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Hint- This question is solved by making truth table of $p,{\text{ }}q,{\text{ }}r,{\text{ }} \sim r,{\text{ }}\left( {p \wedge q} \right){\text{ and }}\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$.

Now given that,

$\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$

Now we have to find the truth values of $p,q{\text{ and }}r$

We will find that by using truth table,

Now the possible combinations are ${2^3} = 8$ (Because variables are three.)

Also we know that,

$

\sim {\text{ means }}NOT \\

\wedge {\text{ means }}AND \\

\vee {\text{ means }}OR \\

$

$

p{\text{ }}q{\text{ }}r{\text{ }} \sim r{\text{ }}\left( {p \wedge q} \right){\text{ }}\left( {p \wedge q} \right) \vee \left( { \sim r} \right) \\

T{\text{ }}T{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}T \\

F{\text{ }}T{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

T{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

T{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}T{\text{ }}T \\

F{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

T{\text{ }}F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

F{\text{ }}F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

$

Now it is given that $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$ and we have to find the values of $p,q{\text{ and }}r$

when $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

Now, from the truth table, we will see the respective values of $p,q{\text{ and }}r$ for $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

Therefore, the values of $p,q{\text{ and }}r$ are $F,F,T$ respectively.

Thus, the correct option is (E).

Note- Whenever we face such types of questions the key concept is that we should solve it by using a truth table. In this question we find the values of $p,q{\text{ and }}r$ by making the truth table and then we see the respective values of $p,q{\text{ and }}r$ for $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

Now given that,

$\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$

Now we have to find the truth values of $p,q{\text{ and }}r$

We will find that by using truth table,

Now the possible combinations are ${2^3} = 8$ (Because variables are three.)

Also we know that,

$

\sim {\text{ means }}NOT \\

\wedge {\text{ means }}AND \\

\vee {\text{ means }}OR \\

$

$

p{\text{ }}q{\text{ }}r{\text{ }} \sim r{\text{ }}\left( {p \wedge q} \right){\text{ }}\left( {p \wedge q} \right) \vee \left( { \sim r} \right) \\

T{\text{ }}T{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}T \\

F{\text{ }}T{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

T{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}F{\text{ }}F \\

T{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}T{\text{ }}T \\

F{\text{ }}T{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

T{\text{ }}F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

F{\text{ }}F{\text{ }}F{\text{ }}T{\text{ }}F{\text{ }}T \\

$

Now it is given that $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$ and we have to find the values of $p,q{\text{ and }}r$

when $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

Now, from the truth table, we will see the respective values of $p,q{\text{ and }}r$ for $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

Therefore, the values of $p,q{\text{ and }}r$ are $F,F,T$ respectively.

Thus, the correct option is (E).

Note- Whenever we face such types of questions the key concept is that we should solve it by using a truth table. In this question we find the values of $p,q{\text{ and }}r$ by making the truth table and then we see the respective values of $p,q{\text{ and }}r$ for $\left( {p \wedge q} \right) \vee \left( { \sim r} \right)$ has truth values $F$.

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