Question

Let $f\left(x\right)=\{\begin{array}{cc}1+x,& 0\le x\le 2\\ 3-x,& 2<x\le 3\end{array}$ then $f\left\{f\right(x\left)\right\}=$

• A. $\{\begin{array}{cc}2+x,& 0\le x\le 1\\ 2-x,& 1<x\le 2\\ 4-x,& 2<x\le 3\end{array}$
• B. $\{\begin{array}{cc}2+x,& 0\le x\le 2\\ 4-x,& 2<x\le 3\end{array}$
• C. $\{\begin{array}{cc}2+x,& 0\le x\le 2\\ 2-x,& 2<x\le 3\end{array}$
• D. none of these

Answer 1

The correct option is A $\{\begin{array}{cc}2+x,& 0\le x\le 1\\ 2-x,& 1<x\le 2\\ 4-x,& 2<x\le 3\end{array}$

$f\left(x\right)=\{\begin{array}{cc}1+x,& 0\le x\le 2\\ 3-x,& 2<x\le 3\end{array}f\left\{f\right(x\left)\right\}=\{\begin{array}{cc}1+f\left(x\right),& 0\le f\left(x\right)\le 2\\ 3-f\left(x\right),& 2<f\left(x\right)\le 3\end{array}=\{\begin{array}{cc}1+(1+x),& 0\le 1+x\le 2\&0\le x\le 2\\ 3-(1+x),& 2<1+x\le 3\&0\le x\le 2\\ 1+(3-x),& 0\le 3-x\le 2\&2<x\le 3\\ 3-(3-x),& 2<3-x\le 3\&2<x\le 3\end{array}=\{\begin{array}{cc}2+x,& 0\le x\le 1\\ 2-x,& 1<x\le 2\\ 4-x,& 2<x\le 3\end{array}$