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If f:R→R is a function such that f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3) ∀ x∈R, then f(2)−f(1)=

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If f:RR is a function such that f(x)=x3+x2f(1)+xf′′(2)+f′′′(3) xR, then f(2)f(1)=




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