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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
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 :
  • ✓
     $2 \mathrm{x}^2-5 \mathrm{x}+2=0$
  • B
     $\mathrm{x}^2-2 \mathrm{x}-8=0$
  • C
     $2 x^2-5 x-2=0$
  • D
    $x^2+2 x-8=0$
✓

Answer

Correct option: A.
 $2 \mathrm{x}^2-5 \mathrm{x}+2=0$
a
$ \int_\alpha^{\log _e 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 _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$

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