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w1f5gbWwTUKeqvXF | maths | mathematical-reasoning | logical-connectives | Consider :
<br/><b>Statement − I :</b> $$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$$ is a fallacy.
<br/><b>Statement − II :</b>$$\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)$$ is a tautology. | [{"identifier": "A", "content": "Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I"}, {"identifier": "B", "content": "Statement -I is True; Statement -II is False."}, {"identifier": "C", "content": "Statement -I is False; Statement -II is True"}, {"identifier": "D", "content": "Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I "}] | ["A"] | null | <p>Statement - I : $$(p \wedge \sim q) \wedge ( \sim p \wedge q)$$</p>
<p><style type="text/css">
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.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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.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-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 608px">
<colgroup>
<col style="width: 35px">
<col style="width: 43px">
<col style="width: 83px">
<col style="width: 83px">
<col style="width: 108px">
<col style="width: 92px">
<col style="width: 164px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim p$$</th>
<th class="tg-baqh">$$ \sim q$$</th>
<th class="tg-baqh">$$p \wedge \sim q$$<br></th>
<th class="tg-baqh">$$ \sim p \wedge q$$<br></th>
<th class="tg-baqh">$$(p \wedge \sim q) \wedge ( \sim p \wedge q)$$<br></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
</tbody>
</table></p>
<p>Hence, statement-I is a fallacy.</p>
<p>Statement - II : $$(p \to q) \leftrightarrow ( \sim q \to \sim p)$$</p>
<p><style type="text/css">
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<table class="tg" style="undefined;table-layout: fixed; width: 717px">
<colgroup>
<col style="width: 35px">
<col style="width: 43px">
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<col style="width: 83px">
<col style="width: 108px">
<col style="width: 112px">
<col style="width: 253px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim p$$</th>
<th class="tg-baqh">$$ \sim q$$</th>
<th class="tg-baqh">$$p \to q$$<br></th>
<th class="tg-baqh">$$ \sim q \to \sim p$$<br></th>
<th class="tg-baqh">$$(p \to q) \leftrightarrow ( \sim q \to \sim p)$$<br></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>Hence, statement - II is a tautology.</p>
<p>Now, statement-II does not depend on statement-I.</p> | mcq | jee-main-2013-offline |
eFr6UeIkqoSt0gX8 | maths | mathematical-reasoning | logical-connectives | The statement $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is : | [{"identifier": "A", "content": "equivalent to $${ \\sim p \\leftrightarrow q}$$"}, {"identifier": "B", "content": "a tautology"}, {"identifier": "C", "content": "a fallacy"}, {"identifier": "D", "content": "equivalent to $${p \\leftrightarrow q}$$"}] | ["D"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
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font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 664px">
<colgroup>
<col style="width: 35px">
<col style="width: 43px">
<col style="width: 83px">
<col style="width: 83px">
<col style="width: 108px">
<col style="width: 92px">
<col style="width: 102px">
<col style="width: 118px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim p$$</th>
<th class="tg-baqh">$$ \sim q$$</th>
<th class="tg-baqh">$$p \leftrightarrow q$$</th>
<th class="tg-baqh">$$p \leftrightarrow \sim q$$</th>
<th class="tg-baqh">$$ \sim p \leftrightarrow q$$</th>
<th class="tg-baqh">$$ \sim (p \leftrightarrow \sim q)$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>From the truth table we can say that given statement is equivalent to $$p \leftrightarrow q$$.</p> | mcq | jee-main-2014-offline |
z8kycnQh0D6zJpWr | maths | mathematical-reasoning | logical-connectives | The negation of $$ \sim s \vee \left( { \sim r \wedge s} \right)$$ is equivalent to : | [{"identifier": "A", "content": "$$s \\vee \\left( {r \\vee \\sim s} \\right)$$"}, {"identifier": "B", "content": "$$s \\wedge r$$"}, {"identifier": "C", "content": "$$s \\wedge \\sim r$$"}, {"identifier": "D", "content": "$$s \\wedge \\left( {r \\wedge \\sim s} \\right)$$"}] | ["B"] | null | <img src="https://gateclass.static.cdn.examgoal.net/2OAgJSCgoisOIuHQi/0RUPnVokAM5r5pw4wQPIququqMxsm/6RBqA2leE55pxx7XCRommU/uploadfile.jpg" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2015 (Offline) Mathematics - Mathematical Reasoning Question 118 English Explanation">
<br>From the table you can see (s $$\wedge$$ r) and negation of ~s $$\vee$$ (~r $$\wedge$$ s) have same truth value. So they are equivalent. | mcq | jee-main-2015-offline |
csz1CrCo9hs1gIxJ | maths | mathematical-reasoning | logical-connectives | The Boolean expression
<br/><br/>$$\left( {p \wedge \sim q} \right) \vee q \vee \left( { \sim p \wedge q} \right)$$ is equivalent to : | [{"identifier": "A", "content": "$${ \\sim p \\wedge q}$$"}, {"identifier": "B", "content": "$${p \\wedge q}$$"}, {"identifier": "C", "content": "$$p \\vee q$$"}, {"identifier": "D", "content": "$$p \\vee \\sim q$$"}] | ["C"] | null | <img src="https://gateclass.cdn.examgoal.net/W3iUUKAK091ok1iYR/8HSPWtDptuwd7CcOkhxrHgDo4hy7f/t0aUb5WYW03FsXNGzU2q58/uploadfile.jpg" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Mathematical Reasoning Question 119 English Explanation"><br>
From the table you can see for both p $$\vee$$ q and (p $$\wedge$$ ~q) $$\vee$$ q $$\vee$$ (~p $$\wedge$$ q) truth table matches so they are equivalent. | mcq | jee-main-2016-offline |
IeXfuwNghZ9rPd4A | maths | mathematical-reasoning | logical-connectives | The following statement
<br/><br/>$$\left( {p \to q} \right) \to \left[ {\left( { \sim p \to q} \right) \to q} \right]$$ is : | [{"identifier": "A", "content": "equivalent to $${ \\sim p \\to q}$$"}, {"identifier": "B", "content": "equivalent to $${p \\to \\sim q}$$"}, {"identifier": "C", "content": "a fallacy"}, {"identifier": "D", "content": "a tautology"}] | ["D"] | null | We have<br><br>
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.tg .tg-s6z2{text-align:center}
.tg .tg-hgcj{font-weight:bold;text-align:center}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-hgcj">p</th>
<th class="tg-hgcj">q</th>
<th class="tg-hgcj">~p</th>
<th class="tg-hgcj">p$$\to$$q</th>
<th class="tg-hgcj">~p$$\to$$q</th>
<th class="tg-hgcj">(~p$$\to$$q)$$\to$$q</th>
<th class="tg-hgcj">(p$$\to$$q)$$\to$$((~p$$\to$$q)$$\to$$q)</th>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
</tr>
<tr>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
</tr>
<tr>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
</tr>
</tbody></table>
<br>
$$ \therefore $$ It is tautology. | mcq | jee-main-2017-offline |
O1CrpwejP1w9RrxfWoFIc | maths | mathematical-reasoning | logical-connectives | The proposition $$\left( { \sim p} \right) \vee \left( {p \wedge \sim q} \right)$$ is equivalent to : | [{"identifier": "A", "content": "p $$ \\vee $$ ~ q"}, {"identifier": "B", "content": "p $$ \\to $$ ~ q"}, {"identifier": "C", "content": "p $$ \\wedge $$ ~ q"}, {"identifier": "D", "content": "q $$ \\to $$ p"}] | ["B"] | null | <img src="https://gateclass.static.cdn.examgoal.net/5zTbH8zwu0wJZaFke/tyBqLcGYS9Pq5EOlU5qNL491PuOUh/gEmh8rTirgAroJSmV6GPRz/uploadfile.jpg" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Mathematical Reasoning Question 107 English Explanation"> | mcq | jee-main-2017-online-8th-april-morning-slot |
zJmXVh9oZFlkQlvs | maths | mathematical-reasoning | logical-connectives | The Boolean expression
<br/><br/>$$ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$$ is equvalent to : | [{"identifier": "A", "content": "$${ \\sim q}$$"}, {"identifier": "B", "content": "$${ \\sim p}$$"}, {"identifier": "C", "content": "p"}, {"identifier": "D", "content": "q"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266818/exam_images/u0czhffenck7fqfdrxfd.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Mathematical Reasoning Question 121 English Explanation">
<br><br>From the table you can see ~p and
<br> ~(p $$ \vee $$ q) $$ \vee $$ (~p $$ \wedge $$ q) are equivalent. | mcq | jee-main-2018-offline |
CARdLkhYP3rek0ixIlsbT | maths | mathematical-reasoning | logical-connectives | If (p $$ \wedge $$ $$ \sim $$ q) $$ \wedge $$ (p $$ \wedge $$ r) $$ \to $$ $$ \sim $$ p $$ \vee $$ q is false, then the truth values of $$p, q$$ and $$r$$ are, respectively : | [{"identifier": "A", "content": "F, T, F"}, {"identifier": "B", "content": "T, F, T"}, {"identifier": "C", "content": "T, T, T"}, {"identifier": "D", "content": "F, F, F"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264718/exam_images/klwq1gy6wqlgukcr0vja.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Morning Slot Mathematics - Mathematical Reasoning Question 110 English Explanation"><br>
From the truth table you can see (p $$ \wedge $$ ~q) $$ \wedge $$ (p $$ \wedge $$ r) $$\to$$ ~p $$\vee$$ q is <b>false</b> only when values of (p, q, r) is (T, F, T). | mcq | jee-main-2018-online-15th-april-morning-slot |
9i272wuPEKyF1Q1C6pQQl | maths | mathematical-reasoning | logical-connectives | If p $$ \to $$ ($$ \sim $$ p$$ \vee $$ $$ \sim $$ q) is false, then the truth values of p and q are respectively : | [{"identifier": "A", "content": "F, F"}, {"identifier": "B", "content": "T, F"}, {"identifier": "C", "content": "F, T"}, {"identifier": "D", "content": "T, T"}] | ["D"] | null | <table class="tg">
<tbody><tr>
<th class="tg-hgcj">p</th>
<th class="tg-amwm">q</th>
<th class="tg-amwm">~p</th>
<th class="tg-amwm">~q</th>
<th class="tg-amwm">~p $$ \vee $$ ~q</th>
<th class="tg-amwm">p $$ \to $$ (~p $$\vee$$ ~q)</th>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
</tbody></table><br>
From the truth table,<br><br>
p $$ \to $$ (~p $$\vee$$ ~q) is false only when p and q both are true. | mcq | jee-main-2018-online-16th-april-morning-slot |
SWEwhQ3jZbQCEVRMtlh3A | maths | mathematical-reasoning | logical-connectives | The logical statement
<br/><br/>[ $$ \sim $$ ( $$ \sim $$ p $$ \vee $$ q) $$ \vee $$ (p $$ \wedge $$ r)] $$ \wedge $$ ($$ \sim $$ q $$ \wedge $$ r) is equivalent to : | [{"identifier": "A", "content": "( $$ \\sim $$ p $$ \\wedge $$ $$ \\sim $$ q) $$ \\wedge $$ r"}, {"identifier": "B", "content": "$$ \\sim $$ p $$ \\vee $$ r"}, {"identifier": "C", "content": "(p $$ \\wedge $$ r) $$ \\wedge $$ $$ \\sim $$ q"}, {"identifier": "D", "content": "(p $$ \\wedge $$ $$ \\sim $$ q) $$ \\vee $$ r"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263729/exam_images/sjp2t7hj2rwdsx9ukbot.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Mathematical Reasoning Question 102 English Explanation"> | mcq | jee-main-2019-online-9th-january-evening-slot |
EUw7UhsPl8bIWjmhL43rsa0w2w9jxadlhgq | maths | mathematical-reasoning | logical-connectives | The Boolean expression ~(p $$ \Rightarrow $$ (~q)) is equivalent to : | [{"identifier": "A", "content": "p $$ \\wedge $$ q"}, {"identifier": "B", "content": "q $$ \\Rightarrow $$ ~p"}, {"identifier": "C", "content": "p $$ \\vee $$ q"}, {"identifier": "D", "content": "(~p) $$ \\Rightarrow $$ q"}] | ["A"] | null | $$ \sim \left( {p \to \sim q} \right) \Rightarrow \sim \left( { \sim p \vee \sim q} \right) \Rightarrow p \wedge q$$ | mcq | jee-main-2019-online-12th-april-evening-slot |
FAPJ14zebD0AktYldM3rsa0w2w9jx66727a | maths | mathematical-reasoning | logical-connectives | If the truth value of the statement p $$ \to $$ (~q $$ \vee $$ r) is false (F), then the truth values of the statements p, q, r are
respectively : | [{"identifier": "A", "content": "T, F, T"}, {"identifier": "B", "content": "F, T, T"}, {"identifier": "C", "content": "T, T, F"}, {"identifier": "D", "content": "T, F, F"}] | ["C"] | null | p $$ \to $$ ( ~ q $$ \vee $$ r) $$ \equiv $$ ~ p $$ \wedge $$ ( ~ q $$ \vee $$ r) $$ \equiv $$ ( ~ p $$ \vee $$ ~ q) $$ \vee $$ r $$ \equiv $$ ~ ( p $$ \wedge $$ q ) $$ \vee $$ r<br><br>
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<table class="tg">
<tbody><tr>
<th class="tg-hgcj">p</th>
<th class="tg-hgcj">q</th>
<th class="tg-amwm">r</th>
<th class="tg-amwm">(p $$ \wedge $$ q) $$ \vee $$ r </th>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-s6z2">T</td>
<td class="tg-s6z2">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-s6z2">F</td>
<td class="tg-s6z2">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody></table> | mcq | jee-main-2019-online-12th-april-morning-slot |
ArQYSzpz7JMKTYCSCM3rsa0w2w9jx2f318i | maths | mathematical-reasoning | logical-connectives | The negation of the Boolean expression ~ s $$ \vee $$ (~r $$ \wedge $$ s) is equivalent to : | [{"identifier": "A", "content": "~ s $$ \\wedge $$ ~ r"}, {"identifier": "B", "content": "r"}, {"identifier": "C", "content": "s $$ \\vee $$ r"}, {"identifier": "D", "content": "s $$ \\wedge $$ r"}] | ["D"] | null | ~ s $$ \vee $$ ( ~ r $$ \wedge $$ s)<br><br>
$$ \equiv $$ (~s $$ \vee $$ ~r) $$ \wedge $$ (~s $$ \vee $$ s)<br><br>
$$ \equiv $$ ( ~s $$ \vee $$ ~r) $$ \wedge $$ t<br><br>
$$ \equiv $$ ~s $$ \vee $$ ~ r $$ \equiv $$ ~ (s $$ \wedge $$ r)<br><br>
$$ \therefore $$ Negation of ~s $$ \vee $$ (~r $$ \wedge $$ s) is s $$ \wedge $$ r | mcq | jee-main-2019-online-10th-april-evening-slot |
L27qJHKqTbCWQYecUh3rsa0w2w9jwxnnauw | maths | mathematical-reasoning | logical-connectives | Which one of the following Boolean expressions is a tautology? | [{"identifier": "A", "content": "(p $$ \\vee $$ q) $$ \\wedge $$ (~ p $$ \\vee $$ ~ q)"}, {"identifier": "B", "content": "(p $$ \\vee $$ q) $$ \\vee $$ ( p $$ \\vee $$ ~ q)"}, {"identifier": "C", "content": "(p $$ \\wedge $$ q) $$ \\vee $$ ( p $$ \\wedge $$ ~ q)"}, {"identifier": "D", "content": "(p $$ \\vee $$ q) $$ \\wedge $$ ( p $$ \\vee $$ ~ q)"}] | ["B"] | null | (p $$ \vee $$ q) $$ \vee $$ (p $$ \vee $$ ~ q)<br><br>
= p $$ \vee $$ (q $$ \vee $$ p) $$ \vee $$ ~ q<br><br>
= (p $$ \vee $$ p) $$ \vee $$ (q $$ \vee $$ ~ q)<br><br>
= p $$ \vee $$ T<br><br>
= T so first statement is tautology | mcq | jee-main-2019-online-10th-april-morning-slot |
BEuYoTmWnNTkZjURIi18hoxe66ijvwvxo14 | maths | mathematical-reasoning | logical-connectives | If p $$ \Rightarrow $$ (q $$ \vee $$ r) is false, then the truth values of p,
q, r are respectively :- | [{"identifier": "A", "content": "F, F, F"}, {"identifier": "B", "content": "T, F, F"}, {"identifier": "C", "content": "F, T, T"}, {"identifier": "D", "content": "T, T, F"}] | ["B"] | null | p $$ \to $$ (q $$ \vee $$ r)
<br><br>= $$ \sim $$ p $$ \vee $$ (q $$ \vee $$ r)
<br><br>$$ \sim $$ p $$ \vee $$ (q $$ \vee $$ r) is false when
<br><br>(1) $$ \sim $$ p = False $$ \Rightarrow $$ P = True
<br><br>and (2) (q $$ \vee $$ r) is false $$ \Rightarrow $$ q and r both are false. | mcq | jee-main-2019-online-9th-april-evening-slot |
fjCe9jjDKWd14AIB6hdqt | maths | mathematical-reasoning | logical-connectives | For any two statements p and q, the negation of
the expression
<br/>p $$ \vee $$ (~p $$ \wedge $$ q) is : | [{"identifier": "A", "content": "p$$ \\leftrightarrow $$q"}, {"identifier": "B", "content": "~p$$ \\wedge $$~q"}, {"identifier": "C", "content": "p$$ \\wedge $$q"}, {"identifier": "D", "content": "~p$$ \\vee $$~q"}] | ["B"] | null | $$ \sim $$[p $$ \vee $$ (~p $$ \wedge $$ q)]
<br><br>= $$ \sim $$ p $$ \wedge $$ ($$ \sim $$($$ \sim $$ p $$ \vee $$ $$ \sim $$q)
<br><br>= $$ \sim $$ p $$ \wedge $$ ( p $$ \vee $$ $$ \sim $$q)
<br><br>= ( $$ \sim $$ p $$ \wedge $$ p ) $$ \vee $$ ($$ \sim $$ p $$ \wedge $$ $$ \sim $$q)
<br><br>= ($$ \sim $$ p $$ \wedge $$ $$ \sim $$q) [ as $$ \sim $$ p $$ \wedge $$ p is always false) | mcq | jee-main-2019-online-9th-april-morning-slot |
hY5nw44745zszMuFPd7ND | maths | mathematical-reasoning | logical-connectives | Which one of the following statements is not a
tautology? | [{"identifier": "A", "content": "(p $$ \\wedge $$ q) $$ \\to $$ (~ p) $$ \\vee $$ q"}, {"identifier": "B", "content": "(p $$ \\wedge $$ q) $$ \\to $$ p"}, {"identifier": "C", "content": "( p $$ \\vee $$ q) $$ \\to $$ ( p $$ \\vee $$ (~q))"}, {"identifier": "D", "content": "p $$ \\to $$ ( p $$ \\vee $$ q)"}] | ["C"] | null | Here you have to check all the options.
