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Mathematics Question on limits and derivatives

Let ∫log⁡ea4dxex−1=π6.\int_{\log_e a}^{4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}. Then eαe^\alpha and e−αe^{-\alpha} are the roots of the equation:

A

2x2−5x+2=02x^2 - 5x + 2 = 0

B

x2−2x−8=0x^2 - 2x - 8 = 0

C

2x2−5x−2=02x^2 - 5x - 2 = 0

D

x2+2x−8=0x^2 + 2x - 8 = 0

Answer

2x2−5x+2=02x^2 - 5x + 2 = 0

Explanation

Solution

Given: ∫log⁡eαlog⁡e4dxex−1=π6.\int_{\log_e \alpha}^{\log_e 4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}.

Let: ex−1=t2  ⟹  exdx=2tdtanddx=2tdtt2+1.e^x - 1 = t^2 \implies e^x dx = 2t dt \quad \text{and} \quad dx = \frac{2t dt}{t^2 + 1}.

Substituting into the integral: ∫dxex−1=∫2tdt(t2+1)⋅t=∫2dtt2+1=2tan⁡−1t.\int \frac{dx}{\sqrt{e^x - 1}} = \int \frac{2t dt}{(t^2 + 1) \cdot t} = \int \frac{2 dt}{t^2 + 1} = 2 \tan^{-1} t.

Reverting the substitution: 2tan⁡−1(ex−1)∣log⁡eαlog⁡e4.2 \tan^{-1} (\sqrt{e^x - 1}) \Big|_{\log_e \alpha}^{\log_e 4}.

Evaluating at the limits: 2[tan⁡−1(elog⁡e4−1)−tan⁡−1(eα−1)]=π6.2 \left[ \tan^{-1} \left( \sqrt{e^{\log_e 4} - 1} \right) - \tan^{-1} \left( \sqrt{e^\alpha - 1} \right) \right] = \frac{\pi}{6}. Simplifying: 2[tan⁡−1(3)−tan⁡−1(eα−1)]=π6.2 \left[ \tan^{-1} (\sqrt{3}) - \tan^{-1} (\sqrt{e^\alpha - 1}) \right] = \frac{\pi}{6}.

Dividing by 2: tan⁡−1(3)−tan⁡−1(eα−1)=π12.\tan^{-1} (\sqrt{3}) - \tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{12}.

Since tan⁡−1(3)=π3\tan^{-1} (\sqrt{3}) = \frac{\pi}{3}, we have: π3−tan⁡−1(eα−1)=π12.\frac{\pi}{3} - \tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{12}.

Rearranging: tan⁡−1(eα−1)=π4.\tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{4}.

Thus: eα−1=1  ⟹  eα−1=1  ⟹  eα=2.\sqrt{e^\alpha - 1} = 1 \implies e^\alpha - 1 = 1 \implies e^\alpha = 2.

Therefore: e−α=12.e^{-\alpha} = \frac{1}{2}.

The roots eα=2e^\alpha = 2 and e−α=12e^{-\alpha} = \frac{1}{2} satisfy the equation: x2−(2+12)x+1=0  ⟹  2x2−5x+2=0.x^2 - \left( 2 + \frac{1}{2} \right) x + 1 = 0 \implies 2x^2 - 5x + 2 = 0.

Therefore: 2x2−5x+2=0.2x^2 - 5x + 2 = 0.