Question

If ${\log }_{[2\mid x-2\mid /3]}{2}^{(1-x^2)}\ge 0$, then $x\in$

• A. $x\in [-1,2)\cup (2,\frac{7}{2})$
• B. $x\in [-1,-\frac{1}{2})$
• C. $x\in [-1,\frac{1}{2})$
• D. $x\in \left\{\right[-1,0.5)\cup [1,2)\cup (2,\frac{7}{2}\left)\right\}$

Answer 1

The correct option is D $x\in \left\{\right[-1,0.5)\cup [1,2)\cup (2,\frac{7}{2}\left)\right\}$

Given: $\frac{\log 2^{(1-x^2)}}{\log \left(2|x-2|/3\right)}\ge 0$

$\frac{\log 2^{(1-x^2)}}{\log \left(\frac{2|x-2|}{3}\right)}$

Note $x\ne 2$ as $\log 0\to -\infty$

Also $x\ne 0.5$ as $\log 1=0$

$x\ne 3.5$

Numerator is positive for $-1\le x\le 1$

Denominator is positive for$x<0.5$ and $x>3.5$

Numerator is negative for $x\le -1$ and $x\ge 1$

Denominator is negative for$3.5>x>0.5$

$\therefore$ Taking intersections we get $x\in \left\{\right[-1,0.5)\cup [1,2)\cup (2,3.5\left)\right\}$