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Question
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.
Solution
f: R → R is given by,
`f(x) = |x| = [(x,if x>= 0), (-x, if x < 0)]`
It is seen that `f(-1) = |-1| = 1, f(1) = |1| = 1`
∴f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = |x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x) =|x| = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
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