1.

Let `f(x0=|x-1|dot`Then`f(x^2)=(f(x))^2`(b) `f(x+y)=f(x)+f(y)``f(|x|)-|f(x)|`(d) none of theseA. `f(x^(2))={f(x)}^(2)`B. `f(x+y)=f(x)+f(y)`C. `f(|x|)=|f(x)|`D. None of the above

Answer» Correct Answer - D
Given, `f(x)=|x-1|`
` therefore f(x^(2))=|x^(2)-1|`
`and {f(x)}^(2)=(x-1)^(2)`
`rArr f(x^(2)) ne (f(x))^(2)`, hence (a) is false.
Also, `f(x+y)=|x+y-1|`
` and f(x)=|x-1|`,
`f(y)=|y-1|`
`rArr f(x+y) ne f(x) +f(y),` hence (b) is false.
`f(|x|)=||x|-1|`
` and |f(x) |=||x-1||=|x-1|`
` therefore f(|x|) ne |f(x)|,` hence (c) is false.


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