If `f(x)=|x-1|" and "g(x)=f(f(f(x)))`, then for `xgt2,g'(x)` is equal to

Options

A. 1 for all `xgt2`
B. 1 for `2ltxlt3`
C. `-1" for "2ltxlt3`
D. none of these

Answer

C

Explanation

We have, `f(x)=|x-1|`
`:." "g(x)=f(f(f(x)))`
`implies" "g(x)=f(f(|x-1|))`
`implies" "g(x)=f(||x-1|-1|)`
`implies" "g(x)=|||x-1|-1|-1|`
`implies" "g(x)={{:(||x-2|-1|","," if "xge1),(||-x|-1|","," if "xlt1):}`
`implies" "g(x)={{:(|x-3|" ,"," if "xge2),(|2-x-1|","," if "1lexlt2),(|x-1|" ,"," if "0lexlt1),(|-x-1|" ,"," if "xlt0):}`
`implies" "g(x)={{:(x-3" ,"," if "xge3),(3-x" ,"," if "2lexlt3),(x-1" ,"," if "1lexlt2),(1-x" ,"," if "0lexlt1),(x+1" ,"," if "-1lexlt0),(-x-1","," if "xlt-1):}`
`implies" "g(x)={{:(1" ,"," if "xgt3),(-1" ,"," if "2ltxlt3),(1" ,"," if "1ltxlt2),(-1" ,"," if "0ltxlt1),(1" ,"," if "-1ltxlt0),(-1" ,"," if "xlt-1):}`
ALITER For `xgt2`, we have
`f(x)=|x-1|=x-1`
`:." "f(f(x))=f(x-1)=|(x-1)-1|`
`implies" "f(f(x))=|x-2|=x-2" "[becausexgt2]`
`implies" "f(f(f(x)))=f(x-2)=|x-2-1|={{:(x-3",",xge3),(3-x",",2lexlt3):}`
`implies" "g(x)={{:(x-3",",xge3),(3-x",",2lexlt3):}`
`implies" "g'(x)={{:(1" ,",xgt3),(-1",",2ltxlt3):}`