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# Test: Differential Equation (Competition Level) - 1

## 30 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Differential Equation (Competition Level) - 1

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This mock test of Test: Differential Equation (Competition Level) - 1 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Differential Equation (Competition Level) - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Differential Equation (Competition Level) - 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Differential Equation (Competition Level) - 1 exercise for a better result in the exam. You can find other Test: Differential Equation (Competition Level) - 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

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

Solution:

We have, IF = sin x

QUESTION: 2

Solution:

QUESTION: 3

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

Solution:

Clearly, it is a differential equation of degree 2.

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

Solution:

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.

QUESTION: 5

Solution of the differential equation

Solution:

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

QUESTION: 6

Consider the differential equation

Then x is given by:

Solution:

QUESTION: 7

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

Solution:

QUESTION: 8

then the solution of the equation is

Solution:

Put y = vx

Put log v = z

QUESTION: 9

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

Solution:

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

QUESTION: 10

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

Solution:

QUESTION: 11

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

Solution:

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

QUESTION: 12

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

Solution:

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.

QUESTION: 13

The degree of the differential equation satisfying

Solution:

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

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

Solution:

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

On differentiating again,

QUESTION: 15

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

Solution:

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

QUESTION: 16

Solution of the equation

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

Solution:

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

∴ The given equation becomes

Hence, (A) is the correct answer.

QUESTION: 17

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

Solution:

Hence, (A) is the correct answer.

QUESTION: 18

Solution of the equation

Solution:

Hence, (B) is the correct answer

QUESTION: 19

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

Solution:

Hence, (A) is the correct answer.

QUESTION: 20

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

Solution:

QUESTION: 21

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

Solution:

Hence, (A) is the correct answer

QUESTION: 22

Solution of the equation

Solution:

Hence, (C) is the correct answer.

QUESTION: 23

The solution of the differential equation,  given

Solution:

Put x2 y2 = z

Now, given expression transforms to

Hence, (A) is the correct answer

QUESTION: 24

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

Solution:

QUESTION: 25

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

Solution:
QUESTION: 26

Solution of the differential equation

Solution:

Hence, the differential equation becomes

Hence, (A) is the correct answer.

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

Solution:

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

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

Solution:

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

Solution:

The slope of the ray = y/x

Hence, (B) is the correct answer.

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

Solution:

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