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Question
Let A = R - {3} and B = R - {1}. Consider the function f : A → B defined by f(x) = `(x-2)/(x-3).`Show that f is one-one and onto and hence find f-1.
[CBSE 2012, 2014]
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Solution
We have,
A = R {3} and B = R - {1}
The function f : A → B defined by f(x) = `(x-2)/(x-3)`
Let x,y ∈ A such that f (x) = f (y). Then,
`(x-2)/(x-3) = (y-2)/(y-3)`
⇒ xy −3x − 2y + 6 = xy − 2x − 3y + 6
⇒ −x = −y
⇒ x = y
∴ f is one − one .
Let y ∈ B. Then, y ≠ 1.
The function f is onto if there exists x ∈ A such that f (x) = y . Now,
f (x) = y
⇒ `(x-2)/(x-3) = y`
⇒ x - 2 = xy - 3y
⇒ x - xy = 2 - 3y
⇒ x (1 - y ) = 2 - 3y
⇒ ` x = (2 - 3y)/ (1 - y ) in A [ y ≠ 1]`
Thus, for any y ∈ B, there exists `(2-3y)/(1-y)` ∈ A such that
`f ((2 - 3y) / (1-y)) = (((2-3y)/(1-y))-2)/(((2-3y)/(1 - y))-3 `= `(2-3y - 2 +2y)/(2-3y -3 +3y) = (-y)/(-1) = y`
∴ f is onto.
So, f is one−one and onto fucntion.
Now,
` As , x = (2 - 3y)/(1- y)`
`so , f^-1 (x) = (2- 3x)/(1 - x) = (3x - 2)/(x-1)`
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