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Question

If f(ab)+f(bc)+f(ca)=2f(a+b+c) where a,b,cR and ab+bc+ca=0, then which of the following is/are possible ?

A
f(x)=c1x2+c2x4, ci0
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B
f(x)=c0+c1x2+c2x4, ci0
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C
f(x)=c1x2+c2x4+c3x6, ci0
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D
f(x) is always an even function
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Solution

The correct options are
A f(x)=c1x2+c2x4, ci0
D f(x) is always an even function
f(ab)+f(bc)+f(ca)=2f(a+b+c)
Put a=b=c=0
f(0)=0

Put b=c=0, we have f(a)+f(a)=2f(a)
f(a)=f(a)
So, f is an even function.

Let f(x)=c0+c1x2+c2x4+...+cmx2m
We know that
ab+bc+ca=0
a=bcb+c
Put b=(1+3)x and c=(13)x
a=x

f(3x)+f(23x)+f(3x)=2f(3x)
Therefore, comparing the coefficient of x2m, we get
2(3)2m+(23)2m=232m
23m+12m=232m
2+4m=23m
This holds for m=1 and m=2,
When m=3
43=64233=54
So, 2+4m>23m
Therefore,
2+4m>4m>23m for m3
Therefore, f(x) must be of the form
c1x2+c2x4

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