JEE Main 2020 (Online) 8th January Evening Slot | Differential Equations Question 100 | Mathematics | JEE Main - ExamSIDE.com

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1

JEE Main 2020 (Online) 8th January Evening Slot

MCQ (Single Correct Answer)

+4

-1

The differential equation of the family of curves, x² = 4b(y + b), b $$ \in $$ R, is

A

x(y')² = x – 2yy'

B

x(y')² = 2yy' – x

C

xy" = y'

D

x(y')² = x + 2yy'

2

JEE Main 2020 (Online) 8th January Morning Slot

MCQ (Single Correct Answer)

+4

-1

Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to

A

$$ - {{\sqrt 3 } \over 2}$$

B

None of those

C

$${{1 \over {\sqrt 2 }}}$$

D

$$-{{1 \over {\sqrt 2 }}}$$

3

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

+4

-1

Let y = y(x) be the solution curve of the differential equation,
$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is:

A

2 + e

B

-e

C

2

D

2 - e

4

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

+4

-1

Let x^k + y^k = a^k, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:

A

$${1 \over 3}$$

B

$${2 \over 3}$$

C

$${4 \over 3}$$

D

$${3 \over 2}$$

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