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प्रश्न
Solve for x, |x + 1| + |x| > 3.
उत्तर
On L.H.S. of the given inequality, We have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = –1, 0 as critical points. These critical points divide the real line in three parts as (– `oo`, – 1), [–1, 0), [0, `oo`).
Case I: When `–oo < x < –1`
|x + 1| + |x| > 3
⇒ –x – 1 – x > 3
⇒ x < –2
Case II: When –1 ≤ x < 0,
|x + 1| + |x| > 3
⇒ x + 1 – x > 3
⇒ 1 > 3 ....(Not possible)
Case III: When 0 ≤ x < `oo`,
|x + 1| + |x| > 3
⇒ x + 1 + x > 3
⇒ x > 1
Combining the results of cases (I), (II) and (III), We get x ∈ `(–oo , –2) ∪ (1, oo)`.
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