Q. 125.0( 1 Vote )

# If f : Q <span la

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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