Let `f : A to B` be a function defined as `f(x)=(2x+3)/(x-3)`, where A=R-{3} and B=R-{2}. Is the function f one-one and onto ? Is f invertible ? If ye

Let `y = f(x) = (2x+3)/(x-3).......................... (1)`
Let `x_1, x_2 in A = R - {3}`
Let `f(x_1) = f(x_2)`
`rArr (2x_1+3)/(x_1-3) = ( 2x_2 +3)/(x_2 -3)`
`rArr (2x_1 + 3)(x_2 -3) = (2 x_2+3) (x_1-3)`
`rArr (2x_1x_2 - 6x_1 + 3x_2 - 9) = ( 2x_1x_2- 6x_2 + 3x_1 -9)`
`rArr - 6x_1 + 3x_2 = -6x_2 + 3x_1`
`rArr 9x_1 = 9x_2`
`rArr x_1 = x_2 `
Now `f(x_1)= f(x_2) rArr x_1 = x_2 `
so `f(x)` is one-one
For onto
`y = (2x+3)/( x-3)`
`rArr xy-3y= 2x+3`
`rArr xy - 2x= 3y +3`
`rArr x(y-2) =3(y+1)`
`rArr x= (3(y+1))/((y-2)) ............. (2)`
equation (2) is defined for all real values of y except 2 i.e `y in R -{2}` which is same as given set B = R-{R}
(co-domain =range)
Also `y= f(x)`
`f(x) = f((3(y+1))/((y-2)))`
`" " = (2[(3(y+1))/((y-2))] +3)/((3(y+1))/((y-2))-3) ` `(` since `f(x) = (2+3)/(x-3))`
`(2(3y+3)+ 3(y-2))/(3y+3 -3y+6) = (9y)/(9) =y`
Thus for every `y in B`, there exists `x in A` such that `f(x) =y`
Thus function is onto.
Since `f(x)` is one-one and onto so `f(x)` is invertible.
Inverse is given by `x = f^(-1)(y) = (3(y+1))/((y-2))`