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Question

Let f(x) be a polynomial function of second degree . Given that f(1)=f(1) and a,b,c are in A.P. Also f(a),f(b),f(c) are in A.P., then find the values of a,b,c

A
f(x) of second degree cannot exist.
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B
a=2,b=0,c=1
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C
a=3,b=5,c=0
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D
none of the above
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Solution

The correct option is A f(x) of second degree cannot exist.
Let f(x)=αx2+βx+γ=0,α0
f(1)=f(1) gives us β=0
f(a)=αa2+γ
f(b)=αb2+γ
f(c)=αc2+γ
Given 2b=a+c.....(1)
f(c)f(b)=f(b)f(a) (Given f(a),f(b),f(c) are in A.P.)
c2b2=b2a2
(cb)(c+b)=(ba)(b+a)
c+b=b+a or c=b=a

In both cases, a=c=b

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