Question

Let $f\left(x\right)$ be positive, continuous, and differentiable on the interval $(a,b)$ and $\underset{x\to a^{+}}{lim}f\left(x\right)=1,\underset{x\to b^{-}}{lim}f\left(x\right)=3^{1/4}.$ If $f^{\prime }\left(x\right)\ge f^3\left(x\right)+\frac{1}{f\left(x\right)}$, then the greatest value of $b-a$ is

• A. $\frac{\pi }{48}$
• B. $\frac{\pi }{36}$
• C. $\frac{\pi }{24}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is C $\frac{\pi }{24}$

$f^{\prime }\left(x\right)\ge f^3\left(x\right)+\frac{1}{f\left(x\right)}\Rightarrow \frac{f\left(x\right)f^{\prime }\left(x\right)}{1+f^4\left(x\right)}\ge 1$
Integrating on the interval $(a,b),$ we get
$\underset{a}{\overset{b}{\int }}\frac{f\left(x\right)f^{\prime }\left(x\right)}{1+f^4\left(x\right)}dx\ge \underset{a}{\overset{b}{\int }}dx$

Put $f^2\left(x\right)=t\Rightarrow 2f\left(x\right)f^{\prime }\left(x\right)dx=dt$
${\left[\frac{1}{2}{\tan }^{-1}{f}^2\right(x\left)\right]}_a^b\ge b-a\Rightarrow \frac{1}{2}\left[\underset{x\to b^{-}}{lim}{\tan }^{-1}{f}^2\right(x)-\underset{x\to a^{+}}{lim}{\tan }^{-1}{f}^2(x\left)\right]\ge b-a\Rightarrow \frac{1}{2}[{\tan }^{-1}\sqrt{3}-{\tan }^{-1}1]\ge b-a\Rightarrow \frac{\pi }{24}\ge b-a$