Question : If $x$ = ${\frac{1}{\sqrt{2}+1}}$, then the value of $(x^{2}+2x–1)$ is:
Option 1: $\sqrt[2]{2}$
Option 2: 4
Option 3: 0
Option 4: 2
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Correct Answer: 0
Solution :
Given:
$x= {\frac{1}{\sqrt{2}+1}}$
Rationalising the denominator, we get,
⇒ $x ={\frac{{\sqrt{2}-1}}{(\sqrt{2}+1)×{(\sqrt{2}-1)}}}=\frac{\sqrt{2}-1}{2-1} = {\sqrt{2}-1}$
Putting this value in the expression $(x^{2}+2x-1)$, we get,
$=[(\sqrt{2}-1)^{2}+2(\sqrt{2}-1)-1]$
$=2-2\sqrt{2}+1+2\sqrt{2}-2-1 $
$= 0$
Hence, the correct answer is 0.
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