Q70 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics

Consider the following two propositions: $\mathrm{P}_1: \sim(p \rightarrow \sim q)$ $P_2:(p \wedge \sim q) \wedge((\sim p) \vee q)$ If the proposition $p \rightarrow((\sim p) \vee q)$ is evaluated as FALSE, then :

  • A
    $P_1$ is TRUE and $P_2$ is FALSE
  • B
    $\mathrm{P}_1$ is FALSE and $\mathrm{P}_2$ is TRUE
  • Correct answer
    C
    Both $P_1$ and $P_2$ are FALSE
  • D
    Both $P_1$ and $P_2$ are TRUE

Solution

Given propositions $\mathrm{P}_1: \sim(\mathrm{p} \rightarrow \sim \mathrm{q})$ $\mathrm{P}_2:(p \wedge \sim q) \wedge((\sim p) \vee q)$ Required table is shown below. \(\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline \mathbf{p} & \mathbf{q} & \sim p & \sim q & \sim p \vee q & p \rightarrow(\sim(p \vee q) & p \rightarrow \sim q & \sim(p \rightarrow \sim q) & p \wedge \sim q & p_2 \\ \hline \text { T } & \text { T } & \text { F } & \text { F } & \text { T } & \text { T } & \text { F } & \text { T } & \text { F } & \text { F } \\ \hline \text { T } & \text { F } & \text { F } & \text { T } & \text { F } & \text { F } & \text { T } & \text { F } & \text { T } & \text { F } \\ \hline \text { F } & \text { T } & \text { T } & \text { F } & \text { T } & \text { T } & \text { T } & \text { F } & \text { F } & \text { F } \\ \hline \text { F } & \text { F } & \text { T } & \text { T } & \text { T } & \text { T } & \text { T } & \text { F } & \text { F } & \text { F } \\ \hline \end{array}\) \(p \rightarrow(\sim p \vee q)\) is \(F\) when \(p\) is true \(q\) is false From table \(P_1 ~\&~ P_2\) both are false

Hence, the correct answer is option: 'C'.

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