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Question: Let f(x) the n positive function differentiable on [0, a] such that f(0) = 1 and f (1) = 3<sup>1/6</...

Let f(x) the n positive function differentiable on [0, a] such that f(0) = 1 and f (1) = 31/6. If f'(x) ³ (f(x))4 + (f(x))–2, then the maximum value of a is.

A

p/6

B

p/12

C

p/24

D

p/36

Answer

p/36

Explanation

Solution

3≤3f2f1f6+13 \leq \frac{3f^{2}f^{1}}{f^{6} + 1} integrating from 0 to a

̃ 3a £ tan–13\sqrt{3} – tan–11 =π12\frac{\pi}{12} ̃ a £ π36\frac{\pi}{36}