The correct option is C [2,) We must have x 2 ≥ 0 for √(x 2) to get defined, so, x ≥ 2 Now, √(x 2) ≥ 1 ∀ x ∈ [2,∞) as square roots are always non negative He ...

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Standard XI

Mathematics

Squaring an Inequality

Solve: √x-2≥-...

Question

Solve: $\sqrt{x - 2}$ ≥ -1

[2,3)

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[3,)

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[2,)

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(,3]

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Solution

The correct option is C
[2,)

We must have x-2 ≥ 0 for $\sqrt{x - 2}$ to get defined, so, x ≥ 2

Now, $\sqrt{x - 2} \geq - 1 \forall x \in [2, \infty)$ as square roots are always non-negative

Hence x ≥ 2

Note: Some students solve it by squaring on both sides for which x-2 ≥ 1 or x ≥ 3 clearly interval [2,3) is lost.

Suggest Corrections

Solve: √x−2 ≥ -1

The correct option is C [2,) We must have x-2 ≥ 0 for √x−2 to get defined, so, x ≥ 2 Now, √x−2≥−1∀x∈[2,∞) as square roots are always non-negative Hence x ≥ 2 Note: Some students solve it by squaring on both sides for which x-2 ≥ 1 or x ≥ 3 clearly interval [2,3) is lost.

Solve: $\sqrt{x - 2}$ ≥ -1