JEE Main 2020 (Online) 5th September Morning Slot | Indefinite Integrals Question 33 | Mathematics

If
$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$

where c is a constant of integration, then g(0) is equal to :

e²

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-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):

$$\sec x - {{x\tan x} \over {x\sin x + \cos x}} + C$$

$$\sec x + {{x\tan x} \over {x\sin x + \cos x}} + C$$

$$\tan x - {{x\sec x} \over {x\sin x + \cos x}} + C$$

$$\tan x + {{x\sec x} \over {x\sin x + \cos x}} + C$$

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-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 :

$$ - {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}$$

$$ {\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}$$

$$ - {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}$$

$${\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}$$

JEE Main 2020 (Online) 3rd September Evening Slot

MCQ (Single Correct Answer)

-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 :

(x + 1, -$${\sqrt x }$$)

(x + 1, $${\sqrt x }$$)

(x - 1, -$${\sqrt x }$$)

(x - 1, $${\sqrt x }$$)

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