Mathematical Reasoning & Induction · JEE Main previous year questions

The statement $B\Rightarrow ((\sim A)\vee B)$ is equivalent to :

1. A.$B\Rightarrow ((\sim A)\Rightarrow B)$
2. B.$A\Rightarrow (A\iff B)$
3. C.$A\Rightarrow ((\sim A)\Rightarrow B)$
4. D.$B\Rightarrow (A\Rightarrow B)$

The statement $(p\wedge (\sim q))\Rightarrow (p\Rightarrow (\sim q))$ is

1. A.a tautology
2. B.equivalent to $(\sim p)\vee (\sim q)$
3. C.a contradiction
4. D.$p\vee q$

Let $\Delta ,\nabla \in \{\wedge ,\vee \}$ be such that $(\mathrm{p}\to \mathrm{q})\Delta (\mathrm{p}\nabla \mathrm{q})$ is a tautology. Then

1. A.$\Delta =\vee ,\nabla =\vee$
2. B.$\Delta =\vee ,\nabla =\wedge$
3. C.$\Delta =\wedge ,\nabla =\wedge$
4. D.$\Delta =\wedge ,\nabla =\vee$

Let $\Omega$ be the sample space and $A\subseteq \Omega$ be an event. Given below are two statements : (S1) : If P(A) = 0, then $A=\varphi$ (S2) : If P(A) = 1, then $A=\Omega$ Then :

1. A.both (S1) and (S2) are true
2. B.both (S1) and (S2) are false
3. C.only (S2) is true
4. D.only (S1) is true

The compound statement $(\sim (P\wedge Q))\vee ((\sim P)\wedge Q)\Rightarrow ((\sim P)\wedge (\sim Q))$ is equivalent to

1. A.$((\sim P)\vee Q)\wedge (\sim Q)$
2. B.$(\sim Q)\vee P$
3. C.$((\sim P)\vee Q)\wedge ((\sim Q)\vee P)$
4. D.$(\sim P)\vee Q$

Let p and q be two statements. Then $\sim (p\wedge (p\Rightarrow \sim q))$ is equivalent to

1. A.$(\sim p)\vee q$
2. B.$p\vee (p\wedge (\sim q))$
3. C.$p\vee (p\wedge q)$
4. D.$p\vee ((\sim p)\wedge q)$

The statement $(p\wedge q)\Rightarrow (p\wedge r)$ is equivalent to :

1. A.$q\Rightarrow (p\wedge r)$
2. B.$p\Rightarrow (p\wedge r)$
3. C.$(p\wedge r)\Rightarrow (p\wedge q)$
4. D.$(p\wedge q)\Rightarrow r$

The statement $(p\Rightarrow q)\vee (p\Rightarrow r)$ is NOT equivalent to

1. A.$(p\wedge (\sim r))\Rightarrow q$
2. B.$(\sim q)\Rightarrow ((\sim r)\vee p)$
3. C.$p\Rightarrow (q\vee r)$
4. D.$(p\wedge (\sim q))\Rightarrow r$

Let the operations $\ast ,\odot \in \{\wedge ,\vee \}$. If $(p\ast q)\odot (p\odot \sim q)$ is a tautology, then the ordered pair $(\ast ,\odot )$ is :

1. A.$(\vee ,\wedge )$
2. B.$(\vee ,\vee )$
3. C.$(\wedge ,\wedge )$
4. D.$(\wedge ,\vee )$

Let p : Ramesh listens to music. q : Ramesh is out of his village. r : It is Sunday. s : It is Saturday. Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as

1. A.$((\sim q)\wedge (r\vee s))\Rightarrow p$
2. B.$(q\wedge (r\vee s))\Rightarrow p$
3. C.$p\Rightarrow (q\wedge (r\vee s))$
4. D.$p\Rightarrow ((\sim q)\wedge (r\vee s))$

$(p\wedge r)\iff (p\wedge (\sim q))$ is equivalent to $(\sim p)$ when $r$ is

1. A.$p$
2. B.$\sim p$
3. C.$q$
4. D.$\sim q$

If the truth value of the statement $(P\wedge (\sim R))\to ((\sim R)\wedge Q)$ is $\mathrm{F}$, then the truth value of which of the following is $\mathrm{F}$ ?

