UGC NET CSE | June 2023 | Part 2: 13

Question

 

Consider universe positive integer $X=\{1 \leq n \leq 8\}$, proposition $P=$ " $n$ is an even integers", $Q="(3 \leq n \leq 7) \wedge(n \neq 6)$ ". Then truth set of $P$ $\leftrightarrow \mathrm{Q}$ is

• A. $\{1,4\}$
• B. $\{2,6\}$
• C. $\{3,4,5\}$
• D. $\{1\}$

Answer 1

Answer is A ;

here X ={ 1,2,3,4,5,6,7,8}

p(n) is true if n is even where n belongs to [ 1,8 ] .

Q(n) is true when 3<=n<=7 AND n!=6 .

now p<-->q called as A  is true when both are true AND both are false

p is true for { 2,4,6,8} and false for {1,3,5,7}

q is true for {3,4,5,7} and false for {1,2,6,8}

A : p↔q is {1,4} as for 1 both are false and for 4 both are true .