Let f:R→R be such that for all x∈R, 21+x+21-x, f(x) and 3x+3-x are in A.P. Then the minimum value of f(x) is:
2
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0
f(x)=21-x+21+x+3x+3-x2=2x+12x+123x+13x≥2+1⇒f(x)≥3