Jawaharlal Nehru Technological University Hyderabad
Jawaharlal Nehru Technological University Hyderabad MCA 2021 MATHEMATICAL FOUNDATIONS OF COMPUTER SCIENCE Previous Question Paper (July/August - 2021 - R20)
Answer any five questions
All questions carry equal marks
Main
Prove that the following statement is valid: (p ∧ q) → (¬p ∨ q) ∧ r ∧ ¬s → ¬r ∴ ¬s
Answer
To prove the validity of the argument, we identify the premises and the conclusion. The premises are: 1) (p ∧ q) → (¬p ∨ q), 2) r, and 3) ¬s → ¬r. The conclusion is ¬s.
We can use the method of proof by contradiction. Assume the negation of the conclusion is true: s. If s is true, we examine the premises:
- From premise 3 (¬s → ¬r) and our assumption s, we cannot directly deduce r. However, using contraposition, ¬s → ¬r is equivalent to r → s.
- Given premise 2 is r, and we have r → s, by Modus Ponens, s must be true. This is consistent with our assumption.
- Now consider the premise (p ∧ q) → (¬p ∨ q). This is a tautology because (p ∧ q) → (¬p ∨ q) is equivalent to ¬(p ∧ q) ∨ (¬p ∨ q), which simplifies to (¬p ∨ ¬q) ∨ (¬p ∨ q) = ¬p ∨ (¬q ∨ q) = ¬p ∨ T = T.
- Since the premises are satisfied and the conclusion ¬s leads to a contradiction when combined with the requirement that the premises must hold for all truth values, we check the validity. If s is false, ¬s is true. If s is true, then ¬s is false. Since the premises r and ¬s → ¬r imply r → s, and we are given r, we must have s. Thus, the argument is valid.
Find the conjunctive normal form of (q ∧ p) ∨ (¬q ∧ p) ∨ ¬q
Answer
To find the Conjunctive Normal Form (CNF) of the expression (q ∧ p) ∨ (¬q ∧ p) ∨ ¬q, we first simplify the expression using Boolean algebra laws:
Factor out p from the first two terms:
Since (q ∨ ¬q) is a tautology (T):
The expression p ∨ ¬q is already in Conjunctive Normal Form, as it is a single clause consisting of a disjunction of literals.
Write the following sentences in the symbolic form: i) Arjun is a student, ii) All students like easy courses, iii) Sociology is an easy course.
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Prove that the following argument is valid: No mathematicians are fools. No one who is not a fool is an administrator. Sita is a Mathematician. ∴ Sita is not an administrator.
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Let A = {0, 1, 2, 3, 4}. Show that the relation R = {(0, 0), (0, 4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)} is an equivalence relation.
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Let X={1,2,3} and f, g, h be functions from X to X given by f={(1,2), (2,3), (3,1)}, g={(1,2), (2,1), (3,3)}, h={(1,1), (2,2), (3,1)}. Find: i) f ∘ g, ii) f ∘ h ∘ g.
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A={1,2,3,4} is a Relation R from A to A. R={(1,1),(1,2),(2,3),(3,4)}. S={(3,1),(4,4),(2,4),(1,4)}. Determine R ∘ S, S ∘ R, R², S².
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If f(x) = x+2, g(x) = x-2, h(x) = 3x, then find: i) g ∘ f, ii) f ∘ h, iii) h ∘ g.
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Using the principle of mathematical induction, prove that 1 × 2 + 3 × 4 + 5 × 6 + …. + (2n - 1) × 2n = (n(n+1)(4n-1))/3
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Prove that for any positive integer number n, n³ + 2n is divisible by 3.
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Use the mathematical induction to prove that 3ⁿ > n² for n a positive integer greater than 2.
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Using the principle of mathematical induction, prove that 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ..... + 1/{n(n + 1)} = n/(n + 1)
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There are Three boxes I, II and III. Box I contains 4 Red, 5 Blue and 6 White balls. Box II contains 3 Red, 4 Blue and 5 White balls. Box III contains 5 Red, 10 Blue and 5 White balls. One box is chosen and one ball is drawn from it. What is the probability that i) Red ball is chosen ii) Blue ball is chosen iii) White ball is chosen.
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Solve the recurrence relation aₙ - 10aₙ₋₁ + 12aₙ₋₂ = 0, a₀=0, a₁=3.
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Prove that a graph G is a tree with n vertices if and only if it has (n-1) edges.
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Construct the minimum spanning tree for the following graph using Prim’s algorithm.
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