IIT JAM Mathematics MCQ Test 4


30 Questions MCQ Test Mock Test Series for IIT JAM Mathematics | IIT JAM Mathematics MCQ Test 4


Description
This mock test of IIT JAM Mathematics MCQ Test 4 for Mathematics helps you for every Mathematics entrance exam. This contains 30 Multiple Choice Questions for Mathematics IIT JAM Mathematics MCQ Test 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Mathematics MCQ Test 4 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this IIT JAM Mathematics MCQ Test 4 exercise for a better result in the exam. You can find other IIT JAM Mathematics MCQ Test 4 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

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

Solution:

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

QUESTION: 2

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

Solution:

Given x = cos t, y = sin t

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

Now

QUESTION: 3

The arc length of   Assume that y is positive

Solution:

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.

QUESTION: 4

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

Solution:

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 

QUESTION: 5

Let E   Then 

Solution:

QUESTION: 6

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

Solution:

Hence, the given integral 

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

Solution:

dA = | J | ds, Where 

QUESTION: 8

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

Solution:

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.

QUESTION: 9

Evaluate 

Solution:

QUESTION: 10

Changing the order of the integration  gives

Solution:

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

QUESTION: 11

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

Solution:

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

QUESTION: 12

Let  then, which of the follwing is not true ?

Solution:

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.

QUESTION: 13

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

The y(1) is equal to,

Solution:

the soln is

Given  y(0) = 1

 

 

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,

Solution:

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.

QUESTION: 15

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

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

Solution:

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
QUESTION: 16

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

Solution:

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
QUESTION: 17

if  then

Solution:

By the definition of

again IInd condition

*Multiple options can be correct
QUESTION: 18

Which of the following is / are true ?

Solution:

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
QUESTION: 19

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

Solution:

Given cuive is 

x2 + y2 = a2

about x-axis

similarly about  y- axis

*Multiple options can be correct
QUESTION: 20

Which of the following is /are correct ?

Solution:

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
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.


Solution:

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
QUESTION: 22

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


Solution:

= 0.053

*Answer can only contain numeric values
QUESTION: 23

The value of  is ___________.


Solution:

*Answer can only contain numeric values
QUESTION: 24

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


Solution:

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
QUESTION: 25

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


Solution:

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
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).


Solution:

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
QUESTION: 27

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


Solution:

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
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 _________.


Solution:

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
QUESTION: 29

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


Solution:

given at

Now at x = e, sind y,

 

*Answer can only contain numeric values
QUESTION: 30

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


Solution:

we have 

given