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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