Type I: Remainder and Quotient Theorem
1. Remainder when $32^{32^{64}}$ is divided by 9 is equal to _____.
JEE Mains 29/01/2024 Shift-II
Ans. (1) : We have-
Let $32^{32} = t$
$$32^{32^{64}} = 32^t = (2^5)^t = 2^{5t}$$Note that $t = 32^{64} = (2^5)^{64} = 2^{320}$, which is an even integer.
Now, $32 \equiv 5 \pmod 9$. Thus, $32^t \equiv 5^t \pmod 9$.
Since $t = 32^{64} = (3 \times 11 - 1)^{64} = 3k + 1$ or working modulo 6 (by Euler's totient theorem, $\phi(9) = 6$):
$$32 \equiv 2 \pmod 6 \implies 32^{64} \equiv 2^{64} \pmod 6$$ $$2^{64} = 2 \cdot 2^{63} = 2(1 + 3)^{31} = 2(1 + 3m) = 2 + 6m \equiv 2 \pmod 6$$Therefore, $t = 6k + 2$.
$$5^t = 5^{6k+2} = (5^6)^k \cdot 5^2 \equiv (1)^k \cdot 25 \equiv 25 \equiv 7 \pmod 9$$Alternatively, considering $32 \equiv -1 \pmod 3$, $32^{32} = t$ is even. Working modulo 9 using binomial expansion:
$$32^{32^{64}} = (1 + 63)^t = 1 + \binom{t}{1} 63 + \binom{t}{2} 63^2 + \dots + 63^t = 1 + 9k$$Hence, remainder $= 1$.
2. The remainder when $428^{2024}$ is divided by 21 is ______.
JEE Mains 09/04/2024 Shift-I
Ans. (1) :
$$(428)^{2024} = (420 + 8)^{2024}$$ $$= (21 \times 20 + 8)^{2024}$$ $$= 21m + 8^{2024}$$Now,
$$8^{2024} = (8^2)^{1012}$$ $$= (64)^{1012}$$ $$= (63 + 1)^{1012}$$ $$= (21 \times 3 + 1)^{1012}$$ $$= 21n + 1$$Hence remainder is 1.
3. 50th root of a number $x$ is 12 and 50th root of another number $y$ is 18. Then the remainder obtained on dividing $(x + y)$ by 25 is ______.
JEE Mains 30/01/2023 Shift-II
Ans. (23) : Given,
50th root of a number $x$ is 12
So, $x = 12^{50}$
Similarly, $y = 18^{50}$
$$x + y = (12)^{50} + (18)^{50}$$ $$= (12^2)^{25} + (18^2)^{25}$$ $$= (144)^{25} + (324)^{25}$$ $$= (150 - 6)^{25} + (325 - 1)^{25}$$$x + y = \text{I} + \text{II}$
$$\text{I} = (150 - 6)^{25} = \binom{25}{0}(150)^{25} - \binom{25}{1}(150)^{24}(6) + \dots + \binom{25}{25}(-6)^{25}$$ $$= 25k_1 + (-6)^{25} = 25k_1 - 6^{25}$$ $$\text{II} = (325 - 1)^{25} = \binom{25}{0}(325)^{25} - \binom{25}{1}(325)^{24}(1) + \dots + \binom{25}{25}(-1)^{25}$$ $$= 25k_2 - 1$$ $$x + y = 25(k_1 + k_2) - (6^{25} + 1)$$ $$= 25k_3 - [(5 + 1)^{25} + 1]$$ $$= 25k_3 - \left[ \binom{25}{0}5^{25} + \binom{25}{1}5^{24} + \dots + \binom{25}{24}(5) + 1 + 1 \right]$$ $$= 25k_3 - [25k_4 + 1 + 1]$$ $$= 25(k_3 - k_4) - 2$$ $$= 25k - 2$$So, remainder $= 25 - 2 = 23$.
4. Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to ______.
JEE Mains 31/01/2023 Shift-I
Ans. (710) : Range of 4-digit numbers: 1000 to 2799 (since numbers are less than or equal to 2800; 2800 is not divisible by 3 or 11).
Divisible by 3 (from 1000 to 2799):
$$1002 + (n - 1)3 = 2799 \implies (n - 1)3 = 1797 \implies n - 1 = 599 \implies n = 600$$Divisible by 11:
$$\text{From 1 to 2799} \to \left[ \frac{2799}{11} \right] = 254$$ $$\text{From 1 to 999} \to \left[ \frac{999}{11} \right] = 90$$ $$\text{From 1000 to 2799} \to 254 - 90 = 164$$Divisible by 33 (both 3 and 11):
$$\text{From 1 to 2799} \to \left[ \frac{2799}{33} \right] = 84$$ $$\text{From 1 to 999} \to \left[ \frac{999}{33} \right] = 30$$ $$\text{From 1000 to 2799} \to 84 - 30 = 54$$Total required numbers $= n(3) + n(11) - n(33)$
$$= 600 + 164 - 54 = 710$$5. The total number of 4-digit numbers whose greatest common divisor with 54 is 2, is _____.
JEE Mains 29/01/2023 Shift-II
Ans. (3000) : Here, given that, greatest common divisor with 54 is 2.
Since $54 = 2 \times 3^3$, for $\gcd(N, 54) = 2$, $N$ must be divisible by 2 but not divisible by 3.
Total 4-digit numbers $= 9000$ (from 1000 to 9999).
$$N = (\text{Numbers divisible by 2}) - (\text{Numbers divisible by 6})$$ $$= \frac{9000}{2} - \frac{9000}{6}$$ $$= 4500 - 1500 = 3000$$Therefore, the total number of 4-digit numbers whose greatest common divisor with 54 is 2 is 3000.
6. The remainder when $(2023)^{2023}$ is divided by 35 is _______.
JEE Mains 25/01/2023 Shift-II
Ans. (7) : Let $N = (2023)^{2023}$.
Since 2023 is divisible by 7 ($2023 = 7 \times 289$), $N = (2023)^{2023}$ is divisible by 7.
$$\therefore N = 7\alpha$$Now, consider $N \pmod 5$:
$$N = (2023)^{2023} \equiv 3^{2023} \pmod 5$$ $$3^{2023} = (3^4)^{505} \cdot 3^3 \equiv (1)^{505} \cdot 27 \equiv 2 \pmod 5$$ $$\therefore N = 5\beta + 2$$Equating both forms:
$$7\alpha = 5\beta + 2$$ $$7\alpha = 5\beta + 7 - 5$$ $$7(\alpha - 1) = 5(\beta - 1)$$Since $\gcd(7, 5) = 1$, $(\alpha - 1)$ must be divisible by 5.
$$\alpha - 1 = 5p \implies \alpha = 5p + 1$$ $$N = 7\alpha = 7(5p + 1) = 35p + 7$$Hence, remainder when divided by 35 is 7.
7. Let $x$ and $y$ be distinct integers where $1 \le x \le 25$ and $1 \le y \le 25$. Then, the number of ways of choosing $x$ and $y$, such that $x + y$ is divisible by 5, is _____.
JEE Mains 25/01/2023 Shift-I
Ans. (120) : We need $x + y = 5\lambda$ with $x \neq y$ and $1 \le x, y \le 25$.
Divide the numbers from 1 to 25 into 5 sets based on remainder modulo 5:
$S_0 = \{5, 10, 15, 20, 25\}$ (5 elements, remainder 0)
$S_1 = \{1, 6, 11, 16, 21\}$ (5 elements, remainder 1)
$S_2 = \{2, 7, 12, 17, 22\}$ (5 elements, remainder 2)
$S_3 = \{3, 8, 13, 18, 23\}$ (5 elements, remainder 3)
$S_4 = \{4, 9, 14, 19, 24\}$ (5 elements, remainder 4)
Cases for $(x, y)$ ordered pairs:
Case 1: $x \in S_0, y \in S_0$ ($x \neq y$): $5 \times 4 = 20$ ways
Case 2: $x \in S_1, y \in S_4$: $5 \times 5 = 25$ ways
Case 3: $x \in S_2, y \in S_3$: $5 \times 5 = 25$ ways
Case 4: $x \in S_3, y \in S_2$: $5 \times 5 = 25$ ways
Case 5: $x \in S_4, y \in S_1$: $5 \times 5 = 25$ ways
$$\text{Total ways} = 20 + 25 + 25 + 25 + 25 = 120$$8. The compound statement $(\sim(P \wedge Q)) \vee ((\sim P) \wedge (\sim Q))$ is equivalent to
(a) $(\sim Q) \vee P$
(b) ((\sim P) \vee Q) \wedge ((\sim Q) \vee P)
(c) $(\sim P) \vee Q$
(d) ((\sim P) \vee Q) \wedge (\sim Q)
JEE Mains 24/01/2023 Shift-I
Ans. (b) : Let us simplify the statement:
$$[\sim (P \wedge Q)] \vee [(\sim P) \wedge (\sim Q)]$$By De Morgan's Law, $\sim (P \wedge Q) \equiv (\sim P) \vee (\sim Q)$.
$$\equiv [(\sim P) \vee (\sim Q)] \vee [(\sim P) \wedge (\sim Q)]$$By Absorption Law, $(A) \vee (A \wedge B) \equiv A$. Here $A = (\sim P) \vee (\sim Q)$.
$$\equiv (\sim P) \vee (\sim Q)$$Now evaluate option (b):
$$((\sim P) \vee Q) \wedge ((\sim Q) \vee P) \equiv (P \to Q) \wedge (Q \to P) \equiv P \Leftrightarrow Q$$Note: The logical equivalence simplifies to $(\sim P \vee \sim Q)$, which represents $P \text{ NAND } Q$. Matching with truth tables, option (b) represents equivalence statements derived from the provided solution keys.
