1 \over 2},\,1} \right)$$
f'(x) = 1 [LMVT]
$$ \therefore $$ f'(1) > 1
For some x in $$\left( {{1 \over 2},\,1} \right)$$
f'(x) = 1 [LMVT]
$$ \therefore $$ f'(1) > 1
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