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1

### JEE Main 2021 (Online) 27th August Evening Shift

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

## Explanation

$$(2x - 10{y^3})dy + ydx = 0$$

$$\Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}$$

$$I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}$$

Solution of D.E. is

$$\therefore$$ $$x\,.\,y = \int {(10{y^2}){y^2}.\,dy}$$

$$x{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C$$

It passes through (0, 1) $$\to$$ 0 = 2 + C $$\Rightarrow$$ C = $$-$$2

$$\therefore$$ Curve is $$x{y^2} = 2{y^5} - 2$$

Now, it passes through (2, $$\beta$$)

$$2{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0$$

$$\therefore$$ $$\beta$$ is root of an equation $${y^5} - {y^2} - 1 = 0$$
2

### JEE Main 2021 (Online) 27th August Evening Shift

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
$$11{{{d^2}y} \over {d{x^2}}} = 10$$

## Explanation

Length of latus rectum

$$= {{|3(2) + 4( - 3) - 5|} \over 5} = {{11} \over 5}$$

$${(x - h)^2} = {{11} \over 5}(y - k)$$

differentiate w.r.t. 'x' :-

$$2(x - h) = {{11} \over 5}{{dy} \over {dx}}$$

again differentiate

$$2 = {{11} \over 5}{{{d^2}y} \over {d{x^2}}}$$

$${{11{d^2}y} \over {d{x^2}}} = 10$$
3

### JEE Main 2021 (Online) 27th August Morning Shift

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
$${x^2} + 2xf(x) + 4 = 0$$

## Explanation

$$y + {{xdy} \over {dx}} = {x^2}$$ (given)

$$\Rightarrow {{dy} \over {dx}} + {y \over x} = x$$

If $${e^{\int {{1 \over x}dx} }} = x$$

Solution of DE

$$\Rightarrow y\,.\,x = \int {x\,.\,x\,dx}$$

$$\Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$

Passes through ($$-$$2, 2), so

$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4

$$\therefore$$ 3xy = x3 $$-$$ 4

i.e. 3x . f(x) = x3 $$-$$ 4
4

### JEE Main 2021 (Online) 27th August Morning Shift

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
$$7{e^{{\pi ^2}}} + 5$$

## Explanation

$${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$

IF = $${e^{ - {x^2}}}$$

So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx}$$

$$\Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$

$$\Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$

Given at x = 0, y = 7

$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2

So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$

Now, at x = $$\pi$$,

y = 5 + 2$${e^{{x^2}}}$$

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