<br><br>We know, p $$ \to $$ q = $$ \sim $$ p $$ \vee $$ q
<br><br>(A) (p $$ \wedge $$ q) $$ \to $$ (~ p) $$ \vee $$ q
<br><br>= $$ \sim $$ (p $$ \wedge $$ q) $$ \vee $$ ((~ p) $$ \vee $$ q)
<br><br>= ($$ \sim $$ p $$ \vee $$ $$ \sim $$ q) $$ \vee $$ ((~ p) $$ \vee $$ q)
<br><br>= $$ \sim $$ p $$ \vee $$ (q $$ \vee $$ $$ \sim $$ q)
<br><br>= $$ \sim $$ p $$ \vee $$ t
<br><br>[(q $$ \vee $$ $$ \sim $$ q) = t (tautology)]
<br><br>= t
<br><br>(B) (p $$ \wedge $$ q) $$ \to $$ p
<br><br>= $$ \sim $$ (p $$ \wedge $$ q) $$ \vee $$ p
<br><br>= $$ \sim $$ p $$ \vee $$ $$ \sim $$ q $$ \vee $$ p
<br><br>= t $$ \vee $$ $$ \sim $$ q
<br><br>= t
<br><br>(C) ( p $$ \vee $$ q) $$ \to $$ ( p $$ \vee $$ (~q))
<br><br>= $$ \sim $$ ( p $$ \vee $$ q) $$ \vee $$ ( p $$ \vee $$ (~q))
<br><br>= ($$ \sim $$ p $$ \wedge $$ $$ \sim $$ q) $$ \vee $$ ( p $$ \vee $$ (~q))
<br><br>= $$ \sim $$ q $$ \wedge $$ (p $$ \vee $$ $$ \sim $$p)
<br><br>= $$ \sim $$ q $$ \wedge $$ t
<br><br>= $$ \sim $$ q (not tautology)
<br><br>(D) p $$ \to $$ ( p $$ \vee $$ q)
<br><br>= $$ \sim $$ p $$ \vee $$ ( p $$ \vee $$ q)
<br><br>= t $$ \vee $$ q
<br><br>= t | mcq | jee-main-2019-online-8th-april-evening-slot |
u0jxs6AIyc81iqYkTi5ac | maths | mathematical-reasoning | logical-connectives | The expression $$ \sim $$ ($$ \sim $$ p $$ \to $$ q) is logically equivalent to : | [{"identifier": "A", "content": "p $$ \\wedge $$ q"}, {"identifier": "B", "content": "$$ \\sim $$ p $$ \\wedge $$ $$ \\sim $$ q"}, {"identifier": "C", "content": "p $$ \\wedge $$ $$ \\sim $$ q"}, {"identifier": "D", "content": "$$ \\sim $$ p $$ \\wedge $$ q"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265973/exam_images/zxx1e4fhowtletn8zjkm.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Mathematical Reasoning Question 96 English Explanation"> | mcq | jee-main-2019-online-12th-january-evening-slot |
9p3XSt1mtpsZig0ryRz1S | maths | mathematical-reasoning | logical-connectives | The Boolean expression ((p $$ \wedge $$ q) $$ \vee $$ (p $$ \vee $$ $$ \sim $$ q)) $$ \wedge $$ ($$ \sim $$ p $$ \wedge $$ $$ \sim $$ q) is equivalent to : | [{"identifier": "A", "content": "p $$ \\wedge $$ q"}, {"identifier": "B", "content": "p $$ \\wedge $$ ($$ \\sim $$ q)"}, {"identifier": "C", "content": "p $$ \\vee $$ ($$ \\sim $$ q)"}, {"identifier": "D", "content": "($$ \\sim $$ p) $$ \\wedge $$ ($$ \\sim $$ q)"}] | ["D"] | null | ((p $$ \wedge $$ q) $$ \vee $$ (p $$ \vee $$ $$ \sim $$ q)) $$ \wedge $$ ($$ \sim $$ p $$ \wedge $$ $$ \sim $$ q)
<br><br>$$ \equiv $$ $$\left( {\left( {\left( {p \vee \left( {p \vee \sim q} \right)} \right)} \right) \wedge \left( {q \vee \left( {p \vee \sim q} \right)} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {\left( {p \vee \sim q} \right) \wedge \left( {q \vee \sim q \vee p} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {\left( {p \vee \sim q} \right) \wedge \left( {t \vee p} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {\left( {p \vee \sim q} \right) \wedge t} \right) \wedge \left( { \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {p \vee \sim q} \right) \wedge \left( { \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {p \wedge \sim p \wedge \sim q} \right) \vee \left( { \sim q \wedge \sim p \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$\left( {f \wedge \sim q} \right) \vee \left( { \sim q \wedge \sim p} \right)$$
<br><br>$$ \equiv $$ $$f' \vee \left( { \sim q \wedge \sim p} \right)$$
<br><br>$$ \equiv $$ $$\left( { \sim q \wedge \sim p} \right)$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
xwfoVcp6WOGDPwSwrCee3 | maths | mathematical-reasoning | logical-connectives | If q is false and p $$ \wedge $$ q $$ \leftrightarrow $$ r is true, then which one of the following statements is a tautology ? | [{"identifier": "A", "content": "P $$ \\wedge $$ r"}, {"identifier": "B", "content": "(p $$ \\vee $$ r) $$ \\to $$ (p $$ \\wedge $$ r)"}, {"identifier": "C", "content": "p $$ \\vee $$ r"}, {"identifier": "D", "content": "(p $$ \\wedge $$ r) $$ \\to $$ (p $$ \\vee $$ r)"}] | ["D"] | null | Given q is F and (p $$ \wedge $$ q) $$ \leftrightarrow $$ r is T
<br><br>$$ \Rightarrow $$ p $$ \wedge $$ q is F which implies that r is F
<br><br>$$ \Rightarrow $$ q is F and r is F
<br><br>$$ \Rightarrow $$ (p $$ \wedge $$ r) is always F
<br><br>$$ \Rightarrow $$ (p $$ \wedge $$ r) $$ \to $$ (p $$ \vee $$ r) is tautology. | mcq | jee-main-2019-online-11th-january-morning-slot |
SzEiL7YZ0iQAZgnQKyN4R | maths | mathematical-reasoning | logical-connectives | If the Boolean expression
<br/>(p $$ \oplus $$ q) $$\wedge$$ (~ p $$ \odot $$ q) is equivalent
<br/> to p $$\wedge$$ q, where $$ \oplus , \odot \in \left\{ { \wedge , \vee } \right\}$$, then the
<br/>ordered pair $$\left( { \oplus , \odot } \right)$$ is : | [{"identifier": "A", "content": "$$\\left( { \\vee , \\wedge } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { \\vee , \\vee } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { \\wedge , \\vee } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { \\wedge , \\wedge } \\right)$$"}] | ["C"] | null | <img src="https://gateclass.cdn.examgoal.net/qTCfGCsgnLJmurfJO/mTRBMelO4xBQhIbdNCj3SfmvgSIpr/yQ9Y64EkKKeaEzp8Qw7Tue/uploadfile.jpg" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Mathematical Reasoning Question 103 English Explanation">
<br><br>Given that, (p $$ \oplus $$ q) $$\wedge$$ (~ p $$ \odot $$ q) $$ \equiv $$ p $$\wedge$$ q.
<br><br>From the truth table you can see this equivalence only hold when ordered pair $$\left( { \oplus , \odot } \right)$$ is = $$\left( { \wedge , \vee } \right)$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
lOLfccfYozsLG7HBHYjgy2xukfurii1o | maths | mathematical-reasoning | logical-connectives | The negation of the Boolean expression p $$ \vee $$ (~p $$ \wedge $$ q) is equivalent to : | [{"identifier": "A", "content": "$$p \\wedge \\sim q$$"}, {"identifier": "B", "content": "$$ \\sim $$$$p \\vee \\sim q$$"}, {"identifier": "C", "content": "$$ \\sim p \\wedge q$$"}, {"identifier": "D", "content": "$$ \\sim p \\wedge \\sim q$$"}] | ["D"] | null | p $$ \vee $$ (~p $$ \wedge $$ q)
<br><br>= (p $$ \vee $$ ~p) $$ \wedge $$ (p $$ \vee $$ q)
<br><br>= (T) $$ \wedge $$ (p $$ \vee $$ q)
<br><br>= (p $$ \vee $$ q)
<br><br>Now negation of (p $$ \vee $$ q) is
<br><br>$$ \sim $$(p $$ \vee $$ q) = $$ \sim p \wedge \sim q$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
FrleKOpX9kV4yTWIC0jgy2xukfqcz1nw | maths | mathematical-reasoning | logical-connectives | The statement
<br/>$$\left( {p \to \left( {q \to p} \right)} \right) \to \left( {p \to \left( {p \vee q} \right)} \right)$$ is : | [{"identifier": "A", "content": "a tautology"}, {"identifier": "B", "content": "a contradiction"}, {"identifier": "C", "content": "equivalent to (p $$ \\vee $$ q) $$ \\wedge $$ ($$ \\sim $$ p)"}, {"identifier": "D", "content": "equivalent to (p $$ \\wedge $$ q) $$ \\vee $$ ($$ \\sim $$ q)"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267009/exam_images/ug8ifnvkt1ne5r5zuiz4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Mathematics - Mathematical Reasoning Question 74 English Explanation">
<br><br>So it is tautology. | mcq | jee-main-2020-online-5th-september-evening-slot |
x4gljq07Pu1OCLNlpwjgy2xukfjjl99r | maths | mathematical-reasoning | logical-connectives | The negation of the Boolean expression x $$ \leftrightarrow $$ ~ y is equivalent to : | [{"identifier": "A", "content": "$$\\left( {x \\wedge y} \\right) \\vee \\left( { \\sim x \\wedge \\sim y} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { \\sim x \\wedge y} \\right) \\vee \\left( { \\sim x \\wedge \\sim y} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {x \\wedge y} \\right) \\wedge \\left( { \\sim x \\vee \\sim y} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {x \\wedge \\sim y} \\right) \\vee \\left( { \\sim x \\wedge y} \\right)$$"}] | ["A"] | null | We know, p $$ \leftrightarrow $$ q $$ \equiv $$ $$\left( {p \to q} \right) \wedge \left( {q \to p} \right)$$
<br><br>$$ \therefore $$ x $$ \leftrightarrow $$ ~ y $$ \equiv $$ $$\left( {x \to \sim y} \right) \wedge \left( { \sim y \to x} \right)$$
<br><br>[ Also $$p \to q \equiv \sim p \vee q$$ ]
<br><br>$$ \Rightarrow $$ x $$ \leftrightarrow $$ ~ y $$ \equiv $$ $$\left( { \sim x \vee \sim y} \right) \wedge \left( {y \vee x} \right)$$
<br><br>$$ \therefore $$ $$ \sim \left( {x \leftrightarrow \sim y} \right) \equiv \left( {x \wedge y} \right) \vee \left( { \sim y \wedge \sim x} \right)$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
hySFpWbwwA3tcTNjAMjgy2xukf8zn82s | maths | mathematical-reasoning | logical-connectives | Given the following two statements:<br/><br/>
$$\left( {{S_1}} \right):\left( {q \vee p} \right) \to \left( {p \leftrightarrow \sim q} \right)$$ is a tautology<br/><br/>
$$\left( {{S_2}} \right): \,\,\sim q \wedge \left( { \sim p \leftrightarrow q} \right)$$ is a fallacy. Then: | [{"identifier": "A", "content": "both (S<sub>1</sub>) and (S<sub>2</sub>) are not correct"}, {"identifier": "B", "content": "only (S<sub>1</sub>) is correct"}, {"identifier": "C", "content": "only (S<sub>2</sub>) is correct"}, {"identifier": "D", "content": "both (S<sub>1</sub>) and (S<sub>2</sub>) are correct"}] | ["A"] | null | <b>Truth table for S<sub>1</sub> :
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<th class="tg-amwm">p</th>
<th class="tg-amwm">q</th>
<th class="tg-amwm">~q</th>
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<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
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<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
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<td class="tg-baqh">T</td>
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<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
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</tbody>
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<br>$$ \therefore $$ S<sub>1</sub>
is not correct.
<br><br><b>Truth table for S<sub>2</sub> :</b>
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<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
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<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
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<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
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<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
</tbody>
</table>
<br><br>$$ \therefore $$ S<sub>2</sub>
is also not correct. | mcq | jee-main-2020-online-4th-september-morning-slot |
nOq66cqiaklAbEp9Ztjgy2xukf3yrsm1 | maths | mathematical-reasoning | logical-connectives | Let p, q, r be three statements such that the
truth value of <br/>(p $$ \wedge $$ q) $$ \to $$ ($$ \sim $$q $$ \vee $$ r) is F. Then the
truth values of p, q, r are respectively : | [{"identifier": "A", "content": "T, F, T"}, {"identifier": "B", "content": "F, T, F"}, {"identifier": "C", "content": "T, T, T"}, {"identifier": "D", "content": "T, T, F"}] | ["D"] | null | Given, (p $$ \wedge $$ q) $$ \to $$ ($$ \sim $$q $$ \vee $$ r) is false.<br><br>This statement is false when<br><br>p $$ \wedge $$ q = T<br><br>and ($$ \sim $$q $$ \vee $$ r) = F<br><br>Now, p $$ \wedge $$ q = T when<br><br>both p and q are True.<br><br>As q = T<br><br>$$ \therefore $$ $$ \sim $$q = F<br><br>Now, ($$ \sim $$q $$ \vee $$ r) = F when<br><br>r = F as $$ \sim $$q is already false.<br><br>$$ \therefore $$ p = T<br><br>q = T<br><br>r = F | mcq | jee-main-2020-online-3rd-september-evening-slot |
FXaLJaXBdLHMOgP6Umjgy2xukf0yjac8 | maths | mathematical-reasoning | logical-connectives | The proposition p
$$ \to $$ ~ (p
$$ \wedge $$ ~q) is equivalent
to :
| [{"identifier": "A", "content": "($$ \\sim $$p) $$ \\vee $$ q"}, {"identifier": "B", "content": "q"}, {"identifier": "C", "content": "($$ \\sim $$p) $$ \\wedge $$ q"}, {"identifier": "D", "content": "($$ \\sim $$p) $$ \\vee $$ ($$ \\sim $$q)"}] | ["A"] | null | $$p \to \, \sim (p \wedge \sim q)$$<br><br>$$ \equiv p \to ( \sim p \vee q)$$<br><br>$$ \equiv \, \sim p \vee ( \sim p \vee q)$$<br><br>$$ \equiv \, \sim p \vee q$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
nFlqKR4iPQURZOKVmP7k9k2k5khoedh | maths | mathematical-reasoning | logical-connectives | If p $$ \to $$ (p $$ \wedge $$ ~q) is false, then the truth values
of p and q are respectively : | [{"identifier": "A", "content": "T, T"}, {"identifier": "B", "content": "T, F"}, {"identifier": "C", "content": "F, T"}, {"identifier": "D", "content": "F, F"}] | ["A"] | null | p $$ \to $$ (p $$ \wedge $$ ~q) will be false only when p is true and (p $$ \wedge $$ ~q) is false.
