Question

Consider the polynomial function $f(x)=\frac{x^7}{7}-\frac{x^6}{6}+\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}-\frac{x^2}{2}+x$
Statement-1: The equation $f(x)=0$ can not have two or more roots
Statement-2: Rolles theorem is not applicable for $y=f(x)$ on any interval $[a,b]$ where $a,b\in R$

• A. Statement-1 is true, statement- 2 is true and statement-2 is correct explanation for statement- 1 .
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (A)

Let $f(x)=0$ has two roots say $x=r_1$ and $x=r_2$ where $r_1,r_2\in [a,b]$
$\Rightarrow f\left(r_1\right)=f\left(r_2\right)$
$\text{ hence }\exists \text{ there must exist some }c\in \left(r_1,r_2\right)\text{ where }{f}^{\prime }(c)=0$
$\text{ but }{f}^{\prime }(x)=x^6-x^5+x^4-x^3+x^2-x+1$
$\text{ for }x\ge 1,f^{\prime }(x)=\left(x^6-x^5\right)+\left(x^4-x^3\right)+\left(x^2-x\right)+1>0$
$\text{ for }x\le 1,f^{\prime }(x)=(1-x)+\left(x^2-x^3\right)+\left(x^4-x^5\right)+x^6>0$
$\text{ hence }{f}^{\prime }(x)>0\text{ for all }x$
$\therefore \text{ Rolles theorem fails }\Rightarrow f(x)=0\text{ can not have two or more roots. }$