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All questions of Practice Test for IIT Jam Mathematics for Mathematics Exam
is equal to _______________
A group (G, *) has 10 elements. The minimum number of elements of G, which are their own inverse is- a)2
- b)1
- c)9
- d)0
Correct answer is option 'B'. Can you explain this answer?
A group (G, *) has 10 elements. The minimum number of elements of G, which are their own inverse is
a)
2
b)
1
c)
9
d)
0
|
Chirag Verma answered |
Since,in a group there must be an identity element.Also it is its own inverse. Therefore,minimum number of elements of G,which are their own inverse=1
Find the number of element of order 10 in Z30‘
Correct answer is '4'. Can you explain this answer?
Find the number of element of order 10 in Z30‘
|
|
Chirag Verma answered |
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
is
is equal to ..........
___________ is the order of {3} in the multiplicative group of integers modulo 5.
Correct answer is '4'. Can you explain this answer?
___________ is the order of {3} in the multiplicative group of integers modulo 5.
|
|
Chirag Verma answered |
The identity of the multiplicative group of integers modulo 5 is {1}.
Here
{3}.5{3} = 4 * {1} [•, 3.3 0 4 (mod 5)]
∴ {3}.5{3}.5{3} = {4}.5{3} = {2} * {1} [∵, 4.30 2 (mod 5)]
∴{3}.5{3}.5{3}.5{3} = {2}.5{3} = {1} [∵, 2.30 (mod 5)]
Hence, the order of {3} is 4.
Which one is a group under multiplication modulo 11?- a){1,8}
- b){9,1}
- c){10,1}
- d){1,3,5,7,8}
Correct answer is option 'C'. Can you explain this answer?
Which one is a group under multiplication modulo 11?
a)
{1,8}
b)
{9,1}
c)
{10,1}
d)
{1,3,5,7,8}
|
|
Chirag Verma answered |
By the definition of operation table of a group.
Given axa = b in a group G , where a,b ∈ G then x is equal to- a)a-1b
- b)a-1b-1
- c)a-1b-1b-1
- d)a-1ba-1
Correct answer is option 'D'. Can you explain this answer?
Given axa = b in a group G , where a,b ∈ G then x is equal to
a)
a-1b
b)
a-1b-1
c)
a-1b-1b-1
d)
a-1ba-1
|
|
Chirag Verma answered |
If G be a group and a,b ∈ G
then
Find the directional derivative of φ = x2yz + 4xz2 at (1, - 2 , - 1 ) in the direction 2i - j - 2k.
Correct answer is '12.34'. Can you explain this answer?
Find the directional derivative of φ = x2yz + 4xz2 at (1, - 2 , - 1 ) in the direction 2i - j - 2k.
|
|
Chirag Verma answered |
The unit vector in the direction of 2i - j - 2k is
Then the required directional derivative is
Since this is positive,increasing in this direction.
is ______.
The system of linear equations(4d - 1) x + y + z = 0, - y + z = 0, (4d - 1 )z = 0 has a non-trivial solution, if the value of d is ___ .
Correct answer is '0.25'. Can you explain this answer?
The system of linear equations
(4d - 1) x + y + z = 0, - y + z = 0, (4d - 1 )z = 0 has a non-trivial solution, if the value of d is ___ .
|
|
Chirag Verma answered |
The system of homogeneous linear equations has a non-trivial solution, if
Which of the following is correct?- a)Every subgroup of an abelian group is normal
- b)Every subgroup of a cyclic group is normal
- c)Intersection of any two normal subgroup is a normal subgroup
- d)If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.
Correct answer is option 'A,B,C'. Can you explain this answer?
Which of the following is correct?
a)
Every subgroup of an abelian group is normal
b)
Every subgroup of a cyclic group is normal
c)
Intersection of any two normal subgroup is a normal subgroup
d)
If N is a normal subgroup of G and H is any subgroup of G then NH is normal subgroup of G.
|
|
Chirag Verma answered |
Every subgroup of an abelian group is normal as for every x ∈ G and h ∈ H
Every subgroup of cyclic group is normal and intersection of two normal subgroups is normal subgroup again.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.
If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.
then- a)<G, •> is not group
- b)<G. •> is a group
- c)<G,•> is a sub group
- d)<G, •> is not a sub group.
Correct answer is option 'B'. Can you explain this answer?
