Given,
x∣x−1∣+∣x+2∣+a=0
Now taking, Case I: x<−2, we get
−x2+x−x−2+a=0
⇒a=x2+2
And y=x2+2 is decreasing ∀x∈(−∞,−2)
Now taking Case II: −2≤x<1 we get,
−x2+x+x+2+a=0
⇒a=x2−2x−2
And y=x2−2x−2
So, dxdy=2x−1≤0∀x∈[−2,1)
Hence, y is decreasing ∀x∈[−2,1)
Now taking Case III: x≥1 we get,
x2−x+x+2+a=0
⇒a=−(x2+2)
And y=−(x2+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