Mathematical Reasoning & Induction · JEE Main previous year questions

2023 24 questions

1. Q172023 · 13 Apr 2023 · Shift 1

   The negation of the statement $((A\wedge (B\vee C))\Rightarrow (A\vee B))\Rightarrow A$ is

   1. A.equivalent to $B\vee \sim C$
   2. B.equivalent to $\sim A$
   3. C.equivalent to $\sim C$
   4. D.a fallacy
   Show answer

   Answer: B

2. Q192023 · 13 Apr 2023 · Shift 2

   The statement $(p\wedge (\sim q))\vee ((\sim p)\wedge q)\vee ((\sim p)\wedge (\sim q))$ is equivalent to _________.

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

   Answer: A

3. Q152023 · 12 Apr 2023 · Shift 1

   Among the two statements (S1) : $(p\Rightarrow q)\wedge (p\wedge (\sim q))$ is a contradiction and (S2) : $(p\wedge q)\vee ((\sim p)\wedge q)\vee (p\wedge (\sim q))\vee ((\sim p)\wedge (\sim q))$ is a tautology

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

   Answer: C

4. Q192023 · 11 Apr 2023 · Shift 1

   Let $f:[2,4]\to \mathbb{R}$ be a differentiable function such that $(x{\log }_{e}x)f^{\prime }(x)+({\log }_{e}x)f(x)+f(x)\ge 1,x\in [2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$. Consider the following two statements : (A) : $f(x)\le 1$, for all $x\in [2,4]$ (B) : $f(x)\ge \frac{1}{8}$, for all $x\in [2,4]$ Then,

   1. A.Neither statement (A) nor statement (B) is true
   2. B.Only statement (A) is true
   3. C.Only statement (B) is true
   4. D.Both the statements (A) and (B) are true
   Show answer

   Answer: D

5. Q292023 · 11 Apr 2023 · Shift 1· Numerical answer

   The number of ordered triplets of the truth values of $p,q$ and $r$ such that the truth value of the statement $(p\vee q)\wedge (p\vee r)\Rightarrow (q\vee r)$ is True, is equal to _________.

   Show answer

   Answer: 7

6. Q42023 · 11 Apr 2023 · Shift 2

   For $a\in \mathbb{C}$, let $\mathrm{A}=\{z\in \mathbb{C}:\mathrm{Re}(a+\bar{z})>\mathrm{Im}(\bar{a}+z)\}$ and $\mathrm{B}=\{z\in \mathbb{C}:\mathrm{Re}(a+\bar{z})<\mathrm{Im}(\bar{a}+z)\}$. Then among the two statements : (S1) : If $\mathrm{Re}(a),\mathrm{Im}(a)>0$, then the set A contains all the real numbers (S2) : If $\mathrm{Re}(a),\mathrm{Im}(a)<0$, then the set B contains all the real numbers,

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

   Answer: A

7. Q162023 · 11 Apr 2023 · Shift 2

   The converse of $((\sim p)\wedge q)\Rightarrow r$ is

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

   Answer: C

8. Q172023 · 10 Apr 2023 · Shift 1

   The negation of the statement $(p\vee q)\wedge (q\vee (\sim r))$ is :

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

   Answer: A

9. Q132023 · 10 Apr 2023 · Shift 2

   The statement $\sim [p\vee (\sim (p\wedge q))]$ is equivalent to :

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

   Answer: C

10. Q202023 · 8 Apr 2023 · Shift 1

    Negation of $(p\Rightarrow q)\Rightarrow (q\Rightarrow p)$ is :

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

    Answer: B

11. Q182023 · 8 Apr 2023 · Shift 2

    The negation of $(p\wedge (\sim q))\vee (\sim p)$ is equivalent to :

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

    Answer: A

12. Q192023 · 6 Apr 2023 · Shift 1

    Statement $(\text{P}\Rightarrow \text{Q})\wedge (\text{R}\Rightarrow \text{Q})$ is logically equivalent to :

    1. A.$(\text{P}\Rightarrow \text{R})\wedge (\text{Q}\Rightarrow \text{R})$
    2. B.$(\text{P}\Rightarrow \text{R})\vee (\text{Q}\Rightarrow \text{R})$
    3. C.$(\text{P}\wedge \text{R})\Rightarrow \text{Q}$
    4. D.$(\text{P}\vee \text{R})\Rightarrow \text{Q}$
    Show answer

    Answer: D

13. Q62023 · 6 Apr 2023 · Shift 2

    Among the statements : (S1) : $2023^{2022}-1999^{2022}$ is divisible by 8 (S2) : $13(13{)}^n-12n-13$ is divisible by 144 for infinitely many $n\in \mathbb{N}$

    1. A.both (S1) and (S2) are incorrect
    2. B.only (S1) is correct
    3. C.only (S2) is correct
    4. D.both (S1) and (S2) are correct
    Show answer

    Answer: D

14. Q182023 · 6 Apr 2023 · Shift 2

    Among the statements (S1) : $(p\Rightarrow q)\vee ((\sim p)\wedge q)$ is a tautology (S2) : $(q\Rightarrow p)\Rightarrow ((\sim p)\wedge q)$ is a contradiction

    1. A.neither (S1) and (S2) is True
    2. B.only (S2) is True
    3. C.both (S1) and (S2) are True
    4. D.only (S1) is True
    Show answer

    Answer: A

15. Q182023 · 1 Feb 2023 · Shift 1

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

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

    Answer: A

16. Q192023 · 31 Jan 2023 · Shift 1

    (S1) $(p\Rightarrow q)\vee (p\wedge (\sim q))$ is a tautology (S2) $((\sim p)\Rightarrow (\sim q))\wedge ((\sim p)\vee q)$ is a contradiction. Then

    1. A.only (S2) is correct
    2. B.both (S1) and (S2) are correct
    3. C.only (S1) is correct
    4. D.both (S1) and (S2) are wrong
    Show answer

    Answer: C

17. Q172023 · 31 Jan 2023 · Shift 2

    The number of values of $r\in \{p,q,\sim p,\sim q\}$ for which $((p\wedge q)\Rightarrow (r\vee q))\wedge ((p\wedge r)\Rightarrow q)$ is a tautology, is :

    1. A.2
    2. B.1
    3. C.4
    4. D.3
    Show answer

    Answer: A

18. Q112023 · 30 Jan 2023 · Shift 1

    Among the statements : (S1) $((p\vee q)\Rightarrow r)\iff (p\Rightarrow r)$ (S2) $((p\vee q)\Rightarrow r)\iff ((p\Rightarrow r)\vee (q\Rightarrow r))$

    1. A.only (S1) is a tautology
    2. B.neither (S1) nor (S2) is a tautology
    3. C.both (S1) and (S2) are tautologies
    4. D.only (S2) is a tautology
    Show answer

    Answer: B

19. Q162023 · 29 Jan 2023 · Shift 2

    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.$B\Rightarrow (A\Rightarrow B)$
    Show answer

    Answer: A, C, D

20. Q162023 · 25 Jan 2023 · Shift 1

    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.$p\vee q$
    Show answer

    Answer: A

21. Q172023 · 25 Jan 2023 · Shift 2

    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.$\Delta =\wedge ,\nabla =\vee$
    Show answer

    Answer: A

22. Q72023 · 24 Jan 2023 · Shift 1

    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
    Show answer

    Answer: A

23. Q172023 · 24 Jan 2023 · Shift 1

    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.$(\sim P)\vee Q$
    Show answer

    Answer: C

24. Q192023 · 24 Jan 2023 · Shift 2

    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.$p\vee ((\sim p)\wedge q)$
    Show answer

    Answer: A