Question

A real valued differentiable function defined on $[1,\infty )$ where $f\left(1\right)=1.$
If $f^{\prime }\left(x\right)=\frac{1}{x^2+f^2\left(x\right)}$ then the maximum value of $\left[f\right(x\left)\right]$ is
(where $[.]$ is greatest integer function)

• A. $2$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is B $1$

Since, $f^{\prime }\left(x\right)>0$
$\therefore f\left(x\right)$ is increasing function.
$\because f\left(1\right)=1,\therefore f\left(x\right)>1$
$\Rightarrow \frac{1}{x^2+f^2\left(x\right)}<\frac{1}{x^2+1}$
$f\left(t\right)-f\left(1\right)=\underset{1}{\overset{t}{\int }}\frac{1}{x^2+f^2\left(x\right)}dx$
$\Rightarrow f\left(t\right)-f\left(1\right)<\underset{1}{\overset{\infty }{\int }}\frac{1}{x^2+1}dx$
$\Rightarrow f\left(t\right)-f\left(1\right)<{\tan }^{-1}x{|}_1^{\infty }$
$\Rightarrow f\left(t\right)-\frac{\pi }{4}<1$
$\Rightarrow f\left(x\right)-\frac{\pi }{4}<1\Rightarrow f\left(x\right)<1+\frac{\pi }{4}$
Hence $\left[f\right(x\left)\right]=1$