A) (1, 2)
B) (0, 1)
C) (2, 3)
D) (3, 4)
Correct Answer: B
Solution :
Let\[f(x)=a{{x}^{2}}+bx+c\] Again, let\[f(x)=\int{f(x)}\,\,dx\] \[=\frac{a}{3}{{x}^{3}}+\frac{b{{x}^{2}}}{2}+cx\] At \[x=0\] \[f(0)=0\] Again, at\[x=1\] \[f(1)=\frac{a}{3}+\frac{b}{2}+c\] \[=\frac{2a+3b+6c}{6}=0\] Hence, one root lies between (0, 1).You need to login to perform this action.
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