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