Question

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

• 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

$f\left(4\sqrt{2}\right)$

Explanation for the correct option:

STEP 1: Given: $f\left(x\right)=x^2-3$, $x\in \mathrm{ℝ}$.

Now,

$\begin{array}{rcl}\left(fof\right)\left(x\right)& =& f\left(f\left(x\right)\right)\\ & =& f\left(x^2-3\right)\\ & =& {\left(x^2-3\right)}^2-3\\ & =& x^4+9-6x^2-3\\ & =& x^4-6x^2+6\end{array}$

STEP 2 using above formula to get $\begin{array}{rcl}& & \left(fofof\right)\left(x\right)\end{array}$

$\begin{array}{rcl}\left(fofof\right)\left(x\right)& =& f\left(f\left(f\left(x\right)\right)\right)\\ & =& \left(fof\right)\left(f\left(x\right)\right)\\ & =& {\left(x^2-3\right)}^4-6{\left(x^2-3\right)}^2+6\end{array}$

STEP 3 When $x=-1$, then $\begin{array}{rcl}& & \left(fofof\right)\end{array}\begin{array}{rcl}& & \left(x\right)\end{array}$is

$\begin{array}{rcl}\left(fofof\right)\left(-1\right)& =& {\left({\left(-1\right)}^2-3\right)}^4-6{\left({\left(-1\right)}^2-3\right)}^2+6\\ & =& {\left(1-3\right)}^4-6{\left(1-3\right)}^2+6\\ & =& {\left(-2\right)}^4-6{\left(-2\right)}^2+6\\ & =& 16-24+6\\ & =& -2\end{array}$

STEP 4 When $x=0$, then $\begin{array}{rcl}& & \left(fofof\right)\left(x\right)\end{array}$ is

$\begin{array}{rcl}\left(fofof\right)\left(0\right)& =& {\left({\left(0\right)}^2-3\right)}^4-6{\left({\left(0\right)}^2-3\right)}^2+6\\ & =& {\left(-3\right)}^4-6{\left(-3\right)}^2+6\\ & =& 81-54+6\\ & =& 33\end{array}$

STEP 5 When $x=1$, then $\begin{array}{rcl}& & \left(fofof\right)\left(x\right)\end{array}$ is

$\begin{array}{rcl}\left(fofof\right)\left(1\right)& =& {\left({\left(1\right)}^2-3\right)}^4-6{\left({\left(1\right)}^2-3\right)}^2+6\\ & =& {\left(1-3\right)}^4-6{\left(1-3\right)}^2+6\\ & =& {\left(-2\right)}^4-6{\left(-2\right)}^2+6\\ & =& 16-24+6\\ & =& -2\end{array}$

Thus, from STEP 3, 4 and 5, we get

$\begin{array}{rcl}\left(fofof\right)\left(-1\right)& =& -2\\ \left(fofof\right)\left(0\right)& =& 33\\ \left(fofof\right)\left(1\right)& =& -2\end{array}$

STEP 6 Substituting values from STEP 3, 4 and 5 in given expression:

$\begin{array}{rcl}\left(fofof\right)\left(-1\right)+\left(fofof\right)\left(0\right)+\left(fofof\right)\left(1\right)& =& -2+33-2\\ & =& 29\end{array}$

Now it is given that $f\left(x\right)=x^2-3$ for $x\in \mathrm{ℝ}$, thus

$\begin{array}{rcl}f\left(4\sqrt{2}\right)& =& {\left(4\sqrt{2}\right)}^2-3\\ & =& 32-3\\ & =& 29\end{array}$

$\begin{array}{rcl}\therefore \mathbf{\left(}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{\right)}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{+}\mathbf{\left(}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{\right)}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{o}\mathbf{f}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{\right)}& \mathbf{=}& \mathbf{29}\\ & \mathbf{=}& \mathit{f}\mathbf{\left(}\mathbf{4}\sqrt{\mathbf{2}}\mathbf{\right)}\end{array}$

Hence, option (A) is correct.