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 between

• A. $0$ and $1$
• B. $1$ and $3$
• C. $0$ and $3$
• D. None

Answer 1

The correct option is C $0$ and $3$

Given $27a+9b+3c+d=0$ .....(1)
Also given $4a{x}^3+3b{x}^2+2cx+d=0$ has at least one root
Hence, Rolle's theorem holds.
$\int f^{\prime }\left(x\right)dx=\int (4a{x}^3+3b{x}^2+2cx+d)dx$
$\Rightarrow f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Now, we will check through options.
Let's take $(a,b)$ as option A, $(0,1)$
$f\left(0\right)=e$
$f\left(1\right)=a+b+c+d+e$
Here, $f\left(0\right)\ne f\left(1\right)$
Hence, Rolle's theorem does not hold for $(0,1)$
Now, option B,$(1,3)$
$f\left(1\right)=a+b+c+d+e$
$f\left(3\right)=81a+b\times 27+9c+3d+e$
$\Rightarrow f\left(3\right)=3(27a+9b+3c+d)+e=e$
Here, $f\left(1\right)\ne f\left(3\right)$
Hence, Rolle's theorem does not hold for $(1,3)$
Now, option C, $(0,3)$
$f\left(0\right)=e$
$f\left(3\right)=81a+b\times 27+9c+3d+e$
$3(27a+9b+3c+d)+e$
$3\times \left(0\right)+e=e$ (by (1))
$f\left(0\right)=f\left(3\right)$
Rolle's theorem holds for $(0,3)$