A) (0,1)
B) (1,3)
C) (2,3)
D) (1,2)
Correct Answer: A
Solution :
Let \[f(x)=a{{x}^{2}}+bx+c\] \[g(x)\int_{{}}^{{}}{f(x)=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx}\] \[\therefore \]\[g(1)=\frac{a}{3}+\frac{b}{2}+c=2a+3b=c=0\] \[g(0)=0=g(1)\] \[\therefore \]g(x) is continuous apply Rolle's theorem \[\Rightarrow \]g(x) = 0 for some value \[x\in (0,1)\Rightarrow g'(x)=f(x)=a{{x}^{2}}+bx+c=0\]has a root in (0, 1)You need to login to perform this action.
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