question_answer

The area (in sq. units) of the region \[A=\{(x,y):{{x}^{2}}\le y\le x+2\}\] is                         [JEE Main 9-4-2019 Morning]

A) \[\frac{10}{3}\]                                  

B) \[\frac{9}{2}\]

C) \[\frac{31}{6}\]                                  

D) \[\frac{13}{6}\]

Correct Answer: B

Solution :

\[{{x}^{2}}\le y\le x+2\]           \[{{x}^{2}}=y;y=x+2\]           \[{{x}^{2}}=x+2\]           \[{{x}^{2}}-x-2=0\]           \[(x-2)(x-1)=0\]           \[x=2,-1\]           Area,\[=\int\limits_{-1}^{2}{(x+2)}-{{x}^{2}}dx=\frac{9}{2}\]