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If $$p,q$$ and $$r$$ are three propositions, then which of the following combination of truth values of $$p,q$$ and $$r$$ makes the logical expression $$\left\{ {(p \vee q) \wedge \left( {( \sim p) \vee r} \right)} \right\} \to \left( {( \sim q) \vee r} \right)$$ false?

JEE · Math · previous-year question

  1. A.$$p = F,q = T,r = F$$correct
  2. B.$$p = T,q = T,r = F$$
  3. C.$$p = T,q = F,r = T$$
  4. D.$$p = T,q = F,r = F$$

Answer

A. $$p = F,q = T,r = F$$

Explanation

.tg {border-collapse:collapse;border-spacing:0;width:100%} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} $\mathrm{p}$ $\mathrm{q}$ $\mathrm{r}$ $(p \vee q) \wedge((\sim p) \vee r)$ $\sim \mathrm{q} \vee \mathrm{r}$ $(1)$ $\mathrm{T}$ $\mathrm{F}$ $\mathrm{T}$ $\mathrm{T}$ $\mathrm{T}$ $(2)$ $\mathrm{T}$ $\mathrm{T}$ $\mathrm{F}$ $\mathrm{F}$ $\mathrm{F}$ $(3)$ $\mathrm{F}$ $\mathrm{T}$ $\mathrm{F}$ $\mathrm{T}$ $\mathrm{F}$ $(4)$ $\mathrm{T}$ $\mathrm{F}$ $\mathrm{F}$ $\mathrm{F}$ $\mathrm{T}$ So, $(p \vee q) \wedge(\sim q \vee r) \rightarrow(\sim p \vee r)$ will be False.

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