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 $
$\text { but } f^{\prime}(x)=x^6-x^5+x^4-x^3+x^2-x+1$
$\text { for } x \geq 1, f^{\prime}(x)=\left(x^6-x^5\right)+\left(x^4-x^3\right)+\left(x^2-x\right)+1>0$
$\text { for } x \leq 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 $
$\therefore \text { Rolles theorem fails } \Rightarrow f(x)=0 \text { can not have two or more roots. }$