JEE Mains 06-Sep-2020 Shift 1 Solved Paper

|x|+|y|≤1
2y2≥|x| For point of intersection
x+y=1⇒x=1−y
y2=

x

2

⇒2y2=x
2y2=1−y ⇒2y2+y−1=0
(2y−1)(y+1)=0
y=

1

2

or −1
Now Area of Δ‌OAB=

1

2

×1×1=

1

2

Area of Region R1=

1

2

×

1

2

×

1

2

=

1

8

Area of Region R2=

1

√2

‌

1 2

∫

0

√xdx=

1

6

Now area of shaded region in first quadrant = Area of Δ‌OAB−R1−R2
=

1

2

−(

1

6

)−(

1

8

)=

5

24

So required area =4(

5

24

)=

5

6