JEE Mathematics Mathematical Reasoning MCQs Set 10

Answer: (a) If you are not a citizen of India, then you are not born in India.

Explanation:
1. Let \( p \): you are born in India and \( q \): you are a citizen of India. The statement is \( p\Rightarrow q \).
2. Its contrapositive is \( \sim q\Rightarrow\sim p \): if you are not a citizen of India, then you are not born in India. Answer: (a)

Teacher's Note:
1. Write the statement as \( p\Rightarrow q \).

Answer: (d) \( (p\vee q)\to(p\vee(\sim q)) \)

Explanation:
1. Options (a), (b) and (c) simplify to \( T \): for example \( (p\wedge q)\to p\equiv(\sim p\vee\sim q)\vee p\equiv T \).
2. For (d), take \( p=F \) and \( q=T \): \( p\vee q=T \) but \( p\vee\sim q=F \), so the implication is false. It is not a tautology. Answer: (d)

Teacher's Note:
1. Test a suspicious row of the truth table.

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

Explanation:
1. \( p\vee(\sim p\wedge q)\equiv(p\vee\sim p)\wedge(p\vee q)\equiv p\vee q \).
2. Its negation is \( \sim(p\vee q)\equiv\sim p\wedge\sim q \). Answer: (a)

Teacher's Note:
1. Simplify using the distributive law.

Answer: (b) T, F, F

Explanation:
1. An implication is false only when the antecedent is true and the consequent is false.
2. So \( p=T \) and \( q\vee r=F \), which needs \( q=F \) and \( r=F \). Answer: (b)

Teacher's Note:
1. When is an implication false?

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

Explanation:
1. (a) \( \equiv p\vee(q\vee\sim q)\equiv p\vee T\equiv T \), a tautology.
2. (b) \( \equiv p \), (c) \( \equiv p \) and (d) is false when \( p=q \) (both \( T \) or both \( F \)), so none of them is a tautology. Answer: (a)

Teacher's Note:
1. Simplify each option.

Answer: (a) \( s\wedge r \)

Explanation:
1. \( \sim[\sim s\vee(\sim r\wedge s)]\equiv s\wedge\sim(\sim r\wedge s)\equiv s\wedge(r\vee\sim s) \).
2. \( \equiv(s\wedge r)\vee(s\wedge\sim s)\equiv s\wedge r \). Answer: (a)

Teacher's Note:
1. Apply De Morgan's law.
2. Use the distributive law.

Answer: (a) T, T and F

Explanation:
1. The implication is false when \( p=T \) and \( \sim q\vee r=F \).
2. That needs \( \sim q=F \) (so \( q=T \)) and \( r=F \). So \( p,q,r=T,T,F \). Answer: (a)

Teacher's Note:
1. Make the consequent false.

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

Explanation:
1. \( p\Rightarrow\sim q\equiv\sim p\vee\sim q \).
2. Negating: \( \sim(\sim p\vee\sim q)\equiv p\wedge q \). Answer: (a)

Teacher's Note:
1. Rewrite the implication.

Answer: (a) \( (\wedge,\vee) \)

Explanation:
1. For \( (\wedge,\vee) \): \( (p\wedge q)\wedge(\sim p\vee q)\equiv p\wedge q \), since \( p\wedge q \) already makes \( \sim p\vee q \) true.
2. For \( (\wedge,\wedge) \) we get \( F \); for \( (\vee,\wedge) \) we get \( \sim p\wedge q \); for \( (\vee,\vee) \) we get \( q \). None of these equals \( p\wedge q \). Answer: (a)

Teacher's Note:
1. Test each pair.

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

Explanation:
1. \( \sim(\sim p\vee q)\equiv p\wedge\sim q \), so the expression is \( [(p\wedge\sim q)\vee(p\wedge r)]\wedge(\sim q\wedge r) \).
2. \( \equiv p\wedge[(\sim q\vee r)\wedge(\sim q\wedge r)]\equiv p\wedge(\sim q\wedge r)\equiv(p\wedge r)\wedge\sim q \). Answer: (c)

Teacher's Note:
1. Use De Morgan's law and the distributive law.

Answer: (b) \( P\vee(\sim Q\wedge R) \)

Explanation:
1. \( P \) is true, \( Q \) is false (\( 192=2^6\cdot3 \)) and \( R \) is true.
2. (a) \( F\vee F=F \); (b) \( T\vee(T\wedge T)=T \); (c) \( F\vee F=F \); (d) \( F\wedge T=F \). Answer: (b)

Teacher's Note:
1. Find the truth value of each statement.

Answer: (b) \( (p\wedge r)\to(p\vee r) \)

Explanation:
1. \( q=F \) makes \( p\wedge q \) false, so for the biconditional to be true \( r \) must be false.
2. (b) \( (p\wedge r)\to(p\vee r) \) is true for every \( p \) and \( r \) (a tautology). (a) fails if \( p=F \), (c) fails if \( p=T \), and (d) is false as \( r=F \). Answer: (b)

Teacher's Note:
1. Find \( r \) first.

Answer: (b) If the squares of two numbers are equal, then the numbers are equal.