<br><br>So, p = T, q = T | mcq | jee-main-2020-online-9th-january-evening-slot |
hAUxfzeSOEKdF4vGZL7k9k2k5hizmsn | maths | mathematical-reasoning | logical-connectives | Which of the following statements is a tautology? | [{"identifier": "A", "content": "~(p $$ \\wedge $$ ~q) $$ \\to $$ p $$ \\vee $$ q"}, {"identifier": "B", "content": "~(p $$ \\vee $$ ~q) $$ \\to $$ p $$ \\vee $$ q"}, {"identifier": "C", "content": "~(p $$ \\vee $$ ~q) $$ \\to $$ p $$ \\wedge $$ q"}, {"identifier": "D", "content": "p $$ \\vee $$ (~q) $$ \\to $$ p $$ \\wedge $$ q"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265134/exam_images/emajs7ltkurmpe8hmtbv.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263920/exam_images/d5mbdvxcpol7gfsdgwd3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Evening Slot Mathematics - Mathematical Reasoning Question 84 English Explanation"></picture> | mcq | jee-main-2020-online-8th-january-evening-slot |
eNuZyyW4HZ3ebDB5rx7k9k2k5gs3o9i | maths | mathematical-reasoning | logical-connectives | Which one of the following is a tautology? | [{"identifier": "A", "content": "P $$ \\wedge $$ (P $$ \\vee $$ Q)"}, {"identifier": "B", "content": "P $$ \\vee $$ (P $$ \\wedge $$ Q)"}, {"identifier": "C", "content": "Q $$ \\to $$ (P $$ \\wedge $$ (P $$ \\to $$ Q))"}, {"identifier": "D", "content": "(P $$ \\wedge $$ (P $$ \\to $$ Q)) $$ \\to $$ Q"}] | ["D"] | null | <b>Option A :</b>
<br><br><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-wa1i{font-weight:bold;text-align:center;vertical-align:middle}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-wa1i">P</th>
<th class="tg-amwm">Q</th>
<th class="tg-amwm">P $$ \vee $$ Q</th>
<th class="tg-amwm">P $$ \wedge $$ (P $$ \vee $$ Q)</th>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
</tbody></table>
<br><br><b>Option B :</b>
<br><br><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-wa1i{font-weight:bold;text-align:center;vertical-align:middle}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-wa1i">P</th>
<th class="tg-amwm">Q</th>
<th class="tg-amwm">P $$ \wedge $$ Q</th>
<th class="tg-amwm">P $$ \vee $$ (P $$ \wedge $$ Q)</th>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
</tbody></table>
<br><br><b>Option C :</b>
<br><br>Q $$ \to $$ (P $$ \wedge $$ (P $$ \to $$ Q))
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ (P $$ \wedge $$ (P $$ \to $$ Q))
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ (P $$ \wedge $$ (~P $$ \vee $$ Q))
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ (P $$ \wedge $$ ~P) $$ \vee $$ (P $$ \wedge $$ Q))
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ (ƒ $$ \vee $$ (P $$ \wedge $$ Q))
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ (P $$ \wedge$$ Q)
<br><br>$$ \equiv $$ (~Q $$ \vee $$ P) $$ \wedge $$ (~Q $$ \vee $$ Q)
<br><br>$$ \equiv $$ (~ Q $$ \vee $$ P) $$ \wedge $$ t
<br><br>$$ \equiv $$ ~ Q $$ \vee $$ P
<br><br><b>Option D :</b>
<br><br>(P $$ \wedge $$ (P $$ \to $$ Q)) $$ \to $$ Q
<br><br>$$ \equiv $$ (P $$ \wedge $$ (~P $$ \vee $$ Q)) $$ \to $$ Q
<br><br>$$ \equiv $$ ((P $$ \wedge $$ ~P) $$ \vee $$ (P $$ \wedge $$ Q)) $$ \to $$ Q
<br><br>$$ \equiv $$ (ƒ $$ \vee $$ (P $$ \wedge $$ Q)) $$ \to $$ Q
<br><br>$$ \equiv $$ P $$ \wedge $$ Q $$ \to $$ Q
<br><br>$$ \equiv $$ ~ (P $$ \wedge $$ Q) $$ \vee $$ Q
<br><br>$$ \equiv $$ (~ P) $$ \vee $$ (~Q) $$ \vee $$ Q
<br><br>$$ \equiv $$ (~ P) $$ \vee $$ t
<br><br>$$ \equiv $$ t
| mcq | jee-main-2020-online-8th-january-morning-slot |
9I2l8OlGVOosJWMkWO7k9k2k5fmm2cv | maths | mathematical-reasoning | logical-connectives | Let A, B, C and D be four non-empty sets. The contrapositive statement of "If A $$ \subseteq $$ B and B $$ \subseteq $$ D,
then A $$ \subseteq $$ C" is : | [{"identifier": "A", "content": "If A \u2288 C, then A \u2288 B or B \u2288 D"}, {"identifier": "B", "content": "If A \u2288 C, then A \u2288 B and B $$ \\subseteq $$ D"}, {"identifier": "C", "content": "If A $$ \\subseteq $$ C, then B $$ \\subset $$ A or D $$ \\subset $$ B"}, {"identifier": "D", "content": "If A \u2288 C, then A $$ \\subseteq $$ B and B $$ \\subseteq $$ D"}] | ["A"] | null | Contrapositive of p $$ \to $$ q is ~q $$ \to $$ ~p
<br><br>If A $$ \subseteq $$ B and B $$ \subseteq $$ D,
then A $$ \subseteq $$ C means
<br><br>(A $$ \subseteq $$ B) $$ \wedge $$ (B $$ \subseteq $$ D) $$ \to $$ (A $$ \subseteq $$ C)
<br><br>Contrapositive is
<br><br>$$ \sim $$(A $$ \subseteq $$ C) $$ \to $$ $$ \sim $$(A $$ \subseteq $$ B) $$ \vee $$ $$ \sim $$(B $$ \subseteq $$ D)
<br><br>$$ \Rightarrow $$ A ⊈ C $$ \to $$ (A ⊈ B) $$ \vee $$ (B ⊈ D) | mcq | jee-main-2020-online-7th-january-evening-slot |
L5tV8VTiGQ5Joiaa5M7k9k2k5e4j5it | maths | mathematical-reasoning | logical-connectives | The logical statement (p $$ \Rightarrow $$ q) $$\Lambda $$ ( q $$ \Rightarrow $$ ~p) is equivalent to : | [{"identifier": "A", "content": "q"}, {"identifier": "B", "content": "$$ \\sim $$p"}, {"identifier": "C", "content": "p"}, {"identifier": "D", "content": "$$ \\sim $$q"}] | ["B"] | null | (p $$ \Rightarrow $$ q) $$\Lambda $$ ( q $$ \Rightarrow $$ ~p)
<br><br>$$ \equiv $$ $$\left( { \sim p \vee q} \right) \wedge \left( { \sim q \vee \sim p} \right)$$
<br><br>$$ \equiv $$ $$ \sim p \vee \left( {q \wedge \sim q} \right)$$
<br><br>$$ \equiv $$ $$ \sim $$p
<br><br>As $${q \wedge \sim q}$$ is a fallacy. | mcq | jee-main-2020-online-7th-january-morning-slot |
rkDqKdV7f4jhoxxZQhjgy2xukezbqrn0 | maths | mathematical-reasoning | logical-connectives | Which of the following is a tautology ? | [{"identifier": "A", "content": "$$\\left( { \\sim p} \\right) \\wedge \\left( {p \\vee q} \\right) \\to q$$"}, {"identifier": "B", "content": "$$\\left( {q \\to p} \\right) \\vee \\sim \\left( {p \\to q} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {p \\to q} \\right) \\wedge \\left( {q \\to p} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { \\sim q} \\right) \\vee \\left( {p \\wedge q} \\right) \\to q$$"}] | ["A"] | null | ~ p $$ \wedge $$ (p $$ \vee $$ q) $$ \to $$ q<br><br>
$$ \equiv $$ (~ p $$ \wedge $$ p) $$ \vee $$ (~ p $$ \wedge $$ q) $$ \to $$ q<br><br>
$$ \equiv $$ C $$ \vee $$ (~ p $$ \wedge $$ q) $$ \to $$ q [C = contradiction]<br><br>
$$ \equiv $$ (~ p $$ \wedge $$ q) $$ \to $$ q<br><br>
$$ \equiv $$ ~ (~ p $$ \wedge $$ q) $$ \vee $$ q<br><br>
$$ \equiv $$ (p $$ \vee $$ ~ q) $$ \vee $$ q<br><br>
$$ \equiv $$ (p $$ \vee $$ q) $$ \vee $$ (~ q $$ \vee $$ q)<br><br>
$$ \equiv $$ (p $$ \vee$$ q) $$ \vee $$ t<br><br>
so ~ p $$ \wedge $$ (p $$ \vee $$ q) $$ \to $$ q is a tautology | mcq | jee-main-2020-online-2nd-september-evening-slot |
e8m7fiFp2ii9uhwxX41klrhnfcs | maths | mathematical-reasoning | logical-connectives | The statement among the following that is a tautology is : | [{"identifier": "A", "content": "$$B \\to \\left[ {A \\wedge \\left( {A \\to B} \\right)} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {A \\wedge \\left( {A \\to B} \\right)} \\right] \\to B$$"}, {"identifier": "C", "content": "$$\\left[ {A \\wedge \\left( {A \\vee B} \\right)} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {A \\vee \\left( {A \\wedge B} \\right)} \\right]$$"}] | ["B"] | null | Given, $$\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$$
<br/><br/>= $$A \wedge \left( { \sim A \vee B} \right) \to B$$
<br/><br/>= $$\left[ {\left( {A \wedge \sim A} \right) \vee \left( {A \wedge B} \right)} \right] \to B$$
<br/><br/>= $$\left( {A \wedge B} \right) \to B$$
<br/><br/>= $${ \sim A \vee \sim B \vee B}$$
<br/><br/>= t
<br/><br/>Hence, $$\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$$ is a tautology. | mcq | jee-main-2021-online-24th-february-morning-slot |
ganiqHSEGa3zsqkGqV1klrl43pu | maths | mathematical-reasoning | logical-connectives | The negation of the statement <br/><br/>$$ \sim p \wedge (p \vee q)$$ is : | [{"identifier": "A", "content": "$$p \\vee \\sim q$$"}, {"identifier": "B", "content": "$$ \\sim p \\vee q$$"}, {"identifier": "C", "content": "$$ \\sim p \\wedge q$$"}, {"identifier": "D", "content": "$$p \\wedge \\sim q$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264386/exam_images/pkgg8abdxwnuxm3mksuq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Mathematical Reasoning Question 70 English Explanation">
<br>$$ \therefore $$ $$ \sim $$ p $$ \wedge $$ (p $$ \vee $$ q) $$ \equiv $$ p $$ \vee $$ $$ \sim $$ q | mcq | jee-main-2021-online-24th-february-evening-slot |
Y4dstDxJfkHb7DfLzh1klrl963n | maths | mathematical-reasoning | logical-connectives | For the statements p and q, consider the following compound statements :<br/><br/>(a) $$( \sim q \wedge (p \to q)) \to \sim p$$<br/><br/>(b) $$((p \vee q) \wedge \sim p) \to q$$<br/><br/>Then which of the following statements is correct? | [{"identifier": "A", "content": "(b) is a tautology but not (a)."}, {"identifier": "B", "content": "(a) and (b) both are not tautologies."}, {"identifier": "C", "content": "(a) and (b) both are tautologies."}, {"identifier": "D", "content": "(a) is a tautology but not (b)."}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263768/exam_images/zyaqwun6l1vaxc3vzpc7.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264303/exam_images/pe5z4lubb8xitalkkhvu.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266697/exam_images/hyq0krwnkwrcah7t3agk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Mathematical Reasoning Question 69 English Explanation"></picture>
<br>(b) is tautologies<br><br>$$ \therefore $$ a & b are both tautologies. | mcq | jee-main-2021-online-24th-february-evening-slot |
UiyDtnS5Vrk1mJy1Gf1kls47g1s | maths | mathematical-reasoning | logical-connectives | The statement A $$ \to $$ (B $$ \to $$ A) is equivalent to : | [{"identifier": "A", "content": "A $$ \\to $$ (A $$\\mathrel{\\mathop{\\kern0pt\\longleftrightarrow}\n\\limits_{}} $$ B)"}, {"identifier": "B", "content": "A $$ \\to $$ (A $$ \\vee $$ B)"}, {"identifier": "C", "content": "A $$ \\to $$ (A $$ \\wedge $$ B)"}, {"identifier": "D", "content": "A $$ \\to $$ (A $$ \\to $$ B)"}] | ["B"] | null | $$A \to (B \to A)$$<br><br>$$ \Rightarrow A \to ( \sim B \vee A)$$<br><br>$$ \Rightarrow \, \sim A \vee ( \sim B \vee A)$$<br><br>$$ \Rightarrow \, \sim B \vee ( \sim A \vee A)$$<br><br>$$ \Rightarrow \, \sim B \vee t$$<br><br>= t (tantology)<br><br>From options :<br><br>(B) $$A \to (A \vee B)$$<br><br>$$ \Rightarrow \, \sim A \vee (A \vee B)$$<br><br>$$ \Rightarrow ( \sim A \vee A) \vee B$$<br><br>$$ \Rightarrow t \vee B$$<br><br>$$ \Rightarrow t$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
fPu8c7tDybZ3josxp31kluvu5ch | maths | mathematical-reasoning | logical-connectives | Let F<sub>1</sub>(A, B, C) = (A $$ \wedge $$ $$ \sim $$ B) $$ \vee $$ [$$\sim$$C $$\wedge$$ (A $$\vee$$ B)] $$\vee$$ $$\sim$$ A and <br/>F<sub>2</sub>(A, B) = (A $$\vee$$ B) $$\vee$$ (B $$ \to $$ $$\sim$$A) be two logical expressions. Then : | [{"identifier": "A", "content": "Both F<sub>1</sub> and F<sub>2</sub> are not tautologies"}, {"identifier": "B", "content": "F<sub>1</sub> and F<sub>2</sub> both are tautologies"}, {"identifier": "C", "content": "F<sub>1</sub> is not a tautology but F<sub>2</sub> is a tautology"}, {"identifier": "D", "content": "F<sub>1</sub> is a tautology but F<sub>2</sub> is not a tautology"}] | ["C"] | null | Truth table for F<sub>1</sub> :
<br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265217/exam_images/nh8hwgezkfdnwxzdht25.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265361/exam_images/y6kerhcssgaioyopzx7w.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267047/exam_images/qymdxeesnhkcyrbboufz.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265554/exam_images/lygyaa4jt0s21gxmkcpu.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263451/exam_images/xunl3mwezb46aboycqqp.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267765/exam_images/svrwznt0tkoqxjzevdb1.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264465/exam_images/uisrv0kdfwaxgjmqyyrp.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265076/exam_images/btgpz9omvap1aqidueis.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265163/exam_images/vluwfg75psmgucvf09o8.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267518/exam_images/h1zyjcoqy8xi3yytdif8.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Mathematical Reasoning Question 66 English Explanation 1"></picture>
$$ \therefore $$ F<sub>1</sub> is not a tautology.
