Mathematical Reasoning

Q1 FREE PREVIEW

The statement \(B\Rightarrow ((~A)∨B)\) is not equivalent to :

[JEE Main 2023, 29 Jan (Shift 2)]

a

\(B\Rightarrow ((~A)\Rightarrow B)\)

b

\(A\Rightarrow ((~A)\Rightarrow B)\)

c

\(A\Rightarrow (A\Leftrightarrow B)\)

d

\(B\Rightarrow (A\Rightarrow B)\)

✓ Correct answer: c)

\(A\Rightarrow (A\Leftrightarrow B)\)

Explanation

\(\begin{matrix}\mathrm{A} & \mathrm{B} & ~\mathrm{A} & ~\mathrm{A}∨\mathrm{B} & \mathrm{B}\Rightarrow ((~\mathrm{A})∨\mathrm{B}) & ~\mathrm{A}\Rightarrow \mathrm{B} & \mathrm{A}\Rightarrow \mathrm{B} & \mathrm{A}\Leftrightarrow \mathrm{B} & 1 & 2 & 3 & 4 \\ \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\ \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T}\end{matrix}\)
1,2 and 4 are correct option.

Q2 FREE PREVIEW

The negation of \( (\sim p \wedge q) \vee(p \wedge \sim q) \) is

a

\( (p \vee \sim q) \vee(\sim p \vee q) \)

b

\( (p \vee \sim q) \wedge(\sim p \vee q) \)

c

\( (p \wedge \sim q) \wedge(\sim p \vee q) \)

d

\( (p \wedge \sim q) \wedge(p \vee \sim q) \)

✓ Correct answer: b)

\( (p \vee \sim q) \wedge(\sim p \vee q) \)

Explanation

\( (p \vee \sim q) \wedge(\sim p \vee q) \)

Q3 FREE PREVIEW

Which of the following Boolean expressions is not a tautology?

[JEE Main 2021, 22 Jul (Shift 2)]

a

\((~p\Rightarrow q)∨(~q\Rightarrow p)\)

b

\((q\Rightarrow p)∨(~q\Rightarrow p)\)

c

\((p\Rightarrow ~q)∨(~q\Rightarrow p)\)

d

\((p\Rightarrow q)∨(~q\Rightarrow p)\)

✓ Correct answer: a)

\((~p\Rightarrow q)∨(~q\Rightarrow p)\)

Explanation

(a) \((~p\Rightarrow q)∨(~q\Rightarrow p)\)

\( (\sim(\sim p) \vee q) \vee(\sim(\sim q) \vee p) \)
\( \equiv(p \vee q) \vee(q \vee p) \)
\( \equiv p \vee q \vee q \vee p \)
\( \equiv p \vee q\)

(b) \((q\Rightarrow p)∨(~q\Rightarrow p)\)

\( (\sim q \vee p) \vee(\sim(\sim q) \vee p) \)

\( \equiv(\sim q \vee p) \vee(q \vee p) \)
\( \equiv \sim q \vee p \vee q \vee p \)
\( \equiv(\sim q \vee q) \vee(p \vee p) \)
\( \equiv \text { True } \vee p \)
\( \equiv \text { True }\)

(c) \((p\Rightarrow ~q)∨(~q\Rightarrow p)\)

\( (\sim p \vee \sim q) \vee(\sim(\sim q) \vee p) \)
\( \equiv(\sim p \vee \sim q) \vee(q \vee p) \)
\( \equiv \sim p \vee \sim q \vee q \vee p \)
\( \equiv(\sim p \vee p) \vee(\sim q \vee q) \)
\( \equiv \text { True } \vee \text { True } \)
\( \equiv \text { True }\)

(d) \((p\Rightarrow q)∨(~q\Rightarrow p)\)

\( (\sim p \vee q) \vee(\sim(\sim q) \vee p) \)
\( \equiv(\sim p \vee q) \vee(q \vee p) \)
\( \equiv \sim p \vee q \vee q \vee p \)
\( \equiv(\sim p \vee p) \vee(q \vee q) \)
\( \equiv \text { True } \vee q \)
\( \equiv \text { True }\)

Q4 FREE PREVIEW

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

[JEE Main 2023, 13 Apr (Shift 2)]

a

\((\sim p) \vee(\sim q)\)

b

\(p \vee(\sim q)\)

c

\((\sim p) \vee q\)

d

\(p \vee q\)

