A) (0, 1)
B) (1, 2)
C) (2, 3)
D) (1, 3)
Correct Answer: A
Solution :
Let \[f(x)=a{{x}^{2}}+bx+c\] \[\Rightarrow \] \[f(x)=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+d\] \[\Rightarrow \] \[f(x)=\frac{2a{{x}^{3}}+3b{{x}^{2}}+6cx+6d}{6}\] \[\Rightarrow \] \[f(1)=\frac{2a+3b+6c+6d}{6}=\frac{6d}{6}=d\] \[(\because 2a+3b+6c=0)\] \[f(0)=\frac{6d}{6}=d\] \[\therefore \] \[f(0)=f(1)\] \[\Rightarrow \]\[f(x)\]will vanish at least once between 0 and 1. \[\therefore \]One of the roots of\[a{{x}^{2}}+bx+c=0\]lies between 0 and 1.
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