If G be an one empty set
(1) a(bc) = (ab)c for all a, b, c ∈ G
(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.
then
then
a)
<G, •> is not group
b)
<G. •> is a group
c)
<G,•> is a sub group
d)
<G, •> is not a sub group.
|
|
Mister Genius answered |
Let (1) and (2) hold. To show G is a group, we need prove existence of identity and
Let a ∈ G be any element inverse.
By (2) the equations ax = a
ya = a
have solutions in G.
Let x = e and y = f be the solutions.
ae = a
fa = a
Let now b e G be any element then again by (2) 
some x, y in G s.t.,
some x, y in G s.t.,ax = b
ya = b.
Now ax = b ⇒ f.(a.x) = f.b
⇒ (f.a).x = f.b
⇒ a.x = f.b
⇒ b = f.b
Again y.a = h
⇒ (y.a).e = b.e
⇒y.(a.e) = b.e
⇒ y.a = be
⇒ b = be
thus we have b = fb ... (i)
b = be ... (ii)
for any b ∈ G
Putting b = e in (i) and b = f in (ii) we get
e = fe
f = fe
⇒ e = f.
Hence ae = a = fa = ea
Hence ae = a = fa = ea
i.e., 
e ∈ G, s.t., ae = ea = a
e ∈ G, s.t., ae = ea = a⇒ e is identity.
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
then aa1 = e,
a2a = e
Now a1 = ea1 = (a2a)a1 = a2(aa1) = a2e = a2.
Hence aa1 = e = a1a for any a ∈ G
i.e., for any a ∈ G, 
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
then value of a for which A2= B, is _____
is given by ______
, then which of the following(s) is/are true for {xn}?
Let G be a group of order 143, then the centre of G is isomorphic to- a)Z
- b)Z11
- c)Z13
- d)Z143
Correct answer is option 'D'. Can you explain this answer?
Let G be a group of order 143, then the centre of G is isomorphic to
a)
Z
b)
Z11
c)
Z13
d)
Z143
|
|
Chirag Verma answered |
O(G) = 143
i.e. O(G) =11 x 13 (p < q)
Here.O(G) = pq , p < q and p X q -1, so G be a cyclic group of order 143 and every cyclic group is an abelian group.
Hence G is an abelian group, if G be an abelian group then centre of G is equal to group G.
G = Z(G) [∵ G is an abelian]
i.e. O[Z(G)] = 143 so Z(G) is isomorphic to Z143.
i.e. O(G) =11 x 13 (p < q)
Here.O(G) = pq , p < q and p X q -1, so G be a cyclic group of order 143 and every cyclic group is an abelian group.
Hence G is an abelian group, if G be an abelian group then centre of G is equal to group G.
G = Z(G) [∵ G is an abelian]
i.e. O[Z(G)] = 143 so Z(G) is isomorphic to Z143.
Let GL(2, Z5) is a group for multiplication of matrix operation such that ad - be ≠ 0 then order of the group GL(2,Z5) is
Correct answer is '480'. Can you explain this answer?
Let GL(2, Z5) is a group for multiplication of matrix operation such that ad - be ≠ 0 then order of the group GL(2,Z5) is
|
|
Chirag Verma answered |
We know that the order of general linear group GL(n, Zp) is
= (Pn-1) (Pn - P) (Pn - P2)_____ (Pn - Pn-1)
Here P = 5 , n = 2
0[GL(2,Z5)] = (52 - 1) (52 - 5)
= 24 x 20
= 480
= (Pn-1) (Pn - P) (Pn - P2)_____ (Pn - Pn-1)
Here P = 5 , n = 2
0[GL(2,Z5)] = (52 - 1) (52 - 5)
= 24 x 20
= 480
Which of the following is a 2 -dimensional subspace of R3 over R ?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Which of the following is a 2 -dimensional subspace of R3 over R ?
a)
b)
c)
d)
|
|
Chirag Verma answered |
Here (a) be a subspace but basis of it is {(0. 1, 0)} so it be a subspace of dimension 1.
(b), (c) are not subspace, so they don't have dim.
Now (d) be a subspace and its basis is {(0, 0. 1) (0, 1. 0)} so it be a subspace with dimension 2.
(b), (c) are not subspace, so they don't have dim.
Now (d) be a subspace and its basis is {(0, 0. 1) (0, 1. 0)} so it be a subspace with dimension 2.
