RD Chapter 2 Functions Ex 2.1 Solutions
Question - 11 : - If f: R → R be thefunction defined by f(x) = 4x3 + 7, showthat f is a bijection.
Answer - 11 : -
Givenf: R → R is a function defined by f(x) = 4x3 +7
Injectivity:
Let x and y beany two elements in the domain (R), such that f(x) = f(y)
⇒ 4x3 +7 = 4y3 + 7
⇒ 4x3 = 4y3
⇒ x3 = y3
⇒ x = y
So, f isone-one.
Surjectivity:
Let y be any element in the co-domain (R), suchthat f(x) = y for some element x in R (domain)
f(x) = y
⇒ 4x3 +7 = y
⇒ 4x3 = y −7
⇒ x3 =(y – 7)/4
⇒ x = ∛(y-7)/4 in R
So, for every elementin the co-domain, there exists some pre-image in the domain. f is onto.
Since, f is both one-to-one and onto, it is a bijection.
Question - 12 : - 
Answer - 12 : -
Question - 13 : -
Answer - 13 : -
Question - 14 : - If A = { 1, 2, 3}, show that a ono-one function f : A → A must be onto.
Answer - 14 : -
Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.
Hence, f has to be onto.
Question - 15 : - If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Answer - 15 : -
Suppose f is notone-one.
Then, there exists twoelements, say 1 and 2 in the domain whose image in the co-domain is same.
Also, the image of 3under f can be only one element.
Therefore, the rangeset can have at most two elements of the co-domain {1, 2, 3}
i.e f is not an onto function, a contradiction.
Hence, f must be one-one.
Question - 16 : -
Answer - 16 : -