Question

Let $f\left(x\right)=-1+\left|x-1\right|,-1\le x\le 3$ and $g\left(x\right)=2-\left|x+1\right|,-2\le x\le 2$. Then $f\circ g\left(x\right)=$

• A. $x+1$
• B. $-1-x$
• C. $x-1$
• D. $$\begin{gathered} x+1if-2\le x\le -1 \\ -1-xif-1\le x\le 0 \\ x-1if0\le x\le 2 \end{gathered}$$

Answer 1

The correct option is D

$$\begin{gathered} x+1if-2\le x\le -1 \\ -1-xif-1\le x\le 0 \\ x-1if0\le x\le 2 \end{gathered}$$

Explanation for the correct answer:

$$\begin{gathered} f\left(x\right)=-1+|x-1| \\ g\left(x\right)=2-|x+1| \\ f\left(g\right(x\left)\right)=-1+|g(x)–1|,-1\le g\left(x\right)\le 3 \\ =-1+|2–|x+1|-1|,-1\le 2–|x+1|\le 3,-2\le x\le 2 \\ =-1+|1–|x+1|,-1\le 2–|x+1|\le 3,-2\le x\le 2 \end{gathered}$$

Now,$-1\le 2–|x+1|\le 3$

$$\begin{gathered} \Rightarrow -3\le –|x+1|\le 1 \\ \Rightarrow -1\le |x+1|\le 3 \\ \Rightarrow 0\le |x+1|\le 3 \\ \Rightarrow -3\le x+1\le 3 \\ \Rightarrow -4\le x\le 2 \end{gathered}$$

$f\circ g\left(x\right)=-1+|1–|x+1\left|\right|,-2\le x\le 2$

That is

$$\begin{gathered} x+1if-2\le x\le -1 \\ -1-xif-1\le x\le 0 \\ x-1if0\le x\le 2 \end{gathered}$$

Hence, the correct option is (D)