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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 $f'$ of $f$ satisfies the equation $f'(x) = \dfrac{{f(x)}}{{{b^2} + {x^2}}}$ . For all $x \in \mathbb{R}$ , then which of the following statements is/are TRUE?A. If $b > 0$ , then $f$ is an increasing functionB. If $b < 0$ , then $f$ is a decreasing functionC. $f(x)f( - x) = 1$ for all $x \in \mathbb{R}$ D. $f(x) - f( - x) = 0$ for all $x \in \mathbb{R}$

Last updated date: 11th Sep 2024
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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: $\int {\dfrac{{f'(x)}}{{f(x)}}} dx = \ln f(x)$
$\int {\dfrac{1}{{{a^2} + {x^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}$
Logarithm property: $\ln 1 = 0$

Complete step by step solution:
Given equation $f'(x) = \dfrac{{f(x)}}{{{b^2} + {x^2}}}$ ……………….(1)
and $f(0) = 1$
Now cross multiplying the equation (1) and we get
$f'(x) = \dfrac{{f(x)}}{{{b^2} + {x^2}}}$
$\Rightarrow \dfrac{{f'(x)}}{{f(x)}} = \dfrac{1}{{{b^2} + {x^2}}}$ ………………………(2)
Integrating the equation (2) and we get
$\Rightarrow \int {\dfrac{{f'(x)}}{{f(x)}}} dx = \int {\dfrac{1}{{{b^2} + {x^2}}}}dx$
Using integration formulas $\int {\dfrac{{f'(x)}}{{f(x)}}} dx = \ln f(x)$ and $\int {\dfrac{1}{{{a^2} + {x^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a}$, we get
$\Rightarrow \ln f(x) = \dfrac{1}{b}{\tan ^{ - 1}}\dfrac{x}{b} + c$……………(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
$\Rightarrow \ln 1 = \dfrac{1}{b}{\tan ^{ - 1}}\dfrac{0}{b} + c$
$\Rightarrow 0 = \dfrac{1}{b}{\tan ^{ - 1}}0 + c$
$\Rightarrow 0 = \dfrac{1}{b} \times 0 + c$
$\Rightarrow c = 0$ …………(4)
Substitute the value $c = 0$ in (3) and we get
$\Rightarrow \ln f(x) = \dfrac{1}{b}{\tan ^{ - 1}}\dfrac{x}{b} + 0$
$\Rightarrow \ln f(x) = \dfrac{1}{b}{\tan ^{ - 1}}\dfrac{x}{b}$
Taking antilog and we get
$\Rightarrow f(x) = {e^{\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}$
Now finding $f( - x)$ , we get
$\Rightarrow f( - x) = {e^{\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{{ - x}}{b}}}$
$\Rightarrow f( - x) = {e^{ - \dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}$
For option A and B,
$f'(x) = \dfrac{{f(x)}}{{{b^2} + {x^2}}}$
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 $f(x) = {e^{\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}$ and $f( - x) = {e^{ - \dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}$
Now $f(x)f( - x)$
$= \left( {{e^{\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}} \right) \times \left( {{e^{ - \dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}}} \right)$
$= {e^{\left( {\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}} \right) - \left( {\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}} \right)}}$
$= {e^0}$
$= 1$
= R.H.S.
Therefore, option C is correct.
For option D,
${e^{\dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}} - {e^{ - \dfrac{1}{b}{{\tan }^{ - 1}}\dfrac{x}{b}}} \ne 0\,\forall \,x \in \mathbb{R}$
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.