Question 14
If $$x=\frac{\sqrt2+1}{\sqrt2-1}$$ and $$y=\frac{\sqrt2-1}{\sqrt2+1}$$, then what is the value of x+y ?
Solution
Given : $$x=\frac{\sqrt2+1}{\sqrt2-1}$$ and $$y=\frac{\sqrt2-1}{\sqrt2+1}$$
To find : $$x+y=$$ $$\frac{\sqrt2+1}{\sqrt2-1}+\frac{\sqrt2-1}{\sqrt2+1}$$
= $$\frac{(\sqrt2+1)^2+(\sqrt2-1)^2}{(\sqrt2-1)(\sqrt2+1)}$$
= $$\frac{(2+1+2\sqrt{2})+(2+1-2\sqrt{2})}{2-1}$$
= $$\frac{(3+3)}{1}=6$$
=> Ans - (A)
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