Q. 125.0( 1 Vote )

If f : Q <span la

Answer :

We have f : Q Q and f(x) = 2x.


Recall that a function is a bijection only if it is both one-one and onto.


First, we will prove that f is one-one.


Let x1, x2ϵ Q (domain) such that f(x1) = f(x2)


2x1 = 2x2


x1 = x2


So, we have f(x1) = f(x2) x1 = x2.


Thus, function f is one-one.


Now, we will prove that f is onto.


Let y ϵ Q (co-domain) such that f(x) = y


2x = y



Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that f(x) = y and hence, function f is onto.


Thus, the function f is a bijection and has an inverse.


We have f(x) = y x = f-1(y)


But, we found f(x) = y


Hence,


Thus,


Now, we have g : Q Q and g(x) = x + 2.


First, we will prove that g is one-one.


Let x1, x2ϵ Q (domain) such that g(x1) = g(x2)


x1 + 2 = x2 + 2


x1 = x2


So, we have g(x1) = g(x2) x1 = x2.


Thus, function g is one-one.


Now, we will prove that g is onto.


Let y ϵ Q (co-domain) such that g(x) = y


x + 2 = y


x = y – 2


Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that g(x) = y and hence, function g is onto.


Thus, the function g is a bijection and has an inverse.


We have g(x) = y x = g-1(y)


But, we found g(x) = y x = y – 2


Hence, g-1(y) = y – 2


Thus, g-1(x) = x – 2


We have (f-1og-1)(x) = f-1(g-1(x))


We found and g-1(x) = x – 2


(f-1og-1)(x) = f-1(x – 2)



We know (gof)(x) = g(f(x)) and gof : Q Q


(gof)(x) = g(2x)


(gof)(x) = 2x + 2


Clearly, gof is a bijection and has an inverse.


Let y ϵ Q (co-domain) such that (gof)(x) = y


2x + 2 = y


2x = y – 2



We have (gof)(x) = y x = (gof)-1(y)


But, we found (gof)(x) = y


Hence,


Thus,


So, it is verified that (gof)-1 = f-1og-1.


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