question_answer

If the area (in sq. units) of the region \[\{(x,y):{{y}^{2}}\le 4x,x+y\le 1,x\ge 0,y\ge O\}\]\[a\sqrt{2}+b,\]is then \[ab\]is equal to : [JEE Main Held on 12-4-2019 Morning]

A) \[\frac{8}{3}\]                        

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

C) \[6\]     

D) \[-\frac{2}{3}\]

Correct Answer: C

Solution :

\[\{(x,y):{{y}^{2}}\le 4x,x+y\le 1,x\ge 0,y\ge 0\}\]                     \[A\int\limits_{0}^{3-2\sqrt{2}}{2\sqrt{x}dx}+\frac{1}{2}\left( 1-\left( 3-2\sqrt{2} \right) \right)\left( 1-(3-2\sqrt{2}) \right)\]           \[=\frac{2[{{x}^{3/2}}]_{0}^{3-2\sqrt{2}}}{3/2}+\frac{1}{2}\left( 2\sqrt{2}-2 \right)\left( 2\sqrt{2}-2 \right)\]           \[=\frac{8\sqrt{2}}{3}+\left( -\frac{10}{3} \right)\]           \[a=\frac{8}{3},b=-\frac{10}{3}\]                  \[a-b=6\]