Question

Let $$b$$ be a non-zero real number. Suppose $$f:\mathbb{R}\to \mathbb{R}$$ is a differentiable function such that $$f(0)=1$$ . If the derivative $$f^{\prime }$$ of $$f$$ satisfies the equation $$f^{\prime }(x)={\displaystyle \frac{f(x)}{b^2+x^2}}$$ . For all $$x\in \mathbb{R}$$ , 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. $$f(x)f(-x)=1$$ for all $$x\in \mathbb{R}$$
D. $$f(x)-f(-x)=0$$ for all $$x\in \mathbb{R}$$

Answer 1

Hint: Firstly, we need to simplify the given function and then find the integration constant by using integration properties. After that we simply the integrating value by using the given value $$f(0)=1$$ . Then we find the condition of increasing or decreasing function and the $$f(-x)$$ to get required answer.

Formula used:
Trigonometric property: $${\tan }^{-1}(-x)=-{\tan }^{-1}x$$
Exponential property: $$e^x\times e^y=e^{x+y}$$
Integration formula: $$\int {\displaystyle \frac{f^{\prime }(x)}{f(x)}}dx=\ln f(x)$$
$$\int {\displaystyle \frac{1}{a^2+x^2}}dx={\displaystyle \frac{1}{a}}{\tan }^{-1}{\displaystyle \frac{x}{a}}$$
Logarithm property: $$\ln 1=0$$

Complete step by step solution:
Given equation $$f^{\prime }(x)={\displaystyle \frac{f(x)}{b^2+x^2}}$$ ……………….(1)
and $$f(0)=1$$
Now cross multiplying the equation (1) and we get
$$f^{\prime }(x)={\displaystyle \frac{f(x)}{b^2+x^2}}$$
$$\Rightarrow {\displaystyle \frac{f^{\prime }(x)}{f(x)}}={\displaystyle \frac{1}{b^2+x^2}}$$ ………………………(2)
Integrating the equation (2) and we get
$$\Rightarrow \int {\displaystyle \frac{f^{\prime }(x)}{f(x)}}dx=\int {\displaystyle \frac{1}{b^2+x^2}}dx$$
Using integration formulas $$\int {\displaystyle \frac{f^{\prime }(x)}{f(x)}}dx=\ln f(x)$$ and $$\int {\displaystyle \frac{1}{a^2+x^2}}dx={\displaystyle \frac{1}{a}}{\tan }^{-1}{\displaystyle \frac{x}{a}}$$, we get
$$\Rightarrow \ln f(x)={\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}+c$$……………(3), where $$c$$ is the constant of integration.
Now, given that $$f(0)=1$$, i.e., $$x=0$$
Substitute $$f(0)=1$$ and we get
$$\Rightarrow \ln 1={\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{0}{b}}+c$$
$$\Rightarrow 0={\displaystyle \frac{1}{b}}{\tan }^{-1}0+c$$
$$\Rightarrow 0={\displaystyle \frac{1}{b}}\times 0+c$$
$$\Rightarrow c=0$$ …………(4)
Substitute the value $$c=0$$ in (3) and we get
$$\Rightarrow \ln f(x)={\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}+0$$
$$\Rightarrow \ln f(x)={\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}$$
Taking antilog and we get
$$\Rightarrow f(x)=e^{{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}$$
Now finding $$f(-x)$$ , we get
$$\Rightarrow f(-x)=e^{{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{-x}{b}}}$$
$$\Rightarrow f(-x)=e^{-{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}$$
For option A and B,
$$f^{\prime }(x)={\displaystyle \frac{f(x)}{b^2+x^2}}$$
Here, the differential function is always positive when $$b>0$$ or $$b<0$$.
Therefore, $$f(x)>0$$ is an increasing function for $$b>0$$
Therefore, option A is correct and B is incorrect.
For option C,
Here $$f(x)=e^{{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}$$ and $$f(-x)=e^{-{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}$$
Now $$f(x)f(-x)$$
$$=\left(e^{{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}\right)\times \left(e^{-{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}\right)$$
$$=e^{\left({\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}\right)-\left({\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}\right)}$$
$$=e^0$$
$$=1$$
= R.H.S.
Therefore, option C is correct.
For option D,
$$e^{{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}-e^{-{\displaystyle \frac{1}{b}}{\tan }^{-1}{\displaystyle \frac{x}{b}}}\ne 0\mathrm{\forall }x\in \mathbb{R}$$
Therefore, option D is incorrect.
Hence, options A and C are correct.

Note: Students need to take care about small properties of logarithm and trigonometry. Like $$\ln 1=0$$ and $${\tan }^{-1}(-x)=-{\tan }^{-1}x$$. We need to take care while we find the function $$f(-x)$$ , we need to substitute $$x$$ with $$-x$$ only in the main function. If we make any mistakes in these steps then we got the wrong solution.