JEE Mathematics Mathematical Reasoning MCQs Set 07

Answer: (b) \( p \Rightarrow q \)

Explanation:
1. 'p only if q' means that p can hold only when q holds, so p cannot be true without q.
2. That is the implication \( p \Rightarrow q \). The reverse implication \( q \Rightarrow p \) is read 'p if q'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) \( p \wedge (\sim q) \)

Explanation:
1. 'x is odd' is p and 'x^2 is not odd' is \( \sim q \).
2. The word 'and' joins them, so the form is \( p \wedge (\sim q) \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) \( \sim p \Rightarrow \sim q \)

Explanation:
1. The converse swaps the parts: \( q \Rightarrow p \). The contrapositive swaps and negates: \( \sim q \Rightarrow \sim p \).
2. The inverse negates both parts but keeps the order: \( \sim p \Rightarrow \sim q \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) a tautology

Explanation:
1. \( p \wedge (\sim p) \) is always false.
2. An implication with a false hypothesis is always true, so the statement is a tautology.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) true

Explanation:
1. 3 + 2 = 5 is true and 1 x 0 = 0 is true.
2. A true statement implying a true statement is true.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) false

Explanation:
1. 20 + 10 = 2 is false while 20 x 10 = 200 is true.
2. A conjunction is true only when both parts are true, so the statement is false.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) \( 3 + 7 = 10 \Leftrightarrow 1 + 2 = 3 \)

Explanation:
1. A biconditional is true when both sides have the same truth value.
2. 3 + 7 = 10 and 1 + 2 = 3 are both true, so the first statement is true. In the others one side is true and the other false (or the pair is mixed), so they are false.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) a contradiction

Explanation:
1. \( p \Rightarrow \sim p \) is equivalent to \( \sim p \), and \( \sim p \Rightarrow p \) is equivalent to p.
2. So the proposition is \( \sim p \wedge p \), which is always false: a contradiction.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) \( p \Leftrightarrow \sim q \)

Explanation:
1. \( p \Leftrightarrow q \) is true when p and q have the same truth value, so its negation is true when they differ.
2. That is exactly the behaviour of \( p \Leftrightarrow \sim q \) (for p = T, q = F both are true; for p = T, q = T both are false).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) A number is divisible by 15 and it is divisible by neither 5 nor 3

Explanation:
1. Negation of \( p \Rightarrow (q \vee r) \) is \( p \wedge \sim(q \vee r) = p \wedge \sim q \wedge \sim r \).
2. So: divisible by 15, and not divisible by 5 and not divisible by 3. (Such a number cannot exist, which shows the original statement is true.)

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) We control population growth but we do not prosper

Explanation:
1. The negation of \( p \Rightarrow q \) is \( p \wedge \sim q \).
2. So the negation is 'we control population growth and we do not prosper'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) If in a triangle ABC, angle B = angle C, then AB = AC

Explanation:
1. The converse of 'if p then q' is 'if q then p'.
2. Here p : AB = AC and q : angle B = angle C, so the converse is 'if angle B = angle C then AB = AC'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) \( (\sim p \vee q) \Rightarrow \sim r \)

Explanation:
1. The inverse of \( a \Rightarrow b \) is \( \sim a \Rightarrow \sim b \).
2. Here \( \sim(p \wedge \sim q) = \sim p \vee q \) by De Morgan's law, so the inverse is \( (\sim p \vee q) \Rightarrow \sim r \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) If the diagonals of a figure are not perpendicular, then the figure is not a rhombus

Explanation:
1. The statement means 'if a figure is a rhombus, then its diagonals are perpendicular'.
2. Its contrapositive is 'if the diagonals are not perpendicular, then the figure is not a rhombus'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) \( (\sim q \vee r) \Rightarrow (p \vee \sim q) \)

Explanation:
1. The contrapositive of \( a \Rightarrow b \) is \( \sim b \Rightarrow \sim a \).
2. \( \sim(q \wedge \sim r) = \sim q \vee r \) and \( \sim(\sim p \wedge q) = p \vee \sim q \), giving \( (\sim q \vee r) \Rightarrow (p \vee \sim q) \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

