Definite Integration Question 415
Question: If $ \int _{\log 2}^{x}{\frac{du}{{{(e^{u}-1)}^{1/2}}}}=\frac{\pi }{6} $ , then $ e^{x}= $
[Orissa JEE 2005]
Options:
A) 1
B) 2
C) 4
D) -1
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int _{\log 2}^{x}{\frac{du}{{{(e^{u}-1)}^{1/2}}}}=\frac{\pi }{6} $
therefore $ \int_1^{\sqrt{e^{x}-1}}{\frac{2t}{1+t^{2}}}\ dt=\frac{\pi }{6} $ as $ e^{u}-1=t^{2} $
therefore $ 2({{\tan }^{-1}}t)_1^{\sqrt{e^{x}-1}}=\frac{\pi }{6} $
therefore $ {{\tan }^{-1}}\sqrt{e^{x}-1}-\frac{\pi }{4}=\frac{\pi }{12} $
therefore $ \sqrt{e^{x}-1}=\tan \frac{\pi }{3} $
therefore $ \sqrt{e^{x}-1}=\sqrt{3} $
therefore $ e^{x}=4 $ .
BETA
