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Statement-2 $: \sim(p \leftrightarrow \sim q)$ is a tautology. (1) Statement–1 is true, Statement–2 is false. (2) Statement–1 is false, Statement–2 is true. (3) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1. (4) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for statement–1. [AIEEE-2009]
$p:$ There is a rational number $x \in S$ such that $x>0$ which of the following statements is the negation of the statement p? (1) There is a rational number $x \in$ S such that $x \leq 0$ (2) There is no rational number $x \in S$ such that $x \leq 0$ (3) Every rational number $x \in S$ satisfies $x \leq 0$ (4) $x \in S$ and $x \leq 0 \Rightarrow x$ is not rational. [AIEEE-2010]
Given $\mathrm{S} \subseteq \mathrm{R}$ and $\mathrm{p}=$ There is a rational number $\mathrm{x} \in \mathrm{S}$ such that $\mathrm{x}>0$ then $\sim \mathrm{p}:$ Any rational number $\mathrm{x} \in \mathrm{S}$ such that $\mathrm{x}$ $\not>$ 0 i.e. $\sim \mathrm{p}:$ Every rational number $\mathrm{x} \in \mathrm{S}$ satisfy $\mathrm{x} \leq 0$
p : Suman is brilliant q : Suman is rich r : Suman is honest The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as :- ( 1)$\sim \mathrm{q} \leftrightarrow \sim \mathrm{p} \wedge \mathrm{r}$ ( 2)$\sim(\mathrm{p} \wedge \sim \mathrm{r}) \leftrightarrow \mathrm{q}$ ( 3)$\sim \mathrm{p} \wedge(\mathrm{q} \leftrightarrow \sim \mathrm{r})$ ( 4)$\sim(\mathrm{q} \leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$ [AIEEE-2011]
Given Statement : $(\mathrm{p} \wedge \sim \mathrm{r}) \Leftrightarrow \mathrm{q}$ Negations of $\mathrm{p} \Leftrightarrow \mathrm{q}$ are $\sim(\mathrm{p} \Leftrightarrow \mathrm{q}), \sim(\mathrm{q} \Leftrightarrow \mathrm{p})$ $\sim \mathrm{p} \Leftrightarrow \mathrm{q}$ and $\sim \mathrm{q} \Leftrightarrow \mathrm{p}$ Hence negations of given statement are $\sim(\mathrm{q} \Leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$ and $\sim(\mathrm{p} \wedge \sim \mathrm{r}) \Leftrightarrow \mathrm{q}$
( 1) $\mathrm{q} \rightarrow[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})]$ (2) $\mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q})$ (3) $\mathrm{p} \vee(\mathrm{p} \wedge \mathrm{q})$ (4) $[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$ [AIEEE-2011]
$[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$ $[\mathrm{p} \wedge(\sim \mathrm{p} \vee \mathrm{q})] \rightarrow \mathrm{q}$ $[(\mathrm{p} \wedge \sim \mathrm{p}) \vee(\mathrm{p} \wedge \mathrm{q})] \rightarrow \mathrm{q}$ $[\mathrm{c} \vee(\mathrm{p} \wedge \mathrm{q})] \rightarrow \mathrm{q}$ $\left\{\begin{array}{c}{\mathrm{p} \wedge \sim \mathrm{p} \equiv \mathrm{c} \equiv \mathrm{contradiction}} \\ {\because \mathrm{c} \vee \mathrm{p} \equiv \mathrm{p}}\end{array}\right.$ $\Rightarrow(\mathrm{p} \wedge \mathrm{q}) \rightarrow \mathrm{q}$ $\Rightarrow \sim(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{q}$ $\Rightarrow(\sim \mathrm{p} \vee \sim \mathrm{q}) \vee \mathrm{q}$ $\Rightarrow \sim \mathrm{p} \vee(\mathrm{q} \vee \sim \mathrm{q})$ $\Rightarrow \sim \mathrm{p} \vee(\mathrm{t}) \equiv$ tautology
