Question

The solution set of $\frac{3x^2-7x+8}{x^2+1}\le 2$ is

• A. $[1,4]$
• B. $[1,2]$
• C. $[2,6]$
  
• D. $[1,6]$

Answer 1

The correct option is D $[1,6]$

$\frac{3x^2-7x+8}{x^2+1}\le 2\Rightarrow \frac{3x^2-7x+8}{x^2+1}-2\le 0\Rightarrow \frac{x^2-7x+6}{x^2+1}\le 0\Rightarrow \frac{(x-6)(x-1)}{x^2+1}\le 0$


$\therefore x\in [1,6]$