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IIT JAM Mathematics Practice Test- 4 - Mathematics MCQ


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30 Questions MCQ Test IIT JAM Mathematics Mock Test Series - IIT JAM Mathematics Practice Test- 4

IIT JAM Mathematics Practice Test- 4 for Mathematics 2024 is part of IIT JAM Mathematics Mock Test Series preparation. The IIT JAM Mathematics Practice Test- 4 questions and answers have been prepared according to the Mathematics exam syllabus.The IIT JAM Mathematics Practice Test- 4 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for IIT JAM Mathematics Practice Test- 4 below.
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IIT JAM Mathematics Practice Test- 4 - Question 1

Compute the integral  along the arc of the parabola x = y2 from (1,-1) to (1,1)

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 1

When the equation of the curve of integration is taken by expressing x as a single valued function of y, we put x = y2

so that

IIT JAM Mathematics Practice Test- 4 - Question 2

Evaluate the integral  taken along the quarter circle x = cos t, y = sin t, joining the same points.

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 2

Given x = cos t, y = sin t

=> dx = - sin t d t , dy = cost dt

Now

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IIT JAM Mathematics Practice Test- 4 - Question 3

The arc length of   Assume that y is positive

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 3

The derivative and root will then be.

Before writing down the length, notice that we  were given x limits and we will need y limits for this ds. With the assumption that y is positive there are easy enough to get. All we need to do is plug x into our equation and solve for y. Doing this gives.

The integral for the arc length is then,

This integral will require the following trig substitution.

The length is then.

IIT JAM Mathematics Practice Test- 4 - Question 4

By changing the order of integration, the integral can be represented as determine the value of A. 

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 4

Here the lim its of y are given by the circle  and the  straigh t line y = x + 2a. The limits of x are given by the straigh t lines x = 0 and x = a

Therefore , the integral extends to all points in the space bounded by the axis of y, the circle A B , the straight line A L and the straight line LM . Draw BH and M K perpendiculars to A L . Now wh en the order o f integration is ch a n g e d , w e take strips parallel to axis o f x instead o f that o f y, thus the integral b reaks up into th ree parts: First corresponding to the area BAH , second to the rectangle BHKM and the third to the triangle M K L , hence

so Here 

IIT JAM Mathematics Practice Test- 4 - Question 5

Let E   Then 

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 5

IIT JAM Mathematics Practice Test- 4 - Question 6

Evaluate  over the positive quadrant of the circle x2 + y2 = 1 .

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 6

Hence, the given integral 

IIT JAM Mathematics Practice Test- 4 - Question 7

Let R be the image of the triangular region S with vertices (0, 0) (0,1) is uv-plane under the transformation x = 2u - 3v and y = u + v.

Then  equals

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 7

dA = | J | ds, Where 

IIT JAM Mathematics Practice Test- 4 - Question 8

 over the semicircle x2 + y2 = ax in the positive quadrant is equal to -

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 8

Let P(r,θ) be any point on the circle x2+y2 = ax

∴ x = r cos θ, y = r sin θ

= r(cos2 θ + sin2 θ) = r

and the equation of the circle

r2(cos2 θ + sin2 θ) = ar cos θ

=> r = a cos θ

The correct answer is C.

IIT JAM Mathematics Practice Test- 4 - Question 9

Evaluate 

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 9

IIT JAM Mathematics Practice Test- 4 - Question 10

Changing the order of the integration  gives

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 10

Here the integration extends to all points of the space bounded by parabola y = 1 - x2 and circle 

The given integral breaks up in two integrals : first corresponding to the area ABCO and second to the area AOCD.
Now solving y = 1 - x2 for x, we have 

Clearly upper and lower limit o f x for the area OABC are  and for area AOCD upper and lower limits of x is  and limits of y is O to -1

Hence, we have

IIT JAM Mathematics Practice Test- 4 - Question 11

An integrating factor for the differential equation (cos y sin 2x) dx + (cos2y - cos2x ) dy = 0 is,

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 11

Compare the given eqn with M dx + N dy = 0 we have M = cos y sin 2x ⇒ - sin y sin 2x

IIT JAM Mathematics Practice Test- 4 - Question 12

Let  then, which of the follwing is not true ?

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 12

we have 

clearly when

and f'(x) < 0

f is decreasing in 

Now at x = -1, sign of f changes from -ive to +ive

=> at x = -1, f has minimum.

=> min f = 0 at x = 4 also f has minimum.

=> only option (c) is not true.

IIT JAM Mathematics Practice Test- 4 - Question 13

Let y(x) be the solution of initial value problem,

The y(1) is equal to,

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 13

the soln is

Given  y(0) = 1

 

 

IIT JAM Mathematics Practice Test- 4 - Question 14

Let F(x) be the particular integral of the differential equation y" + y = (x - cot x) If there exist c e R such that F(c) = c, then c is equal to,

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 14

Then by variation of parameter method, particular integral is given by 

P.l. = cos x f(x) + sin x g(x) ....(1)

Here

= x sin x + cos x - In ( cosecx - cotx) - cos x

= x sin x - In (cosecx - cotx)

So by (1), then P.l. is given by,

F(x) = cos x [ x cos x] + sin x [ x sin x - In ( cosecx - cotx) ]

F(x) = x - sin x In ( cosecx - cotx)

F(0) = Not exirt.

