Question

Let $f\left(x\right)=x^2+20x+90$, then the number of real solutions of the equation f(f(f(x))) = 0 is

• A. 6
• B. 4
• C. 2
• D. \N

Answer 1

The correct option is C 2

$f\left(x\right)=(x+10{)}^2-10$

$f\left(f\right(f\left(x\right))=0\Rightarrow f(f\left(x\right))=\alpha ,\beta ,where\alpha <-10and-10<\beta <0$
$f\left(f\right(x\left)\right)=\beta$, (no $So{l}^n$ from $\alpha )$
$f\left(x\right)=\gamma ,\delta (\delta <-10and-10<\gamma <0)$
$f\left(x\right)=\gamma$
$\underset{\underset{\text{two solutions}}{\downarrow }}{f\left(x\right)=\gamma },\underset{\underset{\text{So no solution}}{\downarrow }}{f\left(x\right)=\delta }(\delta <-10)$
So f(f(f(x))) = 0 has two solutions.