JEE Advanced Level Test: Differential Equation - JEE MCQ

JEE Advanced Level Test: Differential Equation - JEE MCQ

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30 Questions MCQ Test Mathematics (Maths) Class 12 - JEE Advanced Level Test: Differential Equation

JEE Advanced Level Test: Differential Equation for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The JEE Advanced Level Test: Differential Equation questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Differential Equation MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Differential Equation below.
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JEE Advanced Level Test: Differential Equation - Question 1

If sin x is an integrating factor of the differential equation  then P can be

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 1

We have, IF = sin x

JEE Advanced Level Test: Differential Equation - Question 2

Solution of the differential equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 2

JEE Advanced Level Test: Differential Equation - Question 3

The degree of the differential equation of all curves having normal of constant length c, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 3

Clearly, it is a differential equation of degree 2.

JEE Advanced Level Test: Differential Equation - Question 4

The equation of the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 4

The equation of the tangent at any point
This cuts the coordinate axes at
It is given that P (x, y) is the mid point of AB.

Clearly, it represents a rectangular hyperbola.

JEE Advanced Level Test: Differential Equation - Question 5

Solution of the differential equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 5

cos x dy = y (sin x- y) dx

JEE Advanced Level Test: Differential Equation - Question 6

Consider the differential equation

Then x is given by:

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 6

JEE Advanced Level Test: Differential Equation - Question 7

The equation of the curve for which the tangent at P(x, y) cuts the y-axis at (0, y3) is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 7

JEE Advanced Level Test: Differential Equation - Question 8

then the solution of the equation is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 8

Put y = vx

Put log v = z

JEE Advanced Level Test: Differential Equation - Question 9

Find the area of smaller portion of the circle x2 + y2 = 4 cut off by the line x = 1.

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 9

Equation of the circle is x2 + y2 = 4 and equation of the line is x = 1.
Required area = area

JEE Advanced Level Test: Differential Equation - Question 10

The region bounded by the curve = log x and y = 2x , then the area of the region, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 10

JEE Advanced Level Test: Differential Equation - Question 11

The order and degree of the differential equation, of which xy = cex + be-x + x2 is a solution, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 11

We have, xy = cex + be-x + x2....(i)
On differentiating w.r.t. x, we get

On differentiating again,

Hence, the required differentiable equation is

The order of this differential equation is 2 and degree 1.
Hence, (B) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 12

The order of the differential equation whose general solution is given by

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 12

Since, the above relation contains five arbitrary constants, so the order of the differential equation satisfying it, is 5.
Hence, (C) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 13

The degree of the differential equation satisfying

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 13

Put x2 = sin α, y2 = sin β
∴ Given equation reduces to cos α + cos β = a (sin α - sin β)

On differentiating w.r.t. x, we get

which is a differential equation of first order and first degree.
Hence, (A) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 14

The differential equation of the family of curves y = ex (A cos x + B sin x) , where A and B are arbitrary constants, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 14

We have, y = ex (A cos x + B sin x)....(i)
On differentiating w.r.t. x, we get

On differentiating again,

JEE Advanced Level Test: Differential Equation - Question 15

The differential equation of family of parabolas with foci at the origin and axis along the x-axis, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 15

Let the directrix be x = - 2a and latusrectum be 4a. Then, the equation of the parabola is (distance from focus = distance from directrix),

On differentiating w.r.t, we get

On putting this value of a in Eq. (i), the differential equation is

Hence, (A) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 16

Solution of the equation

(x - y) (2dy - dx) = 3dx - 5dy is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 16

We have, (2x - 2y + 5) dy = (x - y + 3) dx

∴ The given equation becomes

Hence, (A) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 17

A curve passes through the point and its slope at any point is given by  Then, the curve has the equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 17

Hence, (A) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 18

Solution of the equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 18

Hence, (B) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 19

A function y = f (x) satisfies the condition f ' (x) sin x + f (x) cos x= 1, f (x) being bounded when

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 19

Hence, (A) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 20

A function  y = f (x) satisfies (0) = 5, then f (x ) is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 20

JEE Advanced Level Test: Differential Equation - Question 21

The solution of  satisfying y (1) = 0 is given by

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 21

Hence, (A) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 22

Solution of the equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 22

Hence, (C) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 23

The solution of the differential equation,  given

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 23

Put x2 y2 = z

Now, given expression transforms to

Hence, (A) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 24

The solution of  sec y satisfying y (1) = 0, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 24

JEE Advanced Level Test: Differential Equation - Question 25

If normal at every point to a curve passes through a fixed point then the curve must be

JEE Advanced Level Test: Differential Equation - Question 26

Solution of the differential equation

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 26

Hence, the differential equation becomes

Hence, (A) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 27

The equation of the curve which passes through the point (2a, a) and for which the sum of the Cartesian sub tangent and the abscissa is equal to the constant a, is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 27

We have,
Cartesian sub tangent + abscissa = constant

Integrating, we get log y + log (x - a) = log c
∴ y(x - a) = c
As the curve passes through the point (2a, a), we have c = a2
∴ The required curve is y (x - a) = a2
Hence, (A) is the correct answer

JEE Advanced Level Test: Differential Equation - Question 28

Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth y, where the constant of proportionality k > 0 depends on the acceleration due to gravity and the geometry of the hole. If t is measured in minutes and  then the time to drain the tank, if the water is 4 meter deep to start with is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 28

JEE Advanced Level Test: Differential Equation - Question 29

A ray of light coming from origin after reflection at the point P(x, y) of any curve becomes parallel to x-axis, if the curve passes through (8, 6) then its equation is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 29

The slope of the ray = y/x

Hence, (B) is the correct answer.

JEE Advanced Level Test: Differential Equation - Question 30

A function y = f (x) has a second order derivative f "(x) = 6 (x - 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x- 5, then the function is

Detailed Solution for JEE Advanced Level Test: Differential Equation - Question 30

We have, f "(x) = 6 (x- 1)
⇒  f '(x) = 3(x -1)2 + c .......(i)
It is given that, y = 3x- 5 is tangent to the curve y = f (x) at the point (2, 1)

On putting, x = 2, f ' (2) = 3 Eq. (i), we get c = 0
∴ f ' (x) = 3(x- 1)2
⇒ f (x) = (x - 1)3 + c1 .....(ii)
The curve y = f (x) passes through (2, 1)
∴ f (2) = 1
On putting, x = 2, f (2) = 1 in Eq. (ii), we get
c1 = 0
On putting c1 = 0 in eq. (ii), we get f (x) = (x- 1)3

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