2023 question paper

Discrete Mathematics

19 questions

  1. 1a. Let A be the set odd positive integers less than 10. Then cardinality of A, ∣A∣ is (i) 5 (ii) 9 (iii) 6 (iv) 420232m

    Module 1: Sets, Relation and Function.

    Let A be the set odd positive integers less than 10. Then cardinality of A, ∣A∣ is

    (i) 5
    (ii) 9
    (iii) 6
    (iv) 4

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    Worked Solution

    Answer

    Correct Option: (i) 5

    Explanation

    A={x∈Z+∣x is odd and x<10}A = \{ x \in \mathbb{Z}^+ \mid x \text{ is odd and } x < 10 \}

    Listing elements:
    A={1,3,5,7,9} A = \{1, 3, 5, 7, 9\}

    Cardinality ∣A∣|A| is the number of distinct elements in AA.

    Counting:
    ∣A∣=5 |A| = 5

    Final Answer

    ∣A∣=5|A| = 5 → Option (i) 5

  2. 1b. If m is the number of objects (pigeons) and n is the number of boxes (pigeonholes), then the function is both one-to-one and onto if (i) m<n (ii) m=n (iii) m>n (iv) none of these20232m

    Module 1: Sets, Relation and Function.

    If m is the number of objects (pigeons) and n is the number of boxes (pigeonholes), then the function is both one-to-one and onto if
    (i) m<n
    (ii) m=n
    (iii) m>n
    (iv) none of these

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  3. 1c. If A \times B = B \times A, (Where A and B are general matrices) then (i) A=I (ii) A=B' (iii) B=A (iv) A'=B20232m

    Module 4: Algebraic Structures & Boolean Algebra.

    If A×B=B×AA \times B = B \times A, (Where A and B are general matrices) then
    (i) A=IA=I
    (ii) A=B′A=B'
    (iii) B=AB=A
    (iv) A′=BA'=B

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  4. 1g. If f(x) = \cos x and g(x) = x^3 then (f \circ g)(x) is (i) (\cos x)^3 (ii) \cos 3x (iii) x^{(\cos x)^3} (iv) \cos x^320232m

    Module 1: Sets, Relation and Function.

    If f(x)=cos⁡xf(x) = \cos x and g(x)=x3g(x) = x^3 then (f∘g)(x)(f \circ g)(x) is
    (i) (cos⁡x)3(\cos x)^3
    (ii) cos⁡3x\cos 3x
    (iii) x(cos⁡x)3x^{(\cos x)^3}
    (iv) cos⁡x3\cos x^3

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  5. 1h. The number of distinguishable permutations of the letters in the word BANANA are (i) 60 (ii) 36 (iii) 20 (iv) 1020232m

    Module 1: Sets, Relation and Function.

    The number of distinguishable permutations of the letters in the word BANANA are
    (i) 60
    (ii) 36
    (iii) 20
    (iv) 10

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  6. 1i. Which of the following pair is not congruent modulo 7? (i) 10, 24 (ii) 25, 56 (iii) -31, -15 (iv) -64, -1520232m

    Module 4: Algebraic Structures & Boolean Algebra.

    Which of the following pair is not congruent modulo 7?
    (i) 10, 24
    (ii) 25, 56
    (iii) -31, -15
    (iv) -64, -15

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  7. 1j. Let N = \{1,2,3,......\} be ordered by divisibility, which of the following subset is totally ordered (i) \{2, 6, 24\} (ii) \{3, 5, 15\} (iii) \{2, 9, 16\} (iv) \{4, 15, 30\}20232m

    Module 1: Sets, Relation and Function.

    Let N={1,2,3,......}N = \{1,2,3,......\} be ordered by divisibility, which of the following subset is totally ordered
    (i) {2,6,24}\{2, 6, 24\}
    (ii) {3,5,15}\{3, 5, 15\}
    (iii) {2,9,16}\{2, 9, 16\}
    (iv) {4,15,30}\{4, 15, 30\}

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  8. 2a. a) Let A = B = \{x | -1 \leq x \leq 1\} for each of the following functions state where it is injective, surjective or bijective i) g(x) = \sin \pi x ii) h(x) = \frac{2x}{3}20237m

    Module 1: Sets, Relation and Function.

    a) Let A=B={x∣−1≤x≤1}A = B = \{x | -1 \leq x \leq 1\} for each of the following functions state where it is injective, surjective or bijective
    i) g(x)=sin⁡πxg(x) = \sin \pi x
    ii) h(x)=2x3h(x) = \frac{2x}{3}

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  9. 2b. Use Cantor's diagonal argument to prove that set F of all functions f: (0,1) \rightarrow R has larger Cardinality than |R|.20237m

    Module 1: Sets, Relation and Function.

    Use Cantor's diagonal argument to prove that set F of all functions f:(0,1)→Rf: (0,1) \rightarrow R has larger Cardinality than ∣R∣|R|.

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  10. 2b. Let f(x) = x+2, g(x) = x-2, h(x) = 3x find (i) f \circ g (ii) f \circ g \circ h20237m

    Module 1: Sets, Relation and Function.

