Question

If $f\left(x\right)=1+x/1!+x^2/2!+x^3/3!+........+x^n/n!$ , then $f\left(x\right)=0$ (n is odd , $n\ge 3$)

• A. can't have any real root
• B. can't have any repeated root
• C. has one positive root
• D. none of these

Answer 1

The correct option is A can't have any real root

If $f\left(x\right)=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...............+\frac{x^n}{n!}$ , Then $f\left(x\right)=0$

$f\left(x\right)=e^x=0$

Which have not any root.