MHT CET · Maths · Mathematical Reasoning
The expression \((p \wedge \sim q) \vee q \vee(\sim p \wedge q)\) is equivalent to
- A \(\sim p \vee q\)
- B \(p \wedge q\)
- C \(\mathrm{p} \vee \mathrm{q}\)
- D \(\mathrm{p} \vee \sim \mathrm{q}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{p} \vee \mathrm{q}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& (p \wedge \sim q) \vee q \vee(\sim p \wedge q) \\
& \equiv[(p \vee q) \wedge(\sim q \vee q)] \vee(\sim p \wedge q)
\end{aligned}
\)
...[Distributive law]
\(
\begin{aligned}
& \equiv[(p \vee q) \wedge T] \vee(\sim p \wedge q) \ldots[\text { Complement law] } \\
& \equiv(p \vee q) \vee(\sim p \wedge q) \\
& \equiv(p \vee q \vee \sim p) \wedge(p \vee q \vee q)
\end{aligned}
\)
...[Distributive law]
\(
\equiv(T \vee q) \wedge(p \vee q)
\)
...[Complement law and Idempotent law]
\(
\begin{aligned}
& \equiv T \wedge(p \vee q) \\
& \equiv p \vee q
\end{aligned}
\)
\(\therefore\) [Identity law]
...[Identity law]
\begin{aligned}
& (p \wedge \sim q) \vee q \vee(\sim p \wedge q) \\
& \equiv[(p \vee q) \wedge(\sim q \vee q)] \vee(\sim p \wedge q)
\end{aligned}
\)
...[Distributive law]
\(
\begin{aligned}
& \equiv[(p \vee q) \wedge T] \vee(\sim p \wedge q) \ldots[\text { Complement law] } \\
& \equiv(p \vee q) \vee(\sim p \wedge q) \\
& \equiv(p \vee q \vee \sim p) \wedge(p \vee q \vee q)
\end{aligned}
\)
...[Distributive law]
\(
\equiv(T \vee q) \wedge(p \vee q)
\)
...[Complement law and Idempotent law]
\(
\begin{aligned}
& \equiv T \wedge(p \vee q) \\
& \equiv p \vee q
\end{aligned}
\)
\(\therefore\) [Identity law]
...[Identity law]
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