<br><br>Truth table for F<sub>2</sub> :
<br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267759/exam_images/ouiidbgpcv8np7qejwco.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265234/exam_images/damobviwczkf76yqzjt6.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264678/exam_images/plyjbuq6g3z6rtk0rmdy.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263998/exam_images/nolswctxgfhwlcc3cjwf.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265436/exam_images/q68zp1j4dgbplvb7valw.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267330/exam_images/dqfhydblnh3cypugufxd.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264972/exam_images/rm6uq5k8twxcdxfuujgh.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266138/exam_images/yohiugp51qucizixpg7h.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263751/exam_images/jqhu9mqnbozimvjnixpz.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265647/exam_images/ofzxonxbghnfr4htgcmr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Mathematical Reasoning Question 66 English Explanation 2"></picture>
<br> $$ \therefore $$ F<sub>2</sub> is a tautology. | mcq | jee-main-2021-online-26th-february-evening-slot |
1p2dl8GpcxkStpiRZt1kmhvyzos | maths | mathematical-reasoning | logical-connectives | Which of the following Boolean expression is a tautology? | [{"identifier": "A", "content": "(p $$ \\wedge $$ q) $$ \\vee $$ (p $$ \\to $$ q)"}, {"identifier": "B", "content": "(p $$ \\wedge $$ q) $$ \\vee $$ (p $$\\vee$$ q)"}, {"identifier": "C", "content": "(p $$ \\wedge $$ q) $$ \\to $$ (p $$ \\to $$ q)"}, {"identifier": "D", "content": "(p $$ \\wedge $$ q) $$ \\wedge $$ (p $$ \\to $$ q)"}] | ["C"] | null | $$\matrix{
p & q & {p \wedge q} & {p \vee q} & {p \to q} & {(p \wedge q) \to (p \to q)} \cr
T & T & T & T & T & T \cr
F & T & F & T & T & T \cr
T & F & F & T & F & T \cr
F & F & F & F & T & T \cr
} $$ | mcq | jee-main-2021-online-16th-march-morning-shift |
8hZxyvOr3nEBB8fFxy1kmjapolf | maths | mathematical-reasoning | logical-connectives | If the Boolean expression (p $$ \Rightarrow $$ q) $$ \Leftrightarrow $$ (q * ($$ \sim $$p) is a tautology, then the boolean expression (p * ($$ \sim $$q)) is equivalent to : | [{"identifier": "A", "content": "q $$ \\Rightarrow $$ p"}, {"identifier": "B", "content": "p $$ \\Rightarrow $$ q"}, {"identifier": "C", "content": "p $$ \\Rightarrow $$ $$ \\sim $$ q"}, {"identifier": "D", "content": "$$ \\sim $$q $$ \\Rightarrow $$ p"}] | ["A"] | null | <p>The Boolean expression</p>
<p>$(p \Rightarrow q) \Leftrightarrow\left(q^*(\sim p)\right)$ is a tautology.</p>
<p>Making the truth table for this</p>
<style type="text/css">
.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}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-c3ow">p</th>
<th class="tg-c3ow">q</th>
<th class="tg-c3ow">$p \rightarrow q$</th>
<th class="tg-c3ow">$q^* \sim p$</th>
<th class="tg-c3ow">$\sim q$</th>
<th class="tg-c3ow">$-q \wedge p$</th>
<th class="tg-c3ow">$\sim(\sim {q} \wedge p)$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
</tr>
<tr>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
</tr>
<tr>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
</tr>
</tbody>
</table><br/>
<p>$\therefore \quad \sim(\sim q \wedge p)=q \vee \sim p=\sim p \vee q$</p>
<p>$\therefore{ }^*$ is equivalent to $v$.</p>
<p>So, $\quad p^* \sim q=p \vee \sim q=\sim q \vee p$</p>
<p>$=q \Rightarrow p$</p> | mcq | jee-main-2021-online-17th-march-morning-shift |
fbDsZRHROBb4k6QKaY1kmkmyoqc | maths | mathematical-reasoning | logical-connectives | If the Boolean expression $$(p \wedge q) \odot (p \otimes q)$$ is a tautology, then $$ \odot $$ and $$ \otimes $$ are respectively given by : | [{"identifier": "A", "content": "$$ \\vee , \\to $$"}, {"identifier": "B", "content": "$$ \\to $$, $$ \\to $$"}, {"identifier": "C", "content": "$$ \\wedge $$, $$ \\vee $$"}, {"identifier": "D", "content": "$$ \\wedge $$, $$ \\to $$"}] | ["B"] | null | $$(p \wedge q)\, \to \,(p \to q)$$<br><br>$$(p \wedge q)\, \to \,( \sim p \vee q)$$<br><br>$$( \sim p \vee \sim q)\, \vee ( \sim p \vee q)$$<br><br>$$ \sim p \vee ( \sim q \vee q) \Rightarrow $$ Tautology<br><br>$$ \Rightarrow \odot \Rightarrow \to $$<br><br>$$ \otimes \Rightarrow \to $$ | mcq | jee-main-2021-online-17th-march-evening-shift |
nzAOiYmH4hkTFVNC2R1kmm3sgyz | maths | mathematical-reasoning | logical-connectives | If P and Q are two statements, then which of the following compound statement is a tautology? | [{"identifier": "A", "content": "((P $$ \\Rightarrow $$ Q) $$ \\wedge $$ $$ \\sim $$ Q) $$ \\Rightarrow $$ (P $$ \\wedge $$ Q)"}, {"identifier": "B", "content": "((P $$ \\Rightarrow $$ Q) $$ \\wedge $$ $$ \\sim $$ Q) $$ \\Rightarrow $$ Q"}, {"identifier": "C", "content": "((P $$ \\Rightarrow $$ Q) $$ \\wedge $$ $$ \\sim $$ Q) $$ \\Rightarrow $$ P"}, {"identifier": "D", "content": "((P $$ \\Rightarrow $$ Q) $$ \\wedge $$ $$ \\sim $$ Q) $$ \\Rightarrow $$ $$ \\sim $$ P"}] | ["D"] | null | <p>LHS of all the options are same i.e.</p>
<p>$$((P \to Q) \wedge \sim Q)$$</p>
<p>$$ \equiv ( \sim P \vee Q) \wedge \sim Q$$</p>
<p>$$ \equiv ( \sim P \wedge \sim Q) \vee (Q \wedge \sim Q)$$</p>
<p>$$ \equiv \sim P \wedge \sim Q$$</p>
<p>(A) $$( \sim P \wedge \sim Q) \to Q$$</p>
<p>$$ \equiv \sim ( \sim P \wedge \sim Q) \vee Q$$</p>
<p>$$ \equiv (P \vee Q) \vee Q \ne $$ Tautology</p>
<p>(B) $$( \sim P \wedge \sim Q) \to P$$</p>
<p>$$ \equiv \sim ( \sim P \wedge \sim Q) \vee \sim P$$</p>
<p>$$ \equiv (P \vee Q) \vee \sim P$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0mtai/207863b7-e6a3-4fcf-9b41-05c186319670/5daac9b0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0mtaj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0mtai/207863b7-e6a3-4fcf-9b41-05c186319670/5daac9b0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0mtaj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Mathematical Reasoning Question 62 English Explanation"></p>
<p>(C) $$( \sim P \wedge \sim Q) \to P$$</p>
<p>$$ \equiv (P \wedge Q) \wedge P \ne $$ Tautology</p>
<p>(D) $$( \sim P \wedge \sim Q) \to (P \vee Q)$$</p>
<p>$$ \equiv (P \wedge Q) \wedge (P \vee Q) \ne $$ Tautology</p> | mcq | jee-main-2021-online-18th-march-evening-shift |
1krpripml | maths | mathematical-reasoning | logical-connectives | The Boolean expression $$(p \wedge \sim q) \Rightarrow (q \vee \sim p)$$ is equivalent to : | [{"identifier": "A", "content": "$$q \\Rightarrow p$$"}, {"identifier": "B", "content": "$$p \\Rightarrow q$$"}, {"identifier": "C", "content": "$$ \\sim q \\Rightarrow p$$"}, {"identifier": "D", "content": "$$p \\Rightarrow \\, \\sim q$$"}] | ["B"] | null | <table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim p$$</th>
<th class="tg-baqh">$$ \sim q$$</th>
<th class="tg-baqh">$$p \wedge \sim q$$</th>
<th class="tg-baqh">$$q \vee \sim p$$</th>
<th class="tg-baqh">$$(p \wedge \sim q) \Rightarrow (q \vee \sim p)$$</th>
<th class="tg-baqh">$$p \Rightarrow q$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table><br><br>$$\therefore$$ $$(p \wedge \sim q) \Rightarrow (q \vee \sim p)$$<br><br>$$ \equiv p \Rightarrow q$$<br><br>So, option (2) is correct | mcq | jee-main-2021-online-20th-july-morning-shift |
1kru3ommn | maths | mathematical-reasoning | logical-connectives | Which of the following Boolean expressions is not a tautology? | [{"identifier": "A", "content": "(p $$\\Rightarrow$$ q) $$ \\vee $$ ($$ \\sim $$ q $$\\Rightarrow$$ p)"}, {"identifier": "B", "content": "(q $$\\Rightarrow$$ p) $$ \\vee $$ ($$ \\sim $$ q $$\\Rightarrow$$ p)"}, {"identifier": "C", "content": "(p $$\\Rightarrow$$ $$ \\sim $$ q) $$ \\vee $$ ($$ \\sim $$ q $$\\Rightarrow$$ p)"}, {"identifier": "D", "content": "($$ \\sim $$ p $$\\Rightarrow$$ q) $$ \\vee $$ ($$\\sim$$ q $$\\Rightarrow$$ p)"}] | ["D"] | null | (1) (p $$\to$$ q) $$\vee$$ ($$\sim$$ q $$\to$$ p)<br><br>= ($$\sim$$ p $$\vee$$ q) $$\vee$$ (q $$\vee$$ p)<br><br>= ($$\sim$$ p $$\vee$$ p) $$\vee$$ q<br><br>= t $$\vee$$ q = t<br><br>(2) (q $$\to$$ p) $$\vee$$ ($$\sim$$ q $$\to$$ p)<br><br>= ($$\sim$$ q $$\vee$$ p) $$\vee$$ (q $$\vee$$ p)<br><br>= ($$\sim$$ q $$\vee$$ q) $$\vee$$ p<br><br>= t $$\vee$$ p = t<br><br>(3) (p $$\to$$ $$\sim$$ q) $$\vee$$ ($$\sim$$ q $$\to$$ p)<br><br>= ($$\sim$$ p $$\vee$$ $$\sim$$ p) $$\vee$$ (q $$\vee$$ p)<br><br>= ($$\sim$$ p $$\vee$$ p) $$\vee$$ ($$\sim$$ q $$\vee$$ q)<br><br>= t $$\vee$$ t = t<br><br>(4) ($$\sim$$ q $$\to$$ q) $$\vee$$ ($$\sim$$ q $$\to$$ p)<br><br>= (p $$\vee$$ q) $$\vee$$ (q $$\vee$$ p)<br><br>= (p $$\vee$$ p) $$\vee$$ (q $$\vee$$ p)<br><br>= p $$\vee$$ q<br><br>Which is not a tautology. | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvyssvv | maths | mathematical-reasoning | logical-connectives | The Boolean expression $$(p \Rightarrow q) \wedge (q \Rightarrow \sim p)$$ is equivalent to : | [{"identifier": "A", "content": "$$ \\sim $$ q"}, {"identifier": "B", "content": "q"}, {"identifier": "C", "content": "p"}, {"identifier": "D", "content": "$$ \\sim $$ p"}] | ["D"] | null | $$(p \to q) \wedge (q \to \sim p)$$<br><br>$$ \equiv ( \sim p \vee q) \wedge ( \sim q \vee \sim p)\{ p \to q \equiv \sim p \vee q\} $$<br><br>$$ \equiv ( \sim p \vee q) \wedge ( \sim p \vee q)$$ {commutative property}<br><br>$$ \equiv \sim p \vee (q \wedge \sim q)$$ {distributive property}<br><br>$$ \equiv \sim p$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1ks07sngu | maths | mathematical-reasoning | logical-connectives | The compound statement $$(P \vee Q) \wedge ( \sim P) \Rightarrow Q$$ is equivalent to : | [{"identifier": "A", "content": "$$P \\vee Q$$"}, {"identifier": "B", "content": "$$P \\wedge \\sim Q$$"}, {"identifier": "C", "content": "$$ \\sim (P \\Rightarrow Q)$$"}, {"identifier": "D", "content": "$$ \\sim (P \\Rightarrow Q) \\Leftrightarrow P \\wedge \\sim Q$$"}] | ["D"] | null | Using Truth Table :<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264245/exam_images/uemdlbsr1t66wwcno3ny.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266789/exam_images/jobctzefoxr8ftpbzqtr.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267081/exam_images/ktfohnrwpak25onaf06y.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266046/exam_images/d18zz8tkcr3s2bkasyjb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Mathematical Reasoning Question 55 English Explanation"></picture> | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbgkdhv | maths | mathematical-reasoning | logical-connectives | If the truth value of the Boolean expression $$\left( {\left( {p \vee q} \right) \wedge \left( {q \to r} \right) \wedge \left( { \sim r} \right)} \right) \to \left( {p \wedge q} \right)$$ is false, then the truth values of the statements p, q, r respectively can be : | [{"identifier": "A", "content": "T F T"}, {"identifier": "B", "content": "F F T"}, {"identifier": "C", "content": "T F F "}, {"identifier": "D", "content": "F T F"}] | ["C"] | null | <table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">r</th>
<th class="tg-baqh">$$\underbrace {p \vee q}_a$$</th>
<th class="tg-baqh">$$\underbrace {q \to r}_b$$</th>
<th class="tg-baqh">$${a \wedge b}$$</th>
<th class="tg-baqh">$${ \sim r}$$</th>
<th class="tg-baqh">$$\underbrace {a \wedge b \wedge ( \sim r)}_c$$</th>
<th class="tg-baqh">$$\underbrace {p \wedge q}_d$$</th>
<th class="tg-baqh">$$c \to d$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table> | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktcznobe | maths | mathematical-reasoning | logical-connectives | Consider the two statements :<br/><br/>(S1) : (p $$\to$$ q) $$ \vee $$ ($$ \sim $$ q $$\to$$ p) is a tautology .<br/><br/>(S2) : (p $$ \wedge $$ $$ \sim $$ q) $$ \wedge $$ ($$\sim$$ p $$\wedge$$ q) is a fallacy.<br/><br/>Then : | [{"identifier": "A", "content": "only (S1) is true."}, {"identifier": "B", "content": "both (S1) and (S2) are false."}, {"identifier": "C", "content": "both (S1) and (S2) are true."}, {"identifier": "D", "content": "only (S2) is true."}] | ["C"] | null | S<sub>1</sub> : ($$\sim$$ p $$ \vee $$ q) $$ \vee $$ (q $$ \vee $$ p) = (q $$ \vee $$ $$\sim$$ p) $$ \vee $$ (q $$ \vee $$ p)<br><br>S<sub>1</sub> = q $$ \vee $$ ($$\sim$$ p $$ \vee $$ p) = qvt = t = tautology<br><br>S<sub>2</sub> : (p $$ \wedge $$ $$\sim$$ q) $$ \wedge $$ ($$\sim$$ p $$\vee$$ q) = (p $$ \wedge $$ $$\sim$$ q) $$ \wedge $$ $$\sim$$ (p $$ \wedge $$ $$\sim$$ q) = C = fallacy | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktekw6qp | maths | mathematical-reasoning | logical-connectives | The statement (p $$ \wedge $$ (p $$\to$$ q) $$\wedge$$ (q $$\to$$ r)) $$\to$$ r is : | [{"identifier": "A", "content": "a tautology"}, {"identifier": "B", "content": "equivalent to p $$\\to$$ $$\\sim$$ r"}, {"identifier": "C", "content": "a fallacy"}, {"identifier": "D", "content": "equivalent to q $$\\to$$ $$\\sim$$ r"}] | ["A"] | null | (p $$ \wedge $$ (p $$\to$$ q) $$\wedge$$ (q $$\to$$ r)) $$\to$$ r<br><br>$$\equiv$$ (p $$\wedge$$ ($$\sim$$ p $$\vee$$ q) $$\vee$$ ($$\sim$$ q $$\vee$$ r)) $$\to$$ r<br><br>$$\equiv$$ ((p $$\wedge$$ q) $$\wedge$$ ($$\sim$$ p $$\vee$$ r)) $$\to$$ r<br><br>$$\equiv$$ (p $$\wedge$$ q $$\wedge$$ r) $$\to$$ r<br><br>$$\equiv$$ $$\sim$$ (p $$\wedge$$ q $$\wedge$$ r) $$\vee$$ r<br><br>$$\equiv$$ ($$\sim$$ p) $$\vee$$ ($$\sim$$ q) $$\vee$$ ($$\sim$$ r) $$\vee$$ r<br><br>$$\Rightarrow$$ tautology | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktg2lye2 | maths | mathematical-reasoning | logical-connectives | The Boolean expression (p $$\wedge$$ q) $$\Rightarrow$$ ((r $$\wedge$$ q) $$\wedge$$ p) is equivalent to : | [{"identifier": "A", "content": "(p $$\\wedge$$ q) $$\\Rightarrow$$ (r $$\\wedge$$ q)"}, {"identifier": "B", "content": "(q $$\\wedge$$ r) $$\\Rightarrow$$ (p $$\\wedge$$ q)"}, {"identifier": "C", "content": "(p $$\\wedge$$ q) $$\\Rightarrow$$ (r $$\\vee$$ q)"}, {"identifier": "D", "content": "(p $$\\wedge$$ r) $$\\Rightarrow$$ (p $$\\wedge$$ q)"}] | ["A"] | null | given statement says<br><br>"if p and q both happen then p and q and r will happen"<br><br>it simply implies "If p and q both happen then 'r' too will happen"<br><br>i.e.<br><br>"if p and q both happen then r and p too will happen<br><br>i.e.<br><br>(p $$\wedge$$ q) $$\Rightarrow$$ (r $$\wedge$$ p) | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktio2w8d | maths | mathematical-reasoning | logical-connectives | Let *, ▢ $$\in$${$$\wedge$$, $$\vee$$} be such that the Boolean expression (p * $$\sim$$ q) $$\Rightarrow$$ (p ▢ q) is a tautology. Then : | [{"identifier": "A", "content": "* = $$\\vee$$, \u25a2 = $$\\vee$$"}, {"identifier": "B", "content": "* = $$\\wedge$$, \u25a2 = $$\\wedge$$"}, {"identifier": "C", "content": "* = $$\\wedge$$, \u25a2 = $$\\vee$$"}, {"identifier": "D", "content": "* = $$\\vee$$, \u25a2 = $$\\wedge$$"}] | ["C"] | null | (p $$\wedge$$ $$\sim$$ q) $$\to$$ (p $$\vee$$ q) is tautology<br><br><style type="text/css">