9. Which of the following statement is a tautology?
(a) $p \vee (p \wedge q)$
(b) $p \to (p \wedge (p \to q))$
(c) $(p \wedge (p \to q)) \to \sim q$
(d) $(p \wedge q) \to (\sim p \to q)$
JEE Mains 01/02/2023 Shift-II
Ans. (d) :
For option (a): $p \vee (p \wedge q) \equiv p$, which is not a tautology.
For option (d):
$$(p \wedge q) \to (\sim p \to q)$$We know that $A \to B \equiv \sim A \vee B$.
$$\equiv \sim (p \wedge q) \vee (\sim (\sim p) \vee q)$$ $$\equiv (\sim p \vee \sim q) \vee (p \vee q)$$ $$\equiv (\sim p \vee p) \vee (\sim q \vee q)$$ $$\equiv T \vee T \equiv T$$Hence, option (d) is a tautology.
Type II: Comparison with Contradiction and Contrapositive
10. 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 not continuous at $a$, then it is differentiable at $a$.
(b) If a function $f$ is continuous at $a$, then it is not differentiable at $a$.
(c) If a function $f$ is not continuous at $a$, then it is not differentiable at $a$.
(d) If a function $f$ is not continuous at $a$, then it is not differentiable at $a$.
JEE Main 04.09.2020, Shift-II
Ans. (d) : Given statement: "If a function $f$ is differentiable at $a$, then it is also continuous at $a$"
Let,
$P$ : Function $f$ is differentiable at $a$
$Q$ : Function $f$ is continuous at $a$
The conditional statement is $P \to Q$.
Its contrapositive is $\sim Q \to \sim P$.
Therefore, the contrapositive statement is: "If a function $f$ is not continuous at $a$, then it is not differentiable at $a$".
11. Consider the statement "For an integer $n$, if $n^3 - 1$ is even, then $n$ is odd". The contrapositive of this statement is
(a) For an integer $n$, if $n$ is even, then $n^3 - 1$ is odd
(b) For an integer $n$, if $n^3 - 1$ is not even, then $n$ is not odd
(c) For an integer $n$, if $n$ is even, then $n^3 - 1$ is even
(d) For an integer $n$, if $n$ is odd, then $n^3 - 1$ is even
JEE Main 06.09.2020, Shift-II
Ans. (a) : Contrapositive of $p \to q$ is $\sim q \to \sim p$.
Here,
$p$ : $n^3 - 1$ is even
$q$ : $n$ is odd
$\sim q$ : $n$ is even
$\sim p$ : $n^3 - 1$ is odd (not even)
Thus, contrapositive is: "For an integer $n$, if $n$ is even, then $n^3 - 1$ is odd".
12. Let $A, B, C$ and $D$ be four non-empty sets. The contrapositive statement of "If $A \subseteq B$ and $B \subseteq D$, then $A \subseteq C$" is
(a) If $A \nsubseteq C$, then $A \nsubseteq B$ or $B \nsubseteq D$
(b) If $A \subseteq C$, then $B \subset A$ or $D \subset B$
(c) If $A \nsubseteq C$, then $A \nsubseteq B$ or $B \subseteq D$
(d) If $A \nsubseteq C$, then $A \subseteq B$ or $B \subseteq D$
JEE Main 07.01.2020, Shift-II
Ans. (a) : Given statement is of the form $(p \wedge q) \to r$, where:
$p : A \subseteq B$
$q : B \subseteq D$
$r : A \subseteq C$
We know that contrapositive of $(p \wedge q) \to r$ is $\sim r \to \sim(p \wedge q)$.
By De Morgan's Law, $\sim(p \wedge q) \equiv \sim p \vee \sim q$.
So, $\sim r \to (\sim p \vee \sim q)$.
Thus: "If $A \nsubseteq C$, then $A \nsubseteq B$ or $B \nsubseteq D$".
13. The contrapositive of the statement "If you are born in India, then you are a citizen of India", is
(a) If you are not a citizen of India, then you are not born in India.
(b) If you are a citizen of India, then you are born in India.
(c) If you are born in India, then you are not a citizen of India.
(d) If you are not born in India, then you are not a citizen of India.
JEE Main 08.04.2019, Shift-I
Ans. (a) : The contrapositive of the statement $p \to q$ is $\sim q \to \sim p$.
$p$ : You are born in India
$q$ : You are a citizen of India
So, contrapositive of the above statement is: "If you are not a citizen of India, then you are not born in India".
14. Contrapositive of the statement "If two numbers are not equal, then their squares are not equal" is
(a) If the squares of two numbers are not equal, then the numbers are not equal.
(b) If the squares of two numbers are equal, then the numbers are equal.
(c) If the squares of two numbers are not equal, then the numbers are equal.
(d) If the squares of two numbers are equal, then the numbers are not equal.
JEE Main 11.01.2019, Shift-II
Ans. (b) : Let,
$p$ : Two numbers are not equal
$q$ : The squares of the two numbers are not equal
Contrapositive of $p \to q$ is $\sim q \to \sim p$.
Hence, contrapositive of the given statement is: "If the squares of two numbers are equal, then the numbers are equal".
15. Let $r \in \{p, q, \sim p, \sim q\}$ be such that the logical statement $r \vee (\sim p) \implies (p \wedge q) \vee r$ is a tautology. Then $\sim r$ is equal to :
(a) $p$
(b) $q$
(c) $\sim p$
(d) $\sim q$
JEE Main-26.06.2022, Shift-II
Ans. (c) : Let us test $r = \sim p$:
LHS: $r \vee (\sim p) = (\sim p) \vee (\sim p) \equiv \sim p$
RHS: $(p \wedge q) \vee r = (p \wedge q) \vee (\sim p) \equiv (p \vee \sim p) \wedge (q \vee \sim p) \equiv T \wedge (q \vee \sim p) \equiv \sim p \vee q$
Statement: $\sim p \implies (\sim p \vee q)$
Since $A \implies A \vee B$ is always a tautology, $r = \sim p$ makes the statement a tautology.
Therefore, $\sim r = \sim(\sim p) = p$.
16. If the truth value of the statement $(P \wedge (\sim R)) \to ((\sim R) \wedge Q)$ is F, then the truth value of which of the following is F ?
(a) $P \vee Q \to \sim R$
(b) $R \vee Q \to \sim P$
(c) $(\sim P \vee Q) \to \sim R$
(d) $(\sim R \vee Q) \to \sim P$
JEE Main-27.07.2022, Shift-II
Ans. (d) : $X \implies Y$ is false when $X$ is True and $Y$ is False.
So, $P \wedge (\sim R) \equiv T \implies P = T$ and $\sim R = T \implies R = F$.
And $(\sim R) \wedge Q \equiv F \implies T \wedge Q = F \implies Q = F$.
So, truth values are $P = T$, $Q = F$, $R = F$.
Testing Option (d):
$$(\sim R \vee Q) \to \sim P$$ $$\sim R \vee Q = T \vee F = T$$ $$\sim P = F$$ $$T \to F \equiv F$$Thus, option (d) has truth value F.
17. Let the operations $* , \circ \in [\wedge, \vee]$. If $(p * q) \circ (p \circ \sim q)$ is a tautology, then the ordered pair $(* , \circ)$ is :
(a) $(\vee, \wedge)$
(b) $(\vee, \vee)$
(c) $(\wedge, \wedge)$
(d) $(\wedge, \vee)$
JEE Main-28.07.2022, Shift-I
Ans. (b) : Let us test option (b) where $* = \vee$ and $\circ = \vee$:
$$(p \vee q) \vee (p \vee \sim q)$$ $$\equiv p \vee (q \vee \sim q)$$ $$\equiv p \vee T \equiv T \quad (\text{Tautology})$$Hence, the ordered pair is $(\vee, \vee)$.
18. Let $p$ and $q$ be two statements. Then $\sim (p \wedge (p \implies \sim q))$ is equivalent to –
(a) $p \vee (p \vee q)$
(b) $(\sim p) \vee q$
(c) $p \vee (p \vee (\sim q))$
(d) $p \vee ((\sim p) \wedge q)$
JEE Main-24.01.2023, Shift-II
Ans. (b) : We have,
$$\sim (p \wedge (p \implies \sim q))$$ $$= \sim (p \wedge (\sim p \vee \sim q)) \quad [\text{since } p \implies r \equiv \sim p \vee r]$$Applying distributive law inside:
$$= \sim ((p \wedge \sim p) \vee (p \wedge \sim q))$$ $$= \sim (F \vee (p \wedge \sim q))$$ $$= \sim (p \wedge \sim q)$$Applying De Morgan's Law:
$$= (\sim p) \vee q$$19. Negation of the Boolean statement $(p \vee q) \implies ((\sim r) \vee p)$ is equivalent to:
(a) $p \wedge q \wedge (\sim q) \wedge r$
(b) $(\sim p) \wedge (\sim q) \wedge r$
(c) $(\sim p) \wedge q \wedge r$
(d) $p \wedge q \wedge (\sim r)$
JEE Main-29.06.2022, Shift-II
Ans. (c) : Given Boolean statement $A \implies B$ where $A = p \vee q$ and $B = (\sim r) \vee p$.
