Answer: Option C
$$\eqalign{
& {\text{Given that,}} \cr
& {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr
& {\text{Then, }}{x^2} + 2x - 1 \cr
& = {x^2} + 2x - 1 + 1 - 1 \cr
& = {x^2} + 2x + 1 - 2 \cr
& = {\left( {x + 1} \right)^2} - 2 \cr
& {\text{Now put the value of }}x \cr
& = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr
& = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times \sqrt2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\sqrt 2 } \right)^2} - 2 \cr
& = 2 - 2 \cr
& = 0 \cr} $$