1-2-x-gt-1-x-1-

Question Number 164568 by leonhard77 last updated on 19/Jan/22

\frac{1}{\sqrt{2 - x}} > \frac{1}{x - 1}

Answered by TheSupreme last updated on 19/Jan/22

x < 2, x \neq 1

{\begin{cases} \frac{1}{x - 1} < 0 \cup {\begin{cases} \frac{1}{x - 1} > 0 \\ \frac{1}{2 - x} > \frac{1}{{(x - 1)}^{2}} \end{cases} \end{cases}

{\begin{cases} x < 1 \cup {\begin{cases} x > 1 \\ x^{2} - x > 0 \end{cases} \\ x < 2 \end{cases}

x < 1 \cup 1 < x < 2

x < 2, x \neq 1

Answered by alephzero last updated on 19/Jan/22

\frac{1}{\sqrt{2 - x}} > \frac{1}{x - 1}

\Rightarrow x \in (- \infty, 1) \cup (1, 2) \dots \dots \dots \dots \dots \dots (i)

\frac{1}{\sqrt{2 - x}} - \frac{1}{x - 1} > 0

\frac{x - 1 - \sqrt{2 - x}}{(x - 1) \sqrt{2 - x}} > 0

(i i) {\begin{cases} x - 1 - \sqrt{2 - x} > 0 \\ (x - 1) \sqrt{2 - x} > 0 \end{cases}

(i i i) {\begin{cases} x - 1 - \sqrt{2 - x} < 0 \\ (x - 1) \sqrt{2 - x} < 0 \end{cases}

(i i) {\begin{cases} x \in (\frac{1 + \sqrt{5}}{2}, + \infty) \\ x \in (1, 2) \cup (2, + \infty) \end{cases}

(i i i) {\begin{cases} x \in (- \infty, \frac{1 + \sqrt{5}}{2}) \\ x \in (- \infty, 1) \end{cases}

\Rightarrow x \in (\frac{1 + \sqrt{5}}{2}, 2) \cup (2, + \infty) \dots \dots \dots . . (i i)

x \in (- \infty, 1) \dots \dots \dots \dots \dots \dots \dots \dots \dots (i i i)

(i) \cup (i i) \cup (i i i) \Rightarrow

\Rightarrow x \in (- \infty, 1) \cup (\frac{1 + \sqrt{5}}{2}, 2)

Or x < 1, \frac{1 + \sqrt{5}}{2} < x < 2