MCQ
$\frac{|x-1|}{x-1} \leq 0$ then $x \in$
- ✓$(-\infty, 1)$
- B$(1, \infty)$
- C$(-1,1)$
- D$\phi$
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$(1,1)(-1,1),(1,-1),(-1,-1),(-2,1)(2,-1),(-1,2)$ and $(-2,-1)$
$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because
$STATEMENT -2$$: \sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.