JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 7)
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?
for all x$$ \in $$R, then which of the following statements is/are TRUE?
If b > 0, then f is an increasing function
If b < 0, then f is a decreasing function
f(x) f($$-$$x) = 1 for all x$$ \in $$R
f(x) $$-$$f($$-$$x) = 0 for all x$$ \in $$R
Explanation
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 $$f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}$$
$$ \therefore $$ $$f'(x) > 0\forall x \in R$$ and $$b \in {R_0}$$.
Therefore, f(x) is an increasing function $$\forall b \in {R_0}$$.
and f(x) f($$-$$x)
$$ = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}\,.\,{e^{ - {1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}} = 1$$
$$f'(x) = {{f(x)} \over {{b^2} + {x^2}}}$$
$$ \Rightarrow \int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over {{b^2} + {x^2}}}} } $$
$$ \Rightarrow {\log _e}|f(x)| = {1 \over b}{\tan ^{ - 1}}\left( {{x \over b}} \right) + c$$
$$ \because $$ 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 $$f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}$$
$$ \therefore $$ $$f'(x) > 0\forall x \in R$$ 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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