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Question
In the following case, state whether the function is one-one, onto or bijective. Justify your answer.
f : R → R defined by f(x) = 1 + x2
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Solution
f : R → R defined by f(x) = 1 + x2
Let x1, x2 ∈ R such that f(x1) = f(x2)
⇒ `1 + x_1^2 = 1 + x_2^2`
⇒ `x_1^2 = x_2^2`
⇒ x1 = ±x2
∴ f(x1) = f(x2) does not imply that x1 = x2.
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 + x2 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 and hence not bijective.
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