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Question Number 18653 by Tinkutara last updated on 26/Jul/17 | ||
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{inequality},\:\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{4} \\ $$ | ||
Answered by ajfour last updated on 26/Jul/17 | ||
$$\mathrm{If}\:\mathrm{x}<−\mathrm{1} \\ $$$$−\left(\mathrm{x}−\mathrm{1}\right)−\left(\mathrm{x}+\mathrm{1}\right)<\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2x}>−\mathrm{4}\:\:\mathrm{or}\:\:\mathrm{x}>\:−\mathrm{2} \\ $$$$\mathrm{hence}\:\mathrm{x}\in\left(−\mathrm{2}\:,−\mathrm{1}\right) \\ $$$$\mathrm{If}\:−\mathrm{1}\leqslant\mathrm{x}<\mathrm{1} \\ $$$$−\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{x}+\mathrm{1}<\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2}<\mathrm{4}\: \\ $$$$\mathrm{so}\:\:\mathrm{x}\in\left[−\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{If}\:\mathrm{1}\leqslant\mathrm{x} \\ $$$$\mathrm{x}−\mathrm{1}+\mathrm{x}+\mathrm{1}<\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2x}<\mathrm{4}\:\:\mathrm{or}\:\:\:\mathrm{x}<\mathrm{2} \\ $$$$\:\:\mathrm{x}\in\left[\mathrm{1},\mathrm{2}\right) \\ $$$$\mathrm{Taking}\:\mathrm{union}\:\mathrm{of}\:\mathrm{the}\:\mathrm{three}\:\mathrm{cases}, \\ $$$$\:\:\:\:\:\mathrm{x}\in\left(−\mathrm{2},\mathrm{2}\right)\:. \\ $$ | ||
Commented by Tinkutara last updated on 26/Jul/17 | ||
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{ajfour}\:\mathrm{Sir}! \\ $$ | ||
Commented by ajfour last updated on 26/Jul/17 | ||