The solution set of 16- x 2- 0 is A- -4- -4- -B- -4-4-C- -4-4D- -16-16

The correct option is B [ 4, 4]16 x^2 ≥ 0 ⇒ x^2 16 ≤ 0 ⇒ (x+4)(x 4) ≤ 0 ∴ x ∈ [ 4, 4] ...

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Inequalities Involving Modulus Function

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Question

The solution set of $16 - x^{2} \geq 0$ is

[- 4, 4]

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(- \infty, - 4] \cup [4, \infty)

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(- 4, 4)

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(- 16, 16)

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