1
JEE Main 2021 (Online) 20th July Evening Shift
+4
-1
Let $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$$, where $$f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R$$. Then which one of the following is correct?
A
g(1) = g(0)
B
$$\sqrt 2 g(1) = g(0)$$
C
$$g(1) = \sqrt 2 g(0)$$
D
g(1) + g(0) = 0
2
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
Let a be a positive real number such that $$\int_0^a {{e^{x - [x]}}} dx = 10e - 9$$ where [ x ] is the greatest integer less than or equal to x. Then a is equal to:
A
$$10 - {\log _e}(1 + e)$$
B
$$10 + {\log _e}2$$
C
$$10 + {\log _e}3$$
D
$$10 + {\log _e}(1 + e)$$
3
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
The value of the integral $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx}$$ is equal to:
A
$${1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}$$
B
$$2{\log _e}2 + {\pi \over 4} - 1$$
C
$${\log _e}2 + {\pi \over 2} - 1$$
D
$$2{\log _e}2 + {\pi \over 2} - {1 \over 2}$$
4
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
Let g(x) = $$\int_0^x {f(t)dt}$$, where f is continuous function in [ 0, 3 ] such that $${1 \over 3}$$ $$\le$$ f(t) $$\le$$ 1 for all t$$\in$$ [0, 1] and 0 $$\le$$ f(t) $$\le$$ $${1 \over 2}$$ for all t$$\in$$ (1, 3]. The largest possible interval in which g(3) lies is :
A
$$\left[ { - 1, - {1 \over 2}} \right]$$
B
$$\left[ { - {3 \over 2}, - 1} \right]$$
C
[1, 3]
D
$$\left[ {{1 \over 3},2} \right]$$
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