Q.

Let α, β and γ be three positive real numbers.

Let fx=αx5+βx3+γx, xR and g : RRbe such that gfx=x for all xR.

If a1, a2, a3, ..... an be in arithmetic progression with mean zero, then the value of 

fg1ni=1nfai is equal to:

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a

27

b

0

c

9

d

3

answer is A.

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Detailed Solution

Consider a case when α=β=0 then fx=γx,  and gfx=xgγx=x gx=xγ 1ni=1nfaiyna1+a2+.....+an =0 fg0f0 0

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