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Let f(x)=|x-1|dot Then (a)f(x^2)=(f(x))^...

Let `f(x)=|x-1|dot` Then (a)`f(x^2)=(f(x))^2` (b) `f(x+y)=f(x)+f(y)` (c)`f(|x|)=|f(x)|` (d) none of these

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Given that,
`f(x)=|x-1|`
Since, `abs(x^2-1)!=|x-1|^2`
`f(x^2) !=(f(x)^2)`
...
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