Question

The set values of $x$ satisfying inequality $\sqrt{x^2-x-6}<2x-3$ is

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

Answer 1

The correct option is D $[3,\infty )$

$\sqrt{x^2-x-6}\ge 0$

$(x-3)(x+2)\ge 0$

$x\le -2,x\ge 3$

$2x-3>0$

$x>\frac{3}{2}$

$\sqrt{x^2-x-6}<2x-3$

$\Rightarrow x\ge 3$

$\therefore x[3,\infty )$