Let ( y - C)2 = Cx be the primitive of 
The no. of integral curves which will pass through (1, 2) is,- a)1
- b)2
- c)3
- d)0
Correct answer is option 'D'. Can you explain this answer?
Let ( y - C)2 = Cx be the primitive of The no. of integral curves which will pass through (1, 2) is,
The no. of integral curves which will pass through (1, 2) is,a)
1
b)
2
c)
3
d)
0
|
Sravya Mehta answered |
We have y(1) = 2
We have no real value of C.
We have no curve which passing through (1, 2).
We have no curve which passing through (1, 2).
Which one of the following is a group?- a)(N, *) where a * b = a for all a, b ∈ N
- b)(Z, *) where a + b = a - b for all a, b ∈ Z
- c)(Q, *) where a * b = ab/2 for all a, b ∈ Q
- d)(R, *) where a + b = a + b + 1 for all a, b ∈ R
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is a group?
a)
(N, *) where a * b = a for all a, b ∈ N
b)
(Z, *) where a + b = a - b for all a, b ∈ Z
c)
(Q, *) where a * b = ab/2 for all a, b ∈ Q
d)
(R, *) where a + b = a + b + 1 for all a, b ∈ R
|
|
Chirag Verma answered |
(N, *) 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) (a * b) * c≠a*(b*c)
Hence the property of associativity is not satisfied.
=> (Z, *) is not a group.
(C) (Q, *) where 
for all a, b ∈ Q. This is not a group since inverse of element 0 ∈ Q does not exist.
for all a, b ∈ Q. This 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
⇒ a + e + 1 = a
⇒ e + 1 = a - a
⇒ e + 1 =0
⇒ e = - 1
Since - 1 ∈ Q, so e ∈ Q, So (R, *) is a group with identity - 1 .
The area between the curves y = xex and y = xe-x and the line x = 1 is __________ ( Write upto Four decimal Places)
Correct answer is '0.7357'. Can you explain this answer?
The area between the curves y = xex and y = xe-x and the line x = 1 is __________ ( Write upto Four decimal Places)
|
|
Chirag Verma answered |
The line x = 1 meets the curves in (1,e) and B(1, 1/e). The second curve is lower cuive as ordinate of B is less than ordinate of A.
Both the curves pass through origin.
Both the curves pass through origin.
= 0.7357
{0, 1, 2, 3, 4, 5} is a group under addition modulo 6, then the order of 1 is ___________.
Correct answer is '6'. Can you explain this answer?
{0, 1, 2, 3, 4, 5} is a group under addition modulo 6, then the order of 1 is ___________.
|
|
Chirag Verma answered |
(1)6 = 1 + 1 + 1 + 1 + 1 +1 = 6 (mod 6) = 0
∴ Older of 1 is 6.
∴ Older of 1 is 6.
Find the length of the arc of the semi cubical parabola ay2 = x3 from its vertex to the point (a, a).- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Find the length of the arc of the semi cubical parabola ay2 = x3 from its vertex to the point (a, a).
a)
b)
c)
d)
|
Veda Institute answered |
Differentiating the equation of the curve
Therefore the required length
How many numbers satisfied the equation x ≌ 7 (mod 17), where x in the range 1 < x < 100.- a)4
- b)5
- c)6
- d)7
Correct answer is option 'C'. Can you explain this answer?
How many numbers satisfied the equation x ≌ 7 (mod 17), where x in the range 1 < x < 100.
a)
4
b)
5
c)
6
d)
7
|
Devansh Choudhary answered |
The equation x ≡ 7 (mod 17) can be rewritten as x = 17k + 7, where k is an integer.
To find the range of x, we can substitute the given range 1 ≤ x ≤ 100 into the equation:
1 ≤ 17k + 7 ≤ 100
Subtracting 7 from all parts of the inequality:
-6 ≤ 17k ≤ 93
Dividing by 17:
-6/17 ≤ k ≤ 93/17
Since k must be an integer, we can round down the lower bound and round up the upper bound:
-1 ≤ k ≤ 5
Now we can find the number of values of x within this range:
For k = -1, we have x = 17(-1) + 7 = 0, which is not within the given range.
For k = 0, we have x = 17(0) + 7 = 7.
For k = 1, we have x = 17(1) + 7 = 24.
For k = 2, we have x = 17(2) + 7 = 41.
For k = 3, we have x = 17(3) + 7 = 58.
