Sets Relations And Functions Question 67
Question: If A= {$x|x\in ,N, $ and $ (x^{2}-4) $ $ (x^{2}-5) $ =0} and B= {$x|x\in {l^{+}} $ and $ x(x-1) $ $ (x-2) $ =0} then $ (A\cup B) $ - $ (A\cap B) $ is
Options:
A) {1, 2}
B) {1}
C) {2}
D) $ \phi $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ (x^{2}-4)(x^{2}-5)=0 $
$ \Rightarrow x^{2}-4=0$ or $x^{2}-5=0 $
$ \Rightarrow x=-2+2$ or $x=-\sqrt{5},+\sqrt{5} $
$ So,A={2} $ $ x(x-1)(x-2)=0 $
$ \Rightarrow x=0,1,2 $ $ So,B={1,2} $
$ (A\cup B)-(A\cap B)={1,2}-{2}={1} $
BETA
