Question

If $27a+9b+3c+d=0$, then the equation $4a{x}^3+3b{x}^2+2cx+d=0$ has atleast one real root lying in

• A. $(0,1)$
• B. $(1,3)$
• C. $(0,3)$
• D. $(2,4)$

Answer 1

The correct option is C $(0,3)$

Let,
$f\left(x\right)=\int (4a{x}^3+3b{x}^2+2cx+d)dx=\frac{4a{x}^4}{4}+\frac{3b{x}^3}{3}+\frac{2c{x}^2}{2}+dx$
$f\left(0\right)=0,f\left(3\right)=3(27a+9b+3c+d)=0$
$\Rightarrow$By Rolle's theorem
$\Rightarrow \exists c\in (0,3),f^{\prime }\left(c\right)=0$