1
JEE Main 2020 (Online) 3rd September Evening Slot
+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, $${\sqrt x }$$)
2
JEE Main 2020 (Online) 9th January Evening Slot
+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$$\theta$$)
3
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
If ƒ'(x) = tan–1(secx + tanx), $$- {\pi \over 2} < x < {\pi \over 2}$$,
and ƒ(0) = 0, then ƒ(1) is equal to :
A
$${1 \over 4}$$
B
$${{\pi - 1} \over 4}$$
C
$${{\pi + 1} \over 4}$$
D
$${{\pi + 2} \over 4}$$
4
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
The integral $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}}$$ is equal to :
(where C is a constant of integration)
A
$${1 \over 2}{\left( {{{x - 3} \over {x + 4}}} \right)^{{3 \over 7}}} + C$$
B
$${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}} + C$$
C
$$- {1 \over {13}}{\left( {{{x - 3} \over {x + 4}}} \right)^{{{13} \over 7}}} + C$$
D
-$${\left( {{{x - 3} \over {x + 4}}} \right)^{-{1 \over 7}}} + C$$
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