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.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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<table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim $$ q</th>
<th class="tg-baqh">p $$ \wedge $$ $$ \sim $$ q</th>
<th class="tg-baqh">p $$ \vee $$ q</th>
<th class="tg-baqh">(p $$ \wedge $$ $$ \sim $$ q) $$ \to $$ (p $$ \vee $$ q)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table> | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktk5iipi | maths | mathematical-reasoning | logical-connectives | Negation of the statement (p $$\vee$$ r) $$\Rightarrow$$ (q $$\vee$$ r) is : | [{"identifier": "A", "content": "p $$\\wedge$$ $$\\sim$$ q $$\\wedge$$ $$\\sim$$ r"}, {"identifier": "B", "content": "$$\\sim$$ p $$\\wedge$$ q $$\\wedge$$ $$\\sim$$ 4"}, {"identifier": "C", "content": "$$\\sim$$ p $$\\wedge$$ q $$\\wedge$$ r"}, {"identifier": "D", "content": "p $$\\wedge$$ q $$\\wedge$$ r"}] | ["A"] | null | <p>Negative of (p $$\vee$$ r) $$\Rightarrow$$ (q $$\vee$$ r)</p>
<p>$$ \equiv $$ $$\sim$$ ((p $$\vee$$ r) $$\Rightarrow$$ (q $$\vee$$ r)) $$\equiv$$ (p $$\vee$$ r) $$ \wedge $$ ($$\sim$$ (q $$\vee$$ r))</p>
<p>$$\equiv$$ (p $$\vee$$ r) $$\wedge$$ ($$\sim$$ q $$\wedge$$ $$\sim$$ r) $$\equiv$$ (p $$\vee$$ r) $$\wedge$$ $$\sim$$ r) $$\wedge$$ $$\sim$$ q</p>
<p>$$\equiv$$ ((p $$\wedge$$ $$\sim$$ r)) $$\vee$$ (r $$\wedge$$ $$\sim$$ r) $$\wedge$$ $$\sim$$ q $$\equiv$$ (p $$\wedge$$ $$\sim$$ r) $$\wedge$$ f) $$\wedge$$ $$\sim$$ q</p>
<p>$$\equiv$$ (p $$\wedge$$ $$\sim$$ r) $$\wedge$$ ($$\sim$$ q) $$\equiv$$ p $$\wedge$$ $$\sim$$ q $$\wedge$$ $$\sim$$ r</p> | mcq | jee-main-2021-online-31st-august-evening-shift |
1kto27r9o | maths | mathematical-reasoning | logical-connectives | Which of the following is equivalent to the Boolean expression p $$\wedge$$ $$\sim$$ q ? | [{"identifier": "A", "content": "$$\\sim$$ (q $$\\to$$ p)"}, {"identifier": "B", "content": "$$\\sim$$ p $$\\to$$ $$\\sim$$ q"}, {"identifier": "C", "content": "$$\\sim$$ (p $$\\to$$ $$\\sim$$ q)"}, {"identifier": "D", "content": "$$\\sim$$ (p $$\\to$$ q)"}] | ["D"] | null | <table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim $$ p</th>
<th class="tg-baqh">$$ \sim $$ q</th>
<th class="tg-baqh">p $$ \to $$ q</th>
<th class="tg-baqh">$$ \sim $$ (p $$ \to $$ q)</th>
<th class="tg-baqh">q $$ \to $$ p</th>
<th class="tg-baqh">$$ \sim $$ (q $$ \to $$ p)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
</tbody>
</table><br><br><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.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-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-baqh">p $$ \wedge $$ $$ \sim $$ q</th>
<th class="tg-baqh">$$ \sim $$ p $$ \to $$ $$ \sim $$ q</th>
<th class="tg-baqh">p $$ \to $$ $$ \sim $$ q</th>
<th class="tg-baqh">$$ \sim $$ (p $$ \to $$ $$ \sim $$ q)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
</tbody>
</table><br><br>p $$\wedge$$ $$\sim$$ q $$ \equiv $$ $$\sim$$ (p $$\to$$ q)<br><br>Option (d) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l5450e9z | maths | mathematical-reasoning | logical-connectives | <p>Let $$\Delta$$ $$\in$$ {$$\wedge$$, $$\vee$$, $$\Rightarrow$$, $$\Leftrightarrow$$} be such that (p $$\wedge$$ q) $$\Delta$$ ((p $$\vee$$ q) $$\Rightarrow$$ q) is a tautology. Then $$\Delta$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\wedge$$"}, {"identifier": "B", "content": "$$\\vee$$"}, {"identifier": "C", "content": "$$\\Rightarrow$$"}, {"identifier": "D", "content": "$$\\Leftrightarrow$$"}] | ["C"] | null | $(p \vee q) \Rightarrow q$
<br/><br/>
$$
\begin{aligned}
& \sim(p \vee q) \vee q \\\\
& =(\sim p \wedge \sim q) \vee q \\\\
& =(\sim p \vee q) \wedge(\sim q \vee q) \\\\
& =(\sim p \vee q) \wedge T \\\\
& =\sim p \vee q
\end{aligned}
$$
<br/><br/>
Now $(p \wedge q) \Delta(\sim p \vee q)$
<br/><br/>
$$
\begin{array}{cccccc}
p & q & \sim p & p \wedge q & \sim p \vee q & (p \wedge q) \Delta(\sim p \vee q) \\
T & T & F & T & T & T \\
T & F & F & F & F & T \\
F & T & T & F & T & T \\
F & F & T & F & T & T
\end{array}
$$
<br/><br/>
$\therefore \Delta=\Rightarrow$ | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54tfxa2 | maths | mathematical-reasoning | logical-connectives | <p>Negation of the Boolean statement (p $$\vee$$ q) $$\Rightarrow$$ (($$\sim$$ r) $$\vee$$ p) is equivalent to :</p> | [{"identifier": "A", "content": "p $$\\wedge$$ ($$\\sim$$ q) $$\\wedge$$ r"}, {"identifier": "B", "content": "($$\\sim$$ p) $$\\wedge$$ ($$\\sim$$ q) $$\\wedge$$ r"}, {"identifier": "C", "content": "($$\\sim$$ p) $$\\wedge$$ q $$\\wedge$$ r"}, {"identifier": "D", "content": "p $$\\wedge$$ q $$\\wedge$$ ($$\\sim$$ r)"}] | ["C"] | null | <p>Given,</p>
<p>(p $$\vee$$ q) $$\Rightarrow$$ (($$\sim$$ r) $$\vee$$ p)</p>
<p>Negation is</p>
<p>$$\sim$$ ((p $$\vee$$ q) $$\Rightarrow$$ ($$\sim$$ r) $$\vee$$ p))</p>
<p>= (p $$\vee$$ q) $$\wedge$$ $$\sim$$ (($$\sim$$ r) $$\vee$$ p)</p>
<p>= (p $$\vee$$ q) $$\wedge$$ (r $$\wedge$$ $$\sim$$ p)</p>
<p>[(p $$\wedge$$ $$\sim$$ p) $$\vee$$ (q $$\wedge$$ $$\sim$$ p)] $$\wedge$$ r</p>
<p>= q $$\wedge$$ $$\sim$$ p $$\wedge$$ r</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55jdggk | maths | mathematical-reasoning | logical-connectives | <p>The maximum number of compound propositions, out of p$$\vee$$r$$\vee$$s, p$$\vee$$r$$\vee$$$$\sim$$s, p$$\vee$$$$\sim$$q$$\vee$$s, $$\sim$$p$$\vee$$$$\sim$$r$$\vee$$s, $$\sim$$p$$\vee$$$$\sim$$r$$\vee$$$$\sim$$s, $$\sim$$p$$\vee$$q$$\vee$$$$\sim$$s, q$$\vee$$r$$\vee$$$$\sim$$s, q$$\vee$$$$\sim$$r$$\vee$$$$\sim$$s, $$\sim$$p$$\vee$$$$\sim$$q$$\vee$$$$\sim$$s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to __________.</p> | [] | null | 9 | There are total 9 compound propositions, out of which 6 contain $\sim s$. So if we assign $s$ as false, these 6 propositions will be true.
<br/><br/>
In remaining 3 compound propositions, two contain $p$ and the third contains $\sim r$. So if we assign $p$ and $r$ as true and false respectively, these 3 propositions will also be true.
<br/><br/>
Hence maximum number of propositions that can be true are 9. | integer | jee-main-2022-online-28th-june-evening-shift |
1l567e91x | maths | mathematical-reasoning | logical-connectives | <p>Let p, q, r be three logical statements. Consider the compound statements</p>
<p>$${S_1}:(( \sim p) \vee q) \vee (( \sim p) \vee r)$$ and</p>
<p>$${S_2}:p \to (q \vee r)$$</p>
<p>Then, which of the following is NOT true?</p> | [{"identifier": "A", "content": "If S<sub>2</sub> is True, then S<sub>1</sub> is True"}, {"identifier": "B", "content": "If S<sub>2</sub> is False, then S<sub>1</sub> is False"}, {"identifier": "C", "content": "If S<sub>2</sub> is False, then S<sub>1</sub> is True"}, {"identifier": "D", "content": "If S<sub>1</sub> is False, then S<sub>2</sub> is False"}] | ["C"] | null | <p>$${S_1}:( \sim p \vee q) \vee ( \sim p \vee r)$$</p>
<p>$$ \cong ( \sim p \vee q \vee r)$$</p>
<p>$${S_2}: \sim p \vee (q \vee r)$$</p>
<p>Both are same</p>
<p>So, option (C) is incorrect.</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56rnxj7 | maths | mathematical-reasoning | logical-connectives | <p>Which of the following statement is a tautology?</p> | [{"identifier": "A", "content": "$$(( \\sim q) \\wedge p) \\wedge q$$"}, {"identifier": "B", "content": "$$(( \\sim q) \\wedge p) \\wedge (p \\wedge ( \\sim p))$$"}, {"identifier": "C", "content": "$$(( \\sim q) \\wedge p) \\vee (p \\vee ( \\sim p))$$"}, {"identifier": "D", "content": "$$(p \\wedge q) \\wedge ( \\sim p \\wedge q))$$"}] | ["C"] | null | <p>$$\because$$ (($$\sim$$ q) $$\wedge$$ p) $$\vee$$ (p $$\vee$$ ($$\sim$$ p))</p>
<p>= ($$\sim$$ q $$\wedge$$ p) $$\vee$$ t (t is tautology)</p>
<p>$$\equiv$$ t</p>
<p>$$\therefore$$ option (C) is correct.</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57ouws7 | maths | mathematical-reasoning | logical-connectives | <p>The boolean expression $$( \sim (p \wedge q)) \vee q$$ is equivalent to :</p> | [{"identifier": "A", "content": "$$q \\to (p \\wedge q)$$"}, {"identifier": "B", "content": "$$p \\to q$$"}, {"identifier": "C", "content": "$$p \\to (p \\to q)$$"}, {"identifier": "D", "content": "$$p \\to (p \\vee q)$$"}] | ["D"] | null | <p>Making truth table</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.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-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 972px">
<colgroup>
<col style="width: 49px">
<col style="width: 52px">
<col style="width: 139px">
<col style="width: 187px">
<col style="width: 214px">
<col style="width: 100px">
<col style="width: 89px">
<col style="width: 142px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">p $$ \wedge $$ q</th>
<th class="tg-baqh">$$ \sim $$ p $$ \wedge $$ q</th>
<th class="tg-baqh">($$ \sim $$ (p $$ \wedge $$ q)) $$ \vee $$ q</th>
<th class="tg-baqh">p $$ \vee $$ q</th>
<th class="tg-baqh">p $$ \to $$ q</th>
<th class="tg-baqh">p $$ \to $$ (p $$ \vee $$ q)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh"></td>
<td class="tg-baqh"></td>
<td class="tg-baqh"></td>
<td class="tg-baqh"></td>
<td class="tg-baqh">Tautology</td>
<td class="tg-baqh"></td>
<td class="tg-baqh"></td>
<td class="tg-baqh">Tautology</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ ($$\sim$$ (p $$\wedge$$ q)) $$\vee$$ q $$\equiv$$ p $$\to$$ (p $$\vee$$ q)</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58abzlk | maths | mathematical-reasoning | logical-connectives | <p>Let $$\Delta$$, $$\nabla $$ $$\in$$ {$$\wedge$$, $$\vee$$} be such that p $$\nabla$$ q $$\Rightarrow$$ ((p $$\Delta$$ q) $$\nabla$$ r) is a tautology. Then (p $$\nabla$$ q) $$\Delta$$ r is logically equivalent to :</p> | [{"identifier": "A", "content": "(p $$\\Delta$$ r) $$\\vee$$ q"}, {"identifier": "B", "content": "(p $$\\Delta$$ r) $$\\wedge$$ q"}, {"identifier": "C", "content": "(p $$\\wedge$$ r) $$\\Delta$$ q"}, {"identifier": "D", "content": "(p $$\\nabla$$ r) $$\\wedge$$ q"}] | ["A"] | null | <b>Case-I</b><br/><br/>
If $\nabla$ is same as $\wedge$<br/><br/>
Then $(p \wedge q) \Rightarrow((p \Delta q) \wedge r)$<br/><br/> is equivalent to $\sim(p \wedge q) \vee$ $((p \Delta q) \wedge r)$<br/><br/> is equivalent to $(\sim(p \wedge q) \vee(p \Delta q)) \wedge(\sim(p \wedge$ $q) \vee r)$<br/><br/>
Which cannot be a tautology<br/><br/>
For both $\Delta$ (i.e. $\vee$ or $\wedge$ )<br/><br/>
<b>Case-II</b><br/><br/>
If $\nabla$ is same as $v$<br/><br/>
Then $(p \vee q) \Rightarrow((p \Delta q) \vee r)$ is equivalent to $\sim(p \vee q) \vee(p \Delta q) \vee r$<br/><br/> which can be a tautology if $\Delta$ is also same as $\vee$.<br/><br/>
Hence both $\Delta$ and $\nabla$ are same as $v$.<br/><br/>
Now $(p \nabla q) \Delta r$ is equivalent to $(p \vee q \vee r)$. | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58grngd | maths | mathematical-reasoning | logical-connectives | <p>Let r $$\in$$ {p, q, $$\sim$$p, $$\sim$$q} be such that the logical statement</p>
<p>r $$\vee$$ ($$\sim$$p) $$\Rightarrow$$ (p $$\wedge$$ q) $$\vee$$ r</p>
<p>is a tautology. Then r is equal to :</p> | [{"identifier": "A", "content": "p"}, {"identifier": "B", "content": "q"}, {"identifier": "C", "content": "$$\\sim$$p"}, {"identifier": "D", "content": "$$\\sim$$q"}] | ["C"] | null | <p>Clearly r must be equal to $$\sim$$ p</p>
<p>$$\because$$ $$\sim$$ p $$\vee$$ $$\sim$$ p = $$\sim$$ p</p>
<p>and (p $$\wedge$$ q) $$\vee$$ $$\sim$$ p = p</p>
<p>$$\therefore$$ $$\sim$$ p $$\Rightarrow$$ p = tautology.</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59kxe1t | maths | mathematical-reasoning | logical-connectives | <p>The negation of the Boolean expression (($$\sim$$ q) $$\wedge$$ p) $$\Rightarrow$$ (($$\sim$$ p) $$\vee$$ q) is logically equivalent to :</p> | [{"identifier": "A", "content": "$$p \\Rightarrow q$$"}, {"identifier": "B", "content": "$$q \\Rightarrow p$$"}, {"identifier": "C", "content": "$$ \\sim (p \\Rightarrow q)$$"}, {"identifier": "D", "content": "$$ \\sim (q \\Rightarrow p)$$"}] | ["C"] | null | <p>Let $$S:(( \sim q) \wedge p) \Rightarrow (( \sim p) \vee q)$$</p>
<p>$$ \Rightarrow S:\, \sim (( \sim q) \wedge p) \vee (( \sim p) \vee q)$$</p>
<p>$$ \Rightarrow S:(q \vee ( \sim p)) \vee (( \sim p) \vee q)$$</p>
<p>$$ \Rightarrow S:( \sim p) \vee q$$</p>
<p>$$ \Rightarrow S:p \Rightarrow q$$</p>
<p>So, negation of S will be $$ \sim (p \Rightarrow q)$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5ai3ak1 | maths | mathematical-reasoning | logical-connectives | <p>Consider the following two propositions:</p>
<p>$$P1: \sim (p \to \sim q)$$</p>
<p>$$P2:(p \wedge \sim q) \wedge (( \sim p) \vee q)$$</p>
<p>If the proposition $$p \to (( \sim p) \vee q)$$ is evaluated as FALSE, then :</p> | [{"identifier": "A", "content": "P1 is TRUE and P2 is FALSE"}, {"identifier": "B", "content": "P1 is FALSE and P2 is TRUE"}, {"identifier": "C", "content": "Both P1 and P2 are FALSE"}, {"identifier": "D", "content": "Both P1 and P2 are TRUE"}] | ["C"] | null | Given $p \rightarrow(\sim p \vee q)$ is false <br/><br/>$\Rightarrow \sim p \vee q$ is false and $p$ is true<br/><br/>
Now $p=$ True.<br/><br/>
$\sim T \vee q=F$<br/><br/>
$F \vee q=F \Rightarrow q$ is false<br/><br/>
$P 1: \sim(T \rightarrow \sim F) \equiv \sim(T \rightarrow T) \equiv$ False.<br/><br/>
$$
\begin{aligned}
P 2:(T \wedge \sim F) \wedge(\sim T \vee F) & \equiv(T \wedge T) \wedge(F \vee F) \\\\
& \equiv T \wedge F \equiv \text { False }
\end{aligned}
$$ | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5c1xxom | maths | mathematical-reasoning | logical-connectives | <p>The number of choices for $$\Delta \in \{ \wedge , \vee , \Rightarrow , \Leftrightarrow \} $$, such that <br/><br/>$$(p\Delta q) \Rightarrow ((p\Delta \sim q) \vee (( \sim p)\Delta q))$$ is a tautology, is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["B"] | null | Let $x:(p \Delta q) \Rightarrow(p \Delta \sim q) \vee(\sim p \Delta q)$
<br/><br/>
<b>Case-I</b>
<br/><br/>
When $\Delta$ is same as $v$
<br/><br/>
Then $(p \Delta \sim q) \vee(\sim p \Delta q)$ becomes
<br/><br/>
$(p \vee \sim q) \vee(\sim p \vee q)$ which is always true, so $x$ becomes a tautology.
<br/><br/>
<b>Case-II</b>
<br/><br/>
When $\Delta$ is same as $\wedge$
<br/><br/>
Then $(p \wedge q) \Rightarrow(p \wedge \sim q) \vee(\sim p \wedge q)$
<br/><br/>
If $p \wedge q$ is $T$, then $(p \wedge \sim q) \vee(\sim p \wedge q)$ is $F$
<br/><br/>
so $x$ cannot be a tautology.