For k = 4, we have x = 17(4) + 7 = 75.
For k = 5, we have x = 17(5) + 7 = 92.
Therefore, there are 6 numbers (7, 24, 41, 58, 75, 92) that satisfy the equation x ≡ 7 (mod 17) within the given range 1 ≤ x ≤ 100.
To find the range of x, we can substitute the given range 1 ≤ x ≤ 100 into the equation:
1 ≤ 17k + 7 ≤ 100
Subtracting 7 from all parts of the inequality:
-6 ≤ 17k ≤ 93
Dividing by 17:
-6/17 ≤ k ≤ 93/17
Since k must be an integer, we can round down the lower bound and round up the upper bound:
-1 ≤ k ≤ 5
Now we can find the number of values of x within this range:
For k = -1, we have x = 17(-1) + 7 = 0, which is not within the given range.
For k = 0, we have x = 17(0) + 7 = 7.
For k = 1, we have x = 17(1) + 7 = 24.
For k = 2, we have x = 17(2) + 7 = 41.
For k = 3, we have x = 17(3) + 7 = 58.
For k = 4, we have x = 17(4) + 7 = 75.
For k = 5, we have x = 17(5) + 7 = 92.
Therefore, there are 6 numbers (7, 24, 41, 58, 75, 92) that satisfy the equation x ≡ 7 (mod 17) within the given range 1 ≤ x ≤ 100.
with real constant coefficients, then the least possible value of n is _________.
Let {un} be a sequence of real number's. If ∑|un| is convergent series. Then which of the following's is/are true?- a)
is convergent - b)
divergent - c)
is convergent - d)
is convergent
Correct answer is option 'A,C,D'. Can you explain this answer?
Let {un} be a sequence of real number's. If ∑|un| is convergent series. Then which of the following's is/are true?
a)
is convergent b)
divergentc)
is convergent d)
is convergent |
|
Chirag Verma answered |
Now given ∑|un| Convergent and we know ∑1/n2 is convergent so by comparison test
is Convergent.
If y1 and y2 are two solution of the differential equation 
. Then which of the followings can be given as general soln of this differential eqn.- a)y(x) = c1 y1
- b)y(x) = c2 y2
- c)y(x) = y1 + C(y1 - y2)
- d)None of these
Correct answer is option 'A,B,C'. Can you explain this answer?
If y1 and y2 are two solution of the differential equation . Then which of the followings can be given as general soln of this differential eqn.
. Then which of the followings can be given as general soln of this differential eqn.a)
y(x) = c1 y1
b)
y(x) = c2 y2
c)
y(x) = y1 + C(y1 - y2)
d)
None of these
|
|
Chirag Verma answered |
Option (a), (b) are clearly correct
Now we have
Now from (4) and (5)
If f(x) = (x2 - 1) |x2 - 3x + 2| + cos|x| then the set of point of non-differentiability is,- a){1,0,2}
- b){1,2}
- c){2}
- d){0,2}
Correct answer is option 'C'. Can you explain this answer?
If f(x) = (x2 - 1) |x2 - 3x + 2| + cos|x| then the set of point of non-differentiability is,
a)
{1,0,2}
b)
{1,2}
c)
{2}
d)
{0,2}
|
Ojasvi Malhotra answered |
Point of Non-Differentiability for f(x) = (x^2 - 1) |x^2 - 3x + 2| + cos|x|
- Finding the Points of Non-Differentiability
To find the points of non-differentiability for the given function f(x), we need to consider the points where the function is not differentiable.
- Breakdown of the Function
The given function f(x) consists of three parts:
1. (x^2 - 1)
2. |x^2 - 3x + 2|
3. cos|x|
- Examining Each Part
1. The function (x^2 - 1) is a polynomial function and is differentiable for all real numbers.
2. The absolute value function |x^2 - 3x + 2| will change its behavior at the roots of the expression inside the absolute value.
Solving x^2 - 3x + 2 = 0 gives x = 1 and x = 2. Therefore, the function is non-differentiable at x = 1 and x = 2.
3. The cosine function cos|x| is continuous and differentiable for all real numbers.
- Conclusion
Therefore, the points of non-differentiability for the function f(x) = (x^2 - 1) |x^2 - 3x + 2| + cos|x| are x = 1 and x = 2. Hence, the correct answer is option 'c) {2}'.
