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 $\mathrm{x}$ $\not>$ 0 i.e. $\sim \mathrm{p}:$ Every rational number $\mathrm{x} \in \mathrm{S}$ satisfy $\mathrm{x} \leq 0$
Q. Consider the following statements
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]
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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 statement are $\sim(\mathrm{q} \Leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$ 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) $\sim(p \vee q) \vee(\sim p \wedge q)$ $(\sim p \wedge \sim q) \vee(\sim p \wedge q)$ $\Rightarrow \sim p \wedge(\sim q \vee q)$ $\Rightarrow \sim p \wedge t \equiv \sim p$
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
Really enjoyed the questions
I’m not getting solution of 4 question
Thanks
Do add new question.
Quite helpful
Thanks
Tq
సూపర్
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Thank you so much..these helped me alot ….
I have confusion in this chapter..after doing these bits ..I got clarity……Once again tq
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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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