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Question Number 115882 by Khalmohmmad last updated on 29/Sep/20

Answered by PRITHWISH SEN 2 last updated on 29/Sep/20

sgn(x+1)=1  when x>−1  sgn(x+1)=0  when x+1=0  sgn(x+1)=−1 when x<−1  ∵ denominator ≠0 ∴ sgn(x+1)≠ 1  ∴ domain x∈ (−∞,−1]

$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{1}\:\:\mathrm{when}\:\mathrm{x}>−\mathrm{1} \\ $$$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{0}\:\:\mathrm{when}\:\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=−\mathrm{1}\:\mathrm{when}\:\mathrm{x}<−\mathrm{1} \\ $$$$\because\:\mathrm{denominator}\:\neq\mathrm{0}\:\therefore\:\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)\neq\:\mathrm{1} \\ $$$$\therefore\:\mathrm{domain}\:\mathrm{x}\in\:\left(−\infty,−\mathrm{1}\right] \\ $$

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