Question

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

• 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 $f\left(4\sqrt{2}\right)$

Since, $f\left(x\right)=x^2-3$

$\therefore f(-1)=(-1{)}^2-3=1-3=-2,$

$f\left(0\right)=0-3=-3$ and $f\left(1\right)=1^2-3=-2$

Consider, $(f\circ f\circ f)(-1)$ $=\left(f\right(f\left(f\right)\left)\right)(-1)=\left(f\right(f\left(f\right(-1\left)\right)\left)\right)=\left(f\right(f(-2))=(f\left(\right(-2{)}^2-3\left)\right)=\left(f\right(4-3\left)\right)=\left(f\right(1\left)\right)=-2$.......(1)

Now, consider $(f\circ f\circ f)\left(0\right)$

$=\left(f\right(f\left(f\right)\left)\right)\left(0\right)=\left(f\right(f\left(f\right(0\left)\right)\left)\right)=\left(f\right(f(-3)\left)\right)=\left(f\right((-3{)}^2-3))=(f\left(6\right))=(6{)}^2-3=36-3=33$.......(2)

Again, consider $(f\circ f\circ f)\left(1\right)$

$=\left(f\right(f\left(f\right)\left)\right)\left(1\right)=\left(f\right(f\left(f\right(1\left)\right)\left)\right)=\left(f\right(f(-2)\left)\right)=\left(f\right((-2{)}^2-3))=(f\left(1\right))=-2$.......(3)

Adding (1), (2) and (3), we get

$(f\circ f\circ f)(-1)+(f\circ f\circ f)\left(0\right)+(f\circ f\circ f)\left(1\right)=-2+33-2=29$

Also,

$\left(f\right(4\sqrt{2}\left)\right)=(4\sqrt{2}{)}^2-3=32-3=29$

Hence, A will be correct answer.