The number of real solutions of the equation | vedclass

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(D) The given equation is $x(x^2+3|x|+5|x-1|+6|x-2|) = 0$.This implies either $x = 0$ or $x^2+3|x|+5|x-1|+6|x-2| = 0$.For the second part,let $f(x) =

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$1$

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(D) The given equation is $x(x^2+3|x|+5|x-1|+6|x-2|) = 0$.This implies either $x = 0$ or $x^2+3|x|+5|x-1|+6|x-2| = 0$.For the second part,let $f(x) = x^2+3|x|+5|x-1|+6|x-2|$.Since $x^2 \ge 0$,$3|x| \ge 0$,$5|x-1| \ge 0$,and $6|x-2| \ge 0$,the sum $f(x)$ is always non-negative.Specifically,$f(x) = 0$ only if all terms are zero simultaneously,which is impossible as $x^2=0 \implies x=0$,but at $x=0$,$f(0) = 0^2 + 3(0) + 5|0-1| + 6|0-2| = 0 + 0 + 5 + 12 = 17 
eq 0$.Thus,the only real solution is $x = 0$.Therefore,the number of real solutions is $1$.

The number of real solutions of the equation $x(x^2+3|x|+5|x-1|+6|x-2|)=0$ is

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