Let $\int_{\alpha}^{\log _{e} 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^{\alpha}$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :

(A) $2 \mathrm{x}^{2}-5 \mathrm{x}+2=0$

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Definite Integrals

Let

\int_{α}^{l o g_{e} 4} \frac{dx}{e ^{x} - 1} = \frac{π}{6}

. Then

e^{α}

and

e^{- α}

are the roots of the equation :

Answer: (A)

\int_{α}^{l n 4} \frac{d x}{e ^{x} - 1} = \frac{π}{6}

Substitute

e^{x} - 1 = t^{2}

e^{x} d x = 2 t d t ⟹ d x = \frac{2 t d t}{t ^{2} + 1}

, so the integral becomes

\int \frac{2 d t}{t ^{2} + 1} = 2 tan^{- 1} t

2 [tan^{- 1} e^{x} - 1]_{α}^{l n 4} = \frac{π}{6}

x = ln 4

e^{l n 4} - 1 = 3

tan^{- 1} 3 = π /3

2 (\frac{π}{3} - tan^{- 1} e^{α} - 1) = \frac{π}{6} ⟹ tan^{- 1} e^{α} - 1 = \frac{π}{4} ⟹ e^{α} - 1 = 1 ⟹ e^{α} = 2

e^{α} = 2

e^{- α} = 1/2

; sum

= 5/2

, product

= 1

. Quadratic with these roots:

x^{2} - \frac{5}{2} x + 1 = 0

, i.e.

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

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option