Question

# Let there are n number of polynomial functions fi:R→R, i∈N satisfying the equation f(a+f(a+b))+f(ab)−f(a+b)−bf(a)=a ∀ a,b∈R. 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 n∑i=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+k−0−k2=0 ⇒k(2−k)=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))+2−f(b)−2b=0 ⇒f(f(b))=f(b)+2b−2 Replace b by x, ⇒f(f(x))=f(x)+2x−2 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)=2−x satisfies the above conditions. Hence, the functions are f(x)=x and f(x)=2−x that satisfying eqn (1). f(1)=1 for both the functions. n∑i=1fi(x)=x+(2−x) =2 (constant)

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