- Finding the Points of Non-Differentiability
To find the points of non-differentiability for the given function f(x), we need to consider the points where the function is not differentiable.
- Breakdown of the Function
The given function f(x) consists of three parts:
1. (x^2 - 1)
2. |x^2 - 3x + 2|
3. cos|x|
- Examining Each Part
1. The function (x^2 - 1) is a polynomial function and is differentiable for all real numbers.
2. The absolute value function |x^2 - 3x + 2| will change its behavior at the roots of the expression inside the absolute value.
Solving x^2 - 3x + 2 = 0 gives x = 1 and x = 2. Therefore, the function is non-differentiable at x = 1 and x = 2.
3. The cosine function cos|x| is continuous and differentiable for all real numbers.
- Conclusion
Therefore, the points of non-differentiability for the function f(x) = (x^2 - 1) |x^2 - 3x + 2| + cos|x| are x = 1 and x = 2. Hence, the correct answer is option 'c) {2}'.
If f(x) = xα In x and f(0) = 0, then the value of α for which Rolle’s theorem can be applied in [0,1] is,- a)-2
- b)-1
- c)0
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
If f(x) = xα In x and f(0) = 0, then the value of α for which Rolle’s theorem can be applied in [0,1] is,
a)
-2
b)
-1
c)
0
d)
1/2
|
|
Chirag Verma answered |
The function has to be continuous in [0,1]
If G is a group, Z its center and if G/Z is cyclic then G- a)must be abelian
- b)must be non abelian
- c)must be normal subgroup
- d)must be subgroup
Correct answer is option 'A'. Can you explain this answer?
If G is a group, Z its center and if G/Z is cyclic then G
a)
must be abelian
b)
must be non abelian
c)
must be normal subgroup
d)
must be subgroup
|
|
Chirag Verma answered |
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
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 = Zgm [∵ Z is a normal subgroup of G]
Again
a ∈ Za and Za = Zgm ⇒ a ∈ Z gm.
Now
a ∈ Z g m ⇒ 
z1 ∈ Z such that a ∈ z1 gm
z1 ∈ Z such that a ∈ z1 gmSimilarly, for b ∈ G, b = z2 gn, where z2 ∈ z and n is any integer.
Now
Again
For the three equations
-2x + y + z = a
x - 2y + z = b
x + y - 2z = c- a)have no solution unless a + b + c = 0
- b)when a + b + c = 0, then solution exist.
- c)have solution when a + b + c ≠ 0
- d)when a + b + c ≠ 0, then solution exist
Correct answer is option 'A,B'. Can you explain this answer?
For the three equations
-2x + y + z = a
x - 2y + z = b
x + y - 2z = c
-2x + y + z = a
x - 2y + z = b
x + y - 2z = c
a)
have no solution unless a + b + c = 0
b)
when a + b + c = 0, then solution exist.
c)
have solution when a + b + c ≠ 0
d)
when a + b + c ≠ 0, then solution exist
|
|
Chirag Verma answered |
Take augmented matrix
then Rank A = Rank M = 2,then system has solution otherwise no solution.
So if a + b + c = 0 then solution is exist.
So option A and B are correct.
Total number of non abelian groups of order 23 . 34. 5 is __________.
Correct answer is '15'. Can you explain this answer?
Total number of non abelian groups of order 23 . 34. 5 is __________.
|
Vivek Kumar answered |
Partitions of 3 are three namely
3 = 3,2 + 1, 1 +1 +1.
Partition of 4 are five namely
4 = 4, 3 + 1, 2 + 2, 2 + 1 +1, 1 +1 +1 +1.
Partition of 1 is only one 1 = 1
Thus there are (I) non isomorphic abelian group of order 2.
(II) five non isomorphic abelian group of order 34. and
(III) One abelian group of order 5.
Hence total number of non abelian group of order
23. 34. 5 = 3 x 5 x 1 = 1
=15
Let G be a group of 35 elements. Then the largest possible size of a subgroup of G other than G itself is ______.
Correct answer is '7'. Can you explain this answer?
Let G be a group of 35 elements. Then the largest possible size of a subgroup of G other than G itself is ______.
|
|
Chirag Verma answered |
Group G has 35 elements, i.e. its order is 35
So, possible subgroup sizes can be 1, 5, 7, 35.
Thus the largest possible size of subgroup other than G itself (proper subgroup) is 7.
