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RRB RPF RPSF SI and CONS Model Papers 5 with solutions

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Question : 82 of 119

Marks: +1, -0

The simplified value of

\frac{\sqrt{6}+2}{\sqrt{2}+\sqrt{2+\sqrt{3}}} - \frac{\sqrt{6}-2}{\sqrt{2}-\sqrt{2-\sqrt{3}}}

-\frac{2\sqrt{2}}{2+\sqrt{2}}

is

$2\sqrt{6}$
0
$\sqrt{3}$
$\sqrt{2}$

Solution:

(2)

\frac{\sqrt{6}+2}{\sqrt{2}+\sqrt{2+\sqrt{3}}} - \frac{\sqrt{6}-2}{\sqrt{2}-\sqrt