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Question
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = 1 + x2
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Solution
f : R → R, defined by f(x) = 1 + x2
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
1 + x2=1 + y2
x2 = y2
x = ± y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
1 + x2= y
x2= y − 1
`x = ± sqrt-1 = ± i` is not in R.
So, f is not a surjection and f is not a bijection.
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