Question
Mon June 25, 2012 By: Arvindh Jambunathan
Q. classify as injection ,surjection or bijection 

Q. classify as injection ,surjection or bijection following details question description

Expert Reply
Wed June 27, 2012
Answer 1 . Given : f: Q-{3} ->Q by f(x) = (2x+3) / (x-3)
if f(x) = f(y)
 then => (2x+ 3) /(x-3) = (2 y+ 3) / (y-3 )
        => (2x+3 ) (y-3) =( 2y+3)(x-3)      ( by cross multiplication)
        => 2xy -6x +3y -9 = 2xy -6y +3x -9
         => 9x =9y
         => x = y
       therefore the function f(x) is one -one , hence injective. 
 
 
Now the function to be a surjection also ,
we know that f(x) = (2x+ 3) / (x-3)  , Q -> Q -{3}
now take f(x)=y and take out the values of y in terms of x , we get
 => (2x + 3)  / (x-3) =y
=> 2x +3 = xy - 3y
=> 2x - xy = -3y -3
=> x =( 3* ( y+1 ) ) / (y-2) which is defined for Q -> Q - {2}
 
but its given that the domain of f(x) is Q and  and range of f(y) is signifies that its not defined for value 2 . ( in simpler words , the domain of f(x) should be equal to range of f(y) for  the funtion to be surjective) 
 
hence the function f(x) is injective but not surjective ,hence it is not bijective 
 
Answer 2  Given : f: R -> R by f(x) = 1 +x2
if f(x) = f(y)
 then => 1 + x2 = 1 +y2
        =>x = y or x = -y
         as we are getting two values for a single value of x
        the function f(x) is not one -one , hence not injection .
 
 
Now the function to be a surjection also ,
we know that f(x) = 1 + x2                     
now take f(x)=y and take out the values of y in terms of x , we get
 =>  1 + x2 =y
 => x = ( y - 1 ) 1/2 which is defined for R  -> R 
 => the function is surjective
 
hence the function f(x) is surjective but not injective, and hence it is not bijective .
 
Answer 3
Given :  f : RR by f(x)= 3 - 4x 
if f(x) = f(y)
 then => 3 - 4x = 3 -4y
        => - 4x = - 4y      
        => x = y
       therefore the function f(x) is one -one , hence injective. 
 
 
Now the function to be a surjection also ,
we know that f(x) = 3 - 4x
now take f(x)=y and take out the values of y in terms of x , we get
 => 3  - 4x =y
=> x =(3 - y) / 4 which is defined for R -> R
 => the function is surjective
 
hence the function f(x) is both injective and surjective , hence it is  bijective.
Home Work Help