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Question 106
If √2 = 1.4142, find the value of $$2\sqrt{2} + \sqrt{2} + \frac{1}{2+\sqrt{2}} + \frac{1}{\sqrt{2} + 1}$$
Solution
Expression : $$2\sqrt{2} + \sqrt{2} + \frac{1}{2+\sqrt{2}} + \frac{1}{\sqrt{2} + 1}$$
= $$2\sqrt{2} + \sqrt{2} + [\frac{1}{2+\sqrt{2}}\times(\frac{2-\sqrt2}{2-\sqrt2})] + [\frac{1}{\sqrt{2} + 1}\times(\frac{\sqrt2-1}{\sqrt2-1})]$$
= $$3\sqrt2+\frac{2-\sqrt2}{2}+\frac{\sqrt2-1}{1}$$
= $$\frac{1}{2} [6\sqrt2+2-\sqrt2+2\sqrt2-2]$$
= $$\frac{1}{2}[7\sqrt2]$$
= $$3.5\times1.4142=4.9497$$
=> Ans - (B)
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