IIT JAM Mathematics Practice Test- 19 - Mathematics MCQ

Since 5 has order 6 in Z₃₀ and 3 has order 4in Z₁₂. 
Hence, <5>  <3> is a subgroup of order 24 in Z₃₀ Z_12.

Given that


⇒ n + 2 = 4
⇒ n = 2

By definition of gradient of 

Now,


Now by gauss - divergence theorem

Here 

⇒ T is linear 

⇒ R = 1/e^a

Here sequence in option (A) is not convergent ⇒ it is not cauchy. Sequence in option (C) is convergent which converge to 0.
⇒ It be a cauchy sequence which converge in Q.
Similarly sequence in Option (D) has two subsequences.
 and both are converging to 0. 
But sequence in option (B) will converge to e.
⇒ It be a cauchy sequence which is not converging in Q.

We know that trace (A) 
i.e. 
also know that det (A) = 
Now 

Here A\B = A - B
So (A \ B) U (A ∩ B) = A, which is unbounded.

Given 



Thus for the solution, we must have
c + b - a = 0

By a well-known theorem, we know that " If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is φ(d)".
Where φ (d) denotes Euler's φ - function
Here n = 60 and d = 12 (12 is divisor of 60)
then φ (12) = 1, 5, 7, 11
Hence total no. of elements of order 12 in cyclic group of order 60 is 4.

Since U (10) and U (5) are cyclic group of order 4.
So U (10) ≌ U (5)
Now, Suppose that φ is an isomorphism from U (10) onto U (12). then 

and

If  be any vector function then by the result we know that


by comparison of the result it is clear that 
Hence curl 



[Where V is volume enclosed by closed surface S]

The r^th term  where r varies form 1 to n.
Hence given limit is 
Hence, By the sum of the series by integration, this limit can be written as.

Given that 
then required length of the curve is 
  

which is a parabola with vertex (1, 1).

Here limit is f(X), so we have to consider a horizontal strip

 the bounded region takes values less than 1.}

Hence Area

Given |a| = 4, |b| = 4, |c| = 2
and  


adding (i) (ii) and (iii) we get



= 6

Since (1, a, a²), (1, b, b²) and (1, c, c²) are non-coplanar therefore

and 

xy = c

 : DE of required O.T

Now take


Now take

We know if A and B are n x n matrices, then Rank (A) + Rank (B) - n ≤ Rank (AB)
where n be the order of matrices,
Take A =  B then we have
Rank (A) + Rank (A) - n ≤ Rank (A²)

​

⇒ T cannot be onto
⇒ T not one-one

If T : R⁴ → R³, be a L.T. then this map cannot be one-one, since dim (R⁴) = 4.
So ker (T) ≠ {0}.
(B) We know if V be a V.S. of all (n x n) skew-symmetric matrices then

In this case we cannot find out matrix representation of T w.r. to basis  of M_n(Q) but if F =R₁, Then always char. eqⁿ of T is same as A.
⇒ only (D) is correct.

Clearly f is continuous at  and hence on R
Now 

We know if {an} be a sequence s.t.

Then if ∑u_n is convergent then sequence {a_n} is Cauchy and hence convergent
Now here we have 

at the point (1,2,-8) the normal vector the surface is given by

Here given surface be closed , so we can apply Gauss - divergence theorem, we have