JEE Mathematics Mathematical Reasoning MCQs Set 01

Answer: (b) ~p ∨ q

Explanation:
1. p ⇒ q is false only when p is true and q is false.
2. ~p ∨ q is false in exactly the same case, so they are equivalent.

Teacher's Note:
1. Compare the truth tables.
2. p ⇒ q ≡ ~p ∨ q.

Answer: (a) T, F

Explanation:
1. An implication is false only when the antecedent is true and the consequent is false: p = T and ~p ∨ q = F.
2. With p = T, ~p = F, so ~p ∨ q = F needs q = F.

Teacher's Note:
1. Use the condition for a false implication.
2. Find q from the consequent.

Answer: (c) p, q, r are all true

Explanation:
1. A conjunction is true only if every part is true.
2. So p, q and r are all true.

Teacher's Note:
1. All components of a conjunction must be true.
2. The expression includes p, q and r.

Answer: (a) True

Explanation:
1. ~p = F, so ~p ∨ q = F ∨ F = F.
2. The antecedent (~p ∨ q) ∧ ~r is false, so the implication is true.

Teacher's Note:
1. Evaluate the antecedent first.
2. A false antecedent makes an implication true.

Answer: (d) 3 is not an odd number or 7 is not a rational number

Explanation:
1. By De Morgan's law, \( \sim(p\wedge q)=\sim p\vee\sim q \).
2. So the negation is: "3 is not an odd number or 7 is not a rational number".

Teacher's Note:
1. Negating an AND gives an OR of the negations.
2. Negate each part.

Answer: (b) \( (p\wedge\sim q)\vee(p\vee q) \)

Explanation:
1. De Morgan: \( \sim[(\sim p\vee q)\wedge(\sim p\wedge\sim q)]=\sim(\sim p\vee q)\vee\sim(\sim p\wedge\sim q) \).
2. \( \sim(\sim p\vee q)=p\wedge\sim q \) and \( \sim(\sim p\wedge\sim q)=p\vee q \).
3. The negation is \( (p\wedge\sim q)\vee(p\vee q) \).

Teacher's Note:
1. Apply De Morgan's laws twice.
2. Negate each inner bracket.

Answer: (c) \( (\sim p\vee\sim q)\wedge(\sim q\wedge r) \)

Explanation:
1. \( \sim[(p\wedge q)\vee(q\vee\sim r)]=\sim(p\wedge q)\wedge\sim(q\vee\sim r) \).
2. This equals \( (\sim p\vee\sim q)\wedge(\sim q\wedge r) \).

Teacher's Note:
1. Negation of OR is the AND of negations.
2. ~(q ∨ ~r) = ~q ∧ r.

Answer: (a) p

Explanation:
1. By the absorption law, \( p\vee(p\wedge\sim q)=p \).
2. If p is true the whole is true; if p is false both parts are false.

Teacher's Note:
1. Use the absorption law.
2. Check both truth values of p.

Answer: (c) may be true or false

Explanation:
1. Since q and r are true, \( q\wedge r \) is true, so the whole disjunction is true no matter what the first part is.
2. The truth value of p is not determined: it may be true or false.

Teacher's Note:
1. A disjunction is true if one part is true.
2. Check whether the other part depends on p.

Answer: (b) A number is prime and it is not odd

Explanation:
1. \( \sim(p\Rightarrow q)=p\wedge\sim q \).
2. So the negation is: a number is prime and it is not odd.

Teacher's Note:
1. ~(p ⇒ q) ≡ p ∧ ~q.
2. State the negation in words.

Answer: (a) True

Explanation:
1. r is false, so the antecedent of the implication is false.
2. An implication with a false antecedent is true.

Teacher's Note:
1. Check the antecedent r first.
2. No need to evaluate the bracket.

Answer: (c) \( (p\wedge q)\wedge\sim r \)

Explanation:
1. \( \sim(A\to B)=A\wedge\sim B \) with A = p ∧ q and B = r.
2. So the negation is \( (p\wedge q)\wedge\sim r \).

Teacher's Note:
1. Use ~(A → B) ≡ A ∧ ~B.
2. Substitute A and B.

Answer: (a) \( p\vee(\sim p)\equiv c \)

Explanation:
1. \( p\vee\sim p \) is always true, so it is a tautology t, not a contradiction c: statement (a) is wrong.
2. The identities (b), (c) and (d) are correct.

Teacher's Note:
1. p ∨ ~p is always true.
2. Check the other identities with truth tables.

Answer: (a) \( p\to q \)

Explanation:
1. \( p\to q \) is false only when p is true and q is false, which is exactly when ~p ∨ q is false.
2. So ~p ∨ q ≡ p → q.

Teacher's Note:
1. Compare truth tables.
2. The other options differ from p → q.

Answer: (c) neither tautology nor contradiction

Explanation:
1. Truth table: p = T, q = T: (T) ⇔ F is F. p = T, q = F: F ⇔ F is T.
2. The statement takes both truth values, so it is neither a tautology nor a contradiction.

Teacher's Note:
1. Evaluate for the four combinations of p and q.
2. Find one case of each truth value.

Answer: (a) Tautology

Explanation:
1. If p is false the implication is true. If p is true then p ∨ q is true, so the implication is true.
2. It is true in every case: a tautology.

Teacher's Note:
1. Check both values of p.
2. p ∨ q is true whenever p is true.

Answer: (b) contradiction

Explanation:
1. \( p\to\sim q\equiv\sim(p\wedge q) \).
2. So the statement is \( \sim(p\wedge q)\Leftrightarrow(p\wedge q) \), which says a statement is equivalent to its own negation; it is false for every assignment.

Teacher's Note:
1. p → ~q ≡ ~(p ∧ q).
2. A statement and its negation always have opposite truth values.