1
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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 $${A_1} + {A_2} = 1$$
2
JEE Main 2021 (Online) 24th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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
3
JEE Main 2021 (Online) 24th February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$12\pi + 3\sqrt 3 $$
4
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The area (in sq. units) of the region enclosed
by the curves y = x2 – 1 and y = 1 – x2 is equal to :
A
$${8 \over 3}$$
B
$${4 \over 3}$$
C
$${7 \over 2}$$
D
$${{16} \over 3}$$
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