<br/><br/>
<b>Case-III</b>
<br/><br/>
When $\Delta$ is same as $\Rightarrow$
<br/><br/>
Then $(p \Rightarrow \sim q) \vee(\sim p \Rightarrow q)$ is same at $(\sim p \vee \sim q) \vee$ $(p \vee q)$, which is always true, so $x$ becomes a tautology.
<br/><br/>
<b>Case-IV</b>
<br/><br/>
When $\Delta$ is same as $\Leftrightarrow$
<br/><br/>
Then $(p \Leftrightarrow q) \Rightarrow(p \Leftrightarrow \sim q) \vee(\sim p \Leftrightarrow q)$
<br/><br/>
$p \Leftrightarrow q$ is true when $p$ and $q$ have same truth values, then $p \Leftrightarrow \sim q$ and $\sim p \Leftrightarrow q$ both are false. Hence $x$ cannot be a tautology.
<br/><br/>
So finally $x$ can be $\vee$ or $\Rightarrow$. | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5w0obkl | maths | mathematical-reasoning | logical-connectives | <p>The conditional statement</p>
<p>$$((p \wedge q) \to (( \sim p) \vee r)) \vee ((( \sim p) \vee r) \to (p \wedge q))$$ is :</p> | [{"identifier": "A", "content": "a tautology"}, {"identifier": "B", "content": "a contadiction"}, {"identifier": "C", "content": "equivalent to $$p \\wedge q$$"}, {"identifier": "D", "content": "equivalent to $$( \\sim p) \\vee r$$"}] | ["A"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
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<table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">r</th>
<th class="tg-baqh">$$ \sim $$ p</th>
<th class="tg-baqh">A =<br>p $$ \wedge $$ q</th>
<th class="tg-baqh">B =<br>$$ \sim $$ p $$ \vee $$ r</th>
<th class="tg-baqh">C =<br>(A $$ \to $$ B)</th>
<th class="tg-baqh">D =<br>(B $$ \to $$ A)</th>
<th class="tg-baqh">C $$ \vee $$ D</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ Given statement is a tautology.</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dwvb9t | maths | mathematical-reasoning | logical-connectives | <p>Which of the following statements is a tautology ?</p> | [{"identifier": "A", "content": "$$((\\sim \\mathrm{p}) \\vee \\mathrm{q}) \\Rightarrow \\mathrm{p}$$"}, {"identifier": "B", "content": "$$p \\Rightarrow((\\sim p) \\vee q)$$"}, {"identifier": "C", "content": "$$((\\sim p) \\vee q) \\Rightarrow q$$"}, {"identifier": "D", "content": "$$q \\Rightarrow((\\sim p) \\vee q)$$"}] | ["D"] | null | Truth Table<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97slo37/df9ec0ac-a06c-4e60-b3b5-9f73d35cb598/e7215830-4b5c-11ed-bfde-e1cb3fafe700/file-1l97slo38.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97slo37/df9ec0ac-a06c-4e60-b3b5-9f73d35cb598/e7215830-4b5c-11ed-bfde-e1cb3fafe700/file-1l97slo38.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Mathematical Reasoning Question 34 English Explanation"> | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6gj6vbd | maths | mathematical-reasoning | logical-connectives | <p>The statement $$(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q}$$ is :</p> | [{"identifier": "A", "content": "a tautology"}, {"identifier": "B", "content": "a contradiction"}, {"identifier": "C", "content": "equivalent to $$(p \\Rightarrow q) \\wedge q$$"}, {"identifier": "D", "content": "equivalent to $$(p \\Rightarrow q) \\wedge p$$"}] | ["D"] | null | <p>$$\sim$$ (p $$ \Leftrightarrow $$ $$\sim$$ q) $$\wedge$$ q</p>
<p>= (p $$ \Leftrightarrow $$ q) $$\wedge$$ q</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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.tg .tg-baqh{text-align:center;vertical-align:top}
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<table class="tg">
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">p $$ \leftrightarrow $$ q</th>
<th class="tg-baqh">(p$$ \to $$ q) $$ \wedge $$ q</th>
<th class="tg-baqh">(p $$ \to $$ q)</th>
<th class="tg-baqh">(p $$ \to $$ q) $$ \wedge $$ q</th>
<th class="tg-baqh">(p $$ \to $$ q) $$ \wedge $$ p</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ ($$\sim$$ (p $$ \Leftrightarrow $$ $$\sim$$ q)) $$\wedge$$ q is equivalent to (p $$ \Rightarrow $$ q) $$\wedge$$ p.</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hz70ir | maths | mathematical-reasoning | logical-connectives | <p>Negation of the Boolean expression $$p \Leftrightarrow(q \Rightarrow p)$$ is</p> | [{"identifier": "A", "content": "$$(\\sim p) \\wedge q$$"}, {"identifier": "B", "content": "$$p \\wedge(\\sim q)$$"}, {"identifier": "C", "content": "$$(\\sim p) \\vee(\\sim q)$$"}, {"identifier": "D", "content": "$$(\\sim p) \\wedge(\\sim q)$$"}] | ["D"] | null | <p>$$p \Leftrightarrow (q \Rightarrow p)$$</p>
<p>$$ \sim (p \Leftrightarrow (q \Leftrightarrow p))$$</p>
<p>$$ \equiv p \Leftrightarrow \, \sim (q \Rightarrow p)$$</p>
<p>$$ \equiv p \Leftrightarrow (q \wedge \sim p)$$</p>
<p>$$ \equiv (p \Rightarrow (q \wedge \sim p)) \wedge ((q \wedge \sim p) \Rightarrow p))$$</p>
<p>$$ \equiv ( \sim p \vee (q \wedge \sim p)) \wedge (( \sim q \vee p) \vee p))$$</p>
<p>$$ \equiv (( \sim p \vee q) \wedge \sim p) \wedge ( \sim q \vee p)$$</p>
<p>$$ \equiv \sim p \wedge ( \sim q \vee p)$$</p>
<p>$$ \equiv ( \sim p \wedge \sim q) \vee ( \sim p \wedge p)$$</p>
<p>$$ \equiv ( \sim p \wedge \sim q) \vee c$$</p>
<p>$$ \equiv ( \sim p \wedge \sim q)$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jcwca3 | maths | mathematical-reasoning | logical-connectives | <p>$$(p \wedge r) \Leftrightarrow(p \wedge(\sim q))$$ is equivalent to $$(\sim p)$$ when $$r$$ is</p> | [{"identifier": "A", "content": "$$p$$"}, {"identifier": "B", "content": "$$\\sim p$$"}, {"identifier": "C", "content": "$$q$$"}, {"identifier": "D", "content": "$$\\sim q$$"}] | ["C"] | null | <p>The truth table</p>
<p><style type="text/css">
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.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-c3ow">p</th>
<th class="tg-c3ow">q</th>
<th class="tg-c3ow">$$ \sim $$ p</th>
<th class="tg-c3ow">$$ \sim $$ q</th>
<th class="tg-c3ow">p $$ \wedge $$ q</th>
<th class="tg-c3ow">p $$ \wedge $$ $$ \sim $$ q</th>
<th class="tg-c3ow">p $$ \wedge $$ q $$ \Leftrightarrow $$ p $$ \wedge $$ $$ \sim $$ q</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
</tr>
<tr>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
</tr>
<tr>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
</tr>
<tr>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">T</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">F</td>
<td class="tg-c3ow">T</td>
</tr>
</tbody>
</table></p>
<p>Clearly p $$\wedge$$ q $$\Leftrightarrow$$ p $$\wedge$$ $$\sim$$ q $$\equiv$$ $$\sim$$ p</p>
<p>$$\therefore$$ r = q</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6kl6wn5 | maths | mathematical-reasoning | logical-connectives | <p>If the truth value of the statement $$(P \wedge(\sim R)) \rightarrow((\sim R) \wedge Q)$$ is F, then the truth value of which of the following is $$\mathrm{F}$$ ?</p> | [{"identifier": "A", "content": "$$\\mathrm{P} \\vee \\mathrm{Q} \\rightarrow \\,\\sim \\mathrm{R}$$"}, {"identifier": "B", "content": "$$\\mathrm{R} \\vee \\mathrm{Q} \\rightarrow \\,\\sim \\mathrm{P}$$"}, {"identifier": "C", "content": "$$\\sim(\\mathrm{P} \\vee \\mathrm{Q}) \\rightarrow \\sim \\mathrm{R}$$"}, {"identifier": "D", "content": "$$\\sim(\\mathrm{R} \\vee \\mathrm{Q}) \\rightarrow \\,\\sim \\mathrm{P}$$"}] | ["D"] | null | $\mathrm{X} \Rightarrow \mathrm{Y}$ is a false<br/><br/>
when $X$ is true and $Y$ is false<br/><br/>
So, $\mathrm{P} \rightarrow \mathrm{T}, \mathrm{Q} \rightarrow \mathrm{F}, \mathrm{R} \rightarrow \mathrm{F}$<br/><br/>
(A) $\mathrm{P} \vee \mathrm{Q} \rightarrow \sim \mathrm{R}$ is $\mathrm{T}$<br/><br/>
(B) $\mathrm{R} \vee \mathrm{Q} \rightarrow \sim \mathrm{P}$ is $\mathrm{T}$<br/><br/>
(C) $\sim(\mathrm{P} \vee \mathrm{Q}) \rightarrow \sim \mathrm{R}$ is $\mathrm{T}$<br/><br/>
(D) $\sim(\mathrm{R} \vee \mathrm{Q}) \rightarrow \sim \mathrm{P}$ is $\mathrm{F}$ | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6m5e0bg | maths | mathematical-reasoning | logical-connectives | <p>Let the operations $$*, \odot \in\{\wedge, \vee\}$$. If $$(\mathrm{p} * \mathrm{q}) \odot(\mathrm{p}\, \odot \sim \mathrm{q})$$ is a tautology, then the ordered pair $$(*, \odot)$$ is :</p> | [{"identifier": "A", "content": "$$(\\vee, \\wedge)$$"}, {"identifier": "B", "content": "$$(\\vee, \\vee)$$"}, {"identifier": "C", "content": "$$(\\wedge, \\wedge)$$"}, {"identifier": "D", "content": "$$(\\wedge, \\vee)$$"}] | ["B"] | null | <p>$$ * ,\, \odot \in \{ \wedge ,\, \vee \} $$</p>
<p>Now for $$(p * q) \odot (p \odot \sim q)$$ is tautology</p>
<p>(A) $$( \vee , \wedge ):(p \vee q) \wedge (p \wedge \sim q)$$ not a tautology</p>
<p>(B) $$( \vee , \vee ):(p \vee q) \vee (p \vee \sim q)$$</p>
<p>$$ = P \vee T$$ is tautology</p>
<p>(C) $$( \wedge , \wedge ):(p \wedge q) \wedge (p \wedge \sim q)$$</p>
<p>$$ = (p \wedge p) \wedge (q \wedge \sim q) = p \wedge F$$ not a tautology (Fallasy)</p>
<p>(D) $$( \wedge , \vee ):(p \wedge q) \vee (p \vee \sim q)$$ not a tautology</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6p294z9 | maths | mathematical-reasoning | logical-connectives | <p>The statement $$(p \wedge q) \Rightarrow(p \wedge r)$$ is equivalent to :</p> | [{"identifier": "A", "content": "$$q \\Rightarrow(p \\wedge r)$$"}, {"identifier": "B", "content": "$$p\\Rightarrow(\\mathrm{p} \\wedge \\mathrm{r})$$"}, {"identifier": "C", "content": "$$(\\mathrm{p} \\wedge \\mathrm{r}) \\Rightarrow(\\mathrm{p} \\wedge \\mathrm{q})$$"}, {"identifier": "D", "content": "$$(p \\wedge q) \\Rightarrow r$$"}] | ["D"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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.tg .tg-baqh{text-align:center;vertical-align:top}
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<table class="tg">
<thead>
<tr>
<th class="tg-baqh">$$p$$</th>
<th class="tg-baqh">$$q$$</th>
<th class="tg-baqh">$$r$$</th>
<th class="tg-baqh">$$A = p\, \wedge q$$</th>
<th class="tg-baqh">$$B = p\, \wedge r$$</th>
<th class="tg-baqh">$$A \to B$$</th>
<th class="tg-baqh">$$q \to B$$</th>
<th class="tg-baqh">$$p \to B$$</th>
<th class="tg-baqh">$$B \to A$$</th>
<th class="tg-baqh">$$A \to r$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>$$(p\, \wedge \,q) \Rightarrow (p\, \wedge \,r)$$ is equivalent to $$(p\, \wedge \,q) \Rightarrow r$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6rfes3g | maths | mathematical-reasoning | logical-connectives | <p>The statement $$(p \Rightarrow q) \vee(p \Rightarrow r)$$ is NOT equivalent to</p> | [{"identifier": "A", "content": "$$(p \\wedge(\\sim r)) \\Rightarrow q$$"}, {"identifier": "B", "content": "$$(\\sim q) \\Rightarrow((\\sim r) \\vee p)$$"}, {"identifier": "C", "content": "$$p \\Rightarrow(q \\vee r)$$"}, {"identifier": "D", "content": "$$(p \\wedge(\\sim q)) \\Rightarrow r$$"}] | ["B"] | null | $(\mathrm{A})(p \wedge(\sim r)) \Rightarrow q$
<br/><br/>$$
\begin{aligned}
&\sim(p \wedge \sim r) \vee q \\\\
&\equiv(\sim p \vee r) \vee q \\\\
&\equiv \sim p \vee(r \vee q) \\\\
&\equiv p \rightarrow(q \vee r) \\\\
&\equiv(p \Rightarrow q) \vee(p \Rightarrow r)
\end{aligned}
$$
<br/><br/>(C) $p \Rightarrow(q \vee r)$
<br/><br/>$$
\begin{aligned}
&\equiv \sim p \vee(q \vee r) \\\\
&\equiv(\sim p \vee q) \vee(\sim p \vee r) \\\\
&\equiv(p \rightarrow q) \vee(p \rightarrow r)
\end{aligned}
$$
<br/><br/>(D) $(p \wedge \sim q) \Rightarrow r$
<br/><br/>$$
\equiv p \Rightarrow(q \vee r)
$$
<br/><br/>$$
\begin{aligned}
& \equiv(p \Rightarrow q) \vee(p \Rightarrow r)
\end{aligned}
$$ | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldo5et0w | maths | mathematical-reasoning | logical-connectives | <p>Which of the following statements is a tautology?</p> | [{"identifier": "A", "content": "$$\\mathrm{p\\vee(p\\wedge q)}$$"}, {"identifier": "B", "content": "$$(\\mathrm{p\\wedge(p\\to q))\\to\\,\\sim q}$$"}, {"identifier": "C", "content": "$$\\mathrm{p\\to (p\\wedge (p\\to q))}$$"}, {"identifier": "D", "content": "$$(\\mathrm{p\\wedge q)\\to(\\sim (p)\\to q)}$$"}] | ["D"] | null | $\begin{aligned} & \sim p \rightarrow q \equiv \sim(\sim p) \vee q \equiv p \vee q \\\\ & p \wedge q \rightarrow(\sim p \rightarrow q) \\\\ & \equiv p \wedge q \rightarrow(p \vee q) \\\\ & \equiv \sim(p \wedge q) \vee(p \vee q) \\\\ & \equiv(\sim p \vee \sim q) \vee(p \vee q) \\\\ & \equiv(\sim p \vee(p \vee q)) \vee(\sim q \vee(p \vee q)) \\\\ & \equiv T \vee T \\\\ & \equiv T\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo8e0p7 | maths | mathematical-reasoning | logical-connectives | The number of values of $\mathrm{r} \in\{\mathrm{p}, \mathrm{q}, \sim \mathrm{p}, \sim \mathrm{q}\}$ for which $((\mathrm{p} \wedge \mathrm{q}) \Rightarrow(\mathrm{r} \vee \mathrm{q})) \wedge((\mathrm{p} \wedge \mathrm{r}) \Rightarrow \mathrm{q})$ is a tautology, is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["A"] | null | $((p \wedge q) \Rightarrow(r \vee q)) \wedge((p \wedge r) \Rightarrow q) \equiv T$ (given)
<br/><br/>$\equiv((\sim p \vee \sim q) \vee(r \vee q)) \wedge(\sim p \vee \sim r \vee q)$
<br/><br/>$\equiv((\sim p \vee r) \vee(\sim q \vee q)) \wedge(\sim p \vee \sim r \vee q)$
<br/><br/>$\equiv \sim p \vee \sim r \vee q$
<br/><br/>For above statement to be tautology
<br/><br/>$r$ can be $\sim p$ or $q$
<br/><br/>$\therefore $ Two values of $r$ are possible. | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldon2nkk | maths | mathematical-reasoning | logical-connectives | <p>The negation of the expression $$q \vee \left( {( \sim \,q) \wedge p} \right)$$ is equivalent to</p> | [{"identifier": "A", "content": "$$( \\sim \\,p) \\wedge ( \\sim \\,q)$$"}, {"identifier": "B", "content": "$$( \\sim \\,p) \\vee q$$"}, {"identifier": "C", "content": "$$p \\wedge ( \\sim \\,q)$$"}, {"identifier": "D", "content": "$$( \\sim \\,p) \\vee ( \\sim \\,q)$$"}] | ["A"] | null | $q \vee(\sim q \wedge p)$
<br/><br/>$\Rightarrow(q \vee \sim q) \wedge(q \vee p)$
<br/><br/>$\Rightarrow T \wedge(q \vee p)$
<br/><br/>$\Rightarrow q \vee p$
