Functions Question 743
Let $ f(x)=-1+| x-1 |,\ -1\le x\le 3 $ and $ g(x)=2-| x+1 |,\ -2\le x\le 2, $ then $ (f\circ g)(x) $ , is equal to
Options:
A) $ \begin{cases} x+1-2\le x\le 0 \\ x-10<x\le 2 \\ \end{cases} . $
B) $ \begin{cases} x-1-2\le x\le 0 \\ x+10<x\le 2 \\ \end{cases} . $
C) $ \begin{cases} -1 - x - 2 \le x \le 0 \\ x - 10 < x \le 2 \\ end{cases} . $
D) None of these
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Answer:
Correct Answer: D
Solution:
$ (fog)(x)= \begin{cases} f(x+3),1\le x+3\le 2 \\ f(-x+1),-1\le -x+1\le 1 \\ \end{cases} . $ $ = \begin{cases} \begin{aligned} & f(x+3),1\le x+3\le 2 \\ & f(-x+1),,-1\le -x+1\le 1 \\ \end{aligned} \\ f(-x+1),,1\le -x+1\le 2 \\ \end{cases} . $ $ = \begin{cases} \begin{aligned} & x+1,-2\le x\le -1 \\ & -x-1,,-1\le x\le 0 \\ \end{aligned} \\ x-1,,0\le x\le 2 \\ \end{cases} . $
BETA
