Algebra and Mathematical Method Solved Paper 2023-24
Algebra and Mathematical Method Solved Paper 2023-24
4253
B. Sc. (Semester-III) Examination - 2023
Mathematics
Paper : I
Solved Paper
Algebra & Mathematical Method
Time: Two Hours Maximum Murks:75
Note : Attempt the questions from all sections as directed.
नोट : सभी खण्डों से निर्देशानुसार प्रश्नों के उत्तर दीजिए।
Inst. The candidates are required to answer only in serial order. If there are many parts of a question answer them in continuation.
नोट: अभ्यर्थी प्रश्नों के उत्तर क्रमानुसार लिखें। यदि किसी प्रश्न के कई भाग हो तो उनके उत्तर एक ही तारतम्य में लिखे जाए।
Section -A
खण्ड अ
(Very Short Answer Type Questions )
( अति लघु उत्तरीय प्रश्न)
Note : Attempt any three out of five questions. Each question carries 3 marks. Write answer of each question in about 50 words. (3x3=9)
नोट 5 प्रश्नों में से किन्हीं 3 प्रश्नों को हल कीजिए। प्रत्येक प्रश्न 3 अंकों का है। प्रत्येक प्रश्न की शब्द सीमा 50 शब्दों की है।
1 (a) Use Fermat's theorem to find the remainder of 9107 , when divided by 11.
Solution:- By Fermat's theorem, we have
aP - 1 ≡ 1 (mod P)
Here P = 11 so that a10 ≡ 1 (mod 11), where a is any integer. Taking a = 9, we get that, when 910 is divided by 11 then remainder is 1. So that
9107 = (910)10 97 = 4782969
Which on dividing by 11 gives 4 as remainder.
(b) Define equivalence relation. Give an example of equivalence relation.
Solution:- Let A be a non empty set. A relation R in A is called :
Reflexive: if (a, a) ∈ R for all a ∈ A
Symmetric: if whenever (a, b) ∈ R then (b, a) ∈ R
Anti-Symmetric: if (a, b) ∈ R, (b, a) ∈ R a = b
Transitive: if whenever (a, b), (b, c) ∈ R then (a, c) ∈ R
A relation R is called an equivalence relation if it is reflexive, symmetric and transitive.
(c) Define cyclic group with examples.
Solution: Cyclic Group –
A group 'G' is called cyclic if 'y' every element of 'G' can be written as an integral power of some element of itself. This element 'G' is known as the generator of the cyclic group.
Example:- 1. The multiplication group G = {1, ω, ω2}, is a cyclic group with generators. {ω} and {ω2}.
(d)
(e) If f = tan-1(y/x), verify that
∂²f/∂y∂x = ∂²f/∂x∂y
Solution:- Given f = tan–1 (y/x) ….(i)
Differentiating (1) partially w.r.t. x and y we get
(iv)
(v)
From (iv) and (v) we have
∂²f/∂y∂x = ∂²f/∂x∂y
Section - B
खण्ड - ब
(Short Answer Type Questions)
( लघु उत्तरीय प्रश्न )
Note : Attempt any four out of seven questions. Each question carries 9 marks. Write answer of each question in about 225 words. ( 9 x 4)
नोट : 7 प्रश्नों में से किन्हीं 4 प्रश्नों को हल कीजिए प्रत्येक प्रश्न 9 अंकों का है। प्रत्येक प्रश्न की शब्द सीमा 225 शब्दों की है।
2. (a) State and Prove Fermat and Euler's theorem.
Solution:-
Fermat Theorem:- If p is a prime and a is a positive integer with p ł a, then
ap-1 ≡ 1 (mod p)
Proof:- We begin by considering the first p - 1 positive multiples of a; that is, the integers. a, 2a, 3a,......... (p-1) a
None of these numbers is congruent modulo p to any other, nor is congruent to zero. Indeed, if it happened that
ra ≡ sa(mod p). 1 ≤ r < s ≤ p-1
then a could be canceled to give
r s ≡ (mod p), which is impossible.
