Question

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

• A. $\phi$
• B. {14, 3, 4}
• C. {3}
• D. {4}

Answer 1

Answer: (A)

$\phi$

Answer 2

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. $\therefore \, A = \phi$