1
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
$${\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
2
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to
(where C is a constant of integration):
A
$$\sec x - {{x\tan x} \over {x\sin x + \cos x}} + C$$
B
$$\sec x + {{x\tan x} \over {x\sin x + \cos x}} + C$$
C
$$\tan x - {{x\sec x} \over {x\sin x + \cos x}} + C$$
D
$$\tan x + {{x\sec x} \over {x\sin x + \cos x}} + C$$
3
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
If $$\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$$ = A(x)$${\tan ^{ - 1}}\left( {\sqrt x } \right)$$ + B(x) + C,
where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :
A
(x + 1, -$${\sqrt x }$$)
B
(x + 1, $${\sqrt x }$$)
C
(x - 1, -$${\sqrt x }$$)
D
(x - 1, $${\sqrt x }$$)
4
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
If $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C$$

where C is a constant of integration, then the ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to :
A
(–1, 1 – tan$$\theta $$)
B
(1, 1 + tan$$\theta $$)
C
(–1, 1 + tan$$\theta $$)
D
(1, 1 – tan$$\theta $$)
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