24x2f(x) +f(1-x)=2x-x4 -------(1)
Replace x→1-x
(1-x)2f(1-x) +f(x)=2(1-x)-(1-x)4 ------(2)
Solving (1) and (2)
[x2(1-x)2-1]f(x)=(1-x)2(2x-x4)-2(1-x)+(1-x)4
f(x)=(1-x)2(2x-x4)-2(1-x)+(1-x)4
[x2(1-x)2-1]
62you have made a small mistake in solving the two equations...
See properly
3eure its not for advanced students :P as u have solved the functional eqn buks by bargave for olympiads.
24
eureka123
·2009-10-07 03:24:28
hehe.........
I realised that later....but sinceno one posted ths soln for 30 mins I thought i should do that
1708x2f(x) +f(1-x)=2x-x4 -------(1)
here f(x) must be in the form of 2 degree polynomial
and degree of leading coefficient is -1.
bcz max. power of variable x in R.H S is -1
so f(x) = - (x-a)*(x-b)
put x=0 in eq .(1) we get f(1) = 0 means (x-1) is a factor of f(x)
so f(x) = -(x-1)*(x-b).........(2)
again put x = 1 in eq..(1) we get f(1) +f(0) = 1
again put x = 0 in eq....(2) we get f(0) = -b
so f(1) =0 and f(0) =-b
put these value in f(1) +f(0) = 1
we get b = -1
put value of b in eq....(2)
f(x) = -(x-1)(x+1) =1-x2
ans = (1-x2)
and this solution was given by hsbhatt sir.
62
Lokesh Verma
·2009-10-07 08:55:43
The solution is indeed awesome :)
but I did not understand why f(x) should be a polynomial..? ?
could it not be a function where there is one thing like sin(x) - sin (1-x) which gets canceled?
Am i missing something?
1708
man111 singh
·2009-10-07 08:58:43
No No Sir bcz R.H .S is a polynomial so L.H.S must be a polynomial.
62
Lokesh Verma
·2009-10-07 09:01:52
but that is not guaranteed because they could cancel each other... ?? one -ve term
there could be something like |x| ?? Or somethign still more complex.. say in an exponential form?
srry exponential will not be possible..
19even i have the same doubt !!! the way mavarick is assuming f(x) to be a polynomial is a bit doubtful....if this is the case then we can also suitably divide or multiply the lhs and rhs by suitable quantities and then call it some other polynomial.....
what i want to ask is ,,what is the standard form in these cases to determine the degree of polynomial or the polynomial itself ?
1708
man111 singh
·2009-10-07 09:05:38
but sir here x2 is also present...
62
Lokesh Verma
·2009-10-07 09:11:03
I saw that.. and i was also thinking about it.. but I am still not fully convinced.. (may be I am overlooking something much more fundamental...)
