JEE Advanced 2025 · Paper 2, Mathematics question 16

Substitute $x\to\frac1x$. The limits swap, $\tan^{-1}\frac1x=\frac\pi2-\tan^{-1}x$, and $\frac{dx/x^2}{2/x^2-3/x+2}=\frac{dx}{2x^2-3x+2}$. So
$$\alpha=\int_{1/2}^{2}\frac{\frac\pi2-\tan^{-1}x}{2x^2-3x+2}dx\implies2\alpha=\frac\pi2I,\qquad I=\int_{1/2}^{2}\frac{dx}{2x^2-3x+2}.$$

Evaluate I. Complete the square: $2x^2-3x+2=2\left[(x-\frac34)^2+\frac7{16}\right]$. Then
$$I=\frac{2}{\sqrt7}\left[\tan^{-1}\frac{5}{\sqrt7}+\tan^{-1}\frac{1}{\sqrt7}\right]=\frac{2}{\sqrt7}\tan^{-1}\left(3\sqrt7\right),$$
because $\frac{6/\sqrt7}{1-5/7}=3\sqrt7$.

Answer. $\alpha=\frac{\pi}{2\sqrt7}\tan^{-1}(3\sqrt7)$, so $\frac{2\alpha\sqrt7}{\pi}=\tan^{-1}(3\sqrt7)$ and
$$\sqrt7\tan\left(\frac{2\alpha\sqrt7}{\pi}\right)=\sqrt7\cdot3\sqrt7=21.$$