Let $f\left(x\right)=a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex$  where $a,b,c$ ,$d,e\in R$ and $f\left(x\right)=0$ has a positive root a. Then,

It is given that a is a positive root of  $f\left(x\right)$ and  by 

inspection, we have $f\left(0\right)=0\text{. }$Therefore   $, x=0$ and $x=\alpha$  are the 

roots of $f\left(x\right)=0.$ By Rolle's theorem  $f^{\mathrm{\prime }}\left(x\right)=0$ has a root ${\alpha }_1$

between O and a i.e. $0<{\alpha }_1<\alpha$

So option (a) is correct

 Clearly $f^{\mathrm{\prime }}\left(x\right)=0$  is a fourth degree equation in $x$ and 

imaginary roots always occur in pairs. Since $x={\alpha }_1$

is a root of   $f\left(x\right)=0.$

Therefore  $f^{\mathrm{\prime }}\left(x\right)=0$ will have another real root  ${\alpha }_2$ 2 (say). Now, 

${\alpha }_1$ and ${\alpha }_2$ are real roots of  $f^{\mathrm{\prime }}\left(x\right)=0$ Therefore, by 

Rolle's theorem  $f\left(x\right)=0.$ will have a real root between 

${\alpha }_1$ and  ${\alpha }_2\text{. }$ 

Thus, option (b) is correct. 

We have seen that $x=0,x=\alpha$ are two real roots $f\left(x\right)=0.$ 

$f^{\mathrm{\prime }}\left(x\right)=0$    0 is a fifth degree equation, it will have at least three 

roots. Consequently, by Rolle's theorem $f^{\mathrm{\prime }}\left(x\right)=0$  0 will have

least two real roots.

Thus, option (c) is also correct