1
JEE Main 2021 (Online) 27th August Morning Shift
+4
-1
Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
A
$${x^2} + 2xf(x) - 12 = 0$$
B
$${x^3} + xf(x) + 12 = 0$$
C
$${x^3} - 3xf(x) - 4 = 0$$
D
$${x^2} + 2xf(x) + 4 = 0$$
2
JEE Main 2021 (Online) 26th August Evening Shift
+4
-1
Let y(x) be the solution of the differential equation

2x2 dy + (ey $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
A
0
B
2
C
loge 2
D
loge (2e)
3
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
Let y = y(x) be a solution curve of the differential equation $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$, $$x \in \left( {0,{\pi \over 2}} \right)$$. If $$\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$$, then the value of $$y\left( {{\pi \over 4}} \right)$$ is :
A
$$- {\pi \over 4}$$
B
$${\pi \over 4} - 1$$
C
$${\pi \over 4} + 1$$
D
$${\pi \over 4}$$
4
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Let y = y(x) be the solution of the differential

equation (x $$-$$ x3)dy = (y + yx2 $$-$$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
A
4
B
12
C
8
D
16
EXAM MAP
Medical
NEET