Question

The solution set of $x^2+2\le 3x\le 2x^2-5$, is

• A. $\varphi$
• B. $[1,2]$
• C. $(\infty ,1)\cup [5/2,\infty )$
• D. None of these

Answer 1

The correct option is A $\varphi$

$x^2+2\le 3x\le 2x^2-5$

a. $x^2+2\le 3x$

$x^2-3x+2\le 0$

$x^2-x-2x+2\le 0$

$x(x-1)-2(x-1)\le 0$

$(x-1)(x-2)\le 0$

$x\le 1$ $x\le 2$

b. $3x\le 2x^2-5$

$2x^2-3x-5\ge 0$

$2x^2+2x-5x-5\ge 0$

$2x(x+1)-5(x+1)\ge 0$

$(x+1)(2x-5)\ge 0$

$x\ge -1$ $x\ge \frac{5}{2}$

Answer: A. [$\varphi$].