Question

Solution of the inequality $\left( \frac{1}{3} \right)^{\frac{|x + 2|}{2 - |x|}}$> 9 :

• A. (0, 2)
• B. (2, 4)
• C. (2, 6)
• D. (4, 6)

Answer 1

Answer: (C)

(2, 6)

Answer 2

$3^{- \frac{|x + 2|}{2 - |x|}}$ > 3²

⇒$\frac{- |x + 2|}{2 - |x|}$ > 2 ⇒ $\frac{|x + 2|}{2 - |x|}$ < – 2

Case I : $\underline{x < - 2}$⇒$\frac{- (x + 2)}{2 + x}$< – 2 ⇒ – 1 < – 2

which is not true

Case II : – 2 < x < 0 ⇒ $\frac{x + 2}{2 + x}$ < – 2⇒ 1 < – 2

which is not true

Case III : x ≥ 0 ⇒ $\frac{x + 2}{2 - x}$ < – 2, or $\frac{x + 2}{2 - x}$ + 2 < 0

⇒ $\frac{x + 2 + 4 - 2x}{2 - x}$< 0 or $\frac{(6 - x)(2 - x)}{(2 - x)^{2}}$< 0

⇒ (x – 2) (x – 6) < 0 ⇒ $\underline{2 < x < 6}$.