Question

If $ 2a+3b+6c=0, $ then at least one root of the equation $ a{{x}^{2}}+bx+c=0 $ lies in the interval

• A. (0, 1)
• B. (1, 2)
• C. (2, 3)
• D. (1, 3)

Answer 1

Answer: (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} $ $ \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.