Question

The equation $e^x-x-1=0$ has, apart from x=0

• A. One other real root
• B. Two real roots
• C. No other real root
• D. Infinite number of real roots

Answer 1

The correct option is C No other real root

$e^x=x+1\Rightarrow 1+\frac{1}{1!}+\frac{x^2}{2!}+...=x+1$
$\Rightarrow \frac{x^2}{2!}+\frac{x^3}{3!}+....=0$
On equating the various powers of x, we get
$x^2=0,x^3=0,...x^n=0$
Here, $x=0$ is only real root of given equation.