A) (0, 1)
B) (1, 2)
C) (2, 3)
D) (3, 4)
Correct Answer: A
Solution :
\[f(x)=a{{x}^{2}}+bx+c\] Let \[F(x)=\int{f(x)dx=\frac{a}{3}{{x}^{3}}+\frac{b}{2}{{x}^{2}}+cx}\] Clearly \[F(0)=0\] and \[F(1)=\frac{a}{3}+\frac{b}{2}+c\] \[=\frac{2a+3b+6c}{6}=0\] Þ \[F(0)=F(1)=0\] There exist at least one point c in between 0 and 1 such that \[{F}'(x)=0\] or \[a{{x}^{2}}+bx+c=0\]for some \[x\in (0,\,\,1)\].You need to login to perform this action.
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