Verified Solution • 2025–26 Session moderate

QJEEM23Ma8e2f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8e2f Solution: The set of all for which the equation has exactly one real…

The set of all a∈Ra \in \mathbb{R} for which the equation x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0 has exactly one real root, is :

Multiple Choice Options

  • A (−∞,∞)(-\infty, \infty)
  • B (−6,∞)(-6, \infty)
  • C (−∞,−3)(-\infty,-3)
  • D (−6,−3)(-6,-3)

Step-by-Step Verified Solution

x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0

Case I : If x<-2 then

−x2+x−x−2+a=0-x^2+x-x-2+a=0

a=x2+2a=x^2+2

y=x2+2y=x^2+2 is decreasing ∀x∈(−∞,−2)\forall x \in(-\infty,-2)

Case II : If -2 \leq x<1 then

amp;−x2+x+x+2+a=0amp;a=x2−2x−2\begin{aligned}& -x^2+x+x+2+a=0 \\\\& a=x^2-2 x-2\end{aligned}

y=x2−2x−2y=x^2-2 x-2 is decreasing ∀x∈[−2,1)\forall x \in[-2,1) .

Case III: If x≥1x \geq 1 then

amp;x2−x+x+2+a=0amp;a=−(x2+2)\begin{aligned}& x^2-x+x+2+a=0 \\\\& a=-\left(x^2+2\right)\end{aligned}

y=−(x2+2)y=-\left(x^2+2\right) is decreasing ∀x∈[1,∞)\forall x \in[1, \infty)

∴\therefore Exactly one real root ∀x∈R\forall x \in R

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