A) \[\{x:\left| x \right|=5,x\in N\}\]
B) \[\{x:\left| x \right|=6,x\in Z\}\]
C) \[\{x:{{x}^{2}}+2x+1=0,x\in N\}\]
D) \[\{x:{{x}^{2}}=7,x\in N\}\]
Correct Answer: A
Solution :
[a]a. \[\left| x \right|=5\Rightarrow x=5\,\,[\because x\in N]\] \[\therefore \] Given set is singleton. b. \[\left| x \right|=6\Rightarrow x=-6,\,6\,\,[\because x\in Z]\] \[\therefore \] Given set is not singleton. c. \[{{x}^{2}}+2x+1=0\Rightarrow {{(x+1)}^{2}}=0\] \[\Rightarrow x=-1,-1\] Since, \[-1\notin N,\] \[\therefore \] given set \[=\phi \] d.\[{{x}^{2}}=7\Rightarrow x=\pm \sqrt{7}.\]
You need to login to perform this action.
You will be redirected in
3 sec
