If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ…
If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then
$(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$
is equal to
$f(4 \sqrt{2})$
$f(3 \sqrt{2})$
$f(2 \sqrt{2})$
$f(\sqrt{2})$
Solution
Given, $\quad f(x)=x^2-3$
Now, $\quad f(-1)=(-1)^2-3=-2$
$\begin{aligned}
& \Rightarrow \quad f \circ f(-1)=f(-2)=(-2)^2-3=1 \\
& \Rightarrow \quad f \circ f \circ f(-1)=f(1)=1^2-3=-2
\end{aligned}$
Now,
$f(0)=0^2-3=-3$
$\begin{aligned}
& \Rightarrow \quad f \circ f(0)=f(-3)=(-3)^2-3=6 \\
& \Rightarrow \quad f \circ f \circ f(0)=f(6)=6^2-3=33
\end{aligned}$
Again,
$\begin{aligned}
& f(1)=1^2-3=-2 \\
& \Rightarrow \quad f \circ f(1)=f(-2)=(-2)^2-3=1 \\
& \Rightarrow \quad f \circ f \circ f(1)=(1)^2-3=-2 \\
& \therefore f \circ f \circ f(-1)+f \circ f \circ f(0)+f \circ f \circ f(1) \\
& =-2+33-2=29 \\
&
\end{aligned}$
Now,
$\begin{aligned}
f(4 \sqrt{2}) & =(4 \sqrt{2})^2-3=32-3 \\
& =29
\end{aligned}$
Hence, option (a) is correct.