IIT JAM Mathematics Practice Test- 8 - Mathematics MCQ

Let 0(a) = m, 0(b) = n 

then

Let

Similarly,

Given

Ha ≠ Hb ⇒ aH ≠ bH

then 

aH = bH ⇒ Ha = Hb 

that is 

a^-1 b∈ ⇒ ab^-1 ∈ H.

Let new g ∈ G. g ∈ H be any elements than.

Let G be an abelian group and H a sub group of G x be any element G and h any element of H. We have

= eh(∵ G is abelian x^-1 h = hx^-1)

thus x∈G, x∈G, ⇒ xhx^-1 ∈ H

Hence H is normal in G

Let H be a normal subgroup of G then 

Again, g ∈ G => g-1 ∈ G so

Hence from equations (1) and (2), 

Suppose  e∈  H and for any arbitrary x ∈ N. 

Now HN is a subgroup of G, N is a subgroup of G and N ≤ HN therefore N is a subgroup of HN let hx be an arbitary element of HN st.

again x₁ be any arbitary element of N then.

     ..(1)

Since N Δ G therefore 

Supose that  then we are to prove that N₁ = N₂ we have  but  therefore  that is N1 is equal to some coset of N₂ in G. But two cosets of N₂ in G are either disjoint or identical since e ∈ N₁ and e ∈ N₂ th e re fo re N₁ and N₂ are not disjoint so we must have N₁ = N₂.

We have given that G/Z is a cyclic group, so let Zg is a generator of the cyclic group G/Z, where g ∈ G.
W e now show that G is an abelian group i.e.,
Since a ∈ G, so Za ∈ G/Z. But G/Z is a cyclic group which is gen era ted by Zg. Thus there exists an integer m such that

Za = (Zg)^m = Zg^m [∵ Z is a normal subgroup of G]

Again 

a ∈ Za and Za = Zg^m ⇒ a ∈ Z g^m.

Now

a ∈ Z g m ⇒ z₁ ∈ Z such that a ∈ z₁ g^m

Similarly, for b ∈ G, b = z₂ gⁿ, where z₂ ∈  z and n is any integer.

Now

Again

Since G is non abelian 

Suppose if possible f(xy) = f(x) f(y) then 

which is a contradiction hence f is not an automorphism.

Suppose now that H is any normal subgroup of G. Then, by definition, G/H is abelian if and only if for all x, y in   his is the same as to say that fo r all x, y in  i.e., that H contains every commutator [x, y], (x, y ∈ G). Since these commutators generate G', the last statement is equivalent to saying that  Thus G/H is Abelian if and only if H contains G'.

we have,

Again,

or

'implies f_xy = f_yx Hence option (c) is correct.

Let us approaches (0,0) along the line y = mx which passes through origin. Put y = mx, we get

Which depends on m. Hence does not exists, therefore f(x,y) is discontinuous at (0,0)

Now 

Hence, f(x,y) is discontinuous at (0,0). But both the partial derivative f_x and f_y exists at origin.

Where  

closure property a* b = a ∈ N

Associativity ( a * b ) * c = a * c = a

a * ( b * c ) = a * b = a

a * ( b * c ) = a * b = a

identity a * e = a 

a = a

⇒ there does not exist identity

⇒ (N, *) is not a group.

(B) (Z, *) where a * b = a - b for all a, b ∈Z (Z, *) is a group if all four properties of the group are satisfied. These properties are

 (1) Closure property : If a, b ∈ Z, then a * b = a - b ∈ Z ⇒ closure property is satisfied 

(2) Associativity ⇒ If a, b, c ∈ Z, then ( a * b ) * c = (a -b )*c = a - b - c ...(i) and a*(b*c) = a* (b- c) = a - ( b - c ) = a - b + c ...(ii) 

Hence by Eqs. (i) and (ii)  

Hence the property of associativity is not satisfied. => (Z, *) is not a group.

