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Mathematical Reasoning
hard
ધારોકો $r \in\{p, q, \sim p, \sim q\}$ એવો છ કે જેથી તાર્કિક વિધાન $r \vee(\sim p) \Rightarrow(p \wedge q) \vee r$ : નિત્યસત્ય છે. તો $r=\dots\dots$
A
$p$
B
$q$
C
$\sim p$
D
$\sim q$
(JEE MAIN-2022)
Solution
By options
$(1)$
| $p = r$ | $q$ | $\sim p$ | $r \vee(\sim p )$ | $( p \wedge q )$ | $( p \wedge q ) \vee r$ | $r \vee(\sim P )\;\Rightarrow \;( p \wedge q ) \vee r$ |
| $T$ | $F$ | $F$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $T$ | $F$ | $F$ | $F$ |
| $T$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
$T$ |
$(2)$
| $p$ | $\sim p$ | $r \vee(\sim p )$ | $q = r$ | $( p \wedge q )$ | $( p \wedge q ) \vee r$ | $r \vee(\sim p ) \;\Rightarrow\;( p \wedge q ) \vee r$ |
| $T$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $T$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $F$ | $F$ | $F$ |
$(3)$
| $p$ | $q$ | $r =\sim p$ | $r \vee(\sim p )$ | $( p \wedge q )$ | $( p \wedge q ) \vee r$ | $r \vee(\sim p ) \;\Rightarrow\;( p \wedge q ) \vee r$ |
| $T$ | $T$ | $F$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $T$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $F$ | $F$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | $T$ |
$(4)$
| $\sim p$ | $p$ | $q$ | $r \vee(\sim p )$ | $r =\sim p$ | $( p \wedge q )$ | $( p \wedge q ) \vee r$ | $r \vee(\sim p ) \;\Rightarrow\;( p \wedge q ) \vee r$ |
| $F$ | $T$ | $T$ | $F$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ | $T$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $T$ | $T$ | $F$ | $F$ | $F$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | $T$ |
Now finak answer in option no. $3$
Standard 11
Mathematics
