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Solve for x, |x+3|+xx+2>1. - Mathematics

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प्रश्न

Solve for x, `(|x + 3| + x)/(x + 2) > 1`.

योग

उत्तर

We have `(|x + 3| + x)/(x + 2) > 1`

⇒ `(|x + 3| + x)/(x + 2) - 1 > 0`

⇒ `(|x + 3| - 2)/(x + 2) > 0`

Now two cases arise:

Case I: When x + 3 ≥ 0, i.e., x ≥ –3

Then `(|x + 3| - 2)/(x + 2) > 0`

⇒ `(x + 3 - 2)/(x + 2) > 0`

⇒ `(x + 1)/(x + 2) > 0`

⇒ {(x + 1) > 0 and x + 2 > 0} or {x + 1 < 0 and x + 2 < 0}

⇒ {x > –1 and x > –2} or {x < –1 and x < –2}

⇒ x > –1 or x < –2

⇒ x ∈ (–1, `oo`) or x ∈ (`–oo`, –2)

⇒ x ∈ (–3, –2) ∪ ( –1, `oo`) [Since x ≥ –3]    ....(1)

Case II: When x + 3 < 0 i.e., x < –3

`(|x + 3| - 2)/(x + 2) > 0`

⇒ `(-x - 3 - 2)/(x + 2) > 0`

⇒ `(-(x + 5))/(x + 2) > 0`

⇒ `(x + 5)/(x + 2) < 0`

⇒ (x + 5 < 0 and x + 2 > 0) or (x + 5 > 0 and x + 2 < 0)

⇒ (x < –5 and x > –2) or (x > –5 and x < –2)

It is not possible.

⇒ x ∈ (–5 , –2)    ......(2)

Combining (I) and (II), the required solution is x ∈ (–5 , –2) ∪ (–1, `oo`).

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अध्याय 6: Linear Inequalities - Solved Examples [पृष्ठ १०२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 6 Linear Inequalities
Solved Examples | Q 7 | पृष्ठ १०२

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