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1
JEE Main 2021 (Online) 16th March Evening Shift
+4
-1
If y = y(x) is the solution of the differential equation

$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to :
A
$${1 \over 2}$$loge 2
B
$$\left( {{1 \over {2\sqrt 2 }}} \right)$$ loge 2
C
loge 2
D
$${1 \over 4}$$ loge 2
2
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
3
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
$${1 \over 2}$$
4
JEE Main 2021 (Online) 26th February Evening Shift
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 :
$${e^{({e^{x - 1}})}}$$
$$2{e^{{e^x}}} - 1$$
$$2{e^{{e^x} - 1}} - 1$$
$${e^{{e^x}}} - 1$$