Question-164050

Question Number 164050 by mathlove last updated on 13/Jan/22

Answered by Ar Brandon last updated on 13/Jan/22

x - \frac{1}{8} > 0 \neq 1 \Rightarrow x > \frac{1}{8} \neq \frac{9}{8} \dots . (1)

x^{2} - \frac{1}{9} > 0 \Rightarrow x < - \frac{1}{3} \cup x > \frac{1}{3} \dots (2)

\sin^{- 1} (2 - x) > 0 \Rightarrow 2 - x > 0 \Rightarrow x < 2 \dots (3)

(1), (2), (3)

\Rightarrow S = {x : \frac{1}{3} < x < \frac{9}{8} \cup \frac{9}{8} < x < 2}