Question

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

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

Answer 1

Answer: (A)

Given, $f(x)=x^2-3$
Now, $f(-1)=(-1{)}^2-3=-2$
$\Rightarrow$ fof$(-1)=f(-2)=(-2{)}^2-3=1$
$\Rightarrow$ fofof $(-1)=f(1)=1^2-3=-2$
Now,
$f(0)=0^2-3=-3$
$\Rightarrow$ fof $(0)=f(-3)=(-3{)}^2-3=6$
$\Rightarrow$ fofof $(0)=f(6)=6^2-3=33$
Again,
$f(1)=1^2-3=-2$
$\Rightarrow$ fof $(1)=f(-2)=(-2{)}^2-3=1$
$\Rightarrow$ fofof $(1)=(1{)}^2-3=-2$
$\therefore$ fofof $(-1)+$ foof $(0)+$ fof $(1)$
$=-2+33-2=29$
Now,
$f(4\sqrt{2})=(4\sqrt{2}{)}^2-3=32-3$
$=29$