Given f: Q → Q, defined by f(x) = x3 + 1
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q), such that f(x) = f(y).
f(x) = f(y)
x3 + 1 = y3 + 1
x3 = y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).
f(x) = y
x3+ 1 = y
x =\( 3\sqrt{y-1}\), which may not be in Q.
For example, if y= 8,
x3+ 1 = 8
x3= 7
x =\( 3\sqrt{7}\), which is not in Q.
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
Express R and R-1 as sets of ordered pairs. Determine also
(i) the domain of R‑1
(ii) The Range of R.
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