A) (0, 1)
B) (1, 3)
C) (2, 3)
D) (1, 3)
Correct Answer: A
Solution :
[a] Let \[f'(x)=a{{x}^{2}}+bx+c\] \[\Rightarrow f(x)=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+d\] \[\Rightarrow f(x)=\frac{2a{{x}^{3}}+3b{{x}^{2}}+6cx+6d}{6}\] \[\therefore f(1)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d\] \[(\therefore 2a+3b+6c=0)\] And \[f(0)=d\] So from Rolle's Theorem there exists at least one a in (0, 1). For which \[f'(x)=0.\] Or there is at least one root of \[a{{x}^{2}}+bx+c=0\]in (0, 1).
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