question_answer

Given:     and  \[g(x)={{\left( x-\frac{1}{2} \right)}^{2}},\,\,\,x\in R.\] Then the area (in sq. units) of the region bounded by the curves, \[y=f(x)\] and \[y=g(x)\] between the lines, \[2x=1\] and \[2x=\sqrt{3},\] is: [JEE MAIN Held on 09-01-2020 Evening]

A) \[\frac{\sqrt{3}}{4}-\frac{1}{3}\]

B) \[\frac{1}{3}+\frac{\sqrt{3}}{4}\]

C) \[\frac{1}{2}-\frac{\sqrt{3}}{4}\]

D) \[\frac{1}{2}+\frac{\sqrt{3}}{4}\]

Correct Answer: A

Solution :

Required Area = Area of the Region CMBC = Area of trapezium CLMBC - Area of the region CLMC \[=\int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}{\left[ \left( 1-x \right)-{{\left( x-\frac{1}{2} \right)}^{2}} \right]}\,dx\] \[=\int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}{\left( \frac{3}{4}-{{x}^{2}} \right)}\,dx\] \[=\left[ \frac{3}{4}x-\frac{{{x}^{3}}}{3} \right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\] \[=\frac{\sqrt{3}}{4}-\frac{1}{3}\]