1. A.$P\vee Q\to \sim R$
2. B.$R\vee Q\to \sim P$
3. C.$\sim (P\vee Q)\to \sim R$
4. D.$\sim (R\vee Q)\to \sim P$

The statement $(\sim (p\iff \sim q))\wedge q$ is :

1. A.a tautology
2. B.a contradiction
3. C.equivalent to $(p\Rightarrow q)\wedge q$
4. D.equivalent to $(p\Rightarrow q)\wedge p$

Consider the following statements: P : Ramu is intelligent. Q : Ramu is rich. R : Ramu is not honest. The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as:

1. A.$((P\wedge (\sim R))\wedge Q)\wedge ((\sim Q)\wedge ((\sim P)\vee R))$
2. B.$((P\wedge R)\wedge Q)\vee ((\sim Q)\wedge ((\sim P)\vee (\sim R)))$
3. C.$((P\wedge R)\wedge Q)\wedge ((\sim Q)\wedge ((\sim P)\vee (\sim R)))$
4. D.$((P\wedge (\sim R))\wedge Q)\vee ((\sim Q)\wedge ((\sim P)\vee R))$

The conditional statement $((p\wedge q)\to ((\sim p)\vee r))\vee (((\sim p)\vee r)\to (p\wedge q))$ is :

1. A.a tautology
2. B.a contradiction
3. C.equivalent to $p\wedge q$
4. D.equivalent to $(\sim p)\vee r$

The maximum number of compound propositions, out of $p\vee r\vee s$, $p\vee r\vee \sim s$, $p\vee \sim q\vee s$, $\sim p\vee \sim r\vee s$, $\sim p\vee \sim r\vee \sim s$, $\sim p\vee q\vee \sim s$, $q\vee r\vee \sim s$, $q\vee \sim r\vee \sim s$, $\sim p\vee \sim q\vee \sim s$ that can be made simultaneously true by an assignment of the truth values to $p,q,r$ and $s$, is equal to ________.

Which of the following statement is a tautology?

1. A.$((\sim q)\wedge p)\wedge q$
2. B.$((\sim q)\wedge p)\wedge (p\wedge (\sim p))$
3. C.$((\sim q)\wedge p)\vee (p\vee (\sim p))$
4. D.$(p\wedge q)\wedge (\sim p\wedge q)$

Let $\Delta ,\nabla \in \{\wedge ,\vee \}$ be such that $p\nabla q\Rightarrow ((p\Delta q)\nabla r)$ is a tautology. Then $(p\nabla q)\Delta r$ is logically equivalent to :

1. A.$(p\Delta r)\vee q$
2. B.$(p\Delta r)\wedge q$
3. C.$(p\wedge r)\Delta q$
4. D.$(p\nabla r)\wedge q$

Let $r\in \{p,q,\sim p,\sim q\}$ be such that the logical statement $r\vee (\sim p)\Rightarrow (p\wedge q)\vee r$ is a tautology. Then r is equal to :

1. A.p
2. B.q
3. C.$\sim p$
4. D.$\sim q$

Consider the following two propositions: $P1:\sim (p\to \sim q)$ $P2:(p\wedge \sim q)\wedge ((\sim p)\vee q)$ If the proposition $p\to ((\sim p)\vee q)$ is evaluated as FALSE, then :

1. A.P1 is TRUE and P2 is FALSE
2. B.P1 is FALSE and P2 is TRUE
3. C.Both P1 and P2 are FALSE
4. D.Both P1 and P2 are TRUE

Let $A=\{x\in \mathbb{R}:|x+1|<2\}$ and $B=\{x\in \mathbb{R}:|x-1|\ge 2\}$. Then which one of the following statements is NOT true?

1. A.$A-B=(-1,1)$
2. B.$B-A=\mathbb{R}-(-3,1)$
3. C.$A\cap B=(-3,-1]$
4. D.$A\cup B=\mathbb{R}-[1,3)$

The negation of the Boolean expression $((\sim q)\wedge p)\Rightarrow ((\sim p)\vee q)$ is logically equivalent to :

1. A.$p\Rightarrow q$
2. B.$q\Rightarrow p$
3. C.$\sim (p\Rightarrow q)$
4. D.$\sim (q\Rightarrow p)$

Consider the following statements: A : Rishi is a judge. B : Rishi is honest. C : Rishi is not arrogant. The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

1. A.$\mathrm{B}\to (\mathrm{A}\vee \mathrm{C})$
2. B.$(\sim \mathrm{B})\wedge (\mathrm{A}\wedge \mathrm{C})$
3. C.$\mathrm{B}\to ((\sim \mathrm{A})\vee (\sim \mathrm{C}))$
4. D.$\mathrm{B}\to (\mathrm{A}\wedge \mathrm{C})$

Consider the two statements : (S1) : $(p\to q)\vee (\sim q\to p)$ is a tautology . (S2) : $(p\wedge \sim q)\wedge (\sim p\vee q)$ is a fallacy. Then :

1. A.only (S1) is true.
2. B.both (S1) and (S2) are false.
3. C.both (S1) and (S2) are true.
4. D.only (S2) is true.

Which of the following is the negation of the statement "for all M > 0, there exists $x\in S$ such that $x\ge M$" ?

1. A.there exists M > 0, such that x < M for all $x\in S$
2. B.there exists M > 0, there exists $x\in S$ such that $x\ge M$
3. C.there exists M > 0, there exists $x\in S$ such that x < M
4. D.there exists M > 0, such that $x\ge M$ for all $x\in S$