Question

Solve the polynomial inequality. Express the solution set in in interval notation: ${\frac{(x-1)(3-x)}{x-2}}^2\le 0$

• A. $(-\infty ,-3]\cup (-2,-1)\cup [1,\infty )$
• B. $(-\infty ,1)\cup (2,\infty )$
• C. $(-\infty ,-3)\cup (-1,\infty )$
• D. $(-\infty ,1]\cup [3,\infty )$

Answer 1

The correct option is D $(-\infty ,1]\cup [3,\infty )$

given, ${\frac{(x-1)(3-x)}{x-2}}^2\le 0$

$\therefore x-2\ge 0$

$\therefore x\ge 2$

Also, $(x-1)(3-x{)}^2\le 0$

$\therefore x\le 1$ and $x\ge 3$

Hence the solution is, $(-\infty ,1)\cup (3,\infty )$