Classify the function as injection, surjection or bijectionf: f: R → R, defined by f(x) = 1 + x2

Asked by Sakshi | 1 year ago |  41

##### Solution :-

Given f: R → R, defined by f(x) = 1 + x2

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

1 + x= 1 + y2

x= 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 + x= y

x= y − 1

x = $$± \sqrt{-1} = ± i$$ is not in R.

So, f is not a surjection and f is not a bijection.

Answered by Aaryan | 1 year ago

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