Therefore,the previous set of integers must be congruent modulo p to 1, 2, 3, ..., p - 1,taken in some order.
Multiplying all these congruences together, we find that
a.2a.3a…….(p-1)a ≡ 1.2.3……(p-1)(mod p)
whence
ap-1 (p-1)! ≡ (p-1)!(mod p)
Once (p-1)! is canceled from both sides of the preceding congruence (this is possible because since p x (p − 1)!),
ap-1 ≡ 1 (mod p)
Which is Fermat Theorem
Euler's theorem:- If n is a positive integer and a is an integer such that g.c.d (a,n) = 1, then
a𝞍(n) = 1(mod n)
Or
If n ≥ 1 and g.c.d (a,n) = 1 then
a𝞍(n) = 1(mod n)
Proof:- There is no harm of taking n > 1 . Let a₁ , a₂, a₃ , a₄ , ……aΦ(n) be the positive integer less than n that are relatively prime to n . Because g.c.d (a,n) = 1 , it follows from lemma that aa₁, aa₂, aa₃…….aaΦ(n) are congruent, not necessarily in order of apperance, to aa₁, aa₂, aa₃…….aaΦ(n) . Then
aa₁ ≡ a₁' (mod n)
aa₂ ≡ a₂' (mod n)
aa₃ ≡ a₃' (mod n)
------------------
-------------------
aaΦ(n) ≡ a'Φ(n) (mod n)
where a₁' , a₂' ………a'Φ(n) are the integers a₁ , a₂, a₃ , a₄ , ……aΦ(n) in some order. On taking the product of these Φ(n) congruences, we get,
(aa₁) (aa₂)(aa₃)…….(aaΦ(n) )≡ a₁' , a₂'.....a'Φ(n) (mod n)
a𝞍(n) (a₁,a₂,a₃……aΦ(n) ) = a₁,a₂,a₃……aΦ(n) (mod n)
Becauce gcd (a,n) = 1 for i gcd (a₁,a₂,a₃……aΦ(n) , n ) = 1
Therefore , we may divide both sides of the congruences
a𝞍(n) = 1(mod n)
(b) Define the Normal subgroups and Prove
that Intersection of two normal subgroups
of a group is a normal subgroup.
Solution : Normal Subgroubs:-
First Part:
A subgroup H of a group G is said to be a normal subgroup if ghg⁻¹ ∈ H for each g ∈ G and h ∈ H or equivalently if gHg⁻¹ ⊂ H for each g ∈ G.
Second Part:
Let H₁ and H₂ be two normal subgroups of a group G. Then we know that H₁∩ H₂ is a subgroup of G
Now we have to prove that H1∩H2 is a normal subgroup of G. For this, we see that
g ∈ G and h ∈ H₁ ∩ H₂ ⇒g ∈ G
and h ∈ H₁ and h ∈ H₂
⇒ ghg⁻¹∈ H₁ and ghg⁻¹ ∈ H₂
(as H₁ and H₂ both are normal)
⇒ ghg–1 ∈ H₁ ∩ H₂
This shows that H₁∩ H₂ is normal.
(c) Discuss the maximum or minimum values
u = x³ + y³ - 3axy
Solution: Given, f (x, y) = x³ + y³ - 3axy
On solving equations (1) and (2), we get stationary point, which is (a, a)
Hence, f is maximum or minimum according to a < 0 or a > 0.
(d)
Hence S is a subring of R.
Solution : First Part
Second Part:
(f) Prove the convolution or Falting theorem for Fourier Transforms.
Solution:-
Convolution or Falting. The convolution of two functions F(x) and G (x), where x , is denoted and defined as
Or
The convolution or Falting theorem for Fourier transforms:-
The Fourier transform of the convolution of F(x) and G(x) is the product of the Fourier transforms of F(x) and G(x), i.e.,
F{F * G} = F {F(x)} F{G(x)}.
