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1
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

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
$$\exists \mathrm{x} \in \mathrm{R}$$ such that $$\mathrm{x}^2+1=0$$.
2
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$p, q$$ are true statements and $$r$$ is false statement, then which of the following is correct.

A
$$(p \vee q) \vee r$$ has truth value $$F$$.
B
$$(\mathrm{p} \rightarrow \mathrm{r}) \rightarrow \mathrm{q}$$ has truth value $$\mathrm{F}$$.
C
$$(p \wedge q) \rightarrow r$$ has truth value $$T$$.
D
$$(\mathrm{p} \leftrightarrow \mathrm{q}) \rightarrow \mathrm{r}$$ has truth value $$\mathrm{F}$$.
3
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

p : It rains today

q : I am going to school

r : I will meet my friend

s : I will go to watch a movie.

Then symbolic form of the statement "If it does not rain today or I won't go to school, then I will meet my friend and I will go to watch a movie" is

A
$$\mathrm{\sim(p\vee q)\to (r \vee s)}$$
B
$$\mathrm{(p\wedge q)\to (r \vee s)}$$
C
$$\mathrm{\sim(p\wedge q)\to (r \wedge s)}$$
D
$$\mathrm{(\sim p\wedge q)\to (r \wedge s)}$$
4
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

Negation of $$(p \wedge q) \rightarrow(\sim p \vee r)$$ is

A
$$p \vee q \vee(\sim r)$$
B
$$p \wedge q \wedge r$$
C
$$\sim p \wedge q \wedge r$$
D
$$p \wedge q \wedge(\sim r)$$
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