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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 = y
⇒ `x = y/2 in Q` (domain)
⇒ is onto.
So, f is a bijection and, hence, it is invertible.
Finding f -1:
Let f−1 (x) =y ...(1)
⇒ x = f (y)
⇒ x = 2y
⇒ `y = x/2`
So, ` f^1 (x) = x/2` (from (1))
njectivity 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 =y
⇒ x= 2 -y ∈ Q (domain)
⇒ g is onto.
So, g is a bijection and, hence, it is invertible.
Finding g -1:
Let g−1(x) = y ...(2)
⇒ x = g (y)
⇒ x = y+2
⇒ y = x − 2
So, g−1 (x) = x − 2 (From (2)
Verification of (gof)−1 = f−1 og −1:
f(x) = 2x ; g (x) = x + 2
and `f^-1 (x) = x/2 ; g^-1 (x)= x-2`
`Now, (f^-1 o g^-1) (x) = f^-1 (g^-1)(x)) `
⇒ `(f^-1 o g ^-1)(x) = f^-1 (x-2) `
⇒ `(f ^-1 o g^-1) (x) = (x-2)/2 .......... (3)`
(gof) (x) = g (f(x))
= g (2x)
= 2x + 2
Let (gof)-1 (x) = y ............ (4)
x = (gof) (y)
⇒ x = 2y +2
⇒ 2y = x - 2
⇒ `y= (x-2)/2`
⇒` (gof)^-1 (x) = (x-2)/2` [form (4) ....... (5) ]
from (3) and (5)
⇒ `(gof)^-1 = f^-1 o g^-1`
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