Question

If $f:[-6,6] \rightarrow R$ 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,
$f(4 \sqrt{2})=(4 \sqrt{2})^{2}-3=32-3$
$=29$