If frac{|x-1|}{x-1} leq 0,then x in | vedclass

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(A) Given the inequality $\frac{|x-1|}{x-1} \leq 0$.For the expression to be defined,the denominator must not be zero,so $x - 1 
eq 0$,which implies

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$(-\infty, 1)$

Vedclass · Answer

(A) Given the inequality $\frac{|x-1|}{x-1} \leq 0$.For the expression to be defined,the denominator must not be zero,so $x - 1 
eq 0$,which implies $x 
eq 1$.Case $1$: If $x > 1$,then $|x-1| = x-1$. The expression becomes $\frac{x-1}{x-1} = 1$. Since $1 
ot\leq 0$,there is no solution in this interval.Case $2$: If $x Therefore,the solution set is $x \in (-\infty, 1)$.

If $\frac{|x-1|}{x-1} \leq 0$,then $x \in$

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