Which of the following is a singleton set?

Question

Which of the following is a singleton set? (A)  x: |x| < 1, x ∈ I  (B)  x: |x| = 5, x ∈ I  (C)  x: x2= 1, x ∈ I  (D)  x: x2 + x + 1 = 0, x ∈ R

Tardigrade Admin · Accepted Answer

(a) x: |x| < 1,x ∈ I = x: -1 < x < 1, x ∈ I = 0 (b) x: |x| = 5, x ∈ I = x: x = Â± 5 = Â± 5 (c) x: x2 = 1, x ∈ I = x: x = Â± 1 = Â± 1 (d) x: x2 + x + 1 = 0, x ∈ R = x: x = (-1±√3i/2), x ∈ R = φ

Q. Which of the following is a singleton set?

${x : ∣ x ∣ < 1, x \in I}$

${x : ∣ x ∣ = 5, x \in I}$

${x : x^{2} = 1, x \in I}$

${x : x^{2} + x + 1 = 0, x \in R}$

(a) ${x : ∣ x ∣ < 1, x \in I}$
$= {x : - 1 < x < 1, x \in I} = {0}$
(b) ${x : ∣ x ∣ = 5, x \in I}$
$= {x : x = \pm 5} = {\pm 5}$
(c) ${x : x^{2} = 1, x \in I}$
$= {x : x = \pm 1} = {\pm 1}$
(d) ${x : x^{2} + x + 1 = 0, x \in R}$
$= {x : x = \frac{- 1 \pm 3 i}{2}, x \in R} = ϕ$

Q. Which of the following is a singleton set?

{x:∣x∣<1,x∈I}

{x:∣x∣=5,x∈I}

{x:x2=1,x∈I}

{x:x2+x+1=0,x∈R}

(a) {x:∣x∣<1,x∈I} ={x:−1<x<1,x∈I}={0} (b) {x:∣x∣=5,x∈I} ={x:x=±5}={±5} (c) {x:x2=1,x∈I} ={x:x=±1}={±1} (d) {x:x2+x+1=0,x∈R} ={x:x=2−1±3​i​,x∈R}=ϕ

${x : ∣ x ∣ < 1, x \in I}$

${x : ∣ x ∣ = 5, x \in I}$

${x : x^{2} = 1, x \in I}$

${x : x^{2} + x + 1 = 0, x \in R}$

(a) ${x : ∣ x ∣ < 1, x \in I}$
$= {x : - 1 < x < 1, x \in I} = {0}$
(b) ${x : ∣ x ∣ = 5, x \in I}$
$= {x : x = \pm 5} = {\pm 5}$
(c) ${x : x^{2} = 1, x \in I}$
$= {x : x = \pm 1} = {\pm 1}$
(d) ${x : x^{2} + x + 1 = 0, x \in R}$
$= {x : x = \frac{- 1 \pm 3 i}{2}, x \in R} = ϕ$