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QUESTION: 1

Compute the integral along the arc of the parabola x = y^{2} 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 = y^{2}

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 x^{2} + y^{2} = 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 x^{2} + y^{2} = ax in the positive quadrant is equal to -

Solution:

Let P(r,θ) be any point on the circle x^{2}+y^{2} = ax

∴ x = r cos θ, y = r sin θ

= r(cos^{2} θ + sin^{2} θ) = r

and the equation of the circle

r^{2}(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 - x^{2} 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 - x^{2} 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 sol^{n} 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 ) = 4x^{2} - 3x +1 and then

Solution:

f( x ) ∈ [ - 2,2] Such that

f( x ) = 4x^{2} - 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

y^{2} = 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 x^{2} + y^{2} = a^{2} then

Solution:

Given cuive is

x^{2} + y^{2} = a^{2}

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 = e^{x}, y = e^{-x }and x =1

again whole area of the curve a^{2}y^{2} = a^{2}x^{2} - x^{4},

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

Put x = a sin θ, dx = a cosθ dθ

= 4a^{2}/3

*Answer can only contain numeric values

QUESTION: 21

Evaluate where S is the entire surface of the solid bounded by the cylinder x^{2} + y^{2} =1 and the planes z = 0, z = x + 2.

Solution:

As shown in the figure , the surface consists of three parts: S_{1} the circular b ase in the xy-plane, S_{2}, the elliptic plane section, i . e . , part o f the plane z = x + 2 inside the cylinder x^{2} + y 2 = 1 , and S^{3} , the lateral surface of the cylinder.

On S_{1} we have

z = 0, x^{2} + y^{2} = 1

on S_{2}

2 = x + 2, x^{2} + y^{2} = 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 - x^{2} 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, x^{2} + y^{2} ≤ 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 9x^{2} + 4y^{2} = c_{1} where c_{1} > 0 are given by where c_{2} ∈ R , then the value of β i s ..... ( conect upto two decimal places).

Solution:

Family of ellipse : 9x^{2} + 4y^{2} = c_{1} ..... (1)

.. (2)

family of parabolas' ; y = c_{2}X^{β+1} ...(3)

Diff (3 ) ωy. to x ,

... (4)

Now by (2) and (4) we have

m_{1}m_{2 }= -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) = C_{1} cos 2x + C_{2} sin 2x + 2x sin 2x

y(0) = 0 ⇒ C_{1} = 0

y'(0) = 0 ⇒ 2C_{2} = 0 ⇒ C_{2} = 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) = e^{x}

Now

f(0) + y(0) = e^{0} + e^{0} (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

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