JEE Mathematics Mathematical Reasoning MCQs Set 05

Answer: (b) If your mind is not free then you are not watching television.

Explanation:
1. Contrapositive of p ⇒ q is ~q ⇒ ~p.
2. So: if your mind is not free then you are not watching television.

Teacher's Note:
1. The contrapositive of p ⇒ q is ~q ⇒ ~p and it is always logically equivalent to p ⇒ q.
2. The converse q ⇒ p and the inverse ~p ⇒ ~q are equivalent to each other but not to the original statement.

Answer: (a) (F, T)

Explanation:
1. A statement and its contrapositive have the same truth value.
2. P : T ⇒ F = F, so V₁ = F.
3. Q : T ⇒ T = T, so V₂ = T.

Teacher's Note:
1. The contrapositive of p ⇒ q is ~q ⇒ ~p and it is always logically equivalent to p ⇒ q.
2. The converse q ⇒ p and the inverse ~p ⇒ ~q are equivalent to each other but not to the original statement.

Answer: (b) If I will not go to the doctor then I am feeling well.

Explanation:
1. With p : I am not feeling well and q : I will go to the doctor, the contrapositive ~q ⇒ ~p is: if I will not go to the doctor then I am feeling well.

Teacher's Note:
1. The contrapositive of p ⇒ q is ~q ⇒ ~p and it is always logically equivalent to p ⇒ q.
2. The converse q ⇒ p and the inverse ~p ⇒ ~q are equivalent to each other but not to the original statement.

Answer: (c) There does not exist a number which is equal to its square.

Explanation:
1. The negation of 'there exists' is 'there does not exist'.
2. So: there does not exist a number which is equal to its square.

Teacher's Note:
1. The negation of 'for all' is 'there exists ... not', and the negation of 'there exists' is 'for all ... not'.
2. Negate the property inside the quantifier as well.

Answer: (a) Both statements are true.

Explanation:
1. A ⇔ ~B is true when A and B differ, so its negation is true when A and B agree: ~(A ⇔ ~B) = A ⇔ B, and Statement-1 is true.
2. A ∨ ~(A ∧ ~B) = A ∨ ~A ∨ B = T, so Statement-2 is true.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (b) Statement-1 is true, but Statement-2 is false.

Explanation:
1. p ⇒ q = ~p ∨ q = q ∨ ~p, so Statement-1 is a tautology.
2. ~(~p ∧ q) ∧ (p ∨ q) = (p ∨ ~q) ∧ (p ∨ q) = p, so the biconditional with p is a tautology, not a fallacy; Statement-2 is false.

Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.

Answer: (b) Statement-1 is true, but Statement-2 is false.

Explanation:
1. p ⇔ ~q is true when p and q differ, so its negation is true when they agree, which is p ⇔ q: Statement-1 is true.
2. p ⇔ q is false when p and q differ, so ~(p ⇔ ~q) is not always true; Statement-2 is false.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (a) Both statements are true.

Explanation:
1. A person can have both a voter card and a PAN card, so the 'or' is inclusive: Statement-1 is true.
2. The description of inclusive OR in Statement-2 is the standard definition, so it is also true.

Teacher's Note:
1. An inclusive 'or' allows both parts to be true; an exclusive 'or' allows exactly one.
2. Test whether both component statements could hold together to decide which type of 'or' is used.

Answer: (d) (1), (2), (3) and (4)

Explanation:
1. A logical statement must be true or false.
2. Commands, exclamations, questions and wishes cannot be called true or false, so none of the four types is a logical statement.

Teacher's Note:
1. A statement must be either true or false, but not both; questions, commands and exclamations are not statements.
2. A sentence whose truth depends on an unknown (such as 'tomorrow is Friday') is not a statement.

Answer: (c) (1) and (4) only

Explanation:
1. A number cannot be both rational and irrational, so (1) is exclusive.
2. x = 0 satisfies both x ≥ 0 and x ≤ 0, so (2) and (3) are inclusive.
3. Two lines cannot be both non-intersecting and coincident, so (4) is exclusive.

Teacher's Note:
1. An inclusive 'or' allows both parts to be true; an exclusive 'or' allows exactly one.
2. Test whether both component statements could hold together to decide which type of 'or' is used.

Answer: (b) (2), (3) and (4) only

Explanation:
1. (1): for p = q = T both ~(p ⇒ q) and ~p ∧ ~q are false, so the biconditional is true there; hence (1) is not a fallacy.
2. (2): (p ∧ ~q) forces p = T, q = F, but then ~p ∨ q = F; the statement is always false.
3. (3) is always false, and (4) needs p ∧ ~p, so it is always false.
4. So (2), (3), (4) are fallacies.

Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.

Answer: (c) (1), (2) and (4) only

Explanation:
1. (1), (2) and (4) each deny that every equilateral triangle is equiangular.
2. (3) denies the denial, so it says the same as the original statement and is not a negation.

Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).

Answer: (c) Statement-I is true, Statement-II is false.

Explanation:
1. ~(p ⇔ ~q) = p ⇔ q (true exactly when p and q have the same truth value), so Statement-I is true.
2. p ⇔ q is false when p and q differ, so ~(p ⇔ ~q) is not a tautology and Statement-II is false.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (b) Every rational number x ∈ S satisfies x ≤ 0.

Explanation:
1. The negation of 'there exists x with x > 0' is 'for every x, not (x > 0)'.
2. So: every rational number x ∈ S satisfies x ≤ 0.

Teacher's Note:
1. The negation of 'for all' is 'there exists ... not', and the negation of 'there exists' is 'for all ... not'.
2. Negate the property inside the quantifier as well.

Answer: (b) ~(Q ⇔ (P ∧ ~R))

Explanation:
1. The given statement is (P ∧ ~R) ⇔ Q.
2. Its negation is ~((P ∧ ~R) ⇔ Q), the same as ~(Q ⇔ (P ∧ ~R)).
3. (d) is the original statement itself, so it is not a negation.

Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).

Answer: (a) I will become a teacher and I will not open a school.

Explanation:
1. ~(p ⇒ q) = p ∧ ~q.
2. So: I will become a teacher and I will not open a school.

Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).

Answer: (b) Statement-I is true; Statement-II is true; Statement-II is not a correct explanation of Statement-I.

Explanation:
1. (p ∧ ~q) ∧ (~p ∧ q) contains p ∧ ~p, so it is always false: Statement-I is true.
2. A conditional and its contrapositive are equivalent, so Statement-II is true.
3. Statement-II is about contrapositives and does not explain why Statement-I is a fallacy.

Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.

Answer: (c) equivalent to p ⇔ q

Explanation:
1. p ⇔ ~q is true when p and q differ, so its negation is true when they agree, i.e. p ⇔ q.
2. It is not always true or always false.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (b) p ∨ q

Explanation:
1. q ∨ (~p ∧ q) = q by absorption.
2. (p ∧ ~q) ∨ q = (p ∨ q) ∧ (~q ∨ q) = p ∨ q.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (a) a tautology

Explanation:
1. ~p ⇒ q = p ∨ q, and (p ∨ q) ⇒ q = ~(p ∨ q) ∨ q = ~p ∨ q = p ⇒ q.
2. So the statement is (p ⇒ q) ⇒ (p ⇒ q), which is always true.

Teacher's Note:
1. A tautology is true for every combination of truth values and a contradiction (fallacy) is false for every combination.
2. Simplify with the laws of algebra of statements, or build a short truth table to be certain.

Answer: (d) (∧, ∨)

Explanation:
1. Try (⊕, ⊙) = (∧, ∨): (p ∧ q) ∧ (~p ∨ q) = p ∧ q ∧ (~p ∨ q) = p ∧ q, since q ∧ (~p ∨ q) = q.
2. The other three pairs fail: (∧, ∧) gives (p ∧ q) ∧ (~p ∧ q) = F; (∨, ∨) gives (p ∨ q) ∧ (~p ∨ q) = q; (∨, ∧) gives (p ∨ q) ∧ (~p ∧ q) = ~p ∧ q.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (a) (~p) ∧ (~q)

Explanation:
1. (p ∧ q) ∨ (p ∨ ~q) = p ∨ ~q (since p ∧ q implies p).
2. (p ∨ ~q) ∧ (~p ∧ ~q) = [(p ∧ ~p ∧ ~q)] ∨ [~q ∧ ~p ∧ ~q] = ~p ∧ ~q.

Teacher's Note:
1. Use p ⇒ q = ~p ∨ q and p ⇔ q = (p ⇒ q) ∧ (q ⇒ p) to reduce every expression to ∧, ∨, ~.
2. Absorption laws p ∧ (p ∨ q) = p and p ∨ (p ∧ q) = p shorten many simplifications.

Answer: (d) p : 4 + 3 > 2 and 4 + 3 < 9; ~p : 4 + 3 < 2 or 4 + 3 > 9

Explanation:
1. (a) ~(p ∧ q) = ~p ∨ ~q, correct. (b) ~(p ⇒ q) = p ∧ ~q, correct. (c) ~(p ⇔ q) = (p ∧ ~q) ∨ (~p ∧ q), correct.
2. (d) The negation of '4 + 3 > 2 and 4 + 3 < 9' is '4 + 3 ≤ 2 or 4 + 3 ≥ 9'. Writing '<' and '>' leaves out the equality case, so (d) is incorrect.

Teacher's Note:
1. Use De Morgan's laws: ~(p ∧ q) = ~p ∨ ~q and ~(p ∨ q) = ~p ∧ ~q.
2. The negation of p ⇒ q is p ∧ ~q, and the negation of p ⇔ q is (p ∧ ~q) ∨ (~p ∧ q).