Question

Which of the following is a singleton set?

• A. $\{x:|x|=5,x\in \mathrm{\mathbb{N}}\}$
• B. $\{x:|x|=6,x\in \mathrm{\mathbb{Z}}\}$
• C. $\left\{x:x^2+2x+1=0,x\in \mathrm{\mathbb{N}}\right\}$
• D. $\left\{x:x^2=7,x\in \mathrm{\mathbb{N}}\right\}$

Answer 1

The correct option is A

$\{x:|x|=5,x\in \mathrm{\mathbb{N}}\}$

Explanation for the correct answer

Finding the roster form of $\{x:|x|=5,x\in \mathrm{\mathbb{N}}\}$

Singleton: If a set contains only one element then the set is known as a singleton.

For option (a)

$\{x:|x|=5,x\in \mathrm{\mathbb{N}}\}$

Since $x\in \mathrm{\mathbb{N}}$, then

$x=5$

Therefore, $\{x:|x|=5,x\in \mathrm{\mathbb{N}}\}$ is a singleton set.

Explanation for incorrect answer

For option (b)

$\{x:|x|=6,x\in \mathrm{\mathbb{Z}}\}$

Since, $x\in \mathrm{\mathbb{Z}}$, then

$\left|6\right|=6$ and $\left|-6\right|=6$

Therefore, $x=6,-6$

The roster form of $\{x:|x|=6,x\in \mathrm{\mathbb{Z}}\}$ is $\left\{6,-6\right\}$.

Since the set contains two elements, so it is not a singleton set.

For option (c)

$\left\{x:x^2+2x+1=0,x\in \mathrm{\mathbb{N}}\right\}$

$x^2+2x+1=0$

$\Rightarrow (x+1{)}^2=0$ $\left[\because (a+b{)}^2=a^2+2ab+b^2\right]$

$$\begin{gathered} \Rightarrow x+1=0,x+1=0 \\ \Rightarrow x=-1,-1 \end{gathered}$$

Since, $x\in \mathrm{\mathbb{N}}$

The roster form of $\left\{x:x^2+2x+1=0,x\in \mathrm{\mathbb{N}}\right\}$ is $\varphi$.

Therefore given set is a null set.

For option (d)

$\left\{x:x^2=7,x\in \mathrm{\mathbb{N}}\right\}$

$$\begin{gathered} x^2=7 \\ \Rightarrow x=\pm \sqrt{7} \end{gathered}$$

Since $\sqrt{7}\notin \mathrm{\mathbb{N}}$

The roster form of $\left\{x:x^2=7,x\in \mathrm{\mathbb{N}}\right\}$ is $\varphi$

Therefore given set is a null set.

Hence, the correct option is option (a).