Question
Which of the following is a singleton set ?
A.
$$\left\{ {x:\left| x \right| = 5,\,x\, \in \,N} \right\}$$
B.
$$\left\{ {x:\left| x \right| = 6,\,x\, \in \,Z} \right\}$$
C.
$$\left\{ {x:{x^2} + 2x + 1 = 0,\,x\, \in \,N} \right\}$$
D.
$$\left\{ {x:{x^2} = 7,\,x\, \in \,N} \right\}$$
Answer :
$$\left\{ {x:\left| x \right| = 5,\,x\, \in \,N} \right\}$$
Solution :
$$\eqalign{
& \left( {\text{A}} \right)\,\,\left| x \right| = 5\,\, \Rightarrow x = 5\,\,\left[ {\because \,x\, \in \,N\,} \right] \cr
& \therefore \,{\text{ Given set is singleton}}{\text{.}} \cr
& \left( {\text{B}} \right)\,\,\left| x \right| = 6\,\, \Rightarrow x = - 6,\,6\,\,\left[ {\because \,x\, \in \,Z\,} \right] \cr
& \therefore \,{\text{ Given set is not singleton}}{\text{.}} \cr
& \left( {\text{C}} \right)\,\,{x^2} + 2x + 1 = 0\,\, \cr
& \Rightarrow {\left( {x + 1} \right)^2} = 0 \cr
& \Rightarrow x = - 1,\, - 1 \cr
& {\text{Since, }} - 1\, \notin \,N, \cr
& \therefore \,\,{\text{Given set}} = \phi \cr
& \left( {\text{D}} \right)\,\,{x^2} = 7\,\, \Rightarrow x = \pm \sqrt 7 \cr} $$