1
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
The area bounded by the curve 4y2 = x2(4 $$-$$ x)(x $$-$$ 2) is equal to :
A
$${\pi \over {16}}$$
B
$${\pi \over {8}}$$
C
$${3\pi \over {2}}$$
D
$${3\pi \over {8}}$$
2
JEE Main 2021 (Online) 26th February Evening Shift
+4
-1
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then,
A
$${A_1}:{A_2} = 1:\sqrt 2$$ and $${A_1} + {A_2} = 1$$
B
$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \sqrt 2$$
C
$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \sqrt 2$$
D
$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$
3
JEE Main 2021 (Online) 24th February Evening Shift
+4
-1
The area of the region : $$R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\}$$ is :
A
$$6\sqrt 3$$ square units
B
$$12\sqrt 3$$ square units
C
$$11\sqrt 3$$ square units
D
$$9\sqrt 3$$ square units
4
JEE Main 2021 (Online) 24th February Morning Shift
+4
-1
The area (in sq. units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is :
A
$$12\pi - 3\sqrt 3$$
B
$$24\pi + 3\sqrt 3$$
C
$$24\pi - 3\sqrt 3$$
D
$$12\pi + 3\sqrt 3$$
EXAM MAP
Medical
NEET