The number of real solutions of the equation $x\left(x^{2}+3|x|+5|x-1|+6|x-2|\right)=0$ is
$x=0$ and $x^{2}+3|x|+5|x-1|+6|x-2|=0$
Here all terms are + ve except at $x=0$
So there is no value of $x$
Satisfies this equation
Only solution $\mathrm{x}=0$
No of solution 1 .
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