QJEEAdv20Ma40f4: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma40f4 Solution: Let b be a nonzero real number. Suppose f : R R is a…

Question

Verified Solution • 2025–26 Session moderate

Let b be a nonzero real number. Suppose f : R

\to

R is a differentiable function such that f(0) = 1. If the derivative f' of f satisfies the equation

f'(x) = {{f(x)} \over {{b^2} + {x^2}}}

for all x

\in

R, then which of the following statements is/are TRUE?

Accepted Answer

Given differential equation

f'(x) = {{f(x)} \over {{b^2} + {x^2}}}

\Rightarrow \int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over {{b^2} + {x^2}}}} }

\Rightarrow {\log _e}|f(x)| = {1 \over b}{\tan ^{ - 1}}\left( {{x \over b}} \right) + c

\because

f(0) = 1, so c = 0

\therefore

|f(x)| = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}

\Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}

or

- {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}

\because

f(0) = 1 \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}

and

f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}

\therefore

f'(x) &gt; 0\forall x \in R

and

b \in {R_0}

.

Therefore, f(x) is an increasing function

\forall b \in {R_0}

.

and f(x) f(

-

x)

= {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}\,.\,{e^{ - {1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}} = 1

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