1
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
Out of Syllabus
A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, $$-$$3) from the line 3x + 4y = 5, is given by :
A
$$10{{{d^2}y} \over {d{x^2}}} = 11$$
B
$$11{{{d^2}x} \over {d{y^2}}} = 10$$
C
$$10{{{d^2}x} \over {d{y^2}}} = 11$$
D
$$11{{{d^2}y} \over {d{x^2}}} = 10$$
2
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
If the solution curve of the differential equation (2x $$-$$ 10y3)dy + ydx = 0, passes through the points (0, 1) and (2, $$\beta$$), then $$\beta$$ is a root of the equation :
A
y5 $$-$$ 2y $$-$$ 2 = 0
B
2y5 $$-$$ 2y $$-$$ 1 = 0
C
2y5 $$-$$ y2 $$-$$ 2 = 0
D
y5 $$-$$ y2 $$-$$ 1 = 0
3
JEE Main 2021 (Online) 27th August Morning Shift
+4
-1
Let y = y(x) be the solution of the differential equation

$${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that y(0) = 7. Then y($$\pi$$) is equal to :
A
$$2{e^{{\pi ^2}}} + 5$$
B
$${e^{{\pi ^2}}} + 5$$
C
$$3{e^{{\pi ^2}}} + 5$$
D
$$7{e^{{\pi ^2}}} + 5$$
4
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
$${x^2} + 2xf(x) + 4 = 0$$
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