✓ Correct answer: a)

\((\sim p) \vee(\sim q)\)

Explanation

\(\begin{aligned} & (p \wedge(\sim q) \vee((\sim p) \wedge q) \vee((\sim p) \wedge)(\sim q)) \\ = & (p \wedge(\sim q)) \vee((\sim p) \wedge(q \vee(\sim q))) \\ = & (p \wedge(\sim q)) \vee((\sim p) \wedge t) \\ = & (p \wedge(\sim q)) \vee(\sim p) \\ = & (\sim p) \vee(p \wedge \sim q) \\ = & (\sim p \vee p) \wedge(\sim p \wedge \sim q) \\ = & t \wedge(\sim p \vee \sim q) \\ & =\sim p \vee \sim q\end{aligned}\)

Q5

Consider the following statements:
P : I have fever
Q : I will not take medicine
\(R\) : I will take rest
The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:

[JEE Main 2023, 30 Jan (Shift 2)]

a

\(((\sim P ) \vee \sim Q ) \wedge((\sim P ) \vee R )\)

b

\(((\sim P ) \vee \sim Q ) \wedge((\sim P ) \vee \sim R )\)

c

\(( P \vee Q ) \wedge((\sim P ) \vee R )\)

d

\(( P \vee \sim Q ) \wedge( P \vee \sim R )\)

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Q6

Which of the following statements is a tautology?

[JEE Main 2023, 1 Feb (Shift 2)]

a

\(p \rightarrow(p \wedge(p \rightarrow q))\)

b

\((p \wedge q) \rightarrow(\sim(p) \rightarrow q))\)

c

\((p \wedge(p \rightarrow q)) \rightarrow \sim q\)

d

\(p \vee(p \wedge q)\)

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Q7

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

a

\((\sim(p \wedge q)) \wedge q\)

b

\(\sim(p \wedge q)\)

c

\(\sim(p \vee q)\)

d

\((p \wedge q) \wedge(\sim p)\)

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Q8

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

[JEE Main 2023, 25 Jan (Shift 1)]

a

equivalent to \((\sim p) \vee(\sim q)\)

b

a tautology

c

equivalent to \(p \vee q\)

d

a contradiction

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Q9

The converse of \(((~p)∧q)\Rightarrow r\) is

[JEE Main 2023, 11 Apr (Shift 2)]

a

\((~r)\Rightarrow p∧q\)

b

\((~r)\Rightarrow ((~p)∧q)\)

c

\(((~p)∨q)\Rightarrow r\)

d

\((p∨(~q))\Rightarrow (~r)\)

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Q10

Consider the two statements :

\( \left(S_{1}\right):(p \rightarrow q) \vee(\sim q \rightarrow p) \) is a tautology

\( \left(S_{2}\right):(p \wedge \sim q) \wedge(\sim p \vee q) \) is a fallacy Then,

a

only \( \left(S_{1}\right) \) is true

b

both \( \left(S_{1}\right) \) and \( \left(S_{2}\right) \) are false

c

both \( \left(S_{1}\right) \) and \( \left(S_{2}\right) \) are true

d

only \( \left(S_{2}\right) \) is true

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Q11

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

[JEE Main 2023, 10 Apr (Shift 1)]

a

\(((\sim p) \vee r) \wedge(\sim q)\)

b

\(((\sim p) \vee(\sim q)) \wedge(\sim r)\)

c

\(((\sim p) \vee(\sim q)) \vee(\sim r)\)

d

\((p \vee r) \wedge(\sim q)\)

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Q12

The statement \(\ p \) : For any real numbers \(\ x, y \) if \(\ x=y \), then \(\ 2 x+a=2 y+a \) when \(\ a \in Z \).

a

is true

b

is false

c

its contrapositive is not true

d

None of these

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Q13

The negation of the statement \((p∨q)∧(q∨(~r))\) is

[JEE Main 2023, 10 Apr (Shift 1)]

a

\(((~\mathrm{p})∨\mathrm{r})∧)~\mathrm{q})\)

b

\(((~p)∨(~q))∨(~r)\)

c

\((p∨r)∧(~q)\)

d

\(((~p)∨(~q))∧(~r)\)

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Q14

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

[JEE Main 2023, 1 Feb (Shift 1)]

a

\((\sim p) \wedge(\sim q)\)

b

\(p \wedge(\sim q)\)

c

\((\sim p) \vee(\sim q)\)

d

\((\sim p) \vee q\)

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Q15

Which of the following is equivalent to the Boolean expression \(p \wedge \sim q\) ?

