JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Basic Mathematics

$x(x^2+3|x|+5|x-1|+6|x-2|)=0$ holds when $x=0$ or when the bracket vanishes.

Every term in the bracket — $x^2$, $3|x|$, $5|x-1|$, $6|x-2|$ — is non-negative for every real $x$, so the bracket can only be zero if all four terms are zero simultaneously. That would need $x=0$ (from $x^2$ and $|x|$), $x=1$ (from $|x-1|$) and $x=2$ (from $|x-2|$) all at once, which is impossible; in fact at $x=0$ the bracket equals $0+0+5+12=17\ne 0$. So the bracket never vanishes for any real $x$.

The only real solution of the whole equation is $x=0$ — exactly 1 real solution.