If $2 a+3 b+6 c=0$, then at least one root of the equation $a x^2+b x+c=0$ lies in the interval
If $2 a+3 b+6 c=0$, then at least one root of the equation $a x^2+b x+c=0$ lies in the interval
$(0,1)$
$(1,2)$
$(2,3)$
$(1,3)$
Solution
Let $f^{\prime}(x)=a x^2+b x+c \Rightarrow f(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x+d$ $\Rightarrow f(x)=\frac{1}{6}\left(2 a x^3+3 b x^2+6 c x+6 d\right)$, Now $f(1)=f(0)=d$, then according to Rolle's theorem $\Rightarrow f^{\prime}(x)=a x^2+b x+c=0$ has at least one root in $(0,1)$