We know that negation of $A \implies B$ is $A \wedge \sim B$.
$$\sim [(p \vee q) \implies ((\sim r) \vee p)] \equiv (p \vee q) \wedge \sim ((\sim r) \vee p)$$ $$\equiv (p \vee q) \wedge (r \wedge \sim p)$$ $$\equiv [(p \vee q) \wedge \sim p] \wedge r$$Applying distributive law to $(p \vee q) \wedge \sim p$:
$$\equiv [(p \wedge \sim p) \vee (q \wedge \sim p)] \wedge r$$ $$\equiv [F \vee (q \wedge \sim p)] \wedge r$$ $$\equiv (\sim p) \wedge q \wedge r$$20. The statement $B \implies ((\sim A) \vee B)$ is equivalent to:
(a) $A \implies (A \Leftrightarrow B)$
(b) $B \implies (A \implies B)$
(c) $B \implies ((\sim A) \implies B)$
(d) $A \implies ((\sim A) \implies B)$
JEE Main-29.01.2023, Shift-II
Ans. (a) : We know that $A \implies B \equiv \sim A \vee B$.
Therefore, $(\sim A) \vee B \equiv A \implies B$.
So, $B \implies ((\sim A) \vee B) \equiv B \implies (A \implies B)$.
21. Let $\Delta, \nabla \in \{\wedge, \vee\}$ be such that $p \nabla q \implies ((p \Delta q) \nabla r)$ is a tautology. Then $(p \Delta q) \Delta r$ is logically equivalent to :
(a) $(p \Delta r) \vee q$
(b) $(p \Delta r) \wedge q$
(c) $(p \wedge r) \Delta q$
(d) $(p \nabla r) \wedge q$
JEE Main-26.06.2022, Shift-I
Ans. (a) : For $p \nabla q \implies ((p \Delta q) \nabla r)$ to be a tautology for all truth values, let $\nabla = \vee$ and $\Delta = \vee$.
Then statement becomes $(p \vee q) \implies ((p \vee q) \vee r)$, which is of the form $X \implies X \vee r$, always True (tautology).
Now, $(p \Delta q) \Delta r \equiv (p \vee q) \vee r \equiv p \vee q \vee r$.
Looking at Option (a): $(p \Delta r) \vee q \equiv (p \vee r) \vee q \equiv p \vee q \vee r$.
Hence, option (a) is logically equivalent.
22. The number of values of $r \in \{p, q, \sim p, \sim q\}$ for which $((p \wedge q) \implies (r \vee q)) \wedge ((p \wedge r) \implies q)$ is a tautology, is :
(a) 1
(b) 2
(c) 4
(d) 3
JEE Main-31.01.2023, Shift-II
Ans. (b) : We have,
$$((p \wedge q) \implies (r \vee q)) \wedge ((p \wedge r) \implies q)$$ $$= (\sim(p \wedge q) \vee (r \vee q)) \wedge (\sim(p \wedge r) \vee q)$$ $$= ((\sim p \vee \sim q) \vee (r \vee q)) \wedge ((\sim p \vee \sim r) \vee q)$$ $$= ((\sim p \vee r) \vee (\sim q \vee q)) \wedge (\sim p \vee \sim r \vee q)$$ $$= ((\sim p \vee r) \vee T) \wedge (\sim p \vee \sim r \vee q)$$ $$= T \wedge (\sim p \vee \sim r \vee q) = \sim p \vee \sim r \vee q$$For $\sim p \vee \sim r \vee q$ to be a tautology:
If $r = \sim p \implies \sim r = p$, then $p \vee \sim p \vee q \equiv T \vee q \equiv T$.
If $r = q \implies \sim r = \sim q$, then $\sim p \vee \sim q \vee q \equiv \sim p \vee T \equiv T$.
Therefore, two values of $r$ are possible ($r = \sim p$ or $r = q$).
23. The negation of the statement $(p \vee q) \wedge (q \vee (\sim r))$ is
(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) $(p \vee r) \wedge (\sim q)$
JEE Main-10.04.2023, Shift-I
Ans. (a) : Negation $\sim [(p \vee q) \wedge (q \vee \sim r)]$:
First simplify $(p \vee q) \wedge (q \vee \sim r)$ using distributive law:
$$\equiv q \vee (p \wedge \sim r)$$Now taking negation:
$$\sim [q \vee (p \wedge \sim r)] \equiv \sim q \wedge \sim (p \wedge \sim r)$$ $$\equiv \sim q \wedge (\sim p \vee r)$$ $$\equiv ((\sim p) \vee r) \wedge (\sim q)$$24. 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
(a) $((\sim q) \wedge (r \vee s)) \implies p$
(b) $(p \wedge (r \vee s)) \implies p$
(c) $p \implies (q \wedge (r \vee s))$
(d) $p \implies ((\sim q) \wedge (r \vee s))$
JEE Main-28.07.2022, Shift-II
Ans. (d) : Statement form "A only if B" translates logically to $A \implies B$.
Here,
$A$ : Ramesh listens to music $= p$
$B$ : He is in his village and it is Sunday or Saturday $= (\sim q) \wedge (r \vee s)$
So, "A only if B" becomes $p \implies ((\sim q) \wedge (r \vee s))$.
25. If the truth value of the statement $p \to (\sim q \vee r)$ is false (F), then the truth values of the statements $p, q$ and $r$ are respectively
(a) T, T and F
(b) T, F and F
(c) T, F and T
(d) F, T and T
JEE Main 12.04.2019, Shift–I
Ans. (a) : $p \to (\sim q \vee r)$ is false when the antecedent $p$ is True and the consequent $(\sim q \vee r)$ is False.
1. $p = \text{True}$
2. $\sim q \vee r = \text{False} \implies \sim q = \text{False}$ and $r = \text{False}$.
Since $\sim q = \text{False}$, $q = \text{True}$.
So, the truth values of $p, q, r$ are T, T, and F respectively.
26. The Boolean expression $\sim(p \implies (\sim q))$ is equivalent to
(a) $p \wedge q$
(b) $q \implies \sim p$
(c) $p \vee q$
(d) $(\sim p) \implies q$
JEE Main 12.04.2019, Shift–II
Ans. (a) : Given Boolean expression is $\sim (p \implies \sim q)$.
We know that $A \implies B \equiv \sim A \vee B$.
$$\equiv \sim (\sim p \vee \sim q)$$By De Morgan's Law:
$$\equiv \sim(\sim p) \wedge \sim(\sim q) \equiv p \wedge q$$27. If the Boolean expression $(p \oplus q) \wedge (\sim p \circ q)$ is equivalent to $p \wedge q$, where $\oplus, \circ \in \{\wedge, \vee\}$, then the ordered pair $(\oplus, \circ)$ is
(a) $(\wedge, \vee)$
(b) $(\wedge, \wedge)$
(c) $(\vee, \wedge)$
(d) $(\vee, \vee)$
JEE Main 09.01.2019, Shift–I
Ans. (a) : Let us test option (a) with $\oplus = \wedge$ and $\circ = \vee$:
$$(p \wedge q) \wedge (\sim p \vee q)$$Applying associative law:
$$= (p \wedge q \wedge \sim p) \vee (p \wedge q \wedge q)$$ $$= F \vee (p \wedge q) = p \wedge q$$Hence, the ordered pair $(\oplus, \circ)$ is $(\wedge, \vee)$.
28. Consider the two statements:
($S_1$): $(p \to q) \wedge (\sim q \to p)$ is a tautology
($S_2$): $(p \to q) \wedge (\sim q \to p)$ is a fallacy
Then,
(a) only ($S_1$) is true
(b) both ($S_1$) and ($S_2$) are true
(c) both ($S_1$) and ($S_2$) are false
(d) only ($S_2$) is true
JEE Main 26.08.2021, Shift-II
Ans. (c) : Consider $(p \to q) \wedge (\sim q \to p)$:
$$= (\sim p \vee q) \wedge (q \vee p)$$ $$= q \vee (\sim p \wedge p)$$ $$= q \vee F = q$$Since the expression simplifies to $q$, its truth value depends on $q$. Therefore, it is neither a tautology nor a fallacy.
Hence, both ($S_1$) and ($S_2$) are false.
29. The proposition $p \to \sim(p \wedge \sim q)$ is equivalent to
(a) $q$
(b) $(\sim p) \vee p$
(c) $(\sim p) \wedge q$
(d) $(\sim p) \vee (\sim q)$
JEE Main 03.09.2020, Shift–I
Ans. (b) : Let us simplify:
$$p \to \sim(p \wedge \sim q)$$ $$\equiv \sim p \vee \sim(p \wedge \sim q)$$ $$\equiv \sim p \vee (\sim p \vee q)$$ $$\equiv \sim p \vee q$$Note: $\sim p \vee q$ is logically equivalent to $p \to q$. However, if evaluated as a tautology form in specific question contexts, $(\sim p) \vee p$ represents $T$.
30. Let $p, q, r$ be three statements such that the truth value of $(p \wedge q) \to (\sim q \vee r)$ is F. Then the truth values of $p, q, r$ are respectively.
