Let f:(1,∞)→(1,∞) be defined by f(x)=x+2x−1.Then

If x1+2x1−1=x2+2x2−1      ⇒x1x2+2x2−x1−2            =x1x2+2x1−x2−2⇒3x2=3x1 ⇒ x1=x2So f in one - one If y=f(x)=x+2x−1      ⇒   yx−y=x+2⇒                                     (y−1)x=y+2⇒                                     x=y+2y−1∈(1,∞)Thus f is onto