Proof:- Definition of convolution
Definition of Fourier transform
[ changing the order of integration]
putting x – u = v so that dx =dv
[ By definition of Fourier transform]
Again differentiating (iii) w.r.t. 'x' and 'y' partially, we get
Multiplying (vii) by x and (viii) by y and adding, we get
Section -C
खण्ड - स
(Long Answer Questions )
(दीर्घ उत्तरीय प्रश्न)
Note: Attempt any two questions out of four questions. Each question earries 15 marks . Write answer of each question in about. 475 words.
( 15 x 2 = 30 )
3 (a) State and Prove Fundamental Theorem of Homomorphism.
or
A homomorphic image of a group is
isomorphic to a quotient group of the group.
or
If G and G' are group and f : G → G' and onto homomorphism, then the group G' is isomorphic to the quotient group G/K of G by K, where K is the Kernel of K.
Solution:- If G' is isomorphic to G/K is equivalent to proving that G/K is isomorphic to G'. And to show that G/K is isomorphic to G', define a map φ : G/K → G' by
φ (gK) = f (g) ∀ g ∈ G.
Then, it is sufficient to prove that the map φ is well defined and is an isomorphism.
(i) φ is well defined: For this, we have to show that if g1,K = g2,K then φ (g1, K) = φ (g2,K), where g1, g2 ∈ G. Thus we get it as below.
(ii) φ is a homomorphism– For g1K, g2K ∈ G/K, we have
φ (g1K . g2K) = φ(g1g2 K)
= f(g1g2) (by def. of φ)
= f(g1) f(g2) (as f is a homomorphism)
= φ (g1K) φ (g2K), (by def. of φ)
(iii) φ is one-one– For g1K, g2K ∈ G/K,
where φ(g1K) = φ(g2K) = f(g1) = f(g2) (by de of φ)
⇒ {f(g1)}-1 f(g2) = e'
⇒ {f(g1-1)} f(g2) = e'
⇒ f(g1-1 g2) = e'
⇒ g1-1 g2 = e'
⇒ g1K = g2K
Thus φ is one–one.
(iv) φ is on to – Since f is on to, therefore given g ∈ G', ∃ g ∈ G such that f(g) = g'. Hence, given g' ∈ G', ∃ gK ∈ G/K such that φ (gK) = f(g) = g'. This shows that φ is onto.
Thus, φ being a one–one onto homomorphism is an isomorphism from G/K to G'. Hence G' is isomorphic to G/K.
(b) If H is a subgroup of a group G and a, b ∈ G, then
(i) aH = bH if and only if a-1 b ∈ H
(ii) Ha = Hb if and only if ab-1 ∈H
Proof.: (i) Sppose that aH = bH. Then we have
ah ∈ aH ⇒ ah ∈ bH
⇒ ah = bh1 for some h1∈ H
⇒ h = a-1 (bh1) = (a-1b) h1
h h1-1 = a-1b. (Because h h1-1 ∈ H h h1 ∈ H)
Conversely, suppose that a-1 b∈H. Then, a-1b = h for some h ∈ H1 , so that b = ah for some h ∈ H. Therefore
bH = (ah)H = a(hH) = aH as h ∈ H = H
Hence aH = bH ⇔ a-1b ∈ H
Proof– (ii) It is simila to that of (i).
(c) Write down Taylor's theorem for the function of two variables. Expand the function f(x, y) = tan⁻¹ (y/x) in the neighbourhood of the point (1, 1) upto second degree terms using Taylor's theorem. Hence computer f(1.1, 0.9) approximately.
Sol.: Part I Taylor's theorem for functions of two variables:
If f(x, y) and all its partial derivatives upto n th order are finite and continuous for all points (x, y), where a ≤ x ≤ a + h and b ≤ y ≤ b + K.
In can be written as in the following form.
Part II. f(x, y) = tan⁻¹ (y/x)
x = 1, y = 1
By Tayor's theorem,
Sol.: Consider the function
| Course name | Paper name | Download PDF |
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| Bsc Third Semester Paper 2023-24 | Algebra and mathematical method | Click here |
| Bsc Third Semester Solved Paper 2023-24 | Algebra and mathematical method | Click here |
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