IIT JAM Mathematics Practice Test- 17 - Mathematics MCQ

we know three rectors will be coplanar if their scalar triple product is zero

we know two vector’s are collinear iff on vector is the scalar multiple of other. Here among  None o f vector be the Scalar multiple of other.

Given f (x,y,z ) = x²y + 2xz - 4 (surface) then for any given surface we know that the unit vector is defined as 

∴ Derivative in the direction  

Since  and we know that if the line integral is zero then given curve c is closed.

hence

work done

by equation (2) and (3)

Now use this in (1), we have

= All A,B,C option are correct but (D) is not correct.

The surface element

where

Let V be the volume enclosed by the sphere (surface) then by Gauss’s div. the 

By the stock’s theorem we know that

where C is a closed curve bounded by the surface  

x² +y² = 1

given

then by (1)

we will compute

Let S is the surface of Δ ABC, then the eqnof surface is given by 

Now here to evaluate (1), we can use stokes theorem Then we have

where  is outward going UNV to S  and given by

Now

= i(z) - j (-x) + k(y) = zi + xj + yk

Now take orthogonal projection of S on the xy-plane.

then we have

Then by (2)

Given line integral is

By stoke's th.

Here

then curve

⇒ Bisector of C and  is

and

Hence

and unit vector along 

if  are orthogonal then 

it is not possible for real p and q.
Never orthogonal.

Let

then  will a tangent vector

Given T₁ and T₂ are orthogonal

given

Here S be a closed surface so we will apply gauss divergence theorem,

then we have

since

So  is conservative vector field. 

Hence

work done in moving particle from origin o to D is

and ds = dx/dy

flux across z = c,

Flux across the closed surface

C is the curve of intersection of surfaces 

let x=4cos θ, y=4sinθ Hence,

Let  be the position vector of any point on L (in Cartesian from) 

then

we curve in parametric form as

⇒ L be the arc of given curve from  
 

Now by (1) it is clear that  is conservative so 

Here

now S be an open surface so we will use stoke’s theorem here. 

Here curve C is the intersection of x² + y² - 4x + 4z = 0 and z = 0 x² + y² - 4x = 0 = x² - 4x + 4 + y² = 4 = (x -2)² + y² = 4 be a circle with centre (2,0) and radius 2.
Now on C:

Now we can use Green’s Theorem here so Here

M = y²-x² ; N = x²-y²

If we take,

F. dr = 2xyz (cosx² + 1)dx + z (sinx² + x²) dy + y (sinx² + x²)dz = 2xyz cosx² dx + z sinx² dy + y sinx² dz + 2xyz dx + x² z dy + x²y dz = d(yz sinx²) + d(x²yz)

and φ is the solⁿ of Laplace eqⁿ

 not a solenoidal vector field.

we know

Now

Put these values in (1), we have

Now

If f is irrotational then

So First we find curl 

Now

{ where θ is the angle between 

Hence  is maximum if cos θ = 1 i.e. θ = 0°

Let

since given that 

again

Surface z = x² + y²

⇒ surface S = x² + y² - z = 0 
hence

  (1)

the equation of a plane, passing through a point whose position vector is r₀ and which is perpendicular to the normal N 

(r - r₀₎.N = 0

then the required equation is

⇒ 4(x - 2) - 2(y + 1) - 1 (z - 5) = 0

⇒ 4 x - 8 - 2 y - 2 - z + 5 = 0

⇒ 4x - 2y - z = 5