Question

Solve: ${\log }_{\left[\left(2/3\right)|x-2|\right]}{2}^{(1-x^2)}\ge 0$

• A. $x\in [(-1,\frac{1}{2})\cup [(1,2)\cup (2,\frac{7}{2})]]$
• B. $x\in [(-3,\frac{1}{2})\cup [(1,2)\cup (2,\frac{7}{2})]]$
• C. $x\in [(-1,\frac{15}{2})\cup [(1,2)\cup (2,\frac{7}{2})]]$
• D. None of these

Answer 1

The correct option is A $x\in [(-1,\frac{1}{2})\cup [(1,2)\cup (2,\frac{7}{2})]]$

Given expression is

$log[\frac{2}{3}|x-2|{]}^{2(1-x^2)}\ge 0$

$\to 2(1-x^2)log[\frac{2}{3}|x-2|]\ge 0$

$\to 1-x^2\ge 0to-1\le x\le 1$

also $\frac{2}{3}|x-2|\ge 0$

$x\ge 2,x\le \frac{7}{2}$

$x\ne 2$ Also If we put the value of x from $\frac{1}{2}$ to 2

we get negative value but domain of logrithmic function is positive

So

$x\ni \left[\right(-1,\frac{1}{2})\cup [(1,2)\cup (2,\frac{7}{2})\left]\right]$