The recursion relation to solve x = e^-x using Newton-Raphson method is

A. $${{\text{x}}_{{\text{n}} + 1}} = {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$

B. $${{\text{x}}_{{\text{n}} + 1}} = {{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$

C. $${{\text{x}}_{{\text{n}} + 1}} = \left( {1 + {{\text{x}}_{\text{n}}}} \right)\frac{{{{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}{{1 + {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$

D. $${{\text{x}}_{{\text{n}} + 1}} = \frac{{{\text{x}}_{\text{n}}^2 - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}\left( {1 + {{\text{x}}_{\text{n}}}} \right) - 1}}{{{{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$

Answer: Option C