Let 𝑏 be a nonzero real number. Suppose $$f:R \rightarrow R$$ is a differentiable function such that $$f(0) = 1$$. If the derivative $$f′$$ of $$f$$ 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?

We are given the differential equation
$$f'(x)=\frac{f(x)}{b^{2}+x^{2}},\qquad f(0)=1,\qquad b\neq 0$$

Step 1: Solve the differential equation.
Rewrite as a separable equation:$$\frac{f'(x)}{f(x)}=\frac1{b^{2}+x^{2}}.$$
Integrate both sides:$$\int\frac{f'(x)}{f(x)}\,dx=\int\frac{dx}{b^{2}+x^{2}}.$$
The left integral is $$\ln|f(x)|,$$ while the right integral, using $$\int\frac{dx}{a^{2}+x^{2}}=\frac1a\arctan\!\left(\frac{x}{a}\right),$$ is $$\frac1b\arctan\!\left(\frac{x}{b}\right)+C.$$
Thus$$\ln|f(x)|=\frac1b\arctan\!\left(\frac{x}{b}\right)+C.$$

Exponentiating yields the general solution
$$f(x)=K\exp\!\left[\frac1b\arctan\!\left(\frac{x}{b}\right)\right],$$
where $$K=e^{C}.$$(Since an exponential is always positive, $$|\,\cdot\,|$$ can be dropped.)

Step 2: Use the initial condition.
At $$x=0,\;\arctan(0)=0,$$ so $$f(0)=K\cdot e^{0}=K.$$
Given $$f(0)=1,$$ we get $$K=1.$$
Therefore

$$f(x)=\exp\!\left[\frac1b\arctan\!\left(\frac{x}{b}\right)\right].$$

Step 3: Test each statement.

Option A: If $$b\gt 0,$$ is $$f$$ increasing?
Compute the derivative directly from the differential equation:$$f'(x)=\frac{f(x)}{b^{2}+x^{2}}.$$
Here $$f(x)\gt 0$$ (exponential term) and $$b^{2}+x^{2}\gt 0,$$ so $$f'(x)\gt 0$$ for every real $$x.$$ Hence $$f$$ is indeed increasing. Option A is TRUE.

Option B: If $$b\lt 0,$$ is $$f$$ decreasing?
Even when $$b\lt 0,$$ the denominator $$b^{2}+x^{2}$$ is still positive, and the numerator $$f(x)$$ is positive, so $$f'(x)\gt 0$$ everywhere. Thus $$f$$ is still increasing, not decreasing. Option B is FALSE.

Option C: Verify $$f(x)f(-x)=1.$$\br/> Using the explicit form,

$$f(-x)=\exp\!\left[\frac1b\arctan\!\left(\frac{-x}{b}\right)\right]$$
Since $$\arctan(-u)=-\arctan(u),$$
$$f(-x)=\exp\!\left[-\frac1b\arctan\!\left(\frac{x}{b}\right)\right].$$
Therefore
$$f(x)f(-x)=\exp\!\left[\frac1b\arctan\!\left(\frac{x}{b}\right)\right]\; \exp\!\left[-\frac1b\arctan\!\left(\frac{x}{b}\right)\right]=e^{0}=1.$$
Option C is TRUE.

Option D: Verify $$f(x)-f(-x)=0.$$\br/> From the earlier expressions,
$$f(x)-f(-x)=\exp\!\left[\frac1b\arctan\!\left(\frac{x}{b}\right)\right] -\exp\!\left[-\frac1b\arctan\!\left(\frac{x}{b}\right)\right].$$
For $$x\neq 0$$ the two exponentials are unequal, so their difference is not zero. Hence $$f$$ is not an even function. Option D is FALSE.

Step 4: Collect the correct options.
The statements that hold are Option A and Option C.

Final Answer: Option A (f is increasing when $$b\gt 0$$) and Option C ($$f(x)f(-x)=1$$ for all $$x$$).