Question

The equation $e^x-x-1=0$ has

• A. Only one real root x=0
• B. At least two real roots
• C. Exactly two real roots
• D. Infinitely many real roots

Answer 1

The correct option is A Only one real root x=0

$e^x=x+1\Rightarrow 1+\frac{x}{1!}+\frac{x^2}{2!}+........=x+1$
$\Rightarrow \frac{x^2}{2!}+\frac{x^3}{3!}+......=0$
$x^2=0,x^3=0,........x^n=0$