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

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{14, 3, 4}

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{3}

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{4}

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Solution

The correct option is A

$x^{2}$ = 16 ⇒ x = ± 4

2x = 6 ⇒x=3

There is no value of x which satisfies both the above equations.

Thus, A = ∅.

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