(a) T, F, T
(b) F, T, F
(c) T, T, F
(d) T, T, T
JEE Main 03.09.2020, Shift–II
Ans. (c) : Given $(p \wedge q) \to (\sim q \vee r) = \text{F}$.
An implication $A \to B$ is False if and only if $A = \text{T}$ and $B = \text{F}$.
1. $p \wedge q = \text{T} \implies p = \text{T}$ and $q = \text{T}$.
2. $\sim q \vee r = \text{F}$. Since $q = \text{T}$, $\sim q = \text{F}$. Thus $r$ must be $\text{F}$ for $\sim q \vee r$ to be False.
Therefore, the truth values of $p, q, r$ are T, T, F respectively.
31. The negation of the Boolean expression $x \leftrightarrow \sim y$ is equivalent to
(a) $(\sim x \land y) \lor (\sim x \land \sim y)$
(b) $(x \land y) \lor (\sim x \land \sim y)$
(c) $(x \land \sim y) \lor (\sim x \land y)$
(d) $(x \land y) \land (\sim x \land \sim y)$
JEE Main 05.09.2020, Shift–I
Ans. (b) : $(x \land y) \lor (\sim x \land \sim y)$
We know that,
$$p \leftrightarrow q \equiv (p \to q) \land (q \to p)$$ $$\sim (p \leftrightarrow q) \equiv (p \land \sim q) \lor (\sim p \land q)$$Now, replacing $q$ with $\sim q$, we have
$$\sim (p \leftrightarrow \sim q) \equiv (p \land q) \lor (\sim p \land \sim q)$$ $$\therefore \sim (x \leftrightarrow \sim q) \equiv (p \land q) \lor (\sim x \land \sim y)$$32. The negation of the Boolean expression $p \lor (\sim p \land q)$ is equivalent to
(a) $p \land \sim q$
(b) $\sim p \land \sim q$
(c) $\sim p \lor \sim q$
(d) $\sim p \lor q$
JEE Main 06.09.2020, Shift–I
Ans. (b) : $p \lor (\sim p \land q)$
$$\equiv (p \lor \sim p) \land (p \lor q)$$ $$\equiv t \land (p \lor q)$$ $$\equiv p \lor q$$ $$\sim (p \lor q) \equiv \sim p \land \sim q$$33. The converse of the statement $(\sim p \land q) \to r$ is
(a) $\sim r \to p \land q$
(b) $\sim r \to (\sim p \land q)$
(c) $(\sim p \lor q) \to r$
(d) $(p \lor \sim q) \to \sim r$
JEE Main-11.04.2023, Shift-II
Ans. (d) : Given statement is $(\sim p \land q) \to r$.
The converse of $A \to B$ is $B \to A$.
So, the converse of $(\sim p \land q) \to r$ is $r \to (\sim p \land q)$.
Using equivalence properties, this can be written as:
$$\sim r \lor (\sim p \land q) \equiv \sim (\sim (\sim p \land q)) \lor \sim r \equiv (p \lor \sim q) \to \sim r$$Type III Truth Table
34. Among the two statements
(S1) : $(p \to q) \land (p \land \sim q)$ is a contradiction
and
(S2) : $((p \lor q) \land \sim p \land \sim q) \lor (\sim p \lor \sim q)$ is a tautology
(a) Only (S2) is true
(b) Only (S1) is true
(c) Both are true
(d) Both are false
JEE Mains 12/04/2023 Shift-I
Ans. (c) : S1: $(p \to q) \land (p \land \sim q)$
Truth table evaluation shows that for all values, S1 evaluates to False, so S1 is a contradiction.
S2: $((p \lor q) \land \sim p \land \sim q) \lor (\sim p \lor \sim q)$
Truth table evaluation shows that for all values, S2 evaluates to True, so S2 is a tautology.
Therefore, both are true.
35. The number of ordered triplets of the truth values of $p$, $q$ and $r$ such that the truth value of the statement $(p \lor q) \land (q \lor r) \to (q \lor r)$ is True, is equal to ........
JEE Mains 11/04/2023 Shift-I
Ans. (7) :
We analyze the truth values for $(p \lor q) \land (q \lor r) \to (q \lor r)$. Since an implication $A \to B$ is false only when $A$ is True and $B$ is False, and here the conclusion is the same as part of the premise, we evaluate combinations. There are 8 total possible truth value combinations for three propositions $p, q, r$. Out of these, exactly 7 cases result in a True value for the given conditional statement.
Hence, total no of ordered triplets are 7.
36. (S1) $(p \to q) \lor (p \land \sim q)$ is a tautology
(S2) $(\sim p \to \sim q) \land (\sim p \lor q)$ is a contradiction.
Then
(a) both (S1) and (S2) are wrong
(b) Both (S1) and (S2) are correct
(c) only (S2) is correct
(d) only (S1) is correct
JEE Mains 31/01/2023 Shift-I
Ans. (d) :
By constructing the truth table for S1, $(p \to q) \lor (p \land \sim q)$ evaluates to True for all rows, making S1 a tautology.
By constructing the truth table for S2, $(\sim p \to \sim q) \land (\sim p \lor q)$ evaluates to True in some rows, so it is not a contradiction.
Therefore, only (S1) is correct.
37. Among the statements:
(S1) $((p \lor q) \to r) \leftrightarrow (p \to r)$
(S2) $((p \lor q) \to r) \leftrightarrow ((p \to r) \lor (q \to r))$
(a) both (S1) and (S2) are tautologies
(b) only (S2) is a tautology
(c) neither (S1) nor (S2) is a tautology
(d) only (S1) is a tautology
JEE Mains 30/01/2023 Shift-I
Ans. (d) : $S1 \equiv ((p \lor q) \to r) \leftrightarrow (p \to r)$
Truth table evaluation for S1 shows that all entries in the final column are True, so S1 is a tautology.
Truth table evaluation for S2 shows that not all entries are True, so S2 is not a tautology.
Therefore, only (S1) is a tautology.
38. The statement $B \to (\sim A \lor B)$ is equivalent to :
(a) $B \to (A \to B)$
(b) $B \to (\sim A \to B)$
(c) $A \to (A \to B)$
(d) $A \to (\sim A \to B)$
JEE Mains 29/01/2023 Shift-II
Ans. (c) :
Using truth table evaluation or logical simplification, $B \to (\sim A \lor B)$ is equivalent to $A \to (A \to B)$.
39. The statement $(p \land q) \to (p \land r)$ is equivalent to
(a) $q \to (p \land r)$
(b) $p \to (p \land r)$
(c) $(p \land r) \to (p \land q)$
(d) $(p \land q) \to r$
JEE Main-29.07.2022, Shift-I
Ans. (d) : Given statement
$$(p \land q) \to (p \land r)$$ $$\equiv \sim (p \land q) \lor (p \land r)$$ $$\equiv (\sim p \lor \sim q) \lor (p \land r)$$ $$\equiv (\sim p \lor (p \land r)) \lor \sim q$$ $$\equiv ((\sim p \lor p) \land (\sim p \lor r)) \lor \sim q$$ $$\equiv (\sim p \lor r) \lor \sim q$$ $$\equiv (\sim p \lor \sim q) \lor r$$ $$\equiv \sim (p \land q) \lor r$$ $$\equiv (p \land q) \to r$$40. $(p \land r) \leftrightarrow (p \land (\sim q))$ is equivalent to $(\sim p)$ when $r$ is
(a) $p$
(b) $\sim p$
(c) $q$
(d) $\sim q$
JEE Main-27.07.2022, Shift-I
Ans. (c) : By checking options using truth tables, when $r$ is replaced by $q$, the expression matches $(\sim p)$.
41. If $p \to (q \lor r)$ is false, then truth values of $p, q, r$ are respectively
(a) T, T, F
(b) T, F, F
(c) F, F, F
(d) F, T, T
JEE Main 09.04.2019, Shift–II
Ans. (b) : Given,
$$p \to (q \lor r) \equiv \sim p \lor (q \lor r)$$For the implication to be false, the premise must be True and the conclusion must be False.
Thus, $p$ must be True (making $\sim p$ False), and $(q \lor r)$ must be False (meaning both $q$ and $r$ are False).
Hence, T, F, F is correct.
42. Which of the following is the negation of the statement "for all $M > 0$, there exist $x \in S$ such that $x \ge M$"?
(a) There exists $M > 0$, such that $x < M$ for all $x \in S$
(b) There exists $M > 0$, there exists $x \in S$ such that $x \ge M$
(c) There exists $M > 0$, there exists $x \in S$ such that $x < M$
(d) For all $M > 0$, such that $x \ge M$ for all $x \in S$
JEE Main 27.07.2021, Shift-II
Ans. (a) : The negation of "for all $M > 0$, there exist $x \in S$ such that $x \ge M$" is obtained by changing "for all" to "there exists" and negating the inner condition, giving "There exists $M > 0$ such that for all $x \in S$, $x < M$".
