The set of all a∈R for which the equation x |x-l| + |x + 2|+a = 0 has exactly one real root, is

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Correct answer: B. (b) Given x|x-1|+ |x + 2| +a = 0 Now taking, Case I: x < -2, we get −x^2+x−x−2+a=0 ⇒ a = x^2 + 2 And y = x^2 + 2 is decreasing ∀x∈_(−∞,−2_) Now taking Case II: -2 < x < 1 we get, -x^2 + x + x + 2 + a = 0 ⇒ a = x^2 - 2x - 2 And y = x^2 - 2x - 2 So, d yd x=2 x−1≤0∀x∈[−2,1) Hence, y is decreasing ∀x∈ [-2,1) Now taking Case III: x ≥ 1 we get, x^2-x + x + 2 + a = 0 ⇒ a = -(x^2 + 2) And y = - (x^2 + 2) is decreasing ∀x∈[1,∞) Hence, from all the cases we can say that nature of function is continuously decreasing so, it will cut x-axis only one time, ∴ Exactly one real root ∀a∈R

$The set of all a \in R for which the equation x |x-l| + |x + 2|+a = 0 has exactly one real root, is$

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