Let $b$ be a nonzero real number. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $f(0)=1$. If the derivative $f^{\prime}$ of $f$ satisfies the equation $f^{\prime}(x)=\frac{f(x)}{b^{2}+x^{2}}$ 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; (C) $f(x)f(-x)=1$ for all $x \in \mathbb{R}$

(B) If $b<0$, then $f$ is a decreasing function

(D) $f(x)-f(-x)=0$ for all $x \in \mathbb{R}$

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Differential Equations

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^{'}

f

satisfies the equation

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

for all

x \in R

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

Answer: (A), (C)

From

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

, separating variables and integrating:

ln ∣ f (x) ∣ = \frac{1}{b} tan^{- 1} (\frac{x}{b}) + c

. Using

f (0) = 1

gives

c = 0

, so

f (x) = e^{\frac{1}{b} t a n^{- 1} (x / b)}

. Then

f (- x) = e^{- \frac{1}{b} t a n^{- 1} (x / b)}

, so

f (x) f (- x) = 1

(option C is correct). For

b > 0

tan^{- 1} (x / b)

is increasing, so

f (x)

is increasing (option A is correct). Options B and D are false.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options