Q. If $f:\lbrack - 6,\ 6\rbrack \rightarrow R$ 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) $f(\sqrt{2})$
Explanation
(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, $f(0) = 0^{2} - 3 = - 3$
$\Rightarrow \ \ \ fof\ (0) = f( - 3) = ( - 3)^{2} - 3 = 6$
$\Rightarrow \ \ \ \ fof(0) = f(6) = 6^{2} - 3 = 33$
Again, $f(1) = 1^{2} - 3 = - 2$
$\Rightarrow \ \ \ \ \ \ \ fof(1) = f( - 2) = ( - 2)^{2} - 3 = 1$
$\Rightarrow \ \ \ \ fofof( - 1) + fofof(0) + fofof(1) = - 2 + 33 - 2 = 29$
Now, $f\left( 4\sqrt{2} \right) = \left( 4\sqrt{2} \right)^{2} - 3 = 32 - 3 = 29$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.