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

Answer: (C)

$e^x=x+1$

$\Rightarrow 1+\frac{x}{1!}+\frac{x^2}{2!}+...=x+1$

$\Rightarrow \frac{x^2}{2!}+\frac{x^3}{3!}+...=0$

On equating the various powers of $x$, we get

$x^2=0,x^3=0,....,x^n=0$

Hence, $x=0$ is only one real root of given equation.