Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 1 / b / iv

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- Solution iv

Solution

Not injective; surjective. The equal values

f (0) = f (2) = 1

disprove injectivity. For any

y > 0

, take

x = 1 + y^{- 1/2} > 1

. This lies in the stated domain of a function, and

\frac{1}{( 1 - ∣ x ∣ ) ^{2}} = \frac{1}{( - y ^{- 1/2} ) ^{2}} = y .

Consequently every member of the codomain has a preimage.

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