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Question
Following the case, state whether the function is one-one, onto, or bijective. Justify your answer.
f: R → R defined by f(x) = 1 + x2
Solution
f: R → R is defined as
`f(x) = 1 + x^2`
Let `x_1, x_2 in "R such that " f(x_1) = f(x_2)`
`=> 1 + x_1^2 = 1 + x_2^2`
`=> x_1^2 = x_2^2`
`=> x_1 = +-x_2`
∴ `f(x_1) = f(x_2)` does not imply that `x_1 = x_2`
For instance,
f(1) = f(-1) = 2
∴ f is not one-one.
Consider the element −2 in co-domain R.
It is seen that `f(x) = 1 + x^2` is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, f is neither one-one nor onto.
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