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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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