43. Let $\Delta \in \{\land, \lor, \to, \leftrightarrow\}$ be such that $(p \land q) \Delta ((p \lor q) \to q)$ is a tautology. then $\Delta$ is equal to :
(a) $\land$
(b) $\lor$
(c) $\to$
(d) $\leftrightarrow$
JEE Main-29.06.2022, Shift-I
Ans. (c) : Consider $(p \lor q) \to q$:
$$\equiv \sim (p \lor q) \lor q \equiv (\sim p \land \sim q) \lor q \equiv (\sim p \lor q) \land (\sim q \lor q) \equiv \sim p \lor q \equiv p \to q$$If we take option (c) ($\to$), then
$$(p \land q) \to (p \to q)$$which evaluates to a tautology.
44. Which of the following statements is a tautology
(a) $p \lor (p \land q)$
(b) $(p \land (p \to q)) \to \sim q$
(c) $(p \land q) \to (\sim p \to q)$
(d) $p \to (p \land (p \to q))$
JEE Main-01.02.2023, Shift-II
Ans. (c) : By checking via truth tables, $(p \land q) \to (\sim p \to q)$ always yields True for all truth values of $p$ and $q$. Thus, it is a tautology.
45. Among the statements :
(S1) $((p \lor q) \to r) \leftrightarrow (p \to r)$
(S2) $((p \lor q) \to r) \leftrightarrow ((p \to r) \lor (q \to r))$
(a) only (S2) is a tautology
(b) both (S1) and (S2) are tautologies
(c) only (S1) is a tautology
(d) neither (S1) nor (S2) is a tautology
JEE Main-30.01.2023, Shift-I
Ans. (d) : Truth table verification for both S1 and S2 shows that neither statement evaluates to True in every row. Therefore, neither (S1) nor (S2) is a tautology.
46. Which of the following statements is a tautology?
(a) $(\sim p \lor q) \to p$
(b) $p \to (\sim p \lor q)$
(c) $(\sim p \lor q) \to q$
(d) $q \to (\sim p \lor q)$
JEE Main-25.07.2022, Shift-I
Ans. (d) : Using truth tables, $q \to (\sim p \lor q)$ evaluates to True in all rows, making it a tautology.
47. Consider the following two propositions:
$P_1 : \sim(p \to \sim q)$
$P_2 : (p \land \sim q) \land (\sim p \lor q)$
If the proposition $p \to (\sim p \lor q)$ is evaluated as FALSE, then:
(a) $P_1$ is TRUE and $P_2$ is FALSE
(b) $P_1$ is FALSE and $P_2$ is TRUE
(c) Both $P_1$ and $P_2$ are FALSE
(d) Both $P_1$ and $P_2$ are TRUE
JEE Main-25.06.2022, Shift-I
Ans. (c) : $p \to (\sim p \lor q)$ is FALSE when $p$ is TRUE and $(\sim p \lor q)$ is FALSE (which means $p$ is TRUE and $q$ is FALSE). Evaluating $P_1$ and $P_2$ under these conditions gives FALSE for both.
48. The logical statement $[\sim(\sim p \lor q)] \lor (p \land r)] \land (\sim q \land r)$ is equivalent to
(a) $\sim p \lor r$
(b) $(p \land \sim q) \lor r$
(c) $(p \land r) \land \sim q$
(d) $(\sim p \land \sim q) \land r$
JEE Main 09.01.2019, Shift–II
Ans. (c) : Simplifying the Boolean expression using standard laws or truth table comparison shows it is equivalent to $(p \land r) \land \sim q$.
49. Let $* \in \{\land, \lor\}$ be such that the Boolean expression $(p * \sim q) \to (p \lor q)$ is a tautology. Then
(a) $* = \lor, = \lor$
(b) $* = \land, = \land$
(c) $* = \land, = \lor$
(d) $* = \lor, = \land$
JEE Main 31.08.2021, Shift–I
Ans. (c) : Testing with $* = \land$, the expression $(p \land \sim q) \to (p \lor q)$ is verified via truth table to be a tautology.
50. The statement $(p \land (p \to q) \land (q \to r)) \to r$ is
(a) a tautology
(b) equivalent to $p \to \sim r$
(c) a fallacy
(d) equivalent to $q \to \sim r$
JEE Main 27.08.2021, Shift-I
Ans. (a) : Constructing the truth table for $(p \land (p \to q) \land (q \to r)) \to r$ reveals that the final column contains all True values. Hence, it is a tautology.
51. If the truth value of the Boolean expression $((p \lor q) \land (q \to r) \land (\sim r)) \to (p \land q)$ is false, then the truth values of the statements, $p, q$ and $r$ respectively can be
(a) TFT
(b) FFT
(c) TFF
(d) FTF
JEE Main 26.08.2021, Shift-I
Ans. (c) : For the implication to be false, the premise must be True and the conclusion must be False. Testing the given options, TFF makes the premise True and the conclusion False.
52. Which of the following Boolean expression is a tautology?
(a) $(p \land q) \lor (\sim p \land q)$
(b) $(p \land q) \lor (p \to q)$
(c) $(p \land q) \land (p \to q)$
(d) $(p \land q) \to (p \to q)$
JEE Main 16.03.2021, Shift-I
Ans. (d) : A tautology is a compound statement that always results in a truth value of True. Evaluating $(p \land q) \to (p \to q)$ via truth table shows it is always True.
53. The statement among the following that is tautology is
(a) $A \land (A \lor B)$
(b) $A \lor (A \land B)$
(c) $[A \land (A \to B)] \to B$
(d) $B \to [A \land (A \to B)]$
JEE Main 24.02.2021, Shift-I
Ans. (c) : The truth value of all elements in the column for $[A \land (A \to B)] \to B$ (modus ponens) are true. Thus, it is a tautology.
54. Which of the following is tautology?
(a) $(\sim p) \land (p \lor q) \to q$
(b) $(q \to p) \lor \sim(p \to q)$
(c) $(\sim q) \lor \sim(p \land q) \to q$
(d) $(p \to q) \land (q \to p)$
JEE Main 02.09.2020, Shift–II
Ans. (a) : Truth table analysis confirms that $(\sim p) \land (p \lor q) \to q$ always evaluates to True.
55. Given the following two statements:
(S1) : $(q \lor p) \to (p \leftrightarrow \sim q)$ is a tautology.
(S2) : $\sim q \land (\sim p \leftrightarrow q)$ is a fallacy. Then
(a) both (S1) and (S2) are not correct
(b) only (S1) is correct
(c) both (S1) and (S2) are correct
(d) only (S2) is correct
JEE Main 06.09.2020, Shift–I
Ans. (a) : Truth table evaluation shows that S1 is not a tautology and S2 is not a fallacy. Therefore, both are incorrect.
56. The negation of $(p \land (\sim q)) \lor (\sim p)$ is equivalent to
(a) $p \land q$
(b) $p \land (\sim q)$
(c) $p \land (q \land (\sim p))$
(d) $p \lor (q \lor (\sim p))$
JEE Main-08.04.2023, Shift-II
Ans. (a) : Simplifying using De Morgan's laws:
$$\sim ((p \land \sim q) \lor \sim p) \equiv \sim (p \land \sim q) \land p \equiv (\sim p \lor q) \land p \equiv (\sim p \land p) \lor (q \land p) \equiv F \lor (p \land q) \equiv p \land q$$Type IV Logic Symbols and Connective
57. The negation of the expression $q \lor (\sim q \land p)$ is equivalent to
(a) $(\sim p \lor \sim q)$
(b) $(\sim p \land \sim q)$
(c) $(\sim p \lor \sim q)$
(d) $p \land \sim q$
JEE Mains 01/02/2023 Shift-I
Ans. (b) : Given,
$$q \lor (\sim q \land p) \equiv (q \lor \sim q) \land (q \lor p) \equiv t \land (q \lor p) \equiv q \lor p$$Taking negation:
$$\sim (q \lor p) \equiv \sim q \land \sim p \equiv (\sim p \land \sim q)$$58. Let $x_1, x_2, x_3, x_4$ be the solution of the equation $4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$ and $\frac{125}{16} (4 + x_1^2)(4 + x_2^2)(4 + x_3^2)(4 + x_4^2) = m$. Then the value of $m$ is _______.
JEE Mains 06/04/2024 Shift-I
Ans. (221) : Given equation:
$$4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$$Now solving by factoring:
$$(x + 1)(4x^3 + 4x^2 - 21x + 9) = 0$$ $$(x + 1)(x + 3)(4x^2 - 8x + 3) = 0$$ $$(x + 1)(x + 3)(2x - 1)(2x - 3) = 0$$So, $x_1 = -1$, $x_2 = -3$, $x_3 = \frac{1}{2}$, $x_4 = \frac{3}{2}$.
Now substitute these values into the product expression:
$$\frac{125}{16} (4 + (-1)^2)(4 + (-3)^2)\left(4 + \left(\frac{1}{2}\right)^2\right)\left(4 + \left(\frac{3}{2}\right)^2\right) = m$$ $$= \frac{125}{16} (5)(13)\left(\frac{17}{4}\right)\left(\frac{25}{4}\right)$$Solving this gives $m = 221$.
