A) \[a+b+c=3\]
B) \[a+b+c=1\]
C) \[a+b+c=0\]
D) \[a+b+c=2\]
Correct Answer: C
Solution :
\[\int_{0}^{3}{(3a{{x}^{2}}+2bx+c)dx=\int_{1}^{3}{(3a{{x}^{2}}+2bx+c)dx}}\] Þ \[\int_{0}^{1}{(3a{{x}^{2}}+2bx+c)dx+\int_{1}^{3}{(3a{{x}^{2}}+2bx+c)dx}}\] \[=\int_{1}^{3}{(3a{{x}^{2}}+2bx+c)dx}\] Þ \[\int_{0}^{1}{(3a{{x}^{2}}+2bx+c)dx=0}\] Þ \[\left[ \frac{3a{{x}^{3}}}{3}+\frac{2b{{x}^{2}}}{2}+cx \right]_{0}^{1}=0\Rightarrow a+b+c=0\].
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