Consider the polynomial function  $f\left(x\right)=\frac{x^7}{7}-\frac{x^6}{6}+\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}-\frac{x^2}{2}+x$
Statement I: The equation  $f\left(x\right)=0$ cannot have two or more roots.
Statement II: Rolle’s theorem is not applicable for  $y=f\left(x\right)$ on any interval  $[a,b]$, where  $a,b\in R$

Let  $f\left(x\right)=0$ has two roots say  $x=r_{1}nad x=r_2$, where  $r_1,r_2\in [a,b]$
 $\Rightarrow f\left(r_1\right)=f\left(r_2\right)$
Hence, there must exist some  $c\in (r_1,r_2)$, where  $f\text{'}\left(c\right)=0$
But   $f\text{'}\left(x\right)=x^6-x^5+x^4-x^3+x^2-x+1$
for  $x\ge 1,f\text{'}\left(x\right)=(x^6-x^5)+(x^4-x^3)+(x^2-x)+1>0$
for  $x\le 1,f\text{'}\left(x\right)=(1-x)+(x^2-x^3)+(x^4-x^5)+x^6>0$
Hence,  $f\text{'}\left(x\right)>0$ for all x.
$\therefore$ Rolle’s theorem fails. 
$\Rightarrow f\left(x\right)=0$  cannot have two or more roots.