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Show that the Modulus Function `f : R->R ,`given by `f (x) = | x |`, is neither one-one nor onto, where `| x |`is x, if x is positive or 0 and `| x |`is ` x`, if x is negative. |
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Answer» `f: R to R and f(x) = |x|` Let `x, y in R` and` " " f(x) = f(y)` `rArr " " |x| = |y|` `rArr " " x = pm y` `therefore f` is not one-one. Again `-1 in R` and there does not exist `x in R` for which `f(x)=-1` `therefore f ` is not onto. Therefore, `f` is neither one-one nor onto. |
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