IIT JAM Mathematics Practice Test- 21 - Mathematics MCQ

The function has to be continuous in [0,1]

Let 
The limit of integration are 


on reversing the order of integration 
The limit of x and y becomes

The integral is then 

Given equation can be written as,


the given equation becomes,

Let  f(t) = y(t)

= −f(0) + sF(s)
= sY(0) − y(0)

We have dim U = dim W = 7 and U and W are distinct 
⇒ U + W subspace properly contains U and W.
⇒ dim(U + W) > 7
But also ( U + W ) be a subspace of V
⇒ dim(U + W) ≤ dim V = 10
⇒ (U + W) can have dimensions 8. 9 or 10.
Now dim (U ∩ W) = dim U + dim W - dim (U + W)
= 14 - dim(U + W)

By Gauss- divergence theorem

y = z²

So the given equation can be written as,


Thus equation (1) can be written as

where D = d/dt

So required perpendixular length is

= 7

We have 

Integrate with respect to x,

Region of integration is bounded by curves 

after changing the order of the integration we get
after changing the order of the integration we get

When we calculate minimum of f(x) and maximum of g(x) then we find that f(x) has minimum at x = - b and g(x) has maximum at x = - c 
then minf(x) > maxg(x)

⇒α² = 1 and α + 1 = 5
which is not possible at the same time, 
∴ no real values of α exists.

We know if X ⊂ R then X is connected 
⇔ X be an interval
So if X be a non-empty and non-singleton connected set then X always has rational and irrational elements.
So there is only possibility for X is X be singleton 
Set i.e. X = {a} where a ∈ Q 
⇒ cardinality of X is 1.

We put λ = 0 in the given equation, so that we can get

t = 0 ( 0 + 16) + 1( -1 2 ) + 3 ( 4 - 0 ) = -12 + 12 = 0
⇔ t = 0

Equation of the circle is x² + y²=1
Let x = cosθ, y = sinθ
then dx = -sinθdθ, dy = cosθdθ

and z = 0 ⇒ dz = 0
So total work done is 

Here T is one-one and onto so T^-1 is exist 
Let T ( x , y , z) = ( p , q , r)
⇒ T^-1(p, q r) = (x, y, z)
So Now T(x , y , z) = (p ,q ,r )
⇒ (x - y, y + 3z. x + 2y) = (p, q. r)
x - y = p ...(i)
y + 3z = q ...(ii)
x + 2y = r ...(iii)
equation (1) and equation (3) ⇒ - 3y = p - r

then by equation(1), 

and by equation (2),

(A) We know if T: U → V be a L.T and dim U =m,dim V=n .Then if m>n then then T can not be one-one.
here U = R⁴ so its dim is 4 
and V = R³ so its dim is 3 
Thus T can not be one-one
(B) We know if V be a V .S of all n x n, skew symmetric matrices, then

So here n = 4, so dim V = 4.3/2 = 6
(C) Clearly characteristic polynomial of A will be of degree n. while characteristic polynomial of T will be of degree n². 
So they can not be equal.
Thus option A.B,C can not be true, 
hence option (D) is correct.
That vector space can have basis as {(1,0),(0,1)} or {(1,1),(0,1)} or {(1,1),(1,0)}

Applying L’ Hospital ( from), and using leibnitz's rule
we get

here A is the union of three open intervals. Hence A is open, since a not an interval ⇒ A not connected.
Clearly set of boundary points of is  and also A has no isolated points.

By Property of direct product
 iff ( m,n) = 1 i.e.g.c.d of m & n is 1.

 

Similarly, we can show that 

By the definition of operation table of a group.

By D well known theorem of group theory we know that
“If Z be the centre of the group G and G/Z is cyclic then G is abelian”.

Limits are x = 0 , x = 1 
y = x², y = 2x
After changing the order of the integration, we get

Region of integration is bounded by curves, 
y = x ; y = 2 + x ; x = 0 ; x = 1

One-One: Let x,y ∈ R such that.


So, f is one-one.
Onto: Let α ∈ R (co - domain)such that f(x) = α