Let `f(x)` be defined on `[-2,2]` and be given by
`f(x)={(-1",",-2 le x le 0),(x-1",",1 lt x le 2):} and g(x)=f(|x|) +|f(x)|`.
Then find `g(x)`.

Options

A. [-2,2]`
B. `[-2,0) uu (0,2]`
C. `[-2,1) uu (1,2]`
D. `[-2,0) uu (0,1) uu (1,2]`

Answer

D

Explanation

We have
`f(x)={{:(,-1,-2 le x le 0),(,x-1,0 lt x le2):}`
`therefore |f(x)|={{:(,1,-2 le x le 0),(,|x-1|,0 lt x le2):}`
`={{:(,1,-2 le xle 0),(,1-x,0 lt x lt 1),(,x-1,1 le x le 2):}`
and `|f(|x|)=|x|-1,"for "-2 le x le 2`
`{{:(-x,1,-2le xlt 0),(,x-1,0 le x le 2):}`
`therefore g(x)=|f(x)|+f(|x|)`
`Rightarrow g(x)={{:(,-x,-2 le x lt 0),(,0,0 le x lt 1),(,2(x-1),1 le x le 2):}`
Clearly, g(x) is a continuous at x=0,1
Also,
`("LHD at x=0")=((d)/(dx)(-x))_("at x=0")=-1`
`("RHD at x=0")=((d)/(dx)(0))_("at x=0")=0`
`("LHD at x=1")=((d)/(dx)(0))_("at x=0")=0`
`and ("RHD at x=1")=((d)/(dx)[2(x-1)])_("at x=1")=2`
`therefore ("LHD at x=0") ne ("RHD at x=0")`
`and ("LHD at x=1") ne ("RHD at x=1")`
So, g(x) is not differenetiable at x=0,1.
Now,
`("RHD at x=-2")=((d)/(dx)(-x))_("at x=-2")=-1` exist finitely.
Similarly, ("LHD at x=2")=2, exists finitely
Hence, g(x) is differentiable at all points in the interval [-1,1] except at x=0 and x=1