1
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
$$3\left( {\pi - 2} \right)$$
2
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation,
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi$$) = a and $${{dy} \over {dx}}$$ at x = $$\pi$$ is b, then the ordered pair (a, b) is equal to :
A
(2, 1)
B
$$\left( {2,{3 \over 2}} \right)$$
C
(1, -1)
D
(1, 1)
3
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
Given : $$f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.$$

and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3$$, is :
A
$${1 \over 2} + {{\sqrt 3 } \over 4}$$
B
$${1 \over 2} - {{\sqrt 3 } \over 4}$$
C
$${1 \over 3} + {{\sqrt 3 } \over 4}$$
D
$${{\sqrt 3 } \over 4} - {1 \over 3}$$
4
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
The value of
$$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ is equal to :
A
4$$\pi$$
B
2$$\pi$$
C
$$\pi$$2
D
2$$\pi$$2
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