Question 54

If $$x = \frac{1}{\sqrt{13} - 3}, y = \frac{1}{\sqrt{7} - \sqrt{3}}, z = \frac{1}{\sqrt{2}(\sqrt{3} - 1)}$$, then

Solution

$$x=\ \frac{\ 1}{\sqrt{13\ }-3}=\ \frac{\ \sqrt{13\ }+3}{13-9}=\ \frac{\ \sqrt{13\ }+3}{4}=1.65.$$

$$y=\ \frac{\ 1}{\sqrt{7\ }-\sqrt{\ 3}}=\ \frac{\ \sqrt{7\ }+\sqrt{3\ }}{7-3}=1.09.$$

$$z=\ \frac{\ 1}{\sqrt{6\ }-\sqrt{2\ }}=\ \frac{\ \sqrt{6\ }+\sqrt{2\ }}{4}=0.96.$$

So, $$x>y>z.$$

C is correct choice.

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