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1
JEE Main 2021 (Online) 16th March Evening Shift
+4
-1
Let C1 be the curve obtained by the solution of differential equation
$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C2 be the
solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :
A
$${\pi \over 4}$$ + 1
B
$$\pi$$ + 1
C
$$\pi$$ $$-$$ 1
D
$${\pi \over 2}$$ $$-$$ 1
2
JEE Main 2021 (Online) 16th March Morning Shift
+4
-1
If y = y(x) is the solution of the differential equation,

$${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of the function y(x) over R is equal to:
A
$${1 \over 8}$$
B
8
C
$$-$$$${15 \over 4}$$
D
$${1 \over 2}$$
3
JEE Main 2021 (Online) 26th February Evening Shift
+4
-1
Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}}$$ be a differentiable function for all x$$\in$$R. Then f(x) equals :
A
$${e^{({e^{x - 1}})}}$$
B
$$2{e^{{e^x}}} - 1$$
C
$$2{e^{{e^x} - 1}} - 1$$
D
$${e^{{e^x}}} - 1$$
4
JEE Main 2021 (Online) 26th February Morning Shift
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after $${k \over {{{\log }_e}\left( {{6 \over 5}} \right)}}$$ hours, then $${\left( {{k \over {{{\log }_e}2}}} \right)^2}$$ is equal to :