JEE Main 2024 — 8 April, Shift 2 — Question 15

$\quad \int_{\alpha}^{\log _{e} 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6}$

Let $\mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^{2}$

$e^{x} d x=2 t d t$

$=\int \frac{2 \mathrm{dt}}{\mathrm{t}^{2}+1}$

$=2 \tan ^{-1} \mathrm{t}$

$=\left.2 \tan ^{-1}\left(\sqrt{\mathrm{e}^{\mathrm{x}}-1}\right)\right|_{\alpha} ^{\log _{e}^{4}}$

$=2\left[\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{\mathrm{e}^{\alpha}-1}\right]=\frac{\pi}{6}$

$=\frac{\pi}{3}-\tan ^{-1} \sqrt{\mathrm{e}^{\alpha}-1}=\frac{\pi}{12}$

$\Rightarrow \tan ^{-1} \sqrt{\mathrm{e}^{\alpha}-1}=\frac{\pi}{4}$

$\mathrm{e}^{\alpha}=2 ,\quad \mathrm{e}^{-\alpha}=\frac{1}{2}$

$x^{2}-\left(2+\frac{1}{2}\right) x+1=0$

$2 x^{2}-5 x+2=0$