1
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Truth values of $\mathrm{p} \rightarrow \mathrm{r}$ is F and $\mathrm{p} \leftrightarrow \mathrm{q}$ is F . Then the truth values of $(\sim p \vee q) \rightarrow(p \vee \sim q)$ and $(p \wedge \sim q) \rightarrow(\sim p \wedge q)$ are respectively

A
$\mathrm{T, F}$
B
$\mathrm{F}, \mathrm{T}$
C
$\mathrm{T}, \mathrm{T}$
D
$\mathrm{F, F}$
2
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The statement $\sim(p \leftrightarrow \sim q)$ is

A
equivalent to $\mathrm{p} \leftrightarrow \mathrm{q}$
B
a fallacy
C
a tautology
D
equivalent to $\sim \mathrm{p} \leftrightarrow \mathrm{q}$
3
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

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

A
$\mathrm{p} \wedge(\sim \mathrm{q})$
B
$p \vee(q)$
C
$p \rightarrow(\sim q)$
D
$\mathrm{q} \rightarrow \mathrm{p}$
4
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $S$ be a non-empty subset of $\mathbb{R}$. Consider the following statement:

p : There is a rational number $x \in \mathrm{~S}$ such that $x>0$.

Which of the following statements is the negation of the statement p?

A
There is a rational number $x \in \mathrm{~S}$ such that $x \leq 0$.
B
There is no rational number $x \in \mathrm{~S}$ such that $x \leq 0$.
C
Every rational number $x \in S$ satisfies $x \leq 0$.
D
$x \in \mathrm{~S}$ and $x \leq 0 \Rightarrow x$ is not a rational number.
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