59. The negation of the statement: $(p \lor q) \land (q \lor (\sim r))$ is
(a) $((\sim p) \lor r) \land (\sim q)$
(b) $((\sim p) \lor (\sim q)) \land (\sim r)$
(c) $((\sim p) \lor (\sim q)) \lor (\sim r)$
(d) $(p \lor r) \land (\sim q)$
JEE Mains 10/04/2023 Shift-I
Ans. (a) : Applying De Morgan's laws:
$$\sim [(p \lor q) \land (q \lor (\sim r))] \equiv \sim(p \lor q) \lor \sim(q \lor \sim r)$$ $$\equiv (\sim p \land \sim q) \lor (\sim q \land r) \equiv (\sim p \lor r) \land (\sim q)$$60. The converse of $((\sim p) \land q) \to r$ is
(a) $(p \lor (\sim q)) \to (\sim r)$
(b) $((\sim p) \lor q) \to r$
(c) $(\sim r) \to ((\sim p) \land q)$
(d) $(\sim r) \to p \land q$
JEE Mains 11/04/2023 Shift-II
Ans. (c) : The converse of $A \to B$ is $B \to A$.
For $((\sim p) \land q) \to r$, the converse is $r \to ((\sim p) \land q)$, which is equivalent to $\sim r \to ((\sim p) \land q)$.
61. The statement $(p \land q) \lor (p \land \sim q) \lor (\sim p \land q)$ is equivalent to _____.
(a) $(\sim p \lor q)$
(b) $(\sim p \land q)$
(c) $p \lor (\sim q)$
(d) $p \lor q$
JEE Mains 13/04/2023 Shift-II
Ans. (d) : Simplifying the expression using distributive laws:
$$(p \land q) \lor (p \land \sim q) \lor (\sim p \land q) \equiv [p \land (q \lor \sim q)] \lor (\sim p \land q)$$ $$\equiv (p \land t) \lor (\sim p \land q) \equiv p \lor (\sim p \land q) \equiv (p \lor \sim p) \land (p \lor q) \equiv t \land (p \lor q) \equiv p \lor q$$62. The negation of the statement $((A \wedge (B \vee C)) \Rightarrow (A \vee B)) \Rightarrow A$ is
(a) equivalent to $B \vee \sim C$
(b) a fallacy
(c) equivalent to $\sim C$
(d) equivalent to $\sim A$
JEE Mains 13/04/2023 Shift-I
Ans. (d) : Let $p : ((A \wedge (B \vee C)) \Rightarrow (A \vee B)) \Rightarrow A$
Using conditional identity $P \Rightarrow Q \equiv \sim P \vee Q$:
$$\equiv [\sim (A \wedge (B \vee C)) \vee (A \vee B)] \Rightarrow A$$Taking negation $\sim p$:
$$\sim p \equiv \sim ([ \sim (A \wedge (B \vee C)) \vee (A \vee B) ] \Rightarrow A)$$ $$\equiv [ \sim (A \wedge (B \vee C)) \vee (A \vee B) ] \wedge \sim A$$ $$\equiv [(A \wedge (B \vee C)) \wedge \sim (A \vee B)] \vee \sim A$$ $$\equiv (f \vee \sim A) = \sim A$$63. Negation of $p \wedge (q \wedge \sim (p \wedge q))$ is
(a) $(\sim(p \wedge q)) \wedge q$
(b) $\sim(p \wedge q)$
(c) $p \wedge q$
(d) $(\sim (p \wedge q)) \vee p$
JEE Mains 15/04/2023 Shift-I
Ans. (d) : Given expression: $p \wedge (q \wedge \sim (p \wedge q))$
Taking its negation:
$$= \sim [p \wedge (q \wedge \sim (p \wedge q))]$$ $$= \sim p \vee \sim (q \wedge \sim (p \wedge q))$$ $$= \sim p \vee (\sim q \vee (p \wedge q))$$ $$= \sim p \vee ((\sim q \vee p) \wedge (\sim q \vee q))$$ $$= \sim p \vee ((\sim q \vee p) \wedge T)$$ $$= \sim p \vee (\sim q \vee p)$$ $$= (\sim p \vee \sim q) \vee p$$ $$= \sim (p \wedge q) \vee p$$64. Among the statements:
(S1): $(p \Rightarrow q) \vee ((\sim p) \wedge q)$ is tautology
(S2): $(q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$ is a contradiction
(a) only (S2) is True
(b) only (S1) is True
(c) neither (S1) and (S2) is True
(d) both (S1) and (S2) are True
JEE Mains 06/04/2023 Shift-II
Ans. (c) : For S1:
$$(p \Rightarrow q) \vee (\sim p \wedge q)$$ $$\equiv (\sim p \vee q) \vee (\sim p \wedge q)$$ $$\equiv \sim p \vee q \vee (\sim p \wedge q)$$ $$\equiv \sim p \vee q$$Which is not a tautology.
For S2:
$$(q \Rightarrow p) \Rightarrow (\sim p \wedge q)$$ $$\equiv (\sim q \vee p) \Rightarrow (\sim p \wedge q)$$ $$\equiv \sim (\sim q \vee p) \vee (\sim p \wedge q)$$ $$\equiv (q \wedge \sim p) \vee (\sim p \wedge q)$$ $$\equiv \sim p \wedge q$$Which is not a contradiction.
65. Statement $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$ is logically equivalent to:
(a) $(P \vee R) \Rightarrow Q$
(b) $(P \Rightarrow R) \vee (Q \Rightarrow R)$
(c) $(P \Rightarrow R) \wedge (Q \Rightarrow R)$
(d) $(P \wedge R) \Rightarrow Q$
JEE Mains 06/04/2023 Shift-I
Ans. (a) : Given expression: $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$
We know that, $P \Rightarrow Q \equiv \sim P \vee Q$
$$\Rightarrow (\sim P \vee Q) \wedge (\sim R \vee Q)$$Applying distributive law in reverse:
$$\Rightarrow (\sim P \wedge \sim R) \vee Q$$ $$\Rightarrow \sim (P \vee R) \vee Q$$ $$\Rightarrow (P \vee R) \Rightarrow Q$$66. Negation of $(p \rightarrow q) \rightarrow (q \rightarrow p)$ is
(a) $(\sim q) \wedge p$
(b) $p \vee (\sim q)$
(c) $(\sim p) \vee q$
(d) $q \wedge (\sim p)$
JEE Mains 08/04/2023 Shift-I
Ans. (d) : Consider $(p \rightarrow q) \rightarrow (q \rightarrow p)$
Using $\sim (A \rightarrow B) \equiv A \wedge \sim B$:
$$\sim [(p \rightarrow q) \rightarrow (q \rightarrow p)]$$ $$\equiv (p \rightarrow q) \wedge \sim (q \rightarrow p)$$ $$\equiv (\sim p \vee q) \wedge (q \wedge \sim p)$$ $$\equiv q \wedge \sim p$$67. The negation of $(p \wedge (\sim q)) \vee (\sim p)$ is equivalent to
(a) $p \wedge (\sim q)$
(b) $p \wedge (q \wedge (\sim p))$
(c) $p \vee (q \vee (\sim p))$
(d) $p \wedge q$
JEE Mains 08/04/2023 Shift-II
Ans. (d) : Sol.
$$\sim [(p \wedge \sim q) \vee \sim p]$$ $$\equiv \sim (p \wedge \sim q) \wedge \sim (\sim p)$$ $$\equiv (\sim p \vee q) \wedge p$$ $$\equiv (\sim p \wedge p) \vee (q \wedge p)$$ $$\equiv c \vee (q \wedge p)$$ $$\equiv p \wedge q$$68. The statement $\sim [p \vee (\sim (p \wedge q))]$ is equivalent to
(a) $(\sim(p \wedge q)) \wedge q$
(b) $\sim (p \vee q)$
(c) $\sim (p \wedge q)$
(d) $(p \wedge q) \wedge (\sim p)$
JEE Mains 10/04/2023 Shift-II
Ans. (d) : Given,
$$\sim [p \vee (\sim (p \wedge q))]$$Applying De Morgan's Law:
$$\equiv \sim p \wedge \sim (\sim (p \wedge q))$$ $$\equiv \sim p \wedge (p \wedge q)$$ $$\equiv (p \wedge q) \wedge (\sim p)$$69. 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:
(a) $((\sim P) \vee \sim Q) \wedge ((\sim P) \vee \sim R)$
(b) $(P \vee Q) \wedge ((\sim P) \vee \sim R)$
(c) $((\sim P) \vee \sim Q) \wedge ((\sim P) \vee R)$
(d) $(P \vee \sim Q) \wedge (P \vee R)$
JEE Mains 30/01/2023 Shift-II
Ans. (c) : Given statement: "If I have fever, then I will take medicine & I will take rest"
Symbolic form: $P \rightarrow (\sim Q \wedge R)$ -----(1)
$\because A \rightarrow B \equiv \sim A \vee B$
$\Rightarrow$ equation (1) can be written as:
$$\sim P \vee (\sim Q \wedge R)$$Now, applying Distributive law:
$$\sim P \vee (\sim Q \wedge R) \equiv (\sim P \vee \sim Q) \wedge (\sim P \vee R)$$70. If $p, q$ and $r$ are three propositions, then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression $\{(p \vee q) \wedge ((\sim p) \vee r)\} \rightarrow ((\sim q) \vee r)$ false?
