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Consider the following two propositions:

\begin{aligned} &\mathrm{P} 1: \sim(\mathrm{p} \rightarrow \sim \mathrm{q}) \\ &\mathrm{P} 2:(\mathrm{p} \wedge \sim \mathrm{q}) \wedge((\sim \mathrm{p}) \vee \mathrm{q}) \end{aligned}

If the proposition \mathrm{p} \rightarrow((\sim \mathrm{p}) \vee \mathrm{q}) is evaluated as FALSE, then:

Option: 1

P1 is TRUE and P2 is FALSE


Option: 2

P1 is FALSE and P2 is TRUE


Option: 3

Both P1 and P2 are FALSE


Option: 4

Both P1 and P2 are TRUE


Answers (1)

best_answer
p q \sim p \sim q \sim p\vee q p\rightarrow \left ( \sim p\vee q \right ) p\rightarrow \sim q P_{1}:\sim \left ( p\rightarrow \sim q \right ) p\wedge \sim q P_{2}:\left ( p\wedge \sim q \right )\wedge \left ( \left ( \sim p \right )\vee q \right )
T T F F T T F T F F
T F F T F F T F T F
F T T F T T T F F F
F F T T T T T F F F

 \mathrm{p\rightarrow \left ( \sim p\vee q \right )} is \mathrm{F} when p is true and q is false from table.
Hence, \mathrm{P_{1}\, \&\, P_{2}} both are false

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Gaurav

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