1
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
The area (in sq. units) of the region

A = {(x, y) : (x – 1)[x] $$\le$$ y $$\le$$ 2$$\sqrt x$$, 0 $$\le$$ x $$\le$$ 2}, where [t]

denotes the greatest integer function, is :
A
$${8 \over 3}\sqrt 2 - 1$$
B
$${4 \over 3}\sqrt 2 + 1$$
C
$${8 \over 3}\sqrt 2 - {1 \over 2}$$
D
$${4 \over 3}\sqrt 2 - {1 \over 2}$$
2
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
The area (in sq. units) of the region

{ (x, y) : 0 $$\le$$ y $$\le$$ x2 + 1, 0 $$\le$$ y $$\le$$ x + 1,

$${1 \over 2}$$ $$\le$$ x $$\le$$ 2 } is :
A
$${{79} \over {16}}$$
B
$${{79} \over {24}}$$
C
$${{23} \over {6}}$$
D
$${{23} \over {16}}$$
3
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
Consider a region R = {(x, y) $$\in$$ R : x2 $$\le$$ y $$\le$$ 2x}. if a line y = $$\alpha$$ divides the area of region R into two equal parts, then which of the following is true?
A
3$$\alpha$$2 - 8$$\alpha$$ + 8 = 0
B
$$\alpha$$3 - 6$$\alpha$$3/2 - 16 = 0
C
3$$\alpha$$2 - 8$$\alpha$$3/2 + 8 = 0
D
$$\alpha$$3 - 6$$\alpha$$2 + 16 = 0
4
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
Area (in sq. units) of the region outside

$${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$$ and inside the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :
A
$$6\left( {4 - \pi } \right)$$
B
$$3\left( {4 - \pi } \right)$$
C
$$6\left( {\pi - 2} \right)$$
D
$$3\left( {\pi - 2} \right)$$
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