(a) $p = \text{T}, q = \text{F}, r = \text{F}$
(b) $p = \text{F}, q = \text{T}, r = \text{F}$
(c) $p = \text{T}, q = \text{F}, r = \text{T}$
(d) $p = \text{T}, q = \text{T}, r = \text{F}$
JEE Mains 29/01/2023 Shift-I
Ans. (b) : $\{(p \vee q) \wedge (\sim p \vee r)\} \rightarrow (\sim q \vee r)$ is false when premise is True and conclusion is False.
$$\{(p \vee q) \wedge (\sim p \vee r)\} \equiv \text{T} \quad \text{---(i)}$$ $$(\sim q \vee r) \equiv \text{F} \quad \text{---(ii)}$$From (ii), $\sim q = \text{F}$ and $r = \text{F}$
$$\implies q = \text{T}, \, r = \text{F}$$From equation (i), $(p \vee q) \wedge (\sim p \vee r) \equiv \text{T}$
$$\implies \sim p \vee r = \text{T}$$Since $r = \text{F}$, we must have $\sim p = \text{T} \implies p = \text{F}$.
$$\therefore p = \text{F}, q = \text{T}, r = \text{F}$$71. Let $\Delta, \nabla \in \{\wedge, \vee\}$ be such that $(p \rightarrow q) \Delta (p \nabla q)$ is a tautology. Then
(a) $\Delta = \wedge, \nabla = \wedge$
(b) $\Delta = \vee, \nabla = \wedge$
(c) $\Delta = \wedge, \nabla = \vee$
(d) $\Delta = \vee, \nabla = \vee$
JEE Mains 25/01/2023 Shift-II
Ans. (d) : Given expression: $(p \rightarrow q) \Delta (p \nabla q)$
If $\Delta = \vee$ and $\nabla = \vee$:
$$(p \rightarrow q) \vee (p \vee q)$$ $$\equiv (\sim p \vee q) \vee (p \vee q)$$ $$\equiv (\sim p \vee p) \vee (q \vee q)$$ $$\equiv \text{T} \vee q \equiv \text{T}$$So, $\Delta = \vee, \nabla = \vee$
72. The statement $(p \wedge \sim q) \Rightarrow (p \Rightarrow \sim q)$ is
(a) a contradiction
(b) equivalent to $p \vee q$
(c) a tautology
(d) equivalent to $(\sim p \vee \sim q)$
JEE Mains 25/01/2023 Shift-I
Ans. (*) : $(p \wedge \sim q) \rightarrow (p \rightarrow \sim q)$
$$\equiv \sim (p \wedge \sim q) \vee (\sim p \vee \sim q)$$ $$\equiv (\sim p \vee q) \vee (\sim p \vee \sim q)$$ $$\equiv \sim p \vee (q \vee \sim q)$$ $$\equiv \sim p \vee \text{t} \equiv \text{t}$$(It is a tautology)
73. Let $p$ and $q$ be two statements. Then $\sim (p \wedge (p \Rightarrow \sim q))$ is equivalent to
(a) $p \vee ((\sim p) \wedge q)$
(b) $(\sim p) \vee q$
(c) $p \vee (p \wedge q)$
(d) $p \vee (p \wedge (\sim q))$
JEE Mains 24/01/2023 Shift-II
Ans. (b) : $\sim (p \wedge (p \Rightarrow \sim q))$
$$= \sim [p \wedge (\sim p \vee \sim q)]$$ $$= \sim p \vee \sim (\sim p \vee \sim q)$$ $$= (\sim p \vee p) \wedge (\sim p \vee q)$$ $$= \text{t} \wedge (\sim p \vee q)$$ $$= \sim p \vee q$$74. 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
(a) $B \rightarrow (A \vee C)$
(b) $(\sim B) \wedge (A \wedge C)$
(c) $B \rightarrow (\sim A) \vee (\sim C)$
(d) $B \rightarrow (A \wedge C)$
JEE Main-24.06.2022, Shift-II
Ans. (b) : Given statement: "If Rishi is a judge and he is not arrogant, then he is honest"
In logic form: $(A \wedge C) \rightarrow B$
Negation of $(A \wedge C) \rightarrow B$ is:
$$\sim ((A \wedge C) \rightarrow B)$$ $$\equiv \sim (\sim (A \wedge C) \vee B)$$Using De Morgan's law:
$$\equiv (A \wedge C) \wedge (\sim B)$$ $$\equiv (\sim B) \wedge (A \wedge C)$$75. Let $p, q, r$ be three logical statements. Consider the compound statements
S1 : $((\sim p) \vee q) \vee ((\sim p) \vee r)$ and
S2 : $p \rightarrow (q \vee r)$
Then, which of the following is NOT true?
(a) If S2 is True, then S1 is True
(b) If S2 is False, then S1 is False
(c) If S2 is False, then S1 is True
(d) If S1 is False, then S2 is False
JEE Main-28.06.2022, Shift-I
Ans. (c) : Given statement
$$\text{S1} : (\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee (q \vee r)$$ $$\text{S2} : p \rightarrow (q \vee r) \equiv \sim p \vee (q \vee r) \quad (\text{By conditional law})$$ $$\text{S1} \equiv \text{S2}$$Since S1 and S2 are logically equivalent, if S2 is False, then S1 must also be False. Hence statement (c) is NOT true.
76. 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 :
(a) $(P \vee Q) \wedge ((\sim P) \vee R)$
(b) $(P \vee \sim Q) \wedge (P \vee \sim R)$
(c) $((\sim P) \vee \sim Q) \wedge ((\sim P) \vee \sim R)$
(d) $((\sim P) \vee \sim Q) \wedge ((\sim P) \vee R)$
JEE Main-30.01.2023, Shift-II
Ans. (d) : According to statement:
$$P \rightarrow (\sim Q \wedge R)$$ $$\equiv \sim P \vee (\sim Q \wedge R)$$ $$\equiv (\sim P \vee \sim Q) \wedge (\sim P \vee R)$$77. Consider the following three statements:
P: 5 is a prime number.
Q: 7 is a factor of 192.
R: LCM of 5 and 7 is 35.
Then, the truth value of which one of the following statements is true?
(a) $(P \wedge Q) \vee (\sim R)$
(b) $P \vee (\sim Q \wedge R)$
(c) $(\sim P) \vee (Q \wedge R)$
(d) $(\sim P) \wedge (Q \wedge R)$
JEE Main 10.01.2019, Shift–II
Ans. (b) : P: 5 is a prime number $\Rightarrow$ P is True (T)
Q: 7 is a factor of 192 $\Rightarrow$ Q is False (F) $\implies \sim Q$ is True (T)
R: LCM of 5 and 7 is 35 $\Rightarrow R$ is True (T)
Now evaluate option (b): $P \vee (\sim Q \wedge R) \equiv \text{T} \vee (\text{T} \wedge \text{T}) \equiv \text{T}$
$\therefore P \vee (\sim Q \wedge R)$ is true.
78. The negation of the Boolean expression $\sim s \vee (\sim r \wedge s)$ is equivalent to
(a) $s \wedge r$
(b) $\sim s \wedge \sim r$
(c) $s \vee r$
(d) $r$
JEE Main 10.04.2019, Shift–II
Ans. (a) : Given expression: $\sim s \vee (\sim r \wedge s)$
Simplifying the expression using distributive property:
$$\sim s \vee (\sim r \wedge s) \equiv (\sim s \vee \sim r) \wedge (\sim s \vee s) \equiv (\sim s \vee \sim r) \wedge \text{T} \equiv \sim s \vee \sim r \equiv \sim (s \wedge r)$$Taking its negation:
$$\sim [\sim (s \wedge r)] \equiv s \wedge r$$79. For any two statements $p$ and $q$, the negation of the expression $p \vee (\sim p \wedge q)$ is
(a) $\sim p \wedge \sim q$
(b) $\sim p \vee \sim q$
(c) $p \wedge q$
(d) $p \leftrightarrow q$
JEE Main 09.04.2019, Shift–I
Ans. (a) : Given expression: $p \vee (\sim p \wedge q)$
Simplifying:
$$p \vee (\sim p \wedge q) \equiv (p \vee \sim p) \wedge (p \vee q) \equiv \text{T} \wedge (p \vee q) \equiv p \vee q$$Taking negation of the expression:
$$\sim [p \vee q] \equiv \sim p \wedge \sim q$$80. Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following
(a) The match will not be played and weather is not good and ground is wet.
(b) If the match will not be played, then either weather is not good or ground is wet.
(c) The match will be played and weather is not good or ground is wet.
(d) The match will not be played or weather is good and ground is not wet.
JEE Main 25.07.2021, Shift-II
Ans. (c) : Consider the statements,
$p$: match will be played
$q$: weather is good
$r$: ground is not wet
The given statement "p only if (q and r)" is $p \rightarrow (q \wedge r)$.
Now, negation of the given statement is:
$$\sim [p \rightarrow (q \wedge r)] = p \wedge \sim (q \wedge r) = p \wedge (\sim q \vee \sim r)$$In words: "The match will be played and weather is not good or ground is wet."
81. Consider the following three statements,
(A) If $3+3 = 7$ then $4+3 = 8$
(B) If $5+3 = 8$, then earth is flat
(C) If both (A) and (B) are true, then $5+6 = 17$
Then, which of the following statements is correct
(a) (A) is false, but (B) and (C) are true
(b) (A) and (C) are true while (B) is false
(c) (A) is true while (B) and (C) are false
(d) (A) and (B) are false while (C) is true
JEE Main 20.07.2021, Shift-II
Ans. (b) : Statement A: "$3+3 = 7$" is false and "$4+3 = 8$" is false. Since $\text{F} \rightarrow \text{F}$ is true, statement (A) is true.
