Question

# If f(a−b)+f(b−c)+f(c−a)=2f(a+b+c) where a,b,c∈R 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, ci≠0 D f(x) is always an even functionf(a−b)+f(b−c)+f(c−a)=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=(1−√3)x ⇒a=x f(−√3x)+f(2√3x)+f(−√3x)=2f(3x) Therefore, comparing the coefficient of x2m, we get 2(−√3)2m+(2√3)2m=2⋅32m ⇒2⋅3m+12m=2⋅32m ⇒2+4m=2⋅3m This holds for m=1 and m=2, When m=3 43=642⋅33=54 So, 2+4m>2⋅3m Therefore, 2+4m>4m>2⋅3m for m≥3 Therefore, f(x) must be of the form c1x2+c2x4

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