QJEEM24Maaf38: JEE Main 2024 Mathematics - Class 11 JEEM24Maaf38 Solution: The number of real solutions of the equation is _________.

The given equation is $x(x^2+3|x|+5|x-1|+6|x-2|)=0$ , which can be solved by analyzing it in parts. It can be broken down into: $x=0$ and $x^2+3|x|+5|x-1|+6|x-2|=0$ .

For $x=0$ , it's clear that it is a solution to the equation since it makes the entire expression equal to zero.

Case (I)

x<0

$\begin{aligned}& x^2-3 x-5(x-1)-6(x-2)=0 \\\\& x^2-14 x+17=0\end{aligned}$

$\because$ All roots are positive $\Rightarrow$ no solution

Case (II)

$\begin{aligned}& 0 < x < 1 \\\\& x^2+3 x-5(x-1)-6(x-2)=0 \\\\& x^2-8 x+17=0 \\\\& \because D < 0 \Rightarrow \text { no solution }\end{aligned}$

Case (III)

1 < x < 2

$x^2+3 x+5(x-1)-6(x-2)=0$

$x^2+2 x+7=0$

$\Rightarrow$ no solution

Case (IV)

$\begin{aligned}& x > 2 \\\\& x^2+3 x+5(x-1)+6(x-2)=0 \\\\& x^2+14 x-19=0\end{aligned}$

All roots less than 2

$\Rightarrow$ no solution

Here $x=0$ is only solution.