JEE Main On 16 April 2018 Question 22
Question: If the area of the region bounded by the curves, $ y=x^{2},y=\frac{1}{x} $ and the lines $ y=0 $ and $ x=t(t>1) $ is sq. unit, then t is equal to? [JEE Main 16-4-2018]
Options:
A) $ \frac{4}{3} $
B) $ {e^{2/3}} $
C) $ \frac{3}{2} $
D) $ {e^{3/2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Are bounded by the curves is the region ABCD.
Therefore, area $ =\int_0^{1}{x^{2}dx+\int_1^{t}{\frac{1}{x}dx}} $ $ =[ \frac{x^{3}}{3} ]_0^{1}+[\ln (x)]_1^{t} $
$ =\frac{1}{3}+\ln (t) $ It is given that area enclosed is 1
$ \Rightarrow $ $ \frac{1}{3}+\ln (t)=1 $
$ \Rightarrow $ $ \ln (t)=\frac{2}{3} $
$ \Rightarrow $ $ t={e^{\frac{2}{3}}} $
