Question

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

• A. $(0,1)$
• B. $(1,3)$
• C. $(0,3)$
• D. None of the above

Answer 1

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

Given $27a+9b+3c+d=0\cdots \left(i\right)$
Let $f\left(x\right)=4a{x}^3+3b{x}^2+2cx+d=0$, on integrating
$\int f\left(x\right)dx=\int (4a{x}^3+3b{x}^2+2cx+d)dx$
$\Rightarrow g\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Clearly $g\left(x\right)$ is continuous and differentiable
Now,
$g\left(0\right)=e$,
$g\left(3\right)=81a+b\times 27+9c+3d+e$
$=3(27a+9b+3c+d)+e$
$=3\times \left(0\right)+e=e$ $\left\{\text{from}\right(i\left)\right\}$
$\Rightarrow g\left(0\right)=g\left(3\right)$
$\Rightarrow$ Rolle's theorem is satisfied
$\therefore g^{\prime }\left(x\right)=0$ in $(0,3)$
$\Rightarrow f\left(x\right)=0,$ at least once in $(0,3)$