Maximum value of 'a' for Given Condition Let $f(x) $ be a positive differentiable function on $[0,a]$ such that $f(0)=1$ and $f(a)=3^{\frac{1}{4}}$. If $f'(x)\ge(f(x))^3 +(f(x))^{-1} $,what is the maximum value of a? 
$ a) \frac{\pi}{12} b) \frac{\pi}{36}c) \frac{\pi}{24} d) \frac{\pi}{48}$
From the given conditions i was able to make out that the function is strictly increasing in the interval,but that is about all. And i have no idea how to bring $\pi$ into the mix
I know these kind of questions arent too popular here on SE, but i think this question might be good enough
 A: On the one hand, because$$
f'(x) \geqslant (f(x))^3 + \frac{1}{f(x)}, \quad \forall 0 < x < a
$$
then $f(x) \neq 0$ for $0 < x < a$. Since $f$ is continuous on $[0, a]$ and $f(0) > 0$, then $f(x) > 0$ for $0 < x < a$. Now,\begin{align*}\def\d{\mathrm{d}}
a &= \int_0^a \,\d x \leqslant \int_0^a \frac{f'(x)}{(f(x))^3 + \frac{1}{f(x)}} \,\d x = \int_0^a \frac{f(x)}{(f(x))^4 + 1} f'(x) \,\d x\\
&= \frac{1}{2} \int_{(f(0))^2}^{(f(a))^2} \frac{\d ((f(x))^2)}{((f(x))^2)^2 + 1} = \frac{1}{2} \int_{(f(0))^2}^{(f(a))^2} \frac{\d u}{u^2 + 1}\\
&= \frac{1}{2} \arctan u \Biggr|_{(f(0))^2}^{(f(a))^2} = \frac{1}{2} (\arctan((f(a))^2) - \arctan((f(0))^2))\\
&= \frac{1}{2} (\arctan \sqrt{3} - \arctan 1) = \frac{1}{2} \left(\frac{π}{3} - \frac{π}{4}\right) = \frac{π}{24}.
\end{align*}
On the other hand, if $\displaystyle f(x) = \sqrt{\tan\left(2x + \frac{π}{4}\right)} \, \left(x \in \left[0, \frac{π}{8}\right]\right)$, then\begin{gather*}
2f(x) f'(x) = ((f(x))^2)' = \left(\tan\left(2x + \frac{π}{4}\right)\right)' = \frac{2}{\displaystyle\cos^2\left(2x + \frac{π}{4}\right)},\\
(f(x))^4 + 1 = \left(\tan\left(2x + \frac{π}{4}\right)\right)^2 + 1 = \frac{1}{\displaystyle\cos^2\left(2x + \frac{π}{4}\right)}.
\end{gather*}
Therefore,$$
f(x) f'(x) = (f(x))^4 + 1 \Longrightarrow f'(x) = (f(x))^3 + \frac{1}{f(x)}.
$$
Note that $\displaystyle f(0) = \sqrt{\tan \frac{π}{4}} = 1$, $\displaystyle f\left(\frac{π}{24}\right) = \sqrt{\tan \frac{π}{3}} = 3^{\frac{1}{4}}$, so $\displaystyle a = \frac{π}{24}$ is the maximum.