[JEE Main 2021, 1 Sep (Shift 2)]

a

\(\sim p \rightarrow \sim q\)

b

\(\sim(q \rightarrow p)\)

c

\(\sim(p \rightarrow q)\)

d

\(\sim(p \rightarrow \sim q)\)

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Q16

Contrapositive of the statement: If a function \( f \) is differentiable at \( a \), then it is also continuous at \( a \), is

a

If a function \( f \) is continuous at \( a \), then it is not differentiable at \( a \).

b

If a function \( f \) is not continuous at \( a \), then it is not differentiable at \( a \).

c

If a function \( f \) is not continuous at \( a \), then it is differentiable at \( a \).

d

If a function \( f \) is continuous at \( a \), then it is differentiable at \( a \).

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Q17

Which of the following is tautology?

a

\((\sim p) \wedge(p \vee q) \rightarrow a\)

b

\((p \rightarrow q) \wedge(q \rightarrow p)\)

c

\((\sim q) \vee(p \wedge q) \rightarrow q\)

d

\((q \rightarrow p) \vee \sim(p \rightarrow q)\)

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Q18

The only statement among the following that is a tautology is

a

\( \mathrm{A} \wedge(\mathrm{A} \vee \mathrm{B}) \)

b

\( \mathrm{A} \vee(\mathrm{A} \wedge \mathrm{B}) \)

c

\( [\mathrm{A} \wedge(\mathrm{A} \rightarrow \mathrm{B})] \rightarrow \mathrm{B} \)

d

\( \mathrm{B} \rightarrow[\mathrm{A} \wedge(\mathrm{A} \rightarrow \mathrm{B})] \)

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Q19

Negation of the Boolean expression \(p \Leftrightarrow(q \Rightarrow p)\) is :

a

\((\sim p) \wedge q\)

b

\((\mathrm{p}) \wedge(\sim q)\)

c

\((\sim p) \vee(\sim q)\)

d

\((\sim p) \wedge(\sim q)\)

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Q20

If \(P\) and \(Q\) are two statement, then which of the following compound statement is a tautology?

[JEE Main 2021, 18 Mar (Shift 2)]

a

\(((P \Rightarrow Q) \wedge \sim Q) \Rightarrow Q\)

b

\(((P \Rightarrow Q) \wedge \sim Q) \Rightarrow \sim P\)

c

\(((P \Rightarrow Q) \wedge \sim Q) \Rightarrow P\)

d

\(((P \Rightarrow Q) \wedge \sim Q) \Rightarrow(P \wedge Q)\)

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Q21

If \(p \rightarrow(p \wedge \sim q)\) is false, then the truth values of \(\mathrm{p}\) and \(\mathrm{q}\) are respectively:

a

F,F

b

F,T

c

T,F

d

T,T

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Q22

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

[JEE Main 2023, 10 Apr (Shift 2)]

a

\((\sim(p \wedge q)) \wedge q\)

b

\(\sim(p \wedge q)\)

c

\(\sim(p \vee q)\)

d

\((p \wedge q) \wedge(\sim p)\)

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Q23

Among the statements:
\(\left(S_1\right)((p \vee q) \Rightarrow r) \Leftrightarrow(p \Rightarrow r)\)
\(\left(S_2\right)((p \vee q) \Rightarrow r) \Leftrightarrow((p \Rightarrow r) \vee(q \Rightarrow r))\)

a

Only \(\left(S_1\right)\) is a tautology

b

Neither \(\left(S_1\right)\) nor \(\left(S_2\right)\) is a tautology

c

Only \(\left(S_2\right)\) is a tautology

d

Both \(\left(S_1\right)\) and \(\left(S_2\right)\) are tautologies

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Q24

Negation of the statement \((p \vee r) \Rightarrow(q \vee r)\) is.

