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Question Number 178635 by infinityaction last updated on 19/Oct/22
solution set of  log_x^(2   )  ((x/(∣x∣))−x)≥0
$$\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\:\mathrm{log}_{\mathrm{x}^{\mathrm{2}\:\:\:} } \left(\frac{\mathrm{x}}{\mid\mathrm{x}\mid}−\mathrm{x}\right)\geqslant\mathrm{0} \\ $$
Commented by Frix last updated on 19/Oct/22
x≤−2∨0<x<1
$${x}\leqslant−\mathrm{2}\vee\mathrm{0}<{x}<\mathrm{1} \\ $$
Commented by infinityaction last updated on 20/Oct/22
sir solution
$${sir}\:{solution} \\ $$
Answered by mr W last updated on 20/Oct/22
for x^2 ≥1: (x/(∣x∣))−x≥1  for x>0:  1−x≥1 ⇒x≤0 ⇒contradition!  for x<0:  −1−x≥1 ⇒x≤−2 ✓    for x^2 ≤1: 0<(x/(∣x∣))−x≤1  for x>0:  0<1−x≤1 ⇒x≥0 ⇒1>x>0 ✓  for x<0:  0<−1−x≤1 ⇒−2 ≤x<−1 ⇒contradiction    ⇒solution: x≤−2 or 0<x<1
$${for}\:{x}^{\mathrm{2}} \geqslant\mathrm{1}:\:\frac{{x}}{\mid{x}\mid}−{x}\geqslant\mathrm{1} \\ $$$${for}\:{x}>\mathrm{0}: \\ $$$$\mathrm{1}−{x}\geqslant\mathrm{1}\:\Rightarrow{x}\leqslant\mathrm{0}\:\Rightarrow{contradition}! \\ $$$${for}\:{x}<\mathrm{0}: \\ $$$$−\mathrm{1}−{x}\geqslant\mathrm{1}\:\Rightarrow{x}\leqslant−\mathrm{2}\:\checkmark \\ $$$$ \\ $$$${for}\:{x}^{\mathrm{2}} \leqslant\mathrm{1}:\:\mathrm{0}<\frac{{x}}{\mid{x}\mid}−{x}\leqslant\mathrm{1} \\ $$$${for}\:{x}>\mathrm{0}: \\ $$$$\mathrm{0}<\mathrm{1}−{x}\leqslant\mathrm{1}\:\Rightarrow{x}\geqslant\mathrm{0}\:\Rightarrow\mathrm{1}>{x}>\mathrm{0}\:\checkmark \\ $$$${for}\:{x}<\mathrm{0}: \\ $$$$\mathrm{0}<−\mathrm{1}−{x}\leqslant\mathrm{1}\:\Rightarrow−\mathrm{2}\:\leqslant{x}<−\mathrm{1}\:\Rightarrow{contradiction} \\ $$$$ \\ $$$$\Rightarrow{solution}:\:{x}\leqslant−\mathrm{2}\:{or}\:\mathrm{0}<{x}<\mathrm{1} \\ $$
Commented by infinityaction last updated on 21/Oct/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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