Q. 115.0( 1 Vote )
If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
Answer :
TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.
So, is One – One function
⇔ a≠b
⇒ f(a)≠f(b) for all
⇔ f(a) = f(b)
⇒ a = b for all
Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.
So, is Surjection iff for each
, there exists
such that f(a) = b
Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.
Now, f : R → R, defined by f(x) = 4x3 + 7
To Prove : – f : R → R is bijective defined by f(x) = 4x3 + 7
Check for Injectivity:
Let x,y be elements belongs to R i.e such that
⇒ f(x) = f(y)
⇒ 4x3 + 7 = 4y3 + 7
⇒ x3 = y3
⇒ x = y
Hence, f is One – One function
Check for Surjectivity:
Let y be element belongs to R i.e be arbitrary, then
⇒ f(x) = y
⇒ 4x3 + 7 = y
⇒ 4x3 + 7 – y = 0
Now, we know that for 3 degree equation has a real root
So, let be that root
⇒
⇒
Thus for clearly , there exist
such that f(x) = y
Therefore f is onto
Thus, It is Bijective function
Hence Proved
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