JEE Mathematics Mathematical Reasoning MCQs Set 02

Answer: (c) \((F,T)\)

Explanation:
1. A conditional and its contrapositive always have the same truth value.
2. P: hypothesis true (7 is odd), conclusion false (7 is not divisible by 2), so P is false and \(V_1=F\).
3. Q: hypothesis true and conclusion true, so Q is true and \(V_2=T\).
4. Hence \((V_1,V_2)=(F,T)\).

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Answer: (d) If the area of a square does not increase four times, then its side is not doubled

Explanation:
1. The contrapositive of \(p\to q\) is \(\sim q\to\sim p\).
2. Here \(p\): the side doubles; \(q\): the area increases four times.
3. So the contrapositive is: if the area does not increase four times, then the side is not doubled.

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Answer: (a) If I will come, then it is not raining

Explanation:
1. Let \(p\): it is raining; \(q\): I will not come. The statement is \(p\to q\).
2. Contrapositive \(\sim q\to\sim p\): if I will come, then it is not raining.

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

Explanation:
1. The statement is \((P\wedge\sim R)\leftrightarrow Q\).
2. Since \(\sim(X\leftrightarrow Y)\equiv X\leftrightarrow\ \sim Y\), the negation is \((P\wedge\sim R)\leftrightarrow\ \sim Q\).
3. That is \(\sim Q\leftrightarrow P\wedge\ \sim R\).

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Answer: (b) If I do not go to school, it rains

Explanation:
1. The statement is: if it does not rain, then I go to school, i.e. \(\sim r\to s\).
2. Contrapositive \(\sim s\to r\): if I do not go to school, then it rains.

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

Explanation:
1. \(\sim(p\vee\sim q)=\sim p\wedge q\) and \(\sim(p\vee q)=\sim p\wedge\sim q\).
2. Their disjunction is \(\sim p\wedge(q\vee\sim q)=\sim p\wedge T=\sim p\).

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

Explanation:
1. \(p\Rightarrow(q\vee r)\equiv\ \sim p\vee q\vee r\).
2. \((p\Rightarrow q)\vee(p\Rightarrow r)=(\sim p\vee q)\vee(\sim p\vee r)=\sim p\vee q\vee r\), which matches.
3. Option (a) fails for \(p=T,q=F,r=T\) and (c) fails for \(p=T,q=T,r=F\), where \(p\Rightarrow(q\vee r)\) is true but they are false; (d) fails for \(p=F,q=T,r=F\).

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Answer: (c) 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 is \(\sim q\to\sim p\).
2. That reads: if I will not go to the doctor, then I am feeling well.

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

Explanation:
1. \(p\to(q\to p)=\ \sim p\vee(\sim q\vee p)=T\), a tautology.
2. \(p\to(p\vee q)=\ \sim p\vee p\vee q=T\) as well, so (b) is equivalent.
3. The others reduce to \(\sim p\vee q\), which is not always true.

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Answer: (a) \(Q\wedge R\to P\)

Explanation:
1. If \(m\) is prime and \(m\mid n^2\), then \(m\mid n\) (Euclid's lemma), so \(Q\wedge R\to P\) is true.
2. \(P\wedge Q\to R\) fails for \(m=4,\ n=4\); \(Q\to R\) fails for \(m=4,\ n=2\); \(Q\to P\) fails for \(m=4,\ n=2\).

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

Explanation:
1. \((p\wedge\sim q)\vee q=p\vee q\) (absorption).
2. \((p\vee q)\vee(\sim p\wedge q)=p\vee q\), since \(\sim p\wedge q\) is already contained in \(q\).

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

Explanation:
1. \(\sim[\sim s\vee(\sim r\wedge s)]=s\wedge\sim(\sim r\wedge s)=s\wedge(r\vee\sim s)\).
2. \(s\wedge(r\vee\sim s)=(s\wedge r)\vee(s\wedge\sim s)=s\wedge r\).

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Answer: (d) 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, i.e. \(p\leftrightarrow q\).

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Answer: (c) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I

Explanation:
1. Statement I requires both \(p\) and \(\sim p\) to be true, which is impossible, so it is always false: a fallacy.
2. Statement II holds because a conditional is equivalent to its contrapositive: a tautology.
3. Both are true, but II (contrapositive equivalence) does not explain why I is a contradiction.

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