Question

Let A = R − {3} and B = R − {1}. Consider the function f : A → B defined by . Is f one-one and onto? Justify your answer.

Answer 1

The given function f:A→B is defined by f( x )=( x−2 x−3 ), where, A=R−{ 3 } and B=R−{ 1 }.

Let x,y∈A such that,

f( x )=f( y )

x−2 x−3 = y−2 y−3 ( x−2 )( y−3 )=( y−2 )( x−3 ) xy−3x−2y+6=xy−3y−2x+6

Solve further.

−3x−2y=−3y−2x x=y

So, f( x )=f( y ) implies x=y.

Hence, 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

f( x )=y x−2 x−3 =y x−2=xy−3y

Solve further.

x( 1−y )=−3y+2 x= 2−3y 1−y

So, for any y∈B, there exists 2−3y 1−y 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

So f is onto.

Hence, f is one-one and onto.