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

Explanation:
1. p is false (4 is not prime), q is true and r is true.
2. \( \sim p \vee (q \wedge r) = T \vee T = T \). The others evaluate to F: \( p \wedge q = F \), \( (p \vee q) \wedge \sim r = T \wedge F = F \), \( \sim(q \wedge r) \vee p = F \vee F = F \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) If x has no courage, then x will not win

Explanation:
1. The inverse of 'if p then q' is 'if not p then not q'.
2. So: if x has no courage, then x will not win.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) If \( x = 4 \), then \( 2x + 3y \ne 9 \)

Explanation:
1. The negation of \( x \ne 4 \) is \( x = 4 \), and the negation of \( 2x + 3y = 9 \) is \( 2x + 3y \ne 9 \).
2. The contrapositive swaps and negates: if \( x = 4 \), then \( 2x + 3y \ne 9 \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) \( 5 > 7 \) and \( 6 \ge 4 \)

Explanation:
1. The negation of \( p \Rightarrow q \) is \( p \wedge \sim q \).
2. Not (6 < 4) is \( 6 \ge 4 \), so the negation is '5 > 7 and 6 >= 4'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) a contradiction

Explanation:
1. \( p \wedge \sim q \) is equivalent to \( \sim(p \Rightarrow q) \), and \( \sim p \vee q \) is equivalent to \( p \Rightarrow q \).
2. So the statement is \( \sim(p \Rightarrow q) \wedge (p \Rightarrow q) \), which is always false: a contradiction.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) neither a tautology nor a contradiction

Explanation:
1. Take p = F, q = F, r = T: the first bracket is F and \( \sim r \vee q = F \vee F = F \), so the statement is false.
2. Take p = T, q = T, r = T: \( \sim r \vee q = T \), so the statement is true. It takes both values, so it is neither a tautology nor a contradiction.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) a tautology

Explanation:
1. Suppose \( p \Rightarrow q \) is true and \( r \vee p \) is true. If r is true then \( r \vee q \) is true; if r is false then p is true, hence q is true and \( r \vee q \) is true.
2. So the statement can never be false: it is a tautology.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) a tautology

Explanation:
1. This is the law of the hypothetical syllogism: if p leads to q and q leads to r, then p leads to r.
2. A truth table with 8 rows gives T in every row, so it is a tautology.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) a contradiction

Explanation:
1. By De Morgan's law \( \sim p \wedge \sim q = \sim(p \vee q) \).
2. So the statement is \( (p \vee q) \wedge \sim(p \vee q) \), which is always false: a contradiction.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) \( \{3\} \)

Explanation:
1. Factorise: \( x^2 - 3x - \sqrt2 x + 3\sqrt2 = (x - 3)(x - \sqrt2) \), so the roots are 3 and \( \sqrt2 \).
2. Only 3 is rational, so the truth set over Q is \( \{3\} \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (c) \( \triangle ABC \) is isosceles and the base angles A and B are not equal

Explanation:
1. The negation of \( p \Rightarrow q \) is \( p \wedge \sim q \).
2. So the negation is: the triangle is isosceles and the base angles A and B are not equal.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (b) If x = 4, then x + 2 = 6

Explanation:
1. The converse of 'if p then q' is 'if q then p'.
2. Here p : x + 2 = 6 and q : x = 4, so the converse is 'if x = 4, then x + 2 = 6'.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (a) \( q \Rightarrow (p \wedge r) \)

Explanation:
1. p = T, q = F, r = T. Then \( p \Rightarrow q \) is F, so the first two statements are F, and \( p \wedge q = F \) makes the third F.
2. \( q \Rightarrow (p \wedge r) \) has a false hypothesis (q = F), so it is true.

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.

Answer: (d) \( (p \Rightarrow q) \equiv (\sim q \Rightarrow \sim p) \)

Explanation:
1. The implication is equivalent to its contrapositive.
2. The others are wrong: \( \sim(p \vee q) = \sim p \wedge \sim q \), \( \sim(p \Rightarrow q) = p \wedge \sim q \) and \( \sim(p \wedge q) = \sim p \vee \sim q \).

Teacher's Note:
1. Write each statement in symbols first, then check a truth table row by row or use the standard laws (De Morgan, p implies q equals not p or q).
2. The contrapositive is always equivalent to the original statement, but the converse and inverse are not.