JEE Advanced 2020 · Paper 2, Mathematics question 7

Solve. Separating variables with $f(0)=1$:
$$\ln f=\frac1b\tan^{-1}\frac xb\implies f(x)=e^{\frac1b\tan^{-1}(x/b)}>0.$$

(A)/(B) Since $f>0$, $f'=\dfrac f{b^2+x^2}>0$ for every $b\ne0$. So $f$ is increasing for either sign of $b$: (A) is true and (B) is false.

(C) $\tan^{-1}$ is odd, so $f(x)f(-x)=e^0=1$. True.

(D) $f(x)=f(-x)$ would need $f(x)^2=1$, i.e. $f\equiv1$. False.

(A), (C)