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 + 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.
Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y
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