<br/><br/>Now,
<br/><br/>$\sim(q \vee p)$
<br/><br/>$=\sim q \wedge \sim p$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldpt87pz | maths | mathematical-reasoning | logical-connectives | <p>$$(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))$$ is a tautology</p>
<p>$$(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$$ is a contradiction.</p>
<p>Then</p> | [{"identifier": "A", "content": "only (S2) is correct"}, {"identifier": "B", "content": "both (S1) and (S2) are correct"}, {"identifier": "C", "content": "only (S1) is correct"}, {"identifier": "D", "content": "both (S1) and (S2) are wrong"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lek3mdx9/896912c9-3981-4e34-b46a-5223559fea7c/7c31fad0-b51e-11ed-accc-792fddd82133/file-1lek3mdxa.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lek3mdx9/896912c9-3981-4e34-b46a-5223559fea7c/7c31fad0-b51e-11ed-accc-792fddd82133/file-1lek3mdxa.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Mathematics - Mathematical Reasoning Question 21 English Explanation 1">
<br><br>$\therefore \mathrm{S} 1$ is correct.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lek3nk24/09e41b34-c9d0-4e6b-a3aa-b005fb90770b/9cbe58c0-b51e-11ed-8c61-834212db4998/file-1lek3nk25.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lek3nk24/09e41b34-c9d0-4e6b-a3aa-b005fb90770b/9cbe58c0-b51e-11ed-8c61-834212db4998/file-1lek3nk25.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Mathematics - Mathematical Reasoning Question 21 English Explanation 2">
<br><br>$\therefore \mathrm{S} 2$ is incorrect. | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldr5iqx7 | maths | mathematical-reasoning | logical-connectives | <p>Among the statements :</p>
<p>$$(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$$</p>
<p>$$(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$$</p>
| [{"identifier": "A", "content": "only (S1) is a tautology"}, {"identifier": "B", "content": "neither (S1) nor (S2) is a tautology"}, {"identifier": "C", "content": "both (S1) and (S2) are tautologies"}, {"identifier": "D", "content": "only (S2) is a tautology"}] | ["B"] | null | <p>$${S1}:\left( {(p \vee q) \Rightarrow r} \right) \Leftrightarrow (p \Rightarrow r)$$</p>
<p>$${S2}:\left( {(p \vee q) \Rightarrow r} \right) \Leftrightarrow \left( {(p \Rightarrow r) \vee (q \Rightarrow r)} \right)$$</p>
<p>In $$S1:$$ If $$p=F, q=T, r=F$$ then $$S_1$$ is false</p>
<p>In $$S2:$$ If $$p=T, q=F, r=F$$ then $$S_2$$ is false</p>
<p>$$\therefore$$ Neither S1 nor S2 is a tautology</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsegiln | maths | mathematical-reasoning | logical-connectives | <p>The statement $$B \Rightarrow \left( {\left( { \sim A} \right) \vee B} \right)$$ is equivalent to :</p> | [{"identifier": "A", "content": "$$B \\Rightarrow \\left( {\\left( { \\sim A} \\right) \\Rightarrow B} \\right)$$"}, {"identifier": "B", "content": "$$A \\Rightarrow \\left( {A \\Leftrightarrow B} \\right)$$"}, {"identifier": "C", "content": "$$A \\Rightarrow \\left( {\\left( { \\sim A} \\right) \\Rightarrow B} \\right)$$"}, {"identifier": "D", "content": "$$B \\Rightarrow \\left( {A \\Rightarrow B} \\right)$$"}] | null | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfq8buii/842105db-2a88-4b01-bc04-d2093d4bdf40/d00809a0-cc49-11ed-b18d-8994bf82aa9d/file-1lfq8buij.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lfq8buii/842105db-2a88-4b01-bc04-d2093d4bdf40/d00809a0-cc49-11ed-b18d-8994bf82aa9d/file-1lfq8buij.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Evening Shift Mathematics - Mathematical Reasoning Question 18 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfq8cntz/beed2e58-b43d-4388-abf6-ef1deae3f8c5/e6adeb70-cc49-11ed-b18d-8994bf82aa9d/file-1lfq8cnu0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lfq8cntz/beed2e58-b43d-4388-abf6-ef1deae3f8c5/e6adeb70-cc49-11ed-b18d-8994bf82aa9d/file-1lfq8cnu0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Evening Shift Mathematics - Mathematical Reasoning Question 18 English Explanation 2"> | mcqm | jee-main-2023-online-29th-january-evening-shift |
1ldsv5zz0 | maths | mathematical-reasoning | logical-connectives | <p>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?</p> | [{"identifier": "A", "content": "$$p = F,q = T,r = F$$"}, {"identifier": "B", "content": "$$p = T,q = T,r = F$$"}, {"identifier": "C", "content": "$$p = T,q = F,r = T$$"}, {"identifier": "D", "content": "$$p = T,q = F,r = F$$"}] | ["A"] | null | <style type="text/css">
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<th class="tg-7btt"></th>
<th class="tg-7btt">$\mathrm{p}$</th>
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<th class="tg-7btt">$\mathrm{r}$</th>
<th class="tg-7btt">$(p \vee q) \wedge((\sim p) \vee r)$</th>
<th class="tg-7btt">$\sim \mathrm{q} \vee \mathrm{r}$</th>
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<tbody>
<tr>
<td class="tg-c3ow">$(1)$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-c3ow">$(2)$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
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<td class="tg-c3ow">$(3)$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
</tr>
<tr>
<td class="tg-c3ow">$(4)$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
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So, $(p \vee q) \wedge(\sim q \vee r) \rightarrow(\sim p \vee r)$ will be False. | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldu5bbnv | maths | mathematical-reasoning | logical-connectives | <p>Let $$\Delta ,\nabla \in \{ \wedge , \vee \} $$ be such that $$\mathrm{(p \to q)\Delta (p\nabla q)}$$ is a tautology. Then</p> | [{"identifier": "A", "content": "$$\\Delta = \\vee ,\\nabla = \\vee $$"}, {"identifier": "B", "content": "$$\\Delta = \\vee ,\\nabla = \\wedge $$"}, {"identifier": "C", "content": "$$\\Delta = \\wedge ,\\nabla = \\wedge $$"}, {"identifier": "D", "content": "$$\\Delta = \\wedge ,\\nabla = \\vee $$"}] | ["A"] | null | $(p \longrightarrow q) \Delta(p \nabla q)$
<br/><br/>
$\Rightarrow \left(p^{\prime} \vee q\right) \Delta(p \nabla q)\quad...(i)$
<br/><br/>
$$
\text { If } \Delta=v, \nabla=v
$$
<br/><br/>
(i) becomes $\left(p^{\prime} \vee q\right) \vee(p \vee q) \equiv T$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv1942z | maths | mathematical-reasoning | logical-connectives | <p>The statement $$\left( {p \wedge \left( { \sim q} \right)} \right) \Rightarrow \left( {p \Rightarrow \left( { \sim q} \right)} \right)$$ is</p> | [{"identifier": "A", "content": "a tautology"}, {"identifier": "B", "content": "equivalent to $$\\left( { \\sim p} \\right) \\vee \\left( { \\sim q} \\right)$$"}, {"identifier": "C", "content": "a contradiction"}, {"identifier": "D", "content": "$$p \\vee q$$"}] | ["A"] | null | <p>Making truth table (Let $(p \wedge \sim q) \Rightarrow(p \Rightarrow \sim q)=E$ )</p>
<p><style type="text/css">
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<tr>
<th class="tg-c3ow">$p$</th>
<th class="tg-c3ow">$q$</th>
<th class="tg-c3ow">$\sim p$</th>
<th class="tg-c3ow">$\sim q$</th>
<th class="tg-c3ow">$p \wedge \sim q$</th>
<th class="tg-c3ow">$p \Rightarrow \sim q$</th>
<th class="tg-c3ow">$E$</th>
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<tr>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
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<tr>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{F}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
<td class="tg-c3ow">$\mathrm{T}$</td>
</tr>
</tbody>
</table></p>
<p>$\therefore E$ is a tautology</p> | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwx7ql1 | maths | mathematical-reasoning | logical-connectives | <p>Let p and q be two statements. Then $$ \sim \left( {p \wedge (p \Rightarrow \, \sim q)} \right)$$ is equivalent to</p> | [{"identifier": "A", "content": "$$\\left( { \\sim p} \\right) \\vee q$$"}, {"identifier": "B", "content": "$$p \\vee \\left( {p \\wedge ( \\sim q)} \\right)$$"}, {"identifier": "C", "content": "$$p \\vee \\left( {p \\wedge q} \\right)$$"}, {"identifier": "D", "content": "$$p \\vee \\left( {\\left( { \\sim p} \\right) \\wedge q} \\right)$$"}] | ["A"] | null | <p>Making truth table $(E \equiv \sim(p \wedge(p \Rightarrow \sim q))$</p>
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<th class="tg-0pky">$q$</th>
<th class="tg-0pky">$\sim p$</th>
<th class="tg-0pky">$\sim q$</th>
<th class="tg-0pky">$p \vee q$</th>
<th class="tg-0pky">$p \wedge q$</th>
<th class="tg-0pky">$\begin{aligned}p \Rightarrow p \sim q\end{aligned}$</th>
<th class="tg-0pky">$p \wedge (p \Rightarrow \sim q)$</th>
<th class="tg-0pky">$\mathrm{E}$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
</tr>
</tbody>
</table></p>
<p>$\&$</p>
<p><style type="text/css">
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<th class="tg-0pky">$\sim p \vee q$</th>
<th class="tg-0pky">$p \vee(\sim p \wedge q)$</th>
<th class="tg-0pky">$p \vee(p \wedge q)$</th>
<th class="tg-0pky">$p \vee(p \wedge \sim q)$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
</tr>
<tr>
<td class="tg-0pky">$\mathrm{T}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
<td class="tg-0pky">$\mathrm{F}$</td>
</tr>
</tbody>
</table></p>
<p>$\therefore \sim(p \wedge(p \Rightarrow \sim q))$ is equivalent to $\sim p \vee q$</p>
<p><b>Other Method :</b></p>
<p>$$ \sim \left( {p \vee \left( {p \to \sim q} \right.} \right)$$</p>
<p>$$ = \sim p \vee \sim \left( {p \to \sim q} \right)$$</p>
<p>$$ = \sim p \vee \sim \left( { \sim p \vee \sim q} \right)$$</p>
<p>$$= \sim p \vee \left( {p \wedge q} \right)$$</p>
<p>$$ = \left( { \sim p \vee p} \right) \wedge \left( { \sim p \vee q} \right)$$</p>
<p>$$ = t \wedge \left( { \sim p \vee q} \right)$$</p>
<p>$$ = \sim p \vee q$$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldyb76lo | maths | mathematical-reasoning | logical-connectives | <p>The compound statement $$\left( { \sim (P \wedge Q)} \right) \vee \left( {( \sim P) \wedge Q} \right) \Rightarrow \left( {( \sim P) \wedge ( \sim Q)} \right)$$ is equivalent to</p> | [{"identifier": "A", "content": "$$(( \\sim P) \\vee Q) \\wedge ( \\sim Q)$$"}, {"identifier": "B", "content": "$$( \\sim Q) \\vee P$$"}, {"identifier": "C", "content": "$$(( \\sim P) \\vee Q) \\wedge (( \\sim Q) \\vee P)$$"}, {"identifier": "D", "content": "$$( \\sim P) \\vee Q$$"}] | ["C"] | null | $$
\begin{aligned}
&(\sim(P \wedge Q)) \vee((\sim P) \wedge Q) \Rightarrow((\sim P) \wedge(\sim Q)) \\\\
&(\sim P \vee \sim Q) \vee(\sim P \wedge Q) \Rightarrow(\sim P \wedge \sim Q) \\\\
& \Rightarrow(\sim P \vee \sim Q \vee \sim P) \wedge(\sim P \vee \sim Q \vee Q) \Rightarrow(\sim P \wedge \sim Q) \\\\
& \Rightarrow(\sim P \vee \sim Q) \wedge(T) \Rightarrow(\sim P \wedge \sim Q) \\\\
& \Rightarrow(\sim P \vee \sim Q) \Rightarrow(\sim P \wedge \sim Q) \\\\
& \Rightarrow \sim(\sim P \vee \sim Q) \vee(\sim P \wedge \sim Q) \\\\
& \Rightarrow(P \wedge Q) \vee(\sim P \wedge \sim Q) \Rightarrow(\sim P \vee Q) \wedge(\bullet Q \vee P)
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnwpeal | maths | mathematical-reasoning | logical-connectives | Negation of $p \wedge(q \wedge \sim(p \wedge q))$ is : | [{"identifier": "A", "content": "$(\\sim(p \\wedge q)) \\wedge q$"}, {"identifier": "B", "content": "$(\\sim(p \\wedge q)) \\vee p$"}, {"identifier": "C", "content": "$p \\vee q$"}, {"identifier": "D", "content": "$\\sim(p \\vee q)$"}] | ["B"] | null | <p>The given statement is:</p>
<p>$p \wedge (q \wedge \sim(p \wedge q))$</p>
<p>We want to find the negation of this statement. To do that, we negate the whole expression:</p>
<p>$\sim (p \wedge (q \wedge \sim(p \wedge q)))$</p>
<p>Now, we will use De Morgan's laws to simplify the expression. De Morgan's laws state:</p>
<ol>
<li>$\sim (A \wedge B) \equiv \sim A \vee \sim B$</li>
<li>$\sim (A \vee B) \equiv \sim A \wedge \sim B$</li>
</ol>
<p>Applying the first De Morgan's law to the outer conjunction of the negated expression:</p>
<p>$\sim p \vee \sim (q \wedge \sim (p \wedge q))$</p>
<p>Now we need to address the inner conjunction $(q \wedge \sim (p \wedge q))$. To simplify this part of the expression, we again apply the first De Morgan's law:
</p>
<p>$\sim p \vee (\sim q \vee (p \wedge q))$
</p>
<p>Distribute the disjunction over the inner conjunction:</p>
<p>$$
\sim \mathrm{p} \vee((\sim \mathrm{q} \vee \mathrm{p}) \wedge(\sim \mathrm{q} \vee \mathrm{q}))
$$</p>
<p>Apply the law of excluded middle (A ∨ ¬A = T) on the last term:</p>
<p>$$
\sim \mathrm{p} \vee(\sim \mathrm{q} \vee \mathrm{p})
$$</p>
<p>Apply De Morgan's law on the inner disjunction:
</p>
<p>$$
\sim(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{p}
$$
</p>
<p>Now, comparing this with the provided options, it matches with Option B :</p>
<p>$(\sim(p \wedge q)) \vee p$</p>
<p>So, the negation of the given statement is :</p>
<p>$(\sim(p \wedge q)) \vee p$</p> | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgoxllvb | maths | mathematical-reasoning | logical-connectives | <p>The statement $$(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$$ is equivalent to _________.</p> | [{"identifier": "A", "content": "$$(\\sim p) \\vee(\\sim q)$$"}, {"identifier": "B", "content": "$$p \\vee(\\sim q)$$"}, {"identifier": "C", "content": "$$\\mathrm{p} \\vee \\mathrm{q}$$"}, {"identifier": "D", "content": "$$(\\sim p) \\vee q$$"}] | ["A"] | null | $$
\begin{array}{|c|c|c|c|c|c|c|}
\hline \mathbf{p} & \mathbf{q} & \sim \mathbf{q} & \sim \mathbf{p} & \mathbf{p} \wedge \sim \mathbf{q} & \mathbf{\sim p} \wedge \mathbf{q} & \mathbf{\sim p} \wedge \mathbf{\sim q} \\
\hline T & T & F & F & F & F & F \\
T & F & T & F & T & F & F \\
F & T & F & T & F & T & F \\
F & F & T & T & F & F & T \\
\hline
\end{array}
$$
<br/><br/>Using the truth table, we can see that the given statement is equivalent to $$(\sim p) \vee(\sim q)$$.
