question_answer

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]

A)  \[\frac{4}{3}\]                                   

B) \[{{e}^{2/3}}\]        

C)  \[\frac{3}{2}\]                                   

D) \[{{e}^{3/2}}\]

Correct Answer: B

Solution :

 Are bounded by the curves is the region ABCD.                 Therefore, area \[=\int_{0}^{1}{{{x}^{2}}dx+\int_{1}^{t}{\frac{1}{x}dx}}\] \[=\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{1}+[\ln (x)]_{1}^{t}\] \[=\frac{1}{3}+\ln (t)\] It is given that area enclosed is 1 \[\Rightarrow \]\[\frac{1}{3}+\ln (t)=1\] \[\Rightarrow \]\[\ln (t)=\frac{2}{3}\] \[\Rightarrow \]\[t={{e}^{\frac{2}{3}}}\]