Question

The set $\mathrm{A}=\mathrm{\{}\mathrm{x}:\mathrm{x}$ belongs to $\mathrm{R},{\mathrm{x}}^2=16$ and $2\mathrm{x}=6\mathrm{\}}$ is equal to

• A. $\mathrm{\varphi }$
• B. $\left\{14,3,4\right\}$
• C. $\left\{3\right\}$
• D. $\left\{4\right\}$

Answer 1

The correct option is A

$\mathrm{\varphi }$

Explanation for correct option

Option(A): Given that $\mathrm{A}=\mathrm{\{}\mathrm{x}:\mathrm{x}$ belongs to $\mathrm{R},{\mathrm{x}}^2=16$ and $2\mathrm{x}=6\mathrm{\}}$

Here, $2\mathrm{x}=6$

$\Rightarrow$ $\mathrm{x}=3$

And ${\mathrm{x}}^2=16$

$\Rightarrow$ $\mathrm{x}=\pm 4$ [Taking square root on both sides]

So, there is no value of $\mathrm{x}$, which satisfies both the equations.

Thus, $\mathrm{A}=\mathrm{\varphi }$

Thus option(A) is correct.

Explanation for incorrect options

Option (B): $\left\{14,3,4\right\}$

Here for $\mathrm{x}=14$, $2\mathrm{x}=6$ becomes

$\Rightarrow 2\left(14\right)=28\ne 6$

And ${\mathrm{x}}^2=16$ becomes

$\Rightarrow {\left(14\right)}^2=196\ne 16$

Thus option(B) is incorrect.

Option(C): For $\mathrm{x}=3$, $2\mathrm{x}=6$ becomes

$\Rightarrow 2\left(3\right)=6$

And ${\mathrm{x}}^2=16$ becomes

$\Rightarrow {\left(3\right)}^2=9\ne 16$

Thus option(C) is incorrect.

Option(D): For $\mathrm{x}=4$, $2\mathrm{x}=6$ becomes

$\Rightarrow 2\left(4\right)=8\ne 6$

And ${\mathrm{x}}^2=16$ becomes

$\Rightarrow {\left(4\right)}^2=16$

Thus option(D) is incorrect.

Hence, correct option is option(A) i.e. $\mathrm{\varphi }$