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1
JEE Main 2021 (Online) 20th July Evening Shift
+4
-1
Let y = y(x) satisfies the equation $${{dy} \over {dx}} - |A| = 0$$, for all x > 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $$y(\pi ) = \pi + 2$$, then the value of $$y\left( {{\pi \over 2}} \right)$$ is :
A
$${\pi \over 2} + {4 \over \pi }$$
B
$${\pi \over 2} - {1 \over \pi }$$
C
$${{3\pi } \over 2} - {1 \over \pi }$$
D
$${\pi \over 2} - {4 \over \pi }$$
2
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
3
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
$$10 + {\log _e}(1 + e)$$
4
JEE Main 2021 (Online) 20th July Morning Shift
The value of the integral $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx}$$ is equal to:
$${1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}$$
$$2{\log _e}2 + {\pi \over 4} - 1$$
$${\log _e}2 + {\pi \over 2} - 1$$
$$2{\log _e}2 + {\pi \over 2} - {1 \over 2}$$