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MathsMediumClass 11

Connectives — AND, OR, NOT, implication, biconditional

Mathematical Reasoning

7

JEE Qs

8%

Hard

75

min

Thoroughly memorize the truth table of implication and its logical equivalence (~p v q) as it is the most frequently tested and misunderstood connective.

🧮 Key Formulas

Conjunction (AND): p ^ q. True if and only if p is true and q is true.
Disjunction (OR): p v q. True if and only if p is true or q is true (inclusive OR).
Negation (NOT): ~p. True if and only if p is false.
Implication (IF...THEN): p -> q. False if and only if p is true and q is false. (Logically equivalent to ~p v q)
Biconditional (IF AND ONLY IF): p <-> q. True if and only if p and q have the same truth value. (Logically equivalent to (p -> q) ^ (q -> p))
De Morgan's Law 1: ~(p ^ q) <=> ~p v ~q
De Morgan's Law 2: ~(p v q) <=> ~p ^ ~q

✅ Key Points for JEE

  • 1Mastering the truth tables for all five connectives is fundamental for solving any problem in Mathematical Reasoning, especially for complex compound statements.
  • 2The implication (p -> q) is false in only one specific scenario: when the antecedent (p) is true and the consequent (q) is false. In all other cases, it is true.
  • 3Biconditional (p <-> q) is true when both statements p and q have the same truth value (both true or both false); otherwise, it is false.
  • 4Negation (~p) flips the truth value of statement p. Applying negation to compound statements requires careful use of De Morgan's laws or understanding its effect on the truth values of components.
  • 5Remember that 'OR' in mathematical reasoning is almost always inclusive (p or q or both), unless explicitly stated as exclusive OR (XOR).

⚠️ Common Mistakes

  • ✕Misinterpreting the truth table for implication (p -> q), especially when p is false, often incorrectly assuming it must be false if p is false.
  • ✕Confusing 'OR' with 'exclusive OR'. The standard logical 'OR' is inclusive, meaning p, q, or both can be true for the compound statement to be true.
  • ✕Incorrectly negating compound statements, particularly p -> q. The negation of (p -> q) is p ^ ~q, not ~p -> ~q or ~p v q.
  • ✕Failing to evaluate the truth value of substatements or parts of a compound statement correctly before applying the main connective.

📝 Practice Questions

See all

Q68. Choose the correct answer from the options given below : (1) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (2) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

2025·MCQMedium

Q64. Choose the correct answer from the options given below : (1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

2025·MCQMedium

Q63. Given below are two statements : In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are true (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are false

2025·Assertion ReasoningMedium

Q73.Negation of (p →q) →(q →p) is (1) (p~) ∨p (2) q ∧(~p) (3) (~q) ∧p (4) p ∨(~q)

2023·MCQEasy

Q73.Among the statements (S1) : (p ⇒q) ∨((~p) ∧q) is a tautology (S2) : (q ⇒p) ⇒((~p) ∧q) is a contradiction (1) Neither (S1) and (S2) is True (2) Both (S1) and (S2) are True (3) Only (S2) is True (4) Only (S1) is True

2023·Assertion ReasoningMedium

Q72.Which of the following statements is a tautology? (1) p →(p ∧(p →q)) (2) (p ∧q) →(~(p) →q) (3) (p ∧(p →q)) →~q (4) p ∨(p ∧q)

2023·MCQEasy

NCERT Chapters

  • Class 11 Mathematics Chapter 14: Mathematical Reasoning