Question

If A = R – {b} and B=R-{1} and function f:A→ B is defined by f(x)= (x-a)/(x-b) , a≠b ,then f is​

Answer 1

Given: A = R – {b} and B=R-{1}, f:A→ B is defined by f(x)= (x-a)/(x-b) , a≠b

To find: Type of function?

Solution:

• We have given f(x) = ( x - a ) / ( x - b )

• Now the domain of the function is:  R – { b}and the codomain is: R - { 1 }

• For one-one function, we have f(x1) = f(x2) for x1 = x2

             ( x1 - a ) / ( x1 - b ) = ( x2 - a ) / ( x2 - b )

             ( x1 - a )( x2 - b ) =  ( x2 - a ) ( x1 - b )

             x1x2 - bx1 - ax2 + ab = x1x2 - bx2 - ax1 + ab

             bx1 + ax2 = bx2 + ax1

             bx1 - ax1 =  bx2 -  ax2

             x1 (b-a) = x2(b-a)

             x1 = x2

• So the function is one one.

• For onto function, we have range of f(x) =  R - { 1 }

• So, f(x) =  ( x - a ) / ( x - b )

• Adding and subtracting b in numerator, we get:

             f(x) =  ( x - a + b - b ) / ( x - b )

             f(x) =  ( x - b - a  + b ) / ( x - b )

             f(x) =  ( x - b ) - ( a - b ) / ( x - b )

             f(x) =  ( x - b ) / ( x - b ) - ( a - b ) / ( x - b )

             f(x) =  1 - ( a - b ) / ( x - b )

• Let y =  1 - ( a - b ) / ( x - b )

             y = ( x  -b ) - ( a - b ) / ( x - b )

             y ( x - b ) = x - b - a + b

             y ( x - b ) = x - a

             yx - yb = x - a

             xy - x = by - a

             x ( y - 1 ) = by - a

             x = by - a / ( y - 1 )  belongs to co domain R - { 1 }

• So the function is onto.

Answer:

         So, the given function f(x) = ( x - a ) / ( x - b ) is one one and onto.