Question

If the solution set for $f\left(x\right)<3$ is $(0,\infty )$ and the solution set for $f\left(x\right)>-2$ is $(-\infty ,5)$, then the true solution set for ${\left(f\left(x\right)\right)}^2\ge f\left(x\right)+6$, is

• A. $(-\infty ,+\infty )$
• B. $(-\infty ,0]$
• C. $[0,5]$
• D. $(-\infty ,0]\cup [5,\infty )$

Answer 1

Answer: (D)

$(-\infty ,0]\cup [5,\infty )$

$f(x{)}^2-f(x)-6\ge 0$
Upon factorization we get
$\left(f\right(x)-3)\left(f\right(x)+2)\ge 0$ ...(i)
Now we know that $f\left(x\right)<3$ includes a solution set $(0,\infty )$ and $f\left(x\right)>-2$ implies a solution set $(-\infty ,5)$
Hence from one, we get the solution set of the above inequality as
$(-\infty ,0]\cup [5,\infty )$