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 Let f:RR be such that for all xR, 21+x+21-x, f(x)  and  3x+3-x are in A.P.   Then the  minimum value of f(x) is: 

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detailed solution

Correct option is B

f(x)=21-x+21+x+3x+3-x2=2x+12x+123x+13x≥2+1⇒f(x)≥3


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If a+2b+3c=12,a,b,cR+,then ab2c3 is 


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