[JEE Main 2021, 31 Aug (Shift 2)]

a

\(\sim p \wedge q \wedge \sim r\)

b

\(\sim p \wedge q \wedge r\)

c

\(p \wedge \sim q \wedge \sim r\)

d

\(p \wedge q \wedge r\)

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Q25

Negation of \(p \wedge(q \wedge \sim(p \wedge q))\) is

[JEE Main 2023, 15 Apr (Shift 1)]

a

\(\sim( p \vee q)\)

b

\(p \vee q\)

c

\((\sim(p \wedge q)) \wedge q\)

d

\((\sim(p \wedge q)) \vee p\)

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Q26

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

[JEE Main 2023, 24 Jan (Shift 2)]

a

\(p \vee(p \wedge(\sim q))\)

b

\(p \vee((\sim p) \wedge q)\)

c

\((\sim p) \vee q\)

d

\(p \vee(p \wedge q)\)

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Q27

If P and Q are two statements, then which of the following compound statement is a tautology?

[JEE Main 2021, 18 Mar (Shift 2)]

a

\(((P\Rightarrow Q)∧~Q)\Rightarrow Q\)

b

\(((P\Rightarrow Q)∧~Q)\Rightarrow (P∧Q)\)

c

\(((P\Rightarrow Q)∧~Q)\Rightarrow P\)

d

\(((P\Rightarrow Q)∧~Q)\Rightarrow ~P\)

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Q28

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

[JEE Main 2023, 8 Apr (Shift 2)]

a

\(p \wedge q\)

b

\(p \wedge(\sim q)\)

c

\(p \wedge(q \wedge(\sim p))\)

d

\(p \vee(q \vee(\sim p))\)

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Q29

Statement \((P \Rightarrow Q) \wedge(R \Rightarrow Q)\) is logically equivalent to

[JEE Main 2023, 6 Apr (Shift 1)]

a

\(( P \vee R) \Rightarrow Q\)

b

\(( P \Rightarrow R) \wedge(Q \Rightarrow R)\)

c

\(( P \Rightarrow R) \vee(Q \Rightarrow R)\)

d

\(( P \wedge R) \Rightarrow Q\)

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Q30

Among the statements :

\((\mathrm{S}1)((\mathrm{p}∨\mathrm{q})\Rightarrow \mathrm{r})\Leftrightarrow (\mathrm{p}\Rightarrow \mathrm{r})\)

\((\mathrm{S}2)((\mathrm{p}∨\mathrm{q})\Rightarrow \mathrm{r})\Leftrightarrow ((\mathrm{p}\Rightarrow \mathrm{r})∨(\mathrm{q}\Rightarrow \mathrm{r}))\)

[JEE Main 2023, 30 Jan (Shift 1)]

a

neither (S1) nor (S2) is a tautology

b

only (S2) is a tautology

c

both (S1) and (S2) are tautologies

d

only (S1) is a tautology

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Q31

The negation of the statement \(((A∧(B∨C))\Rightarrow (A∨B))\Rightarrow A\)) is

[JEE Main 2023, 13 Apr (Shift 1)]

a

equivalent to \(~A\)

b

equivalent to \(B∨~C\)

c

a fallacy

d

equivalent to \(~C\)

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Q32

Negation of (p\(\to\)q) \(\to\)(q \(\to\) p) is

[JEE Main 2023, 08 Apr (Shift 1)]

a

~q\(∧\)p

b

pv(~q)

c

(~p)vq

d

q\(∧\)(~p)

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Q33

Negation of \((p \rightarrow q) \rightarrow(q \rightarrow p)\) is

[JEE Main 2023, 8 Apr (Shift 1)]

a

\((\sim p) \vee q\)

b

\(q \wedge(\sim p)\)

c

\((\sim q) \wedge p\)

d

\((\sim(p \wedge q)) \vee p\)

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Q34

The Boolean expression \((p \wedge \sim q) \Rightarrow(q \vee \sim p)\) is equivalent to :

[JEE Main 2021, 20 Jul (Shift 1)]

a

\(q\Rightarrow p\)

b

\(~q\Rightarrow p\)

c

\(p\Rightarrow q\)

d

\(p\Rightarrow ~q\)

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Q35

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

[JEE Main 2023, 08 Apr (Shift 2)]

a

\(p \wedge q\)

b

\(p \wedge(\sim q)\)

c

\(p \wedge(q \wedge(\sim p))\)

d

\(p \vee(q \vee(\sim p))\)

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