Question

lf $f:[-6,6]\to R$ is defined by $f\left(x\right)=x^2-3$ for $x\in R$ then
$\left(fofof\right)(-1)+\left(fofof\right)\left(0\right)+\left(fofof\right)\left(1\right)=$

• A. $f\left(4\sqrt{2}\right)$
• B. $f\left(3\sqrt{2}\right)$
• C. $f\left(2\sqrt{2}\right)$
• D. $f\left(\sqrt{2}\right)$

Answer 1

The correct option is A $f\left(4\sqrt{2}\right)$

$fofof(-1)=fof(-2)=f\left(1\right)=-2$
$fofof\left(1\right)=fof(-2)=f\left(1\right)=-2$
$fofof\left(0\right)=fof(-3)=f\left(0\right)=33$
$\therefore fofof(-1)+fofof\left(1\right)+fofof\left(0\right)=29$
$=32-3$
$=(4\sqrt{2}{)}^2-3$
$\Rightarrow f\left(4\sqrt{2}\right)=(4\sqrt{2}{)}^2-3$