The solution set of x 2-16 - 0 and x 2-9 - 0 isA- - 3-B- -3- -3- -C- -4-3- -3-4-D- -4- -

The correct option is C [ 4, 3] ∪ [3, 4]x^2 16 ≤ 0 ⇒ (x +4)(x 4) ≤ 0 ⇒ x ∈ [ 4, 4] ⋯(1) x^2 9 ≥ 0 ⇒ (x+3)(x 3) ≥ 0 ⇒ x ∈ ( ∞, 3] ∪ [3, ∞) ⋯(2) From (1 ...

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Standard XII

Mathematics

Properties of Inequalities

The solution ...

Question

The solution set of $x^{2} - 16 \leq 0$ and $x^{2} - 9 \geq 0$ is

[4, \infty)

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(- \infty, 3]

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(- \infty, - 3] \cup [3, \infty)

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[- 4, - 3] \cup [3, 4]

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Solution

The correct option is D $[- 4, - 3] \cup [3, 4]$
$x^{2} - 16 \leq 0 \Rightarrow (x + 4) (x - 4) \leq 0 \Rightarrow x \in [- 4, 4] \dots (1)$
$x^{2} - 9 \geq 0 \Rightarrow (x + 3) (x - 3) \geq 0 \Rightarrow x \in (- \infty, - 3] \cup [3, \infty) \dots (2)$
From $(1)$ and $(2)$ ,
$x \in [- 4, - 3] \cup [3, 4]$

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Q. Match the elements of the list I, which contains inequalities, with the elements of list II, which has solution sets.

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Properties of Inequalities

Standard XII Mathematics

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The solution set of x2−16≤0 and x2−9≥0 is

The correct option is D [−4,−3]∪[3,4]x2−16≤0⇒(x+4)(x−4)≤0⇒x∈[−4,4]⋯(1) x2−9≥0⇒(x+3)(x−3)≥0⇒x∈(−∞,−3]∪[3,∞)⋯(2) From (1) and (2), x∈[−4,−3]∪[3,4]

The solution set of $x^{2} - 16 \leq 0$ and $x^{2} - 9 \geq 0$ is