<br/><br/><b>Alternate Method :</b>
<br/><br/>The given statement $$(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$$ can be simplified using the following steps:
<br/><br/>$$
\begin{aligned}
&(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q)) \\\\
&= \left(p \wedge(\sim q)\right) \vee \left((\sim p) \wedge (q \vee (\sim q))\right) \\\\
&= \left(p \wedge(\sim q)\right) \vee \left((\sim p) \wedge t\right) \\\\
&= \left(p \wedge(\sim q)\right) \vee (\sim p) \\\\
&= (\sim p) \vee \left(p \wedge (\sim q)\right) \\\\
&= (\sim p \vee p) \wedge (\sim p \vee \sim q) \\\\
&= t \wedge (\sim p \vee \sim q) \\\\
&= \sim p \vee \sim q
\end{aligned}
$$
<br/><br/>Therefore, the given statement is equivalent to $$(\sim p) \vee (\sim q)$$. | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxp5mt | maths | mathematical-reasoning | logical-connectives | <p>The negation of the statement $$((A \wedge(B \vee C)) \Rightarrow(A \vee B)) \Rightarrow A$$ is</p> | [{"identifier": "A", "content": "equivalent to $$B ~\\vee \\sim C$$"}, {"identifier": "B", "content": "equivalent to $$\\sim A$$"}, {"identifier": "C", "content": "equivalent to $$\\sim C$$"}, {"identifier": "D", "content": "a fallacy"}] | ["B"] | null | $$((A \wedge(B \vee C)) \Rightarrow(A \vee B)) \Rightarrow A$$
<br/><br/>$$ \equiv $$ $$
[\sim(\mathrm{A} \wedge(\mathrm{B} \vee \mathrm{C})) \vee(\mathrm{A} \vee \mathrm{B})] \Rightarrow \mathrm{A}
$$
<br/><br/>$$ \equiv $$ $$
\sim(\sim(A \wedge(B \vee C)) \vee(A \vee B)) \vee A
$$
<br/><br/>$$ \equiv $$ $$
(A \wedge(B \vee C)) \wedge \sim(A \vee B) \vee A
$$
<br/><br/>$$ \equiv $$ $$
(f \vee A)=A
$$
<br/><br/>$$
\therefore \text { Negation of statement }=\sim A
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrejueh | maths | mathematical-reasoning | logical-connectives | <p>Among the two statements</p>
<p>$$(\mathrm{S} 1):(p \Rightarrow q) \wedge(p \wedge(\sim q))$$ is a contradiction and</p>
<p>$$(\mathrm{S} 2):(p \wedge q) \vee((\sim p) \wedge q) \vee(p \wedge(\sim q)) \vee((\sim p) \wedge(\sim q))$$ is a tautology</p> | [{"identifier": "A", "content": "both are false."}, {"identifier": "B", "content": "only (S1) is true."}, {"identifier": "C", "content": "both are true."}, {"identifier": "D", "content": "only (S2) is true."}] | ["C"] | null | $$
\begin{aligned}
S_1: & (p \Rightarrow q) \wedge(p \wedge \sim q) \\\\
& \equiv(\sim p \vee q) \wedge(p \wedge \sim q) \\\\
& \equiv(\sim p \wedge p \wedge \sim q) \vee(q \wedge p \wedge \sim q) \\\\
& \equiv(f \wedge \sim q) \vee(f \wedge p) \\\\
& \equiv f \vee f \equiv f \\\\
S_2: & (p \wedge q) \vee(\sim p \wedge q) \vee(p \wedge \sim q) \vee(\sim p \wedge \sim q) \\\\
& \equiv((p \vee \sim p) \wedge q) \vee((p \vee \sim p) \wedge \sim q) \\\\
& \equiv(t \wedge q) \vee(t \wedge \sim q) \equiv q \vee \sim q \equiv t
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsuhzhe | maths | mathematical-reasoning | logical-connectives | <p>The converse of $$((\sim p) \wedge q) \Rightarrow r$$ is</p> | [{"identifier": "A", "content": "$$((\\sim p) \\vee q) \\Rightarrow r$$"}, {"identifier": "B", "content": "$$(\\sim \\mathrm{r}) \\Rightarrow \\mathrm{p} \\wedge \\mathrm{q}$$"}, {"identifier": "C", "content": "$$(\\mathrm{p} \\vee(\\sim \\mathrm{q})) \\Rightarrow(\\sim \\mathrm{r})$$"}, {"identifier": "D", "content": "$$(\\sim \\mathrm{r}) \\Rightarrow((\\sim \\mathrm{p}) \\wedge \\mathrm{q})$$"}] | ["C"] | null | Converse of $((\sim p) \wedge q) \Rightarrow r$
<br/><br/>$$
\begin{aligned}
& \equiv \mathrm{r} \Rightarrow(\sim \mathrm{p} \wedge \mathrm{q}) \\\\
& \equiv \sim \mathrm{r} \vee(\sim \mathrm{p} \wedge \mathrm{q}) \\\\
& \equiv \sim \mathrm{r} \vee(\mathrm{p} \vee \sim \mathrm{q}) \equiv(\mathrm{p} \vee \sim \mathrm{q}) \Rightarrow \sim \mathrm{r}
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgvpc3a0 | maths | mathematical-reasoning | logical-connectives | <p>The statement $$\sim[p \vee(\sim(p \wedge q))]$$ is equivalent to :</p> | [{"identifier": "A", "content": "$$(\\sim(p \\wedge q)) \\wedge q$$"}, {"identifier": "B", "content": "$$\\sim(p \\vee q)$$"}, {"identifier": "C", "content": "$$(p \\wedge q) \\wedge(\\sim p)$$"}, {"identifier": "D", "content": "$$\\sim(p \\wedge q)$$"}] | ["C"] | null | $$
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline p & q & p \wedge q & \sim(p \wedge q) & \begin{array}{c}
p \vee \sim \\
(p \wedge q)
\end{array} & \begin{array}{l}
\sim[p \vee \sim \\
(p \wedge q)]
\end{array} & p \vee q & \begin{array}{l}
\sim(p \\
\vee q)
\end{array} & \begin{array}{l}
(p \wedge q) \wedge \\
\quad(\sim p)
\end{array} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}
$$
<br/><br/>$$
\therefore \sim[p \vee(\sim(p \wedge q))]=\sim p \wedge(p \wedge q)
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxhachl | maths | mathematical-reasoning | logical-connectives | <p>The negation of the statement $$(p \vee q) \wedge (q \vee ( \sim r))$$ is :</p> | [{"identifier": "A", "content": "$$(( \\sim p) \\vee r)) \\wedge ( \\sim q)$$"}, {"identifier": "B", "content": "$$(p \\vee r) \\wedge ( \\sim q)$$"}, {"identifier": "C", "content": "$$(( \\sim p) \\vee ( \\sim q)) \\vee ( \\sim r)$$"}, {"identifier": "D", "content": "$$(( \\sim p) \\vee ( \\sim q)) \\wedge ( \\sim r)$$"}] | ["A"] | null | The negation of the statement $(p \vee q) \wedge(q \vee(\sim r))$ is
<br/><br/>$$
\begin{aligned}
& =\sim[(p \vee q) \wedge(q \vee(\sim r))] \\\\
& =\sim[(p \vee q) \vee \sim(q \vee(\sim r)] \\\\
& =((\sim p) \wedge \sim q)) \vee((\sim q) \wedge r))
\end{aligned}
$$
<br/><br/>Apply distribution law,
<br/><br/>$$
\begin{aligned}
& =\sim q \wedge((\sim p) \vee r)) \\\\
& =((\sim p) \vee r) \wedge(\sim q)
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgylheex | maths | mathematical-reasoning | logical-connectives | <p>The negation of $$(p \wedge(\sim q)) \vee(\sim p)$$ is equivalent to :</p> | [{"identifier": "A", "content": "$$p \\wedge q$$"}, {"identifier": "B", "content": "$$p \\wedge(\\sim q)$$"}, {"identifier": "C", "content": "$$p \\wedge(q \\wedge(\\sim p))$$"}, {"identifier": "D", "content": "$$p \\vee(q \\vee(\\sim p))$$"}] | ["A"] | null | $$
\begin{aligned}
& (p \wedge(\sim q)) \vee(\sim p) \\\\
& \equiv(p \vee \sim p) \wedge(\sim q \vee \sim p) \\\\
& \equiv \mathrm{T} \wedge(\sim q \vee \sim p) \\\\
& \equiv \sim q \vee \sim p \text { negation } p \wedge q
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgzzynuh | maths | mathematical-reasoning | logical-connectives | <p>Negation of $$(p \Rightarrow q) \Rightarrow(q \Rightarrow p)$$ is :</p> | [{"identifier": "A", "content": "$$(\\sim q) \\wedge p$$"}, {"identifier": "B", "content": "$$q \\wedge(\\sim p)$$"}, {"identifier": "C", "content": "$$p \\vee(\\sim q)$$"}, {"identifier": "D", "content": "$$(\\sim p) \\vee q$$"}] | ["B"] | null | Given: $(p \rightarrow q) \rightarrow(q \rightarrow p)$
<br/><br/>Negation of above statement is :
<br/><br/>$$
\begin{aligned}
& \sim[(p \rightarrow q) \rightarrow(q \rightarrow p)] \\\\
& \equiv \sim[\sim p \rightarrow q \wedge q \rightarrow p] \\\\
& \equiv p \rightarrow q \wedge \sim q \rightarrow p \\\\
& \equiv \sim p \vee q \wedge q \wedge \sim p] \\\\
& \equiv q \wedge(\sim p)
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh22nyhl | maths | mathematical-reasoning | logical-connectives | <p>Statement $$\mathrm{(P \Rightarrow Q) \wedge(R \Rightarrow Q)}$$ is logically equivalent to :</p> | [{"identifier": "A", "content": "$$(P \\Rightarrow R) \\wedge(Q \\Rightarrow R)$$"}, {"identifier": "B", "content": "$$(P \\Rightarrow R) \\vee(Q \\Rightarrow R)$$"}, {"identifier": "C", "content": "$$(P \\wedge R) \\Rightarrow Q$$"}, {"identifier": "D", "content": "$$(P \\vee R) \\Rightarrow Q$$"}] | ["D"] | null | We have, $(P \Rightarrow Q) \wedge(R \Rightarrow Q)$
<br/><br/>$$
\begin{aligned}
& \equiv(\sim P \vee Q) \wedge(\sim R \vee Q) \\\\
& \equiv(\sim P \wedge \sim R) \vee Q \text { (by distributive law) }\\\\
& \equiv \sim(P \vee R) \vee Q \\\\
& \equiv(P \vee R) \Rightarrow Q
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2yajqy | maths | mathematical-reasoning | logical-connectives | <p>Among the statements</p>
<p>(S1) : $$(p \Rightarrow q) \vee((\sim p) \wedge q)$$ is a tautology</p>
<p>(S2) : $$(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)$$ is a contradiction</p> | [{"identifier": "A", "content": "neither (S1) and (S2) is True"}, {"identifier": "B", "content": "only (S2) is True"}, {"identifier": "C", "content": "both $$(\\mathrm{S} 1)$$ and $$(\\mathrm{S} 2)$$ are True"}, {"identifier": "D", "content": "only (S1) is True"}] | ["A"] | null | (S1) : $$(p \Rightarrow q) \vee((\sim p) \wedge q)$$
<br/><br/>$$
\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{P} & \mathrm{Q} & \sim p & \sim p \wedge q & p \Rightarrow q & \begin{array}{c}
(p \Rightarrow q) \vee \\
(\sim p \wedge q)
\end{array} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\end{array}
$$
<br/><br/>Here, for every value of components compound statement (last column) are not true. Hence, $S_1$ is not a tautology.
<br/><br/>$S_2:(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)$ | mcq | jee-main-2023-online-6th-april-evening-shift |
IKF8zEfPChsloJ2J | maths | mathematical-reasoning | logical-statement | Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”,
and r be the statement “x is a rational number iff y is a transcendental number”.
<br/><br/> <b>Statement –1:</b> r is equivalent to either q or p.
<br/><br/><b>Statement –2:</b> r is equivalent to $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ | [{"identifier": "A", "content": "Statement \u2212 1 is false, Statement \u2212 2 is false"}, {"identifier": "B", "content": "Statement \u22121 is false, Statement \u22122 is true"}, {"identifier": "C", "content": "Statement \u22121 is true, Statement \u22122 is true, Statement \u22122 is a correct explanation for Statement \u22121"}, {"identifier": "D", "content": "Statement \u22121 is true, Statement \u22122 is true; Statement \u22122 is not a correct explanation for\nStatement \u22121"}] | ["A"] | null | <p>p : x is an irrational number</p>
<p>q : y is a transcendental number</p>
<p>r : x is a rational number, if y is a transcendental number</p>
<p>$$\Rightarrow$$ r : $$\sim$$ p $$\leftrightarrow$$ q</p>
<p>S<sub>1</sub> : r $$\equiv$$ q $$\vee$$ p</p>
<p>and S<sub>2</sub> : r $$\equiv$$ $$\sim$$ (p $$\leftrightarrow$$ $$\sim$$ q)</p>
<p><style type="text/css">
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overflow:hidden;padding:10px 5px;word-break:normal;}
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font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
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<table class="tg" style="undefined;table-layout: fixed; width: 966px">
<colgroup>
<col style="width: 51px">
<col style="width: 53px">
<col style="width: 83px">
<col style="width: 83px">
<col style="width: 184px">
<col style="width: 92px">
<col style="width: 193px">
<col style="width: 227px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">p</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">$$ \sim p$$</th>
<th class="tg-baqh">$$ \sim q$$</th>
<th class="tg-baqh">r<br>$$ \sim p \leftrightarrow q$$<br></th>
<th class="tg-baqh">Statement I<br>$$q \vee p$$<br></th>
<th class="tg-baqh">$$(p \leftrightarrow \sim q)$$</th>
<th class="tg-baqh">Statement II<br>$$ \sim (p \leftrightarrow \sim q)$$<br></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
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<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table></p>
<p>It is clear from the table that r is not equivalent to either of the statements.</p>
<p>Therefore, both statements are false.</p> | mcq | aieee-2008 |
O0QHhaazXNzgQQym | maths | mathematical-reasoning | logical-statement | Let S be a non-empty subset of R. Consider the following statement:
<br/>P : There is a rational number x ∈ S such that x > 0.
<br/>Which of the following statements is the negation of the statement P? | [{"identifier": "A", "content": "There is no rational number x \u2208 S such that x \u2264 0"}, {"identifier": "B", "content": "Every rational number x \u2208 S satisfies x \u2264 0"}, {"identifier": "C", "content": "x \u2208 S and x \u2264 0 $$ \\Rightarrow $$ x is not rational"}, {"identifier": "D", "content": "There is a rational number x \u2208 S such that x \u2264 0"}] | ["B"] | null | <p>Given that S is a non-empty subset of R.</p>
<p>$$\bullet$$ P : There is a rational number x $$\in$$ S such that x > 0.</p>
<p>Now, we need to find the negation of P. Clearly, P is equivalent to saying that "There is a positive rational number in S.</p>
<p>So, its negation ($$\sim$$ P) is "There is no positive rational number in S".</p>
<p>Thus, for $$\sim$$ P : There exists no positive rational number in S.</p>
<p>$$\bullet$$ $$\Leftrightarrow$$ $$\sim$$ P : Every rational number x $$\in$$ S satisfies x $$\le$$ 0.</p> | mcq | aieee-2010 |
aZuPum2ZGIl6pDOy | maths | mathematical-reasoning | logical-statement | Consider the following statements
<br/>P : Suman is brilliant
<br/>Q : Suman is rich
<br/>R : Suman is honest
<br/>The negation of the statement,
<br/><br/>“Suman is brilliant and dishonest if and only if Suman is rich” can be
expressed as : | [{"identifier": "A", "content": "$$ \\sim \\left[ {Q \\leftrightarrow \\left( {P \\wedge \\sim R} \\right)} \\right]$$"}, {"identifier": "B", "content": "$$ \\sim Q \\leftrightarrow P \\wedge R$$"}, {"identifier": "C", "content": "$$ \\sim \\left( {P \\wedge \\sim R} \\right) \\leftrightarrow Q$$"}, {"identifier": "D", "content": "$$ \\sim P \\wedge \\left( {Q \\leftrightarrow \\sim R} \\right)$$"}] | ["A"] | null | "Suman is brilliant and dishonest" an be expressed as : $${P \wedge \sim R}$$
<br><br>So “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as :
<br><br>$$\left( {P \wedge \sim R} \right) \leftrightarrow Q$$
<br><br>Now negation of this = $$ \sim \left[ {\left( {P \wedge \sim R} \right) \leftrightarrow Q} \right]$$
<br><br>You should know, $$p \leftrightarrow q \equiv q \leftrightarrow p$$
<br><br>So, $$ \sim \left[ {\left( {P \wedge \sim R} \right) \leftrightarrow Q} \right] \equiv \sim \left[ {Q \leftrightarrow \left( {P \wedge \sim R} \right)} \right]$$ | mcq | aieee-2011 |
LQt8RJq61sF54YAA | maths | mathematical-reasoning | logical-statement | The negation of the statement “If I become a teacher, then I will open a school” is : | [{"identifier": "A", "content": "I will become a teacher and I will not open a school"}, {"identifier": "B", "content": "Either I will not become a teacher or I will not open a school"}, {"identifier": "C", "content": "Neither I will become a teacher nor I will open a school"}, {"identifier": "D", "content": "I will not become a teacher or I will open a school"}] | ["A"] | null | <p>Let, p : I become a teacher</p>
<p>q : I will open a school</p>
<p>$$\therefore$$ "If I become a teacher, then I will open a school", is p $$\to$$ q</p>
<p>So, negation of $$(p \to q) = \sim (p \to q) = p \wedge \sim q$$ = I will become a teacher and I will not open a school</p> | mcq | aieee-2012 |
pnAtpDdWdcKtXojQpGxye | maths | mathematical-reasoning | logical-statement | Consider the following two statements :
<br/><br/><b>P :</b> If 7 is an odd number, then 7 is
divisible by 2.
<br/><b>Q :</b> If 7 is a prime number, then 7 is an
odd number
<br/><br/>If V<sub>1</sub> is the truth value of the contrapositive of P and V<sub>2</sub> is the truth value of contrapositive of Q, then the ordered pair (V<sub>1</sub> , V<sub>2</sub>) equals : | [{"identifier": "A", "content": "(T, T)"}, {"identifier": "B", "content": "(T, F)"}, {"identifier": "C", "content": "(F, T)"}, {"identifier": "D", "content": "(F, F)"}] | ["C"] | null | Contrapositive of P : If 7 is not divisible by 2,
then 7 is not an odd number.
<br><br>This statement is false.
<br><br>$$ \therefore $$ V<sub>1</sub> = False (F)
<br><br>Contrapositive of Q : If 7 is not an odd number
,then 7 is not a prime number.
<br><br>This statement is true.
<br><br>$$ \therefore $$ V<sub>2</sub> = True (T)
<br><br>$$ \therefore $$ (V<sub>1</sub>, V<sub>2</sub>) = (F, T). | mcq | jee-main-2016-online-9th-april-morning-slot |
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