JEE Mathematics Mathematical Reasoning MCQs Set 14

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

Explanation:
1. The contrapositive of \( p\to q \) is \( \sim q\to\sim p \).
2. Here p: born in India, q: citizen of India.
3. So: if you are not a citizen of India, then you are not born in India.

Teacher's Note:
1. The contrapositive is always logically equivalent to the original conditional.
2. Negate both parts and swap them.

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

Explanation:
1. Test \( p=F,\ q=T \) in \( (p\vee q)\to(p\vee\sim q) \): the antecedent \( F\vee T=T \), the consequent \( F\vee F=F \).
2. So the implication is false for this row and the statement is not a tautology.
3. The other three are true in every row of the truth table.

Teacher's Note:
1. To show a statement is not a tautology, find a single row of the truth table where it is false.

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

Explanation:
1. By the distributive law, \( p\vee(\sim p\wedge q)\equiv(p\vee\sim p)\wedge(p\vee q)\equiv p\vee q \).
2. Its negation, by De Morgan's law, is \( \sim(p\vee q)\equiv\sim p\wedge\sim q \).

Teacher's Note:
1. Simplify the expression first, then apply De Morgan's law once.

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 is T and \( q\vee r \) is F, which requires both q and r to be F.

Teacher's Note:
1. A false implication means true antecedent and false consequent.

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

Explanation:
1. \( (p\vee q)\vee(p\vee\sim q)\equiv p\vee(q\vee\sim q)\equiv p\vee T\equiv T \).
2. The others reduce to p, p, and \( \sim(p\leftrightarrow q) \) respectively, which are not always true.

Teacher's Note:
1. Group the disjunction so that q and not q appear together; their disjunction is always true.

Answer: (b) T, T and F

Explanation:
1. A false implication requires p true and \( \sim q\vee r \) false.
2. \( \sim q\vee r \) is false only if \( \sim q \) is F and r is F, so q = T and r = F.

Teacher's Note:
1. Make the antecedent true and the consequent false, then solve for q and r.

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

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

Teacher's Note:
1. The negation of an implication a to b is a and not b.

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

Explanation:
1. Try \( \oplus=\wedge \) and \( \odot=\vee \): \( (p\wedge q)\wedge(\sim p\vee q) \).
2. \( =(p\wedge q\wedge\sim p)\vee(p\wedge q\wedge q)=\text{F}\vee(p\wedge q)=p\wedge q \).
3. The other pairs do not reduce to \( p\wedge q \), as a truth table check shows.

Teacher's Note:
1. Test the four possible pairs; the one that cancels the p and not p term works.

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

Explanation:
1. \( \sim(\sim p\vee q)\equiv p\wedge\sim q \).
2. So the bracket is \( (p\wedge\sim q)\vee(p\wedge r)\equiv p\wedge(\sim q\vee r) \).
3. Multiply by \( \sim q\wedge r \): \( p\wedge(\sim q\vee r)\wedge(\sim q\wedge r)\equiv p\wedge(\sim q\wedge r) \), since \( (\sim q\wedge r) \) implies \( (\sim q\vee r) \).
4. So the result is \( p\wedge r\wedge\sim q \).

Teacher's Note:
1. Use De Morgan's law on the first term, then absorption: a statement joined with something it implies simplifies to itself.

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

Explanation:
1. Let p: the numbers are not equal; q: their squares are not equal.
2. The contrapositive of \( p\to q \) is \( \sim q\to\sim p \).
3. \( \sim q \): the squares are equal; \( \sim p \): the numbers are equal.

Teacher's Note:
1. Negate both parts of the conditional and swap them..

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

Explanation:
1. \( (p\wedge\sim q)\vee q\equiv(p\vee q)\wedge(\sim q\vee q)\equiv p\vee q \).
2. Then \( (p\vee q)\vee(\sim p\wedge q) \) is still \( p\vee q \), because \( \sim p\wedge q \) already implies \( p\vee q \).

Teacher's Note:
1. Apply the distributive law, then absorb the last term.

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

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

Teacher's Note:
1. Rewrite the implication as a disjunction, then apply De Morgan's law.

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

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

Teacher's Note:
1. Factor out the common not p using the distributive law.

Answer: (a) \( \text{a tautology} \)

Explanation:
1. Let \( x=p\to q \), \( y=(\sim p\to q)\to q \).
2. If \( p\to q \) is true, then \( \sim p\to q \) true means q is true in the case \( p=F \), and when \( p=T \) q is true; so \( (\sim p\to q)\to q \) is true.
3. Checking all four rows of the truth table gives T every time, so it is a tautology.

Teacher's Note:
1. Construct the four-row truth table; the statement is true in every row.

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

Explanation:
1. \( \sim s\vee(\sim r\wedge s)\equiv(\sim s\vee\sim r)\wedge(\sim s\vee s)\equiv\sim s\vee\sim r \).
2. Its negation is \( s\wedge r \).

Teacher's Note:
1. Simplify before negating: the tautology part drops out.

Answer: (b) \( \text{equivalent to }p\leftrightarrow q \)

Explanation:
1. \( p\leftrightarrow\sim q \) is true exactly when p and q have different truth values.
2. Its negation is true exactly when p and q have the same truth value.
3. That is \( p\leftrightarrow q \).

Teacher's Note:
1. A biconditional with a negated side is the exclusive disjunction.
2. Negating it gives the plain biconditional.

Answer: (b) Both I and II are true

Explanation:
1. I: the conjunction contains \( p\wedge\sim p \), so it is false in every row, a fallacy.
2. II: \( \sim q\to\sim p \) is the contrapositive of \( p\to q \), so the biconditional is always true, a tautology.
3. Both are true.

Teacher's Note:
1. A statement containing both a proposition and its negation in a conjunction is a fallacy.
2. A statement and its contrapositive are always equivalent.

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

Explanation:
1. \( \sim(p\to q)\equiv p\wedge\sim q \).
2. So: I become a teacher and I do not open a school.

Teacher's Note:
1. The only way an if-then promise is broken is when the condition holds and the result fails.

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

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

Teacher's Note:
1. Translate the sentence into symbols and place the negation sign in front of the whole biconditional.

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

Explanation:
1. \( [A\wedge(A\to B)]\to B \) is the rule of modus ponens, which is true in every row.
2. The others fail when, for example, \( A=F \): the second and third reduce to A, and the first fails when \( B=T,\ A=F \).

Teacher's Note:
1. Modus ponens (if A holds and A implies B, then B holds) is a tautology.

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

Explanation:
1. The negation of "there exists x such that property" is "for all x, not property".
2. The negation of \( x>0 \) is \( x\le0 \).

Teacher's Note:
1. Negation turns an existence claim into a universal one and reverses the inequality.

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

Explanation:
1. \( q\to p\equiv\sim q\vee p \), so \( p\to(q\to p) \) is always true (a tautology).
2. \( p\to(p\vee q) \) is also always true, so the two are equivalent.
3. The other three options are not tautologies (for example \( p=T,\ q=F \) makes the third and fourth false).

Teacher's Note:
1. Both sides are tautologies, so they are equivalent.
2. Check the other options with p true and q false.