1. ⇒ (MHT CET 2023 9th May Evening Shift )

Negation of the statement

"The payment will be made if and only if the work is finished in time." Is

A. The work is finished in time and the payment is not made.

B. The payment is made and the work is not finished in time.

C. The work is finished in time and the payment is not made, or the payment is made and the work is finished in time.

D. Either the work is finished in time and the payment is not made, or the payment is made and the work is not finished in time.

Correct answer option is (D)

Let $p$ : Payment will be made

$q$ : Work is finished in time.

Given statement is

$(p \to q) \land (q \to p)$

Negation of above statement is

$(p \land \sim q) \lor (q \land \sim p)$

$∴$ Option (D) is correct.

2. ⇒ (MHT CET 2023 9th May Morning Shift )

The negation of the statement $(p \land q) \to (\sim p \lor r)$ is

A. $p \lor q \lor \sim r$

B. $p \land q \land \sim r$

C. $\sim p \lor q \land r$

D. $\sim p \lor \sim q \lor \sim r$

Correct answer option is (B)

$\begin{aligned} \sim [(p \land q) \to (\sim p \lor r)] \\ \equiv (p \land q) \land \sim (\sim p \lor r) \dots [∵\sim (p \to q) \equiv p \land \sim q] \\ \equiv p \land q \land p \land \sim r \dots [Associative Law] \\ \equiv p \land q \land \sim r ...[Idempotent Law] \end{aligned}$

3. ⇒ (MHT CET 2021 21th September Morning Shift )

Negation of the statement $\forall x \in R, x^{2} + 1 = 0$ is

A. $\exists x \in R$ such that $x^{2} + 1 < 0$ .

B. $\exists x \in R$ such that $x^{2} + 1 \neq 0$ .

C. $\exists x \in R$ such that $x^{2} + 1 \leq 0$ .

D. $\exists x \in R$ such that $x^{2} + 1 = 0$ .

Correct answer option is (C)

Negation of $(\forall x \in R, x^{2} + 1 = 0)$ is $\exists x \in R$ , such that $x^{2} + 1 \neq 0$

4. ⇒ (MHT CET 2021 20th September Evening Shift )

Negation of $(p \land q) \to (\sim p \lor r)$ is

A. $p \lor q \lor (\sim r)$

B. $p \land q \land r$

C. $\sim p \land q \land r$

D. $p \land q \land (\sim r)$

Correct answer option is (D)

$\begin{aligned} \sim [(p \land q) \to (\sim p \lor r)] \\ \equiv\sim [\sim (p \land q) \lor (\sim p \lor r)] \\ \equiv (p \land q) \land \sim (\sim p \lor r) \\ \equiv (p \land q) \land (p \land \sim r) \equiv p \land q \land (\sim r) \end{aligned}$

5. ⇒ (MHT CET 2021 20th September Morning Shift )

The negation of a statement 'x $\in$ A $\cap$ B $\to$ (x $\in$ A and x $\in$ B)' is

A. x $\in$ A $\cap$ B $\to$ (x $\in$ A or x $\in$ B)

B. x $\in$ A $\cap$ B and (x $\notin$ A or x $\notin$ B)

C. x $\in$ A $\cap$ B or (x $\in$ A or x $\in$ B)

D. x $\notin$ A $\cap$ B and (x $\in$ A and x $\in$ B)

Correct answer option is (B)

Let $p : x \in A \cap B$ and $q : x \in A$ and $x \in B$ .

$∴$ Logical form of given statement is $p \to q$ .

Now $p \to q \equiv\sim p \lor q$

$∴ \sim (p \to q) \equiv\sim (\sim p \lor q) \equiv p \land \sim q$ , which is stated as $x \in A \cap B$ and $(x \notin A$ or $x \notin B)$ .

⇒ Mathematical Logic