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1
JEE Main 2020 (Online) 5th September Evening Slot
MCQ (Single Correct Answer)
+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) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
The value of $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ is:
A
$$\pi $$
B
$${{3\pi \over 2}}$$
C
$${{\pi \over 2}}$$
D
$${{\pi \over 4}}$$
3
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
The integral
$$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$
is equal to:
A
$$ - {1 \over {9}}$$
B
$$ - {1 \over {18}}$$
C
$$ {7 \over {18}}$$
D
$${9 \over 2}$$
4
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to :
A
$$ - {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}$$
B
$$ {\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
C
$$ - {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}$$
D
$${\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
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