Question

Suppose $f$ is a real valued differentiable function defined on $[1, \infty)$ with $f(1)=1$ such that $f$ satisfies $f^{\prime}(x)=\frac{1}{x^2+f^2(x)}$ then value of $f(x)$ can not exceed.

• A. $\ln 2$
• B. $1+\frac{\pi}{4}$
• C. $1+\ln \tan 1$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (B)

$f ^{\prime}( x )=\frac{1}{ x ^2+ f ^2( x )}>0 $
$\Rightarrow f ( x )$ is an increasing function $\forall x >1$
Now
$f^{\prime}(x)=\frac{1}{x^2+f^2(x)} \leq \frac{1}{x^2+1} \forall x \geq 1 $
$\left.\int\limits_1^x f^{\prime}(x) d x \leq \int\limits_1^x \frac{1}{\left(x^2+1\right.}\right) d x $
$f(x)-1 \leq \tan ^{-1} x-\frac{\pi}{4}$
$\displaystyle\lim _{x \rightarrow \infty} f(x)-1 \leq \pi / 2-\frac{\pi}{4}$
$\Rightarrow \displaystyle\lim _{x \rightarrow \infty} f(x) \leq 1+\frac{\pi}{4} $