JEE Advanced 2020 — Paper 2 — Question 40

Given $f'(x) = \frac{f(x)}{b^2 + x^2}$. Since $f(0)=1$, we can write $\frac{f'(x)}{f(x)} = \frac{1}{b^2 + x^2}$. Integrate both sides: $\int \frac{f'(x)}{f(x)} \,dx = \int \frac{dx}{b^2 + x^2}$, giving $\ln|f(x)| = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) + C$. Using $f(0)=1$, we get $0 = \frac{1}{b} \tan^{-1}(0) + C = 0 + C$, so $C=0$. Hence $\ln|f(x)| = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)$.

Since $f(0)=1>0$, $f(x) = e^{\frac{1}{b} \tan^{-1}(x/b)}$. Differentiate: $f'(x) = f(x) \cdot \frac{d}{dx}\left[\frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right)\right] = f(x) \cdot \frac{1}{b} \cdot \frac{1}{1+(x/b)^2} \cdot \frac{1}{b} = \frac{f(x)}{b^2 + x^2} > 0$ for all $x$ (since $f(x)>0$ and denominator $>0$).

Thus $f$ is increasing for any nonzero $b$, so A is true and B is false. Compute $f(-x) = e^{\frac{1}{b} \tan^{-1}(-x/b)} = e^{-\frac{1}{b} \tan^{-1}(x/b)} = \frac{1}{f(x)}$.

Hence $f(x) f(-x) = 1$, so C is true. For D, $f(x) - f(-x) = f(x) - \frac{1}{f(x)}$ is not identically zero (e.g., at $x=1$, $f(1) eq \pm 1$), so D is false. Thus the true statements are A and C.