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.