“If I become a teacher, then I will open a school”, is : (1) I will not become a teacher or I will open a school. (2) I will become a teacher and I will not open a school. (3) Either I will not become a teacher or I will not open a school. (4) Neither I will become a teacher nor I will open a school. [AIEEE-2012]
Statement-I: $(\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\sim \mathrm{p} \wedge \mathrm{q})$ is a fallacy. Statement-II : $(\mathrm{p} \rightarrow \mathrm{q}) \leftrightarrow(\sim \mathrm{q} \rightarrow \sim \mathrm{p})$ is a tuatology (1) Statement-I is true, Statement-II is true; statement-II is a correct explanation for Statement-I. (2) Statement-I is true, Statement-II is true; statement-II is not a correct explanation for Statement-I. (3) Statement-I is true, Statement-II is false. (4) Statement-I is false, Statement-II is true. [JEE-MAINS-2013]
Given statement is $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$ As we know $\sim(\mathrm{p} \leftrightarrow \mathrm{q}) \equiv \sim \mathrm{p} \leftrightarrow \mathrm{q}$ or $\mathrm{p} \leftrightarrow \sim \mathrm{q}$ $\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$
(1) equivalent to $p \leftrightarrow q$ (2) equivalent to $\sim p \leftrightarrow q$ (3) a tautology (4) a fallacy [JEE(Main)-2014]
Given statement is $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$ As we know $\sim(\mathrm{p} \leftrightarrow \mathrm{q}) \equiv \sim \mathrm{p} \leftrightarrow \mathrm{q}$ or $\mathrm{p} \leftrightarrow \sim \mathrm{q}$ $\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$
( 1) $\mathrm{s} \vee(\mathrm{r} \vee \sim \mathrm{s})$ (2) $\mathrm{s} \wedge \mathrm{r}$ (3) $\mathrm{s} \wedge \sim \mathrm{r}$ ( 4) $\mathrm{s} \wedge(\mathrm{r} \wedge \sim \mathrm{s})$ [JEE(Main)-2015]
$\square \mathrm{s} \vee(\square \mathrm{r} \wedge \mathrm{s})$ $(\square \mathrm{s} \vee \sim \mathrm{r}) \wedge(\square \mathrm{s} \wedge \mathrm{s})$ $(\square \mathrm{s} \vee \sim \mathrm{r}) \wedge \mathrm{t}$ $(\square \mathrm{s} \vee \sim \mathrm{r})$ $\sim(\square \mathrm{s} \vee \sim \mathrm{r})$ $\mathrm{s} \wedge \mathrm{r}$
(1) $\mathrm{pv} \sim \mathrm{q}$ (2) $\sim \mathrm{p} \wedge \mathrm{q}$ (3) $\mathrm{p} \wedge \mathrm{q}$ (4) $\mathrm{p} \vee \mathrm{q}$ [JEE(Main)-2016]
Given boolean expression is $(\mathrm{p} \wedge \sim \mathrm{q}) \vee \mathrm{q} \vee(\sim \mathrm{p} \wedge \mathrm{q})$ $(\mathrm{p} \wedge \sim \mathrm{q}) \mathrm{Vq}=(\mathrm{p} \vee \mathrm{q}) \wedge(\sim \mathrm{q} \vee \mathrm{q})=(\mathrm{p} \vee \mathrm{q}) \wedge \mathrm{t}=(\mathrm{p} \vee \mathrm{q})$ Now, $(\mathrm{pVq}) \vee(\sim \mathrm{p} \wedge \mathrm{q})=\mathrm{p} \vee \mathrm{q}$
(1) a fallacy (2) a tautology (3) equivalent to $\sim \mathrm{p} \rightarrow \mathrm{q}$ (4) equivalent to $\mathrm{p} \rightarrow \sim \mathrm{q}$ [JEE(Main)-2017]
(1) p (2) q (3) $\sim \mathrm{q}$ ( 4)$\sim \mathrm{p}$ [JEE(Main)-2018]
$\sim(p \vee q) \vee(\sim p \wedge q)$ $(\sim p \wedge \sim q) \vee(\sim p \wedge q)$ $\Rightarrow \sim p \wedge(\sim q \vee q)$ $\Rightarrow \sim p \wedge t \equiv \sim p$
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