JEE Advanced 2025 Paper 2, Mathematics Section 3 Q8: Some Properties of Definite Integrals

Step 1: Symmetry substitution.

Put $x=1/t$, $dx=-dt/t^2$. The limits swap $1/2\leftrightarrow 2$. The denominator transforms as $2/t^2-3/t+2=(2t^2-3t+2)/t^2$, and ${\tan }^{-1}(1/t)={\cot }^{-1}t$ (for $t>0$). Therefore

$$\alpha ={\int }_{1/2}^2\frac{{\cot }^{-1}t}{2t^2-3t+2}dt.$$

Step 2: Add both forms.

Using ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$,

$$2\alpha =\frac{\pi }{2}{\int }_{1/2}^2\frac{dx}{2x^2-3x+2}.$$

Step 3: Complete the square.

$2x^2-3x+2=2\left[{\left(x-\frac{3}{4}\right)}^2+\frac{7}{16}\right]$, so

$$\alpha =\frac{\pi }{8}{\int }_{1/2}^2\frac{dx}{{\left(x-\frac{3}{4}\right)}^2+\frac{7}{16}}=\frac{\pi }{2\sqrt{7}}{\left[{\tan }^{-1}\frac{4x-3}{\sqrt{7}}\right]}_{1/2}^2.$$

Step 4: Evaluate.

$\alpha =\frac{\pi }{2\sqrt{7}}\left[{\tan }^{-1}\frac{5}{\sqrt{7}}-{\tan }^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right]=\frac{\pi }{2\sqrt{7}}{\tan }^{-1}\left(\frac{5/\sqrt{7}+1/\sqrt{7}}{1-5/7}\right)=\frac{\pi }{2\sqrt{7}}{\tan }^{-1}(3\sqrt{7})$.

Step 5: Final computation.

$\frac{2\alpha \sqrt{7}}{\pi }={\tan }^{-1}(3\sqrt{7})$, so $\sqrt{7}\tan \left(\frac{2\alpha \sqrt{7}}{\pi }\right)=\sqrt{7}\cdot 3\sqrt{7}=21$.