1
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
$${{\pi + 2} \over 4}$$
2
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
-$${\left( {{{x - 3} \over {x + 4}}} \right)^{-{1 \over 7}}} + C$$
3
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
If $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c$$

where c is a constant of integration, then $$\lambda f\left( {{\pi \over 3}} \right)$$ is equal to
A
$${9 \over 8}$$
B
2
C
-2
D
$$-{9 \over 8}$$
4
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Let $$a \in \left( {0,{\pi \over 2}} \right)$$ be fixed. If the integral

$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx$$ = A(x) cos 2$$\alpha$$ + B(x) sin 2$$\alpha$$ + C, where C is a

constant of integration, then the functions A(x) and B(x) are respectively :
A
$$x - \alpha$$ and $${\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|$$
B
$$x + \alpha$$ and $${\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|$$
C
$$x + \alpha$$ and $${\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|$$
D
$$x - \alpha$$ and $${\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|$$
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