Find the total number of cyclic subgroups of order 10 in 
Correct answer is '6'. Can you explain this answer?
Find the total number of cyclic subgroups of order 10 in
|
|
Veda Institute answered |
Case - I
|a|= 10 and|b| = 1 or 5. since z100 has a unique cyclic subgroup of order 10 and cyclic group of order 10 has four generators, i.e. there are four choices for a Similarly, there are five choices for b. This gives 20 possibilities for (a,b).
Case - II lal = 2 and|b| = 5 .
Since any finite cyclic group of even order has a unique subgroup of order 2, there is only one choice for a.
So, obviously there are four choices for b.
Hence
has 24 elements of order 10. Because each cyclic subgroup of order 10 has four elements of order 10 and no two of the cyclic subgroups can have an element of order 10 in common. There must be 24/4 = 6 cyclic subgroups of order 10.
|a|= 10 and|b| = 1 or 5. since z100 has a unique cyclic subgroup of order 10 and cyclic group of order 10 has four generators, i.e. there are four choices for a Similarly, there are five choices for b. This gives 20 possibilities for (a,b).
Case - II lal = 2 and|b| = 5 .
Since any finite cyclic group of even order has a unique subgroup of order 2, there is only one choice for a.
So, obviously there are four choices for b.
Hence
has 24 elements of order 10. Because each cyclic subgroup of order 10 has four elements of order 10 and no two of the cyclic subgroups can have an element of order 10 in common. There must be 24/4 = 6 cyclic subgroups of order 10.Let T = { z ∈ c : | z | = 1} then (T, x) is a group which o f the following are not subgroup of (T, x)- a)({1, -1, i, -i}, x)
- b)({1.-1}. X)
- c)
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Let T = { z ∈ c : | z | = 1} then (T, x) is a group which o f the following are not subgroup of (T, x)
a)
({1, -1, i, -i}, x)
b)
({1.-1}. X)
c)
d)
None of these
|
|
Mister Genius answered |
If T = {z ∈ s c : | z | = 1} then (T, x) is said to be a circle group All though the circle group has infinite order it has many finite subgroups as the complex numbers satisfying the equation zn = 1 are called the nth root of unity. They are
Number of generators of group Z16- a)7
- b)8
- c)9
- d)16
Correct answer is option 'B'. Can you explain this answer?
Number of generators of group Z16
a)
7
b)
8
c)
9
d)
16
|
|
Chirag Verma answered |
Since Z16 = {0, 1 , 2 ..... . 15}
The numbers 1, 3, 5, 7, 9, 11, 13,15 are the elements of Z16 that are relatively prime to 16. Each of these elements generates Z16.
So no. of generators = 8
So no. of generators = 8
Let T : V → w be a linear transformation, where V is a finite dimensional vector space. Let B be a subspace of W, then- a)T-1 (B) = φ
- b)T-1 (B) is a subspace of V
- c)dim ( ker T ) ≤ dim (T-1 (B))
- d)dim(T-1 (B)) ≤ dim (ker T)
Correct answer is option 'B,C'. Can you explain this answer?
Let T : V → w be a linear transformation, where V is a finite dimensional vector space. Let B be a subspace of W, then
a)
T-1 (B) = φ
b)
T-1 (B) is a subspace of V
c)
dim ( ker T ) ≤ dim (T-1 (B))
d)
dim(T-1 (B)) ≤ dim (ker T)
|
|
Chirag Verma answered |
Since T is a Linear Transformation ⇒ T(0) = 0 ' ∈w
Hence T(O) = 0 ' ∈ B (∴ B be a subspace of W ).
Hence T(O) = 0 ' ∈ B (∴ B be a subspace of W ).
If the equation of the tangent to the curre y2 = ax3 + b at the point (2,3) is y = 4x - 5 then the value of a + b is ________.
Correct answer is '-5'. Can you explain this answer?
If the equation of the tangent to the curre y2 = ax3 + b at the point (2,3) is y = 4x - 5 then the value of a + b is ________.
|
|
Chirag Verma answered |
also (2,3) lies on y2 = ax3 + b
⇒ 9 = 8a + b ⇒ b = - 7
⇒a + b = -5
The system of equation kx + y + z = 1, x + ky + z = k, and x + y + kz = k2, does not have a solution, if k is equal to _____.
Correct answer is '-2'. Can you explain this answer?
