Let log e 4 d x e x - 1 = 6 . Then e and e - are the roots of the equation: [2024]

Question

$Let \int_{α}^{\log_{e} 4} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6} . Then e^{α} and e^{- α} are the roots of the equation:$ [2024]

Smart Achievers · Accepted Answer

(3)

We have, $\int_{α}^{\log_{e} 4} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6}$ ...(i)

Put $\sqrt{e^{x} - 1} = t \Rightarrow e^{x} = 1 + t^{2}$

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

When $x = α,$ $t = \sqrt{e^{α} - 1},$ when $x = \log_{e} 4$

$t = \sqrt{e^{\log 4} - 1} = \sqrt{3}$

$∴$ From (i), $\int_{\sqrt{e^{α} - 1}}^{\sqrt{3}} \frac{2 t d t}{t (1 + t^{2})} = \frac{π}{6} \Rightarrow 2 {[\tan^{- 1} t]}_{\sqrt{e^{α} - 1}}^{\sqrt{3}} = \frac{π}{6}$

$\Rightarrow 2 (\tan^{- 1} \sqrt{3} - \tan^{- 1} \sqrt{e^{α} - 1}) = \frac{π}{6}$

$\Rightarrow \frac{π}{3} - \tan^{- 1} (\sqrt{e^{α} - 1}) = \frac{π}{12} \Rightarrow \tan^{- 1} (\sqrt{e^{α} - 1}) = \frac{π}{4}$

$\Rightarrow e^{α} - 1 = 1$ $(∵ \tan \frac{π}{4} = 1)$

$\Rightarrow e^{α} = 2 \Rightarrow e^{- α} = \frac{1}{2}$

$∴ Quadratic equation whose roots are e^{α} and e^{- α} is$

$x^{2} - (e^{α} + e^{- α}) x + e^{α} \times e^{- α} = 0$

$\Rightarrow x^{2} - (2 + \frac{1}{2}) x + 1 = 0 \Rightarrow 2 x^{2} - 5 x + 2 = 0$

Solution

Q.
$Let \int_{α}^{\log_{e} 4} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6} . Then e^{α} and e^{- α} are the roots of the equation:$ [2024]

1 $x^{2} + 2 x - 8 = 0$

2 $x^{2} - 2 x - 8 = 0$

3 $2 x^{2} - 5 x + 2 = 0$

4 $2 x^{2} - 5 x - 2 = 0$

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Solution

Q. Let ∫αloge4dxex-1=π6. Then eα and e-α are the roots of the equation: [2024]

1 x2+2x-8=0

2 x2-2x-8=0

3 2x2-5x+2=0