Question : If $\text{a}= {\sqrt{2}+1}$ and $\text{b} = {\sqrt{2}–1}$, then the value of $\frac{1}{a+1}+\frac{1}{b+1}$ will be:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
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Correct Answer: 1
Solution :
Given:
$\text{a}= {\sqrt{2}+1}$ and $\text{b} = {\sqrt{2}–1}$
$\frac{1}{a+1}+\frac{1}{b+1}$
$= \frac{1}{(\sqrt{2}+1)+1}+\frac{1}{(\sqrt{2}–1)+1}$
$= \frac{1}{(\sqrt{2}+2)}+\frac{1}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}}[\frac{1}{(1+\sqrt{2})}+1]$
$=\frac{1}{\sqrt{2}}[\frac{(1–\sqrt{2})}{(1–2)}+1]$
$=\frac{1}{\sqrt{2}}[\sqrt{2}–1+1]$
$=\frac{1}{\sqrt{2}}×\sqrt{2}$
$= 1$
Hence, the correct answer is 1.
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