The system of equation kx + y + z = 1, x + ky + z = k, and x + y + kz = k2, does not have a solution, if k is equal to _____.
|
|
Chirag Verma answered |
we have
thus the augmented matix is given by,
clearly the system has no solution if k = -2
thus the augmented matix is given by,
clearly the system has no solution if k = -2
Let {an} be a sequence of real number's defined as 
If{an} is convergent then it converge to - a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Let {an} be a sequence of real number's defined as If{an} is convergent then it converge to
If{an} is convergent then it converge to a)
b)
c)
d)
|
|
Chirag Verma answered |
Putting an = rn in the given recurrence relation, we get
Let y(x) = c1 f(x) + c2 g(x) be the general solution of (x + 2) y" - (4x + 9)y' + (3x + 7)y = 0, then f(0) + g(0) is equal to _________.
Correct answer is '1.75'. Can you explain this answer?
Let y(x) = c1 f(x) + c2 g(x) be the general solution of (x + 2) y" - (4x + 9)y' + (3x + 7)y = 0, then f(0) + g(0) is equal to _________.
|
|
Chirag Verma answered |
we have
Here
⇒ 1 + P + Q = 0
⇒ f(x) = ex
Now
f(0) + y(0) = e0 + e0 (0 + 3) = 4
or if we take
Solution of the differential equation 
- a)y2 = c1x
- b)y2 = c1 x2+c2x3
- c)y2 = c2x2+2c1x
- d)y2= x(c2,x - 2c1)
Correct answer is option 'D'. Can you explain this answer?
Solution of the differential equation
a)
y2 = c1x
b)
y2 = c1 x2+c2x3
c)
y2 = c2x2+2c1x
d)
y2= x(c2,x - 2c1)
|
|
Chirag Verma answered |
The given equation can be re-written as
Integrating
Putting y2 = v so that 2y (dy/dx) = (dv/dx). We then get
which is linear. Its I.F = 
If the matrix B is obtained from the matrix A by interchanging two rows, then-- a)
- b)
- c)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If the matrix B is obtained from the matrix A by interchanging two rows, then-
a)
b)
c)
d)
None of these
|
|
Chirag Verma answered |
Interchanging two rows (or columns) change the sign of determinant.
is equal to _________, where 
The total number of elements of order 5 in 
_____.
Correct answer is ''. Can you explain this answer?
The total number of elements of order 5 in _____.
_____.|
|
Vivek Kumar answered |
Here we count the number of elements (a,b) in 
with the property that
clearly this requires that either |a| = 5 and |b| = 1 or 5 or |b| = 5 and |a|=1 or 5. Here, we consider three mutually exclusive cases.
Case-I |a| = 5 and.|b| = 5 .Here there are four choices for a(5.10,15.20) and four choices for b. This gives 16 elements of order 5.
Case-ll |a| = 5 and |b| = 1 ,So there are four choices for a and only one for b.
This gives four more elements of order 5.
Case-Ill a| = 1 and |b| = 5. here only one choice for a and four choices for b. so we obtain four more elements of order 5.
Hence
has 24 elem ents of order 5.
with the property thatclearly this requires that either |a| = 5 and |b| = 1 or 5 or |b| = 5 and |a|=1 or 5. Here, we consider three mutually exclusive cases.
Case-I |a| = 5 and.|b| = 5 .Here there are four choices for a(5.10,15.20) and four choices for b. This gives 16 elements of order 5.
Case-ll |a| = 5 and |b| = 1 ,So there are four choices for a and only one for b.
This gives four more elements of order 5.
Case-Ill a| = 1 and |b| = 5. here only one choice for a and four choices for b. so we obtain four more elements of order 5.
Hence
has 24 elem ents of order 5.Changing the order of the integration 
gives- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Changing the order of the integration gives
givesa)
b)
c)
d)
|
|
Chirag Verma answered |
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
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
and for area AOCD upper and lower limits of x is and limits of y is O to -1Hence, we have
How many generators are exists for the cyclic group of order 26.
Correct answer is '12'. Can you explain this answer?
How many generators are exists for the cyclic group of order 26.
|
|
Chirag Verma answered |
Let G = [ a ] be a cyclic group . then generators of this group is of the from am
where m < 26 . [ Co-prime of 26 ]
Hence m = 1, 3 ,5 , 7 , 9 , 1 1 , 15 ,17 ,1 9,2 1 ,23 ,25 total generators are 12.
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