If f(x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt

1.	If f(x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt{2x-4}}}\) for x > 2, then f (11) equals(a) \(\frac{7}{6}\)(b) \(\frac{5}{6}\)(c) \(\frac{6}{7}\)(d) \(\frac{5}{7}\)
Answer» Answer : (c) = \(\frac{6}{7}\) Given, f (x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt{2x-4}}}\) ⇒ f(11) = \(\frac{1}{\sqrt{11+2\sqrt{22-4}}} + \frac{1}{\sqrt{11-2\sqrt{22-4}}}\) = \(\frac{1}{\sqrt{11+2\sqrt{18}}} + \frac{1}{\sqrt{11-2\sqrt{18}}}\) = \(\frac{1}{\sqrt{11+6\sqrt{2}}} + \frac{1}{\sqrt{11-6\sqrt{2}}}\) = \(\frac{1}{\sqrt{(9+2\times 3\times \sqrt{2}+(\sqrt{2})^2)}} +\frac{1}{\sqrt{(9-2\times 3\times \sqrt{2}+(\sqrt{2})^2)}}\) = \(\frac{1}{\sqrt{(3+\sqrt{2})^2}} + \frac{1}{\sqrt{(3-\sqrt{2})^2}}\) = \(\frac{1}{{(3+\sqrt{2})}} +\frac{1}{{(3-\sqrt{2})}}\) = \(\frac{3-\sqrt{2} +3+ \sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\) = \(\frac{6}{9-2}\) = \(\frac{6}{7}\)

If f(x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt{2x-4}}}\) for x > 2, then f (11) equals(a) \(\frac{7}{6}\)(b) \(\frac{5}{6}\)(c) \(\frac{6}{7}\)(d) \(\frac{5}{7}\)

Answer»

Answer : (c) = \(\frac{6}{7}\)

Given, f (x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt{2x-4}}}\)

⇒ f(11) = \(\frac{1}{\sqrt{11+2\sqrt{22-4}}} + \frac{1}{\sqrt{11-2\sqrt{22-4}}}\)

= \(\frac{1}{\sqrt{11+2\sqrt{18}}} + \frac{1}{\sqrt{11-2\sqrt{18}}}\)

= \(\frac{1}{\sqrt{11+6\sqrt{2}}} + \frac{1}{\sqrt{11-6\sqrt{2}}}\)

= \(\frac{1}{\sqrt{(9+2\times 3\times \sqrt{2}+(\sqrt{2})^2)}} +\frac{1}{\sqrt{(9-2\times 3\times \sqrt{2}+(\sqrt{2})^2)}}\)

= \(\frac{1}{\sqrt{(3+\sqrt{2})^2}} + \frac{1}{\sqrt{(3-\sqrt{2})^2}}\)

= \(\frac{1}{{(3+\sqrt{2})}} +\frac{1}{{(3-\sqrt{2})}}\)

= \(\frac{3-\sqrt{2} +3+ \sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\) = \(\frac{6}{9-2}\)

= \(\frac{6}{7}\)