    Let f(x)=x+2f(x) = x+2, g(x)=x−2g(x) = x-2, h(x)=3xh(x) = 3x find
    (i) f∘gf \circ g
    (ii) f∘g∘hf \circ g \circ h

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  12. 3a. a.) find the power set of each of these sets i) \{a, b\} ii) \{\emptyset, \{\emptyset\}\}20237m

    Module 1: Sets, Relation and Function.

    a.) find the power set of each of these sets
    i) {a,b}\{a, b\}
    ii) {∅,{∅}}\{\emptyset, \{\emptyset\}\}

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  13. 4. Determine if the sets are countable or uncountable a.) the set A of all function g: Z_+ \rightarrow Z_+ b.) The set B of all functions f: Z_+ \rightarrow \{0,1\}202314m

    Module 1: Sets, Relation and Function.

    Determine if the sets are countable or uncountable
    a.) the set A of all function g:Z+→Z+g: Z_+ \rightarrow Z_+
    b.) The set B of all functions f:Z+→{0,1}f: Z_+ \rightarrow \{0,1\}

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  14. 5. Prove the following by using the principle of mathematical induction for all n \in N: 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2202314m

    Module 2: Mathematical Induction & Counting Techniques.

    Prove the following by using the principle of mathematical induction for all n∈Nn \in N:
    13+23+33+⋯+n3=(n(n+1)2)21^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2

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    Worked Solution

    Answer

    Prove: 13+23+⋯+n3=(n(n+1)2)21^3 + 2^3 + \dots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2 for all n∈Nn \in \mathbb{N} by Mathematical Induction.

    Step 1: Base Case (n=1n=1)

    LHS =13=1= 1^3 = 1

    RHS =(1⋅22)2=12=1= \left(\dfrac{1\cdot 2}{2}\right)^2 = 1^2 = 1

    LHS = RHS, so holds for n=1n=1.

    Step 2: Induction Hypothesis

    Assume true for n=kn=k:
    P(k):13+23+⋯+k3=(k(k+1)2)2 P(k): 1^3 + 2^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2

    Step 3: Inductive Step (n=k+1n = k+1)

    To prove P(k+1)P(k+1):
    13+⋯+k3+(k+1)3=((k+1)(k+2)2)2 1^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2

    Consider LHS of P(k+1)P(k+1):
    LHS=(13+⋯+k3)+(k+1)3=(k(k+1)2)2+(k+1)3(by hypothesis)=k2(k+1)24+(k+1)3=(k+1)2[k24+(k+1)]=(k+1)2[k2+4k+44]=(k+1)2⋅(k+2)24=((k+1)(k+2)2)2=RHS of P(k+1) \begin{aligned} \text{LHS} &= \left(1^3 + \dots + k^3\right) + (k+1)^3 \\ &= \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 \quad \text{(by hypothesis)}\\ &= \frac{k^2(k+1)^2}{4} + (k+1)^3 \\ &= (k+1)^2 \left[ \frac{k^2}{4} + (k+1) \right] \\ &= (k+1)^2 \left[ \frac{k^2 + 4k +4}{4} \right] \\ &= (k+1)^2 \cdot \frac{(k+2)^2}{4} \\ &= \left( \frac{(k+1)(k+2)}{2} \right)^2 = \text{RHS of } P(k+1) \end{aligned}

    Thus P(k)  ⟹  P(k+1)P(k) \implies P(k+1).

    Step 4: Conclusion

    By the Principle of Mathematical Induction, since base case holds and inductive step proved, the formula holds for all n∈Nn \in \mathbb{N}.

    Corollary / Remark

    Note that 1+2+⋯+n=n(n+1)/21+2+\dots+n = n(n+1)/2, so the theorem elegantly states:
    13+23+⋯+n3=(1+2+⋯+n)2 1^3 + 2^3 + \dots + n^3 = (1+2+\dots+n)^2

  15. 6. State and prove Division algorithm theorem well-ordering principle.202314m

    Module 2: Mathematical Induction & Counting Techniques.

    State and prove Division algorithm theorem well-ordering principle.

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  16. 7a. Check the validity of the following argument all integers are rational numbers. Some integers are powers of 5. There fore, some rational numbers are powers of 520237m

    Module 3: Propositional Logic & Proof Techniques.

    Check the validity of the following argument all integers are rational numbers. Some integers are

    powers of 5. There fore, some rational numbers are powers of 5

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  17. 7b. A grocery store employee is stocking apples. Each apple is a different color. There are 10 apples left in the box and the employee pulls out 2 of them at random. What is the probability that the employee pulls out one pink apple and yellow apple?20237m

    Module 2: Mathematical Induction & Counting Techniques.

    A grocery store employee is stocking apples. Each apple is a different color. There are 10 apples left

    in the box and the employee pulls out 2 of them at random. What is the probability that the employee

    pulls out one pink apple and yellow apple?

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  18. 8. Let \Psi : G \rightarrow H be a homomorphism of groups. Show that if a \in G has order n, then \Psi (a) \in H has order dividing n.202314m

    Module 4: Algebraic Structures & Boolean Algebra.

    Let Ψ:G→H\Psi : G \rightarrow H be a homomorphism of groups. Show that if a∈Ga \in G has order nn, then Ψ(a)∈H\Psi (a) \in H has order dividing nn.

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  19. 9a. Consider the following graph (a) Does a Hamiltonian path exist? If so describe it. If not say why not.20237m

    Module 5: Graphs and Trees

    Consider the following graph

    (a) Does a Hamiltonian path exist? If so describe it. If not say why not.

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  20. 9b. Does an Eulerian path exist? If so describe it. If not say why not.20237m

    Module 5: Graphs and Trees

    Does an Eulerian path exist? If so describe it. If not say why not.

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