Question

Let $b$ be a nonzero real number. Suppose $f:R\to R$ is a differentiable function such that $f\left(0\right)=1.$ If the derivative $f^{\prime }$ of $f$ satisfies the equation $f^{\prime }\left(x\right)=\frac{f\left(x\right)}{b^2+x^2}$ for all $x\in R,$ then which of the following statements is/are TRUE?

• A. If $b>0,$ then $f$ is an increasing function
• B. If $b<0,$ then $f$ is a decreasing function
• C. $f\left(x\right)f(-x)=1$ for all $x\in R$
• D. $f\left(x\right)-f(-x)=0$ for all $x\in R$

Answer 1

Answer: (A), (C)

$f\left(x\right)f(-x)=1$ for all $x\in R$

Given $f^{\prime }\left(x\right)=\frac{f\left(x\right)}{b^2+x^2}$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{b^2+x^2}$
On integrating both sides
$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int \frac{1}{b^2+x^2}dx$
$\Rightarrow \ln \left|f\left(x\right)\right|=\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)+C$
Put $x=0\Rightarrow C=0$
$\therefore f\left(x\right)=\exp \left(\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)[\because f(0)=1]$
$f\left(x\right)>0\forall x\in R$
Since $f^{\prime }\left(x\right)=\frac{f\left(x\right)}{b^2+x^2}>0,$
$\Rightarrow f\left(x\right)$ is increasing function.

$f\left(x\right)f(-x)=\exp \left[\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)-\left(\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)\right]=e^0=1$

$f\left(x\right)-f(-x)=\exp \left[\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right]-\exp \left[-\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right]\ne 0\forall x\in R$