The set A = x: x ∈ R, x2 = 16 and 2x = 6 equals

Question

The set A = x: x ∈ R, x2 = 16 and 2x = 6 equals (A) φ (B) 14, 3, 4 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

We have x2 = 16 ⇒ x = ± 4 Also, 2x = 6 ⇒ x = 3 There is no value of x which satisfies both the above equations. Thus the set A contains no elements. ∴ A = φ

Q. The set A = {x : x $\in$ R, $x^{2}$ = 16 and 2x = 6} equals

$ϕ$

{14, 3, 4}

{3}

{4}

We have $x^{2} = 16 \Rightarrow x = \pm 4$
Also, 2x = 6 $\Rightarrow$ x = 3
There is no value of x which satisfies both the above equations. Thus the set A contains no elements.
$∴ A = ϕ$

Q. The set A = {x : x ∈ R, x2 = 16 and 2x = 6} equals

ϕ

{14, 3, 4}

{3}

{4}

We have x2=16⇒x=±4 Also, 2x = 6 ⇒ x = 3 There is no value of x which satisfies both the above equations. Thus the set A contains no elements. ∴A=ϕ

Q. The set A = {x : x $\in$ R, $x^{2}$ = 16 and 2x = 6} equals

$ϕ$

We have $x^{2} = 16 \Rightarrow x = \pm 4$
Also, 2x = 6 $\Rightarrow$ x = 3
There is no value of x which satisfies both the above equations. Thus the set A contains no elements.
$∴ A = ϕ$