(C) (Q, *) where for all a, b ∈ Q. T his is not a group since inverse of element 0 ∈ Q does not exist.

for identity ⇒ let identity is e, then a * e = a, let a ≠ o .

⇒ if a ≠ 0 then e = 2, Thus if a = 0 then e cannot be 2

⇒ inverse of a = 0 does not exist. Hence it is not a group

(D) (R, *) w here a * b = a + b + 1, a ,b∈ R This is a g roup with identity given as a *

e = a + e + 1 = a

Since - 1 ∈ Q, so e ∈ Q, So (R, *) is a group with identity - 1 .

We are given that

We have directional derivative at a in the direction u as

We have,

But f(0,0) = 0

Hence,

implies f(x,y) is continuous at (0,0)

Now,

where A = 0, B = 0 which does not depends on h and k and

Hence , f(x,y) is continuous as well as differentiable at (0,0).

Since, 7 and 11 are coprime to 30. Hence, 7 and 11 are generator of Z₃₀

The number of distinct cycles of length 3 in 

The number of distinct cycles of length of 2 in

Now, an element of S₅ is of order 6 if and only if it is product of two disjoint cycles of one of length 2 and another of length 3. Further, once a cycle of length 3 the disjoint transposition is uniquely determined. Hence, the number of element of 6 in S₅ is 20. 

Given:

Each of the six players participating in a chess competition will play one match with each other player.

Concept Used:

ⁿC_r = n!/r!(n - r)!

Calculations:

Symbolically, this can be represented as C(n, r) where:

n represents the total number of players.
r represents the number of players in a single match.

In this case, n is 6 (the total number of players) and r is 2 (players in a single match).

So, in the notation of combinations, the number of matches would be represented as C(6,2).

C(6,2) = 6! / [2!(6-2)!] = 30/2 = 15

∴ There will be 15 matches played in the entire competition.

Option -(A)  Order 6 (2 groups: 1 abelian, 1 nonabelian)

Option -(B) We Know that every group of order 2, 3, 4 and 5 are abelian group but 6 order group is abelian not necessary, and If 0(G) = p (prime)

then G must be cyclic and abelian also. Here 0(G) = 439 (prime) then it is an abelian group,

Option - (B) False statement

Option -(C) If G be a group and x∈G then G has more than two solution such that x² = e where e is identity element.

every group of order p² (p -> prime) is always abelian group.
Option - (A) true

In S₃ → total no. of elements are 3 of order 2 i.e. x² = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y₃ = e.
Option -(C) A₃ is proper subgroup of S₃ which is cyclic.

Since, the number of elements of order cl in a cyclic group of order n is φ(d), where d is positive divisor of n. Hence, the number of elements of order 10 in  is

U(n) is a group under multiplication modulo n , and we know that

Hence

Hence

"Every group contains a unique subgroup of each order."

By lagrange's theorem we know that order of the subgroup is divisors of the order of the group.i . e. if a subgroup H of G exists then 0(H) must divides 0(G). Here divisors of 200 are 1 , 2 , 4 , 5 , 8 ,10,20 ,25 , 40, 50, 100 , 200 i.e. G has 12 subgroups of order 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200 and it is clear that a sub group with order 25 is unique.

0(G) = 108

i . e. G has an element 

By a well known theorem, we know that if order of an element is n in a group G then

i.e. O (a⁴²) = 18

Here 0(G) = 65

i.e. G is a finite cyclic group and we know that if G is a finite cyclic group then

where n is order of the group G. 

We can write given series as,

be a spanning set for . Now we have to choose L .l. subset of S. and that will be a basis for .

Let 

By using above 9 eq’s , we have

P = 0 , q = 0, b = -d = -e = f, a = -2d , c = r 

or

⇒ Basis for V is {w₁, w₂}

⇒ dim V = 2

Here not defined in a region which is enclosed by c so we cannot apply Green's tum here. So we will solve given line integral by definition.

So we have

Let surface S : y² + z²

Take orthogonal projection of S on xy-plane