IIT JAM Mathematics Practice Test- 4 - Question 15

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

Satisfying the condition   Then which o f the following is/are

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 15

Given DE is homogeneous, so let y = Vx

put these in given DE, we have

clearly

: bounded But x not bounded.

⇒ y is not bunded on R+ .

⇒ y(x) bounded above but not add below clearly In f {y (x)} = - ∞ and sup (y(x)} = 0

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 4 - Question 16

If f(x) is defined [ -2,2 ] by / ( x ) = 4x2 - 3x +1 and  then

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 16

f( x ) ∈ [ - 2,2] Such that

f( x ) = 4x2 - 3 x + 1 

Now

and

Which shows that g(x) is an odd function if g(x) is an odd function then

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 4 - Question 17

if  then

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 17

By the definition of

again IInd condition

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 4 - Question 18

Which of the following is / are true ?

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 18

We know that 

case-1: Area is bounded by

y2 = x and 2y = x

case-ll: Area is bounded by lines y = x + 2.  , y = 2-x and x = 2

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 4 - Question 19

If the equation of the curve is x2 + y2 = a2 then

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 19

Given cuive is 

x2 + y2 = a2

about x-axis

similarly about  y- axis

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 4 - Question 20

Which of the following is /are correct ?

Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 20

Area bounded by y = ex, y = e-x and  x =1

again whole area of the curve a2y2 = a2x2 - x4,

it is Symmetric about x-axis and y-axis then

Put x = a sin θ, dx = a cosθ dθ

= 4a2/3

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 21

Evaluate  where S is the entire surface of the solid bounded by the cylinder x2 + y2 =1 and the planes z = 0, z = x + 2.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 21

As shown in the figure , the surface consists of three parts: S1 the circular b ase in the xy-plane, S2, the elliptic plane section, i . e . , part o f the plane z = x + 2 inside the cylinder x2 + y 2 = 1 , and S3 , the lateral surface of the cylinder.

On S1 we have

z = 0, x2 + y2 = 1

on S2

2 = x + 2, x2 + y2 = 1

where D is the corresponding domain in the θ z- plane.

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 22

The value of  over the area between the parabola y - x2 and the line y = x is ______.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 22

= 0.053

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 23

The value of  is ___________.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 23

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 24

Evaluate the integral  over the volume enclosed by three coordinate planes and the plane x + y + z = 1.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 24

The given plane x + y + z = 1 meets the co ordinate axes a tA (1 , 0, 0) B ( 0 , 1, 0) and C(0, 0, 1).
Here a column parallel to z-axis is bounded by the plane z = 0 and z = 1 - x - y. Also the region S above which the volume V stands is the triangle OAB bounded by the lines y = 0, x = 0 and x + y = 1 (on xy plane )

Hence the given integral

on simplifying

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 25

Evaluate  over the domain {(x, y ) : x ≥ 0, y ≥ 0, x2 + y2 ≤ 1}.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 25

The region of integration is

Put x = sin θ so that dx = cos θ dθ

When x = 0, θ = 0 and x = 1, θ = π/2

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 26

If the orthogenal trajectories of the family of ellipse 9x2 + 4y2 = c1 where c1 > 0  are given by  where c2 ∈ R , then the value of β i s ..... ( conect upto two decimal places).


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 26

Family of ellipse : 9x2 + 4y2 = c1  ..... (1)

   .. (2)

family of parabolas' ; y = c2Xβ+1 ...(3)

Diff (3 ) ωy. to x ,

  ... (4)

Now by (2) and (4) we have

m1m2 = -1

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 27

Consider the differential equation y" + 4y = 8cos2x , with y(0) = 0 and then y( π ) is equal to _________ .


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 27

We have

(D2 +4) y = 8 cos 2x 

A . E.

⇒ y(x) = C1 cos 2x + C2 sin 2x + 2x sin 2x

y(0) = 0 ⇒ C1 = 0

y'(0) = 0 ⇒ 2C2 = 0 ⇒ C2 = 0

⇒ y(x) = 2x sin 2x

⇒ 

and y (π) = 0

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 28

Let y(x) = c1 f(x) + c2 g(x) be the general solution of (x + 2) y" - (4x + 9)y' + (3x + 7)y = 0, then f(0) + g(0) is equal to _________.


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 28

we have

Here

⇒ 1 + P + Q = 0

⇒ f(x) = ex

Now


f(0) + y(0) = e0 + e0 (0 + 3) = 4 

or if we take

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 29

Consider the differential equation  then y(e) is equal to 


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 29

given at

Now at x = e, sind y,

 

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 4 - Question 30

Let y(x) is a solution of differential equation  equal to. satisfying y(o) = 1. then


Detailed Solution for IIT JAM Mathematics Practice Test- 4 - Question 30

we have 

given

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