Let ∫_α^ _e 4 d x e^x-1 =(π)/(6) . Then e^α and e^ -α are the roots of the equation :

Correct answer: C. 2 x^2-5 x+2=0 aligned & ∫_α^ 4 dx e^ x -1 =(π)/(6) & Let e^ x -1=t^2 & e^ x dx=2 t dt & =∫ 2 dt t^2+1 & =2 ^ -1 t & =.2 ^ -1 ( e^ x -1 )|_α ^ _ e^4 & =2[ ^ -1 3 - ^ -1 e^α-1 ]=(π)/(6) & =(π)/(3)- ^ -1 e^α-1 =(π)/(12) & ^ -1 e^α-1 =(π)/(4) & e^α=2 e^ -α =(1)/(2) & x^2-(2+(1)/(2)) x+1=0 & 2 x^2-5 x+2=0 aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Definite Integration

Let ∫_α^ _e 4 d x e^x-1 =(π)/(6) . Then e^α and e^ -α are the roots...

Let

\int_\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}

. Then

\mathrm{e}^\alpha

and

\mathrm{e}^{-\alpha}

are the roots of the equation :

Options

A $x^2+2 x-8=0$
B $x^2-2 x-8=0$
C $2 x^2-5 x+2=0$
D $2 x^2-5 x-2=0$

Correct answer

C. $2 x^2-5 x+2=0$

Step-by-step solution

\begin{aligned} & \int_\alpha^{\log 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6} \\ & \text{ Let } \mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^2 \\ & \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{t} \mathrm{dt} \\ & =\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 _{\mathrm{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} \\ & \mathrm{x}^2-\left(2+\frac{1}{2}\right) \mathrm{x}+1=0 \\ & 2 \mathrm{x}^2-5 \mathrm{x}+2=0\end{aligned}

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