Question

Consider the function, $f\left(x\right)=x+1,0\le x\le 2,$ then $fof\left(x\right)$ is

• A. $x+2,0\le x\le 1$
• B. $x+2,0\le x\le 2$
• C. $x+2,0\le x\le 4$
• D. $x+2,0\le x\le 3$

Answer 1

The correct option is A $x+2,0\le x\le 1$

$fof\left(x\right)=f\left(f\right(x\left)\right)=f\left(x\right)+1,0\le f\left(x\right)\le 2$

$\Rightarrow fof\left(x\right)=(x+1)+1;0\le x+1\le 2\text{and}0\le x\le 2$

$\Rightarrow fof\left(x\right)=x+2,-1\le x\le 1\text{and}0\le x\le 2$

$\therefore fof\left(x\right)=x+2,x\in [0,1]$