IIT JAM Mathematics Practice Test- 4 - Mathematics MCQ

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

Given x = cos t, y = sin t

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

Now

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.

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 

Hence, the given integral 

dA = | J | ds, Where 

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.

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

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

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.

the solⁿ is

Given  y(0) = 1

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.

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

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

By the definition of

again IInd condition

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

Given cuive is 

x² + y² = a²

about x-axis

similarly about  y- axis

Area bounded by y = e^x, 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

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.

= 0.053

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

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

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

   .. (2)

family of parabolas' ; y = c₂X^β+1 ...(3)

Diff (3 ) ωy. to x ,

  ... (4)

Now by (2) and (4) we have

m₁m₂ = -1

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

we have

Here

⇒ 1 + P + Q = 0

⇒ f(x) = e^x

Now

f(0) + y(0) = e⁰ + e⁰ (0 + 3) = 4 

or if we take

given at

Now at x = e, sind y,

we have 

given