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Question

Let there are n number of polynomial functions fi:RR, iN satisfying the equation
f(a+f(a+b))+f(ab)f(a+b)bf(a)=a a,bR. Then

A
there exists at least one odd function.
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B
there exists at least one even function.
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C
fi(1)=1 iN
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D
ni=1fi(x)=k, where k is a constant.
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Solution

The correct option is D ni=1fi(x)=k, where k is a constant.
f(a+f(a+b))+f(ab)f(a+b)bf(a)=a ...(1)

Put a=b=0, we get
f(0+f(0))+f(0)f(0)=0
f(f(0))=0
Let f(0)=k, then f(k)=0

Put a=0,b=k, we get
f(0+f(k))+f(0)f(k)kf(0)=0
f(k)=0 and f(0)=k
k+k0k2=0
k(2k)=0
k=f(0)=0,2

Case 1 : f(0)=0
Put a=0 in the eqn (1), we get
f(f(b))+f(0)f(b)bf(0)=0
f(f(b))=f(b)
Let f(b)=y, then we get f(y)=y
f(x)=x which is an odd function.

Case 2 : f(0)=2
Put a=0 in the eqn (1), we get
f(f(b))+f(0)f(b)bf(0)=0
f(f(b))+2f(b)2b=0
f(f(b))=f(b)+2b2
Replace b by x,
f(f(x))=f(x)+2x2
From above equation, we observe that f(x) must be a polynomial of degree 1 as any other power will make the LHS and RHS of different powers and will not have any non-trivial solutions.
Now, put x=0, we get
f(f(0))=f(0)2
f(2)=0 (f(0)=2)
Hence, only f(x)=2x satisfies the above conditions.

Hence, the functions are f(x)=x and f(x)=2x that satisfying eqn (1).

f(1)=1 for both the functions.

ni=1fi(x)=x+(2x) =2 (constant)

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