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If truth values of statements $p, q$ are true, and $r$, $s$ are false, then the truth values of the following statement patterns are respectively
$\mathrm{a}: \sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee(\sim \mathrm{q} \vee \mathrm{s})$
$\mathrm{b}:(\sim \mathrm{q} \wedge \sim \mathrm{r}) \leftrightarrow(\mathrm{p} \vee \mathrm{s})$
$c:(\sim p \vee q) \rightarrow(r \wedge \sim s)$
Options:
$\mathrm{a}: \sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee(\sim \mathrm{q} \vee \mathrm{s})$
$\mathrm{b}:(\sim \mathrm{q} \wedge \sim \mathrm{r}) \leftrightarrow(\mathrm{p} \vee \mathrm{s})$
$c:(\sim p \vee q) \rightarrow(r \wedge \sim s)$
Solution:
1217 Upvotes
Verified Answer
The correct answer is:
$F,F,F$
a.
$\begin{aligned}
& \sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee(\sim \mathrm{q} \vee \mathrm{s}) \\
& \equiv \sim(\mathrm{T} \wedge \sim \mathrm{F}) \vee(\sim \mathrm{T} \vee \mathrm{F}) \\
& \equiv(\mathrm{F} \vee \mathrm{F}) \vee(\mathrm{F} \vee \mathrm{F}) \\
& \equiv \mathrm{F} \vee \mathrm{F} \\
& \equiv \mathrm{F}
\end{aligned}$
b.
$\begin{aligned}
& (\sim \mathrm{q} \wedge \sim \mathrm{r}) \leftrightarrow(\mathrm{p} \vee \mathrm{s}) \\
& \equiv(\sim \mathrm{T} \wedge \sim \mathrm{F}) \leftrightarrow(\mathrm{T} \vee \mathrm{F}) \\
& \equiv \mathrm{F} \leftrightarrow \mathrm{T} \\
& \equiv \mathrm{F}
\end{aligned}$
c.
$\begin{aligned}
& (\sim p \vee q) \rightarrow(r \wedge \sim s) \\
& \equiv(\sim T \vee T) \rightarrow(F \wedge \sim F) \\
& \equiv(F \vee T) \rightarrow(F \wedge T) \\
& \equiv T \rightarrow F \\
& \equiv F
\end{aligned}$
$\begin{aligned}
& \sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee(\sim \mathrm{q} \vee \mathrm{s}) \\
& \equiv \sim(\mathrm{T} \wedge \sim \mathrm{F}) \vee(\sim \mathrm{T} \vee \mathrm{F}) \\
& \equiv(\mathrm{F} \vee \mathrm{F}) \vee(\mathrm{F} \vee \mathrm{F}) \\
& \equiv \mathrm{F} \vee \mathrm{F} \\
& \equiv \mathrm{F}
\end{aligned}$
b.
$\begin{aligned}
& (\sim \mathrm{q} \wedge \sim \mathrm{r}) \leftrightarrow(\mathrm{p} \vee \mathrm{s}) \\
& \equiv(\sim \mathrm{T} \wedge \sim \mathrm{F}) \leftrightarrow(\mathrm{T} \vee \mathrm{F}) \\
& \equiv \mathrm{F} \leftrightarrow \mathrm{T} \\
& \equiv \mathrm{F}
\end{aligned}$
c.
$\begin{aligned}
& (\sim p \vee q) \rightarrow(r \wedge \sim s) \\
& \equiv(\sim T \vee T) \rightarrow(F \wedge \sim F) \\
& \equiv(F \vee T) \rightarrow(F \wedge T) \\
& \equiv T \rightarrow F \\
& \equiv F
\end{aligned}$
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