Question
Let $$f\left( x \right) = - 1 + \left| {x - 1} \right|,\, - 1 \leqslant x \leqslant 3$$ and $$ \leqslant g\left( x \right) = 2 - \left| {x + 1} \right|,\, - 2 \leqslant x \leqslant 2,$$ then $$\left( {fog} \right)\left( x \right)$$ is equal to :
A.
\[\left\{ \begin{array}{l}
x + 1\,\,\,\,\, - 2 \le x \le 0\\
x - 1\,\,\,\,\,\,\,\,\,\,0 < x \le 2
\end{array} \right.\]
B.
\[\left\{ \begin{array}{l}
x - 1\,\,\,\,\, - 2 \le x \le 0\\
x + 1\,\,\,\,\,\,\,\,\,\,\,0 < x \le 2
\end{array} \right.\]
C.
\[\left\{ \begin{array}{l}
- 1 - x\,\,\,\,\, - 2 \le x \le 0\\
\,\,\,x - 1\,\,\,\,\,\,\,\,\,\,\,\,0 < x \le 2
\end{array} \right.\]
D.
none of these
Answer :
none of these
Solution :
\[\begin{array}{l}
\left( {fog} \right)\left( x \right) = \left\{ \begin{array}{l}
f\left( {x + 3} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \le x + 3 \le 2\\
f\left( { - x + 1} \right),\,\,\,\, - 1 \le - x + 2 \le 2
\end{array} \right.\\
= \left\{ \begin{array}{l}
f\left( {x + 3} \right),\,\,\,\,\,\,\,\,\,\,\,\,1 \le x + 3 \le 2\\
f\left( { - x + 1} \right),\,\,\, - 1 \le - x + 1 \le 1\\
f\left( { - x + 1} \right),\,\,\,\,\,\,\,\,1 \le - x + 1 \le 2
\end{array} \right.\\
= \left\{ \begin{array}{l}
\,\,\,\,\,x + 1,\,\,\,\,\, - 2 \le x \le - 1\\
- x - 1,\,\,\,\,\, - 1 \le x \le 0\\
\,\,\,\,\,x - 1,\,\,\,\,\,\,\,\,\,\,\,0 \le x \le 2
\end{array} \right.
\end{array}\]