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Grade: 12th pass
        
Hi all
I have functional equation.
Find the function such as f(0)=1/2 and
exists a for any x,y f(x+y)=f(x)(a-y)+f(y)(a-x)
I found that a=1/2 then f(x)=1/2-x
It seems OK, but test failed:
f(x+y)=1/2-x-y and
f(x)(1/2-y)+f(y)(1/2-x)=1/2-x-y+2xy
I can't find the mistake (logical? calculation?)
Ask for help.
Thank you.
11 months ago

Answers : (3)

Pavan Kumar
32 Points
							
hiii, thank you for asking the question man,
                   I don’t completely understand your problem but let me solve the problem and check your solution once again
  f(x+y)=f(x)(a-y)+f(y)(a-x)                        -1
             for  a=1/2 then f(x)=1/2-x
                              also  f(y)=1/2-y
                     substitute in -1
                        we get    f(x+y)=f(x)(1/2-y)+f(y)(1/2-x)
                                                 =(1/2-x)(1/2-y)+(1/2-y)(1/2-x)
                                                  =2(1/2-x)(1/2-y)
                                    f(0)=2*1/2*1/2=1/2
            Hope you got your mistake and correct solution

 


 
11 months ago
George
12 Points
							
Well, if the f(x)=1/2-x is the correct answer, then the left side of equation
f(x+y)=f(x)(1/2-y)+f(y)(1/2-x)
will be f(x+y)=1/2-x-y, but the right side
f(x)(1/2-y)+f(y)(1/2-x)=(1/2-x)(1/2-y)+(1/2-y)(1/2-x)=1/2-x-y+2xy
11 months ago
Pavan Kumar
32 Points
							
Hey ,George at first i thought something else was your problem...this question has some many errors 
             
                                   f(x+y)=f(x)(a-y)+f(y)(a-x)   
                    put y=0 gives
                                        f(x)=f(x)(a)+f(0)(a-x)
                                                  given f(0)=1/2 substitute it
                                                f(x)=f(x)(a)+1/2(a-x)    
                                          so f(x)=    1/2*(a-x)/(1-a)   
now here comes something wierd      f(x+y)=(a-x)(a-y)/(1-a)
                                                              put y=a
                                                              f(x+a)=0 this is wrong or else if  x and y not equal to a
                                                             your contradiction so atleast they have to give a not equal to ½ and 1 
                                     see for the remaining conditions this contradiction will not come
                                                                             
11 months ago
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