Question

The set of real numbers r satisfying
$\frac{3r^2-8r+5}{4r^2-3r+7}>0$ is

• A. the set of all real numbers
• B. the set of all positive real numbers
• C. the set of all real numbers strictly between 1 and 5/3
• D. the set of all real numbers which are either less than 1 or greater than 5/3

Answer 1

The correct option is D the set of all real numbers which are either less than 1 or greater than 5/3

$\frac{3.r^2-8.r+5}{4.r^2-3.r+7}>0$
$\Rightarrow 3r^2-3r-5r+5>0$ and $4r^2-3r+7>0$ as D < 0
$\Rightarrow 3r(r-1)-5(r-1)>0$
$\Rightarrow (3r-5)(r-1)>0$
$(3r-5)(r-1)>0$
$r\epsilon (-\infty ,1)\cup (5/3,\infty )$