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Question

If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)−1 = f−1 og −1.

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Solution

Injectivity of f:
Let x and y be two elements of domain (Q), such that
f(x) = f(y)
2x = 2y
x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y.

2x= yx= y2Q domain

f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f -1:
Let f-1x=y ...1x=fyx=2yy=x2 So, f-1x=x2 from 1

Injectivity of g:
Let x and y be two elements of domain (Q), such that
g(x) = g(y)
x + 2 = y + 2
x = y
So, g is one-one.

Surjectivity of g:
Let y be in the co domain (Q), such that g(x) = y.

x+2=yx= 2-yQ domain

g is onto.
So, g is a bijection and, hence, it is invertible.

Finding g -1:
Let g-1x=y ...2x=gyx=y+2y=x-2So, g-1x=x-2 From 2

Verification of (gof)−1 = f−1 og −1:

fx=2x; gx=x+2and f-1x=x2; g-1x=x-2Now, f-1o g-1x=f-1g-1xf-1o g-1x=f-1x-2f-1o g-1x=x-22 ...3gofx=g f x=g 2x=2x+2Let gof-1x=y .... 4x=gofyx=2y+22y=x-2y=x-22 gof-1x=x-22 [from 4 ... 5]From 3 and 5, gof-1=f-1o g-1

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