Differentiable Function Satisfying Differential Equation Involving Arctan

To determine the correct statement(s), we begin by solving the given differential equation:

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

Given that $f(0) = 1$ and $f(x)$ is a differentiable function on $\mathbb{R}$, $f(x) > 0$ for all $x \in \mathbb{R}$. We can separate the variables as follows:

$\frac{f'(x)}{f(x)} = \frac{1}{b^2 + x^2}$

Integrating both sides with respect to $x$:

$\int \frac{f'(x)}{f(x)} \, dx = \int \frac{dx}{b^2 + x^2}$

$\ln|f(x)| = \frac{1}{b} \arctan\left(\frac{x}{b}\right) + C$

Applying the initial condition $f(0) = 1$:

$\ln(1) = \frac{1}{b} \arctan(0) + C \implies C = 0$

Since $f(0) = 1 > 0$, $f(x)$ remains strictly positive for all $x \in \mathbb{R}$. Thus, the solution to the differential equation is:

$f(x) = \exp\left(\frac{1}{b} \arctan\left(\frac{x}{b}\right)\right)$

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Analysis of Options:

1. Monotonicity of $f(x)$ (Options A and B): The derivative of $f(x)$ is given as:

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

   Since $f(x) > 0$ and $b^2 + x^2 > 0$ for all $x \in \mathbb{R}$ (regardless of whether $b > 0$ or $b < 0$), we have:

   $f'(x) > 0 \quad \text{for all } x \in \mathbb{R}$

   This implies that $f(x)$ is a strictly increasing function on $\mathbb{R}$ for any non-zero real constant $b$.

   • Option (A) is TRUE.
   • Option (B) is FALSE.

2. Symmetry and Product Identities (Options C and D): Evaluating $f(-x)$:

   $f(-x) = \exp\left(\frac{1}{b} \arctan\left(-\frac{x}{b}\right)\right) = \exp\left(-\frac{1}{b} \arctan\left(\frac{x}{b}\right)\right) = \frac{1}{f(x)}$

   Therefore, for all $x \in \mathbb{R}$:

   $f(x) f(-x) = f(x) \cdot \frac{1}{f(x)} = 1$

   • Option (C) is TRUE.

   For Option (D), $f(x) - f(-x) = 0 \implies f(x) = f(-x)$, which holds only when $f(x) = 1$ (i.e., at $x = 0$). Thus, it is not identically zero for all $x \in \mathbb{R}$.

   • Option (D) is FALSE.

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Conclusion:

The correct statements are (A) and (C).