Let A = R – {2} and B = R – {1}. If show that f is one-one and onto. Hence, find f-1.

To Show: that f is one-one and onto


To Find: Inverse of f


[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]


one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x1, x2 A & f(x1), f(x2) B, f(x1) = f(x2) x1= x2 or x1 x2f(x1) f(x2)


onto function: If range = co-domain then f(x) is onto functions.


So, We need to prove that the given function is one-one and onto.


Let x1, x2 Q and f(x) = .So f(x1) = f(x2) = , on solving we get x1=x2


So f(x1) = f(x2) x1= x2, f(x) is one-one


Given co-domain of f(x) is R – {1}


Let y = f(x) =, So x = [Range of f(x) = Domain of y]


So Domain of y = Range of f(x) = R – {1}


Hence, Range of f(x) = co-domain of f(x) = R – {1}.


So, f(x) is onto function


As it is a bijective function. So it is invertible


Invers of f(x) is f-1(y) =


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