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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
If $\int_{\log 2}^x {\frac{{du}}{{{{({e^u} - 1)}^{1/2}}}}} = \frac{\pi }{6}$, then ${e^x} = $
  • A
    $1$
  • B
    $2$
  • ✓
    $4$
  • D
    $-1$
✓

Answer

Correct option: C.
$4$
c
(c) $\int_{\log 2}^x {\frac{{du}}{{{{({e^u} - 1)}^{1/2}}}}} = \frac{\pi }{6}$

==> $\int_1^{\sqrt {{e^x} - 1} } {\frac{{2t}}{{1 + {t^2}}}} \;dt = \frac{\pi }{6}$

as ${e^u} - 1 = {t^2}$

==> $2({\tan ^{ - 1}}t)_1^{\sqrt {{e^x} - 1} } = \frac{\pi }{6}$

==> ${\tan ^{ - 1}}\sqrt {{e^x} - 1} - \frac{\pi }{4} = \frac{\pi }{{12}}$

==> $\sqrt {{e^x} - 1} = \tan \frac{\pi }{3}$ 

==> $\sqrt {{e^x} - 1} = \sqrt 3 $

==> ${e^x} = 4$.

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