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 $f^{\text{'}}$of $f$ satisfies the equation $f^{\text{'}}\left(x\right)=\frac{f\left(x\right)}{\left(b^2+x^2\right)}$. 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. $f\left(x\right)f\left(-x\right)=1$ for all $x\in R$
• D. $f\left(x\right)-f\left(-x\right)=0$ for all $x\in R$

Answer 1

Answer: (A), (C)

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

Explanation for correct option:

Consider the given equation :

$f^{\text{'}}\left(x\right)=\frac{f\left(x\right)}{b^2+x^2}$

Rewrite the above Equation as :

$\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=\frac{1}{b^2+x^2}$

Integrate the above Equation

$\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}dx=\int \frac{1}{b^2+x^2}dx$

$$\begin{gathered} \mathrm{Let},t=f\left(x\right) \\ \therefore dt=f^{\text{'}}\left(x\right)dx \end{gathered}$$

$\ln f\left(x\right)=\frac{1}{b}{\tan }^{-1}\frac{x}{b}+C$

when, $f\left(0\right)=1$

$$\begin{gathered} \ln f\left(x\right)=0+c \\ c=0 \end{gathered}$$

Substitute $0$ for $c$ in the above Equation.

$$\begin{gathered} \ln f\left(x\right)=\frac{1}{b}{\tan }^{-1}\frac{x}{b} \\ f\left(x\right)=e^{\frac{1}{b}{\tan }^{-1}\frac{x}{b}} \end{gathered}$$

$$\begin{gathered} f\left(-x\right)=e^{\frac{1}{b}{\tan }^{-1}\left(\frac{-x}{b}\right)} \\ f\left(-x\right)=e^{-\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)} \end{gathered}$$

Option A: If $b>0$, then $f$ is an increasing function

$f\text{'}\left(x\right)=e^{\frac{1}{b}{\tan }^{-1}\frac{x}{b}}\times \frac{1}{b^2+x^2}\times \frac{1}{b}>0$

$f\left(x\right)$ is increasing function.

$f\left(x\right)>0$ for all $x\in R$

Hence, option A is true.

Option B: If $b<0$, then $f$ is a decreasing function

$f\left(x\right)$ is decreasing function then we get negative value

$f\left(x\right)<0$ for all $x\notin R$

Hence, option B is false.

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

$$\begin{gathered} f\left(x\right).f\left(-x\right)=\left(e^{\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)}\right)\left(e^{-\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)}\right) \\ f\left(x\right).f\left(-x\right)=e^{\left(\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)\left(-\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)} \\ f\left(x\right).f\left(-x\right)=e^0 \\ f\left(x\right).f\left(-x\right)=1 \end{gathered}$$

Hence, option C is true.

Option D: $f\left(x\right)-f\left(-x\right)=0$ for all $x\in R$

$$\begin{gathered} f\left(x\right)-f\left(-x\right)=\left(e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}\right)-\left(e^{\left(-\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}\right) \\ f\left(x\right)-f\left(-x\right)=e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}-\frac{1}{e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}} \\ f\left(x\right)-f\left(-x\right)=\frac{e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}.e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}-1}{e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}} \\ f\left(x\right)-f\left(-x\right)=\frac{e^{2\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}-1}{e^{\left(\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right)}} \end{gathered}$$

Hence, option D is false.

Therefore, the correct answer is option A and C.