Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3) Is f one-one and onto? Justify your answer.
Class 12, NCERT Chapter 1, Exercise 1.2, Q10
A = R - {3}, B = R - {1}
f: A → B is defined as
Let x,y∈A such that f(x)=f(y).
⇒ (x-2)(y-3)=(y-2)(x-3)
⇒ xy-3x-2y+6=xy-3y-2x+6
⇒ -3x-2y=-3y-2x
⇒ 3x-2x=3y-2y
⇒ x=y
∴ f is one-one.
Let y ∈B = R - {1}. 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(1-y)=-3y+2
⇒x=2-3y/1-y ∈A
⇒(x-2)(x-3)=y
⇒x-2=xy-3y
⇒x(1-y)=-3y+2
⇒x=2-3y/1-y ∈A
Thus, for any y ∈ B, there exists x=2-3y/1-y ∈A such that
Hence, function f is one-one and onto.
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