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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 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 \[x \in \mathbb{R}\]

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Last updated date: 09th Apr 2024
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MVSAT 2024
Answer
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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.