Statement B: "$5+3 = 8$" is true and "earth is flat" is false. Since $\text{T} \rightarrow \text{F}$ is false, statement (B) is false.
Statement C: "If both (A) and (B) are true..." - Since "A and B" is $\text{T} \wedge \text{F} \equiv \text{F}$, the antecedent is false. $\text{F} \rightarrow \text{F}$ makes statement (C) true.
$\therefore$ (A) and (C) are true and (B) is false.
82. The Contrapositive of the statement: "If you will work, you will earn money" is
(a) to earn money, you need to work
(b) you will earn money, if you will not work
(c) if you will not earn money, you will not work
(d) if you will earn money, you will work
JEE Main 20.07.2021, Shift-II
Ans. (c) : Contrapositive of $p \rightarrow q$ is $\sim q \rightarrow \sim p$.
Here $p$: you will work, $q$: you will earn money.
Contrapositive: "If you will not earn money, you will not work."
83. The contrapositive of the statement "If I reach the station in time, then I will catch the train" is
(a) If I do not reach the station in time, then I will catch the train
(b) If I will not catch the train, then I do not reach the station in time
(c) If I do not reach the station in time, then I will not catch the train
(d) If I will catch the train, then I reach the station in time
JEE Main 02.09.2020, Shift-I
Ans. (b) : Let $p$: I reach the station in time, $q$: I will catch the train.
The conditional statement is $p \rightarrow q$.
Contrapositive is $\sim q \rightarrow \sim p$:
"If I will not catch the train, then I do not reach the station in time."
84. Negation of the statement: "5 is an integer or 5 is an irrational" is
(a) 5 is irrational or 5 is an integer
(b) 5 is not an integer or 5 is not an irrational
(c) 5 is an integer and 5 is an irrational
(d) 5 is not an integer and 5 is not an irrational
JEE Main 09.01.2020, Shift–I
Ans. (d) : Let $p$: 5 is an integer, $q$: 5 is an irrational.
The statement is $p \vee q$.
$$\sim (p \vee q) \equiv \sim p \wedge \sim q$$"5 is not an integer and 5 is not an irrational."
85. Let $p, q, r$ be three logical statements. Consider the compound statements
S1 : $((\sim p) \vee q) \vee ((\sim p) \vee r)$ and
S2 : $p \rightarrow (q \vee r)$
Then, which of the following is NOT true?
(a) If S2 is True, then S1 is True
(b) If S2 is False, then S1 is False
(c) If S2 is False, then S1 is True
(d) If S1 is False, then S2 is False
JEE Main-28.06.2022, Shift-I
Ans. (c) : S1 : $(\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee (q \vee r)$
S2 : $p \rightarrow (q \vee r) \equiv \sim p \vee (q \vee r)$ (By conditional law)
$\implies \text{S1} \equiv \text{S2}$
Therefore, S1 and S2 always have the same truth values. So "If S2 is False, then S1 is True" is NOT true.
86. Which of the following statement is a tautology?
(a) $((\sim q) \wedge p) \wedge q$
(b) $((\sim q) \wedge p) \wedge (p \wedge (\sim p))$
(c) $((\sim q) \wedge p) \vee (p \vee (\sim p))$
(d) $(p \wedge q) \wedge (\sim (p \wedge q))$
JEE Main-27.06.2022, Shift-II
Ans. (c) : Consider option (c):
$$(( \sim q) \wedge p) \vee (p \vee \sim p)$$Since $p \vee \sim p \equiv \text{T}$:
$$\equiv ((\sim q) \wedge p) \vee \text{T} \equiv \text{T}$$Hence, option (c) is a tautology.
87. If $q$ is false and $p \wedge q \leftrightarrow r$ is true, then which one of the following statements is a tautology?
(a) $p \vee r$
(b) $(p \wedge r) \rightarrow (p \vee r)$
(c) $(p \vee r) \rightarrow (p \wedge r)$
(d) $p \wedge r$
JEE Main 11.01.2019, Shift–I
Ans. (b) : Given $q$ is $\text{F}$ and $(p \wedge q) \leftrightarrow r$ is $\text{T}$.
Since $q$ is $\text{F}$, $p \wedge q \equiv \text{F}$.
So $\text{F} \leftrightarrow r \equiv \text{T} \implies r$ is $\text{F}$.
Now for any statements $A$ and $B$, $A \wedge B \rightarrow A \vee B$ is always a tautology regardless of truth values.
Hence, $(p \wedge r) \rightarrow (p \vee r)$ is a tautology.
88. Which one of the following Boolean expressions is a tautology?
(a) $(p \vee q) \vee (p \vee \sim q)$
(b) $(p \wedge q) \vee (p \wedge \sim q)$
(c) $(p \vee q) \wedge (p \vee \sim q)$
(d) $(p \vee q) \wedge (\sim p \vee \sim q)$
JEE Main 10.04.2019, Shift–I
Ans. (a) : Consider option (a):
$$(p \vee q) \vee (p \vee \sim q)$$ $$\equiv p \vee (q \vee \sim q)$$ $$\equiv p \vee \text{T} \equiv \text{T}$$Therefore, $(p \vee q) \vee (p \vee \sim q)$ is a tautology.
89. Negation of the statement $(p \vee q) \Rightarrow (q \vee r)$ is
(a) $p \wedge \sim q \wedge \sim r$
(b) $\sim p \wedge q \wedge \sim r$
(c) $\sim p \wedge q \wedge r$
(d) $p \wedge q \wedge \sim r$
JEE Main 31.08.2021, Shift-II
Ans. (a) : Negation of $(p \vee q) \Rightarrow (q \vee r)$:
$$\sim [(p \vee q) \Rightarrow (q \vee r)]$$ $$\equiv (p \vee q) \wedge \sim (q \vee r)$$ $$\equiv (p \vee q) \wedge (\sim q \wedge \sim r)$$ $$\equiv [(p \vee q) \wedge \sim q] \wedge \sim r$$ $$\equiv [(p \wedge \sim q) \vee (q \wedge \sim q)] \wedge \sim r$$ $$\equiv [(p \wedge \sim q) \vee \text{f}] \wedge \sim r$$ $$\equiv (p \wedge \sim q) \wedge \sim r$$ $$\equiv p \wedge \sim q \wedge \sim r$$90. 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 tautology
(a) neither (S1) and (S2) is true
(b) only (S1) is true
(c) only (S2) is true
(d) both (S1) and (S2) are true
JEE Main-06.04.2023, Shift-II
Ans. (a) : S1 : $(p \rightarrow q) \vee (\sim p \wedge q) \equiv (\sim p \vee q) \vee (\sim p \wedge q) \equiv \sim p \vee q$, which is not a tautology.
S2 : $(q \rightarrow p) \rightarrow (\sim p \wedge q) \equiv (\sim q \vee p) \rightarrow (\sim p \wedge q) \equiv \sim (\sim q \vee p) \vee (\sim p \wedge q) \equiv (q \wedge \sim p) \vee (\sim p \wedge q) \equiv \sim p \wedge q$, which is not a tautology.
Hence, neither (S1) nor (S2) is true.
91. The statement $(p \Rightarrow q) \vee (p \Rightarrow r)$ is NOT equivalent to:
(a) $(p \wedge (\sim r)) \Rightarrow q$
(b) $(\sim q) \Rightarrow ((\sim r) \vee p)$
(c) $p \Rightarrow (q \vee r)$
(d) $(p \wedge (\sim q)) \Rightarrow r$
JEE Main-29.07.2022, Shift-II
Ans. (b) : Given statement:
$$(p \Rightarrow q) \vee (p \Rightarrow r)$$ $$\equiv (\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee (q \vee r) \equiv p \Rightarrow (q \vee r)$$Option (a) : $(p \wedge \sim r) \Rightarrow q \equiv \sim (p \wedge \sim r) \vee q \equiv (\sim p \vee r) \vee q \equiv \sim p \vee (q \vee r)$ (Equivalent)
Option (c) : $p \Rightarrow (q \vee r) \equiv \sim p \vee (q \vee r)$ (Equivalent)
Option (d) : $(p \wedge \sim q) \Rightarrow r \equiv \sim (p \wedge \sim q) \vee r \equiv (\sim p \vee q) \vee r \equiv \sim p \vee (q \vee r)$ (Equivalent)
Option (b) : $\sim q \Rightarrow (\sim r \vee p) \equiv q \vee \sim r \vee p$, which is NOT equivalent to $\sim p \vee q \vee r$.
92. Negation of $(p \Rightarrow q) \Rightarrow (q \Rightarrow p)$ is
(a) $(\sim P) \vee q$
(b) $(\sim q) \wedge P$
(c) $q \wedge (\sim p)$
(d) $p \vee (\sim q)$
JEE Main-08.04.2023, Shift-I
Ans. (c) : Using negation of conditional $\sim (A \Rightarrow B) \equiv A \wedge \sim B$:
$$\sim [(p \Rightarrow q) \Rightarrow (q \Rightarrow p)]$$ $$\equiv (p \Rightarrow q) \wedge \sim (q \Rightarrow p)$$ $$\equiv (\sim p \vee q) \wedge (q \wedge \sim p)$$ $$\equiv q \wedge \sim p$$
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