Question

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

A
(,+)
B
(,0]
C
[0,5]
D
(,0][5,)

Solution

## The correct option is C $$\displaystyle (- \infty, \:0] \cup [5, \: \infty)$$$$f(x)^{2}-f(x)-6\geq0$$Upon factorization we get$$(f(x)-3)(f(x)+2)\geq 0$$ ...(i)Now we know that $$f(x)<3$$ includes a solution set $$(0,\infty)$$ and $$f(x)>-2$$ implies a solution set $$(-\infty,5)$$Hence from one, we get the solution set of the above inequality as$$(-\infty,0]\cup[5,\infty)$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More