Explanation:
1. The contrapositive of \( p\to q \) is \( \sim q\to\sim p \).
2. \( \sim q \): the squares are equal; \( \sim p \): the numbers are equal. Answer: (b)

Teacher's Note:
1. Negate and swap.

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

Explanation:
1. \( (p\wedge q)\vee(p\vee\sim q)\equiv p\vee\sim q \).
2. \( (p\vee\sim q)\wedge(\sim p\wedge\sim q)\equiv\sim p\wedge\sim q \), since \( \sim p\wedge\sim q \) already forces \( \sim q \) true. Answer: (d)

Teacher's Note:
1. Simplify the first bracket.
2. Check with the truth table.

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

Explanation:
1. \( \sim p\to q\equiv p\vee q \).
2. Negation: \( \sim(p\vee q)\equiv\sim p\wedge\sim q \). Answer: (d)

Teacher's Note:
1. Rewrite the implication.

Answer: (a) \( \sim p \)

Explanation:
1. \( \sim(p\vee q)\equiv\sim p\wedge\sim q \).
2. \( (\sim p\wedge\sim q)\vee(\sim p\wedge q)\equiv\sim p\wedge(\sim q\vee q)\equiv\sim p \). Answer: (a)

Teacher's Note:
1. Apply De Morgan's law.
2. Factor out \( \sim p \).

Answer: (a) a tautology

Explanation:
1. \( (\sim p\to q)\equiv p\vee q \), and \( (p\vee q)\to q\equiv(p\to q) \).
2. So the statement is \( (p\to q)\to(p\to q) \), which is always true. Answer: (a)

Teacher's Note:
1. Simplify the inner implications.
2. Build the truth table.

Answer: (c) \( p\vee q \)

Explanation:
1. \( (p\wedge\sim q)\vee q\equiv(p\vee q)\wedge(\sim q\vee q)\equiv p\vee q \).
2. \( (p\vee q)\vee(\sim p\wedge q)\equiv p\vee q \), since \( \sim p\wedge q \) is already covered by \( q \). Answer: (c)

Teacher's Note:
1. Absorb terms using the distributive law.

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

Explanation:
1. The truth table of \( p\leftrightarrow\sim q \) is \( F,T,T,F \) for \( (T,T),(T,F),(F,T),(F,F) \), so its negation is \( T,F,F,T \).
2. That is exactly the truth table of \( p\leftrightarrow q \). It is neither always true nor always false. Answer: (a)

Teacher's Note:
1. Build the truth table.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( (p\wedge\sim q)\wedge(\sim p\wedge q)\equiv p\wedge\sim p\wedge q\wedge\sim q\equiv F \), so the Assertion is true.
2. \( \sim q\to\sim p \) is the contrapositive of \( p\to q \), so the two are equivalent and the biconditional is always true; the Reason is true.
3. The Reason concerns a different statement and does not explain why the Assertion is a fallacy. Answer: (b)

Teacher's Note:
1. Simplify the Assertion.
2. Recognise the contrapositive.

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

Explanation:
1. With \( p \): I become a teacher and \( q \): I will open a school, the statement is \( p\to q \).
2. \( \sim(p\to q)\equiv p\wedge\sim q \): I become a teacher and I do not open a school. Answer: (a)

Teacher's Note:
1. Translate to symbols.

Answer: (a) \( \sim[Q\leftrightarrow(P\wedge\sim R)] \)

Explanation:
1. The statement is \( (P\wedge\sim R)\leftrightarrow Q \), which is \( Q\leftrightarrow(P\wedge\sim R) \).
2. Its negation is \( \sim[Q\leftrightarrow(P\wedge\sim R)] \). Answer: (a)

Teacher's Note:
1. Translate the sentence.

Answer: (d) \( [A\wedge(A\to B)]\to B \)

Explanation:
1. Options (b) and (c) both reduce to \( A \), which is not always true.
2. For (a), take \( B=T \) and \( A=F \): the consequent is false, so the implication is false.
3. (d) is modus ponens: it is true in every row. Answer: (d)

Teacher's Note:
1. Check the options with \( A=F \) or \( B=T \).

Answer: (c) Every rational number \( x\in S \) satisfies \( x\le0 \)

Explanation:
1. The negation of ''there exists \( x \) with property \( \gt0 \)'' is ''for every \( x \), not \( \gt0 \)''.
2. So every rational number \( x\in S \) satisfies \( x\le0 \). Answer: (c)

Teacher's Note:
1. Negate the quantifier.

Answer: (c) A is true, but R is false.

Explanation:
1. The truth values of \( \sim(p\leftrightarrow\sim q) \) are \( T,F,F,T \), identical to \( p\leftrightarrow q \), so the Assertion is true.
2. These values are not all \( T \), so it is not a tautology and the Reason is false. Answer: (c)

Teacher's Note:
1. Build the truth table.

Answer: (c) \( p\to(p\vee q) \)

Explanation:
1. \( q\to p\equiv\sim q\vee p \), so \( p\to(q\to p)\equiv\sim p\vee\sim q\vee p\equiv T \), a tautology.
2. \( p\to(p\vee q) \) is also always true, with the same truth table; the others can be false (for example \( p=T,q=F \) makes (a), (b), (d) false). Answer: (c)

Teacher's Note:
1. Compare the truth tables.