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Q. Statement-1: $\sim(p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$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]
Q. Let $S$ be a non-empty subset of $R$. Consider the following statement:$p:$ 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]
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Sol. (3)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 $\mathrm{x}$ $\not>$ 0i.e. $\sim \mathrm{p}:$ Every rational number $\mathrm{x} \in \mathrm{S}$ satisfy $\mathrm{x} \leq 0$
Q. Consider the following statementsp : Suman is brilliantq : Suman is richr : Suman is honestThe 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]
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Sol. (2,4)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})$$\sim \mathrm{p} \Leftrightarrow \mathrm{q}$ and $\sim \mathrm{q} \Leftrightarrow \mathrm{p}$Hence negations of given statementare $\sim(\mathrm{q} \Leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$and $\sim(\mathrm{p} \wedge \sim \mathrm{r}) \Leftrightarrow \mathrm{q}$
Q. The only statement among the followings that is a tautology is :( 1) $\mathrm{q} \rightarrow[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})]$(2) $\mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q})$(3) $\mathrm{p} \vee(\mathrm{p} \wedge \mathrm{q})$(4) $[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$ [AIEEE-2011]
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Sol. (4)$[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$$[\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
Q. The negation of the statement“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]
Q. Consider :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]
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Sol. (2)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}$$\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$
Q. The statement $\sim(p \leftrightarrow \sim q)$ is :(1) equivalent to $p \leftrightarrow q$(2) equivalent to $\sim p \leftrightarrow q$(3) a tautology(4) a fallacy [JEE(Main)-2014]
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Sol. (1)Given statement is $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$As we know $\sim(\mathrm{p} \leftrightarrow \mathrm{q}) \equiv \sim \mathrm{p} \leftrightarrow \mathrm{q}$ or $\mathrm{p} \leftrightarrow \sim \mathrm{q}$$\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$
Q. The negation of $\sim \mathrm{s} \vee(\sim \mathrm{r} \wedge \mathrm{s})$ is equivalent to :( 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]
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Sol. (2)$\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})$$\mathrm{s} \wedge \mathrm{r}$
Q. The Boolean Expression (p\wedge\simq) Vq\vee(\simp\wedgeq) is equivalent to :-(1) $\mathrm{pv} \sim \mathrm{q}$(2) $\sim \mathrm{p} \wedge \mathrm{q}$(3) $\mathrm{p} \wedge \mathrm{q}$(4) $\mathrm{p} \vee \mathrm{q}$ [JEE(Main)-2016]
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Sol. (4)Given boolean expression is$(\mathrm{p} \wedge \sim \mathrm{q}) \vee \mathrm{q} \vee(\sim \mathrm{p} \wedge \mathrm{q})$$(\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}$
Q. The following statement $(\mathrm{p} \rightarrow \mathrm{q}) \rightarrow[(\sim \mathrm{p} \rightarrow \mathrm{q}) \rightarrow \mathrm{q}]$ is :(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]
Q. The Boolean expression $\sim(p \vee q) \vee(\sim p \wedge q)$ is equivalent to :(1) p (2) q (3) $\sim \mathrm{q}$ ( 4)$\sim \mathrm{p}$ [JEE(Main)-2018]
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Sol. (4)$\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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Please add questions after 2018
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Thank you for these questions, really helped a lot.
The Level of Questions Are moderate
I want Jee important question
Pls add 2019and 2020 question also of this chapter
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I’m not getting solution of 4 question
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Do add new question.
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In question no 3 (from upward) ,their are two solutions. In them only solution no 4 is right according to my answer their is negation of all statement. In option 2, their is negation of q is not taken.
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