Question 27 Let π: A β B be a function defined as π(π₯)=(2π₯ + 3)/(π₯ β 3) , where A = R β {3} and B = R β {2}. Is the function f oneβone and onto? Is f invertible? If yes, then find its inverse
π(π₯)=(2π₯ + 3)/(π₯ β 3)
Checking one-one
Let π₯_1 , π₯_2 β A
π(π₯_1 )=(2π₯_1+ 3)/(π₯_1β 3)
π(π₯_2 )=(2π₯_2+ 3)/(π₯_2 β 3)
One-one Steps
1. Calculate f(x1)
2. Calculate f(x2)
3. Putting f(x1) = f(x2)
we have to prove x1 = x2
Putting π(π₯_1 ) = π(π₯_2 )
(2π₯_1+ 3)/(π₯_1β 3) = (2π₯_2+ 3)/(π₯_2 β 3)
(2π₯_1+ 3) (π₯_2 β 3) = (2π₯_2+ 3) (π₯_1 β 3)
2π₯_1 "(" π₯_2 β 3")"+ 3 (π₯_2 β 3) = 2π₯_2 "(" π₯_1 β 3")"+ 3(π₯_1 β 3)
2π₯_1 π₯_2 β6π₯_1+ 3π₯_2β9 = 2π₯_2 π₯_1 β6π₯_2+ 3π₯_1 β9
β6π₯_1+ 3π₯_2 = β6π₯_2+ 3π₯_1
3π₯_2+6π₯_2 = 3π₯_1+6π₯_1
9π₯_2= 9π₯_1
π₯_2= π₯_1
If π(π₯_1 ) = π(π₯_2 ) , then π₯_1= π₯_2
β΄ f is one-one
Checking onto
π(π₯)=(2π₯ + 3)/(π₯ β 3)
Let y β B where
y = (2π₯ + 3)/(π₯ β 3)
y(x β 3) = 2x + 3
xy β 3y = 2x + 3
xy β 2x = 3y + 3
x(y β 2) = 3(y + 1)
x = (3(π¦ + 1))/((π¦ β 2))
For y = 2 ,
x is not defined
But it is given that y β R β {2}
Hence , x = (3(π¦ + 1))/((π¦ β 2)) β R β {3}
Now,
Checking for y = f(x)
Putting value of x in f(x)
f(x) = f((3(π¦ + 1))/((π¦ β 2)))
= (2(3(π¦ + 1)/((π¦ β 2) ))+ 3)/((3(π¦ + 1)/((π¦ β 2) ))β 3)
= (((6(π¦ + 1)+3(π¦β2))/((π¦ β 2) )))/((3(π¦ + 1)/((π¦ β 2) ))β 3)
Also, y = f(x)
Putting y in f(x)
Hence, f is onto
= (2(3(π¦ + 1)/((π¦ β 2) ))+ 3)/((3(π¦ + 1)/((π¦ β 2) ))β 3)
= (((6(π¦ + 1) + 3(π¦ β 2))/((π¦ β 2) )))/(((3(π¦ + 1) β 3(π¦ β 2))/((π¦ β 2) )) )
= (6(π¦ + 1) + 3(π¦ β 2))/(3(π¦ + 1) β 3(π¦ β 2))
= (6π¦ + 6 + 3π¦ β 6)/(3π¦ + 3 β 3π¦ + 6)
= 9π¦/9
= y
Thus,
for every y β B, there exists x β A such that
f(x) = y
Hence, f is onto
Since f(x) is one-one and onto,
So, f(x) is invertible
And
Inverse of x = π^(β1) (π¦)
= (3(π¦ + 1))/((π¦ β 2))
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.