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QJEEM23Mad97a: JEE Main 2023 Mathematics - Class 11 JEEM23Mad97a Solution: The number of real roots of the equation , is :

The number of real roots of the equation x∣x∣−5∣x+2∣+6=0x|x|-5|x+2|+6=0 , is :

Multiple Choice Options

  • A 4
  • B 3
  • C 5
  • D 6

Step-by-Step Verified Solution

amp;x∣x∣−5∣x+2∣+6=0amp; Case-I : amp; When xlt;−2 then amp;−x2+5(x+2)+6=0amp;⇒x2−5x−16=0amp;⇒x=5±25+642amp;∴x=5−892 is accepted \begin{aligned}& x|x|-5|x+2|+6=0 \\\\& \text { Case-I : } \\\\& \text { When } x<-2 \text { then } \\\\& -x^2+5(x+2)+6=0 \\\\& \Rightarrow x^2-5 x-16=0 \\\\& \Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2} \\\\& \therefore x=\frac{5-\sqrt{89}}{2} \text { is accepted }\end{aligned}

Case-II :

amp; When −2≤xlt;0 then amp;−x2−5(x+2)+6=0⇒amp;x2+5x+4=0⇒amp;(x+1)(x+4)=0amp;x=−1 is accepted \begin{aligned}& \text { When } -2 \leq x < 0 \text { then } \\\\& -x^2-5(x+2)+6=0 \\\\\Rightarrow & x^2+5 x+4=0 \\\\\Rightarrow & (x+1)(x+4)=0 \\\\& x=-1 \text { is accepted }\end{aligned}

Case-III :

amp; When x≥0 then amp;x2−5(x+2)+6=0⇒amp;x2−5x−4=0amp;x=5±25+162amp;=5±412amp;x=5−412 is accepted ∴amp;3 real roots are possible. \begin{aligned}& \text { When } x \geq 0 \text { then } \\\\& x^2-5(x+2)+6=0 \\\\\Rightarrow & x^2-5 x-4=0 \\\\& x=\frac{5 \pm \sqrt{25+16}}{2} \\\\& =\frac{5 \pm \sqrt{41}}{2} \\\\& x=\frac{5 - \sqrt{41}}{2} \text { is accepted } \\\\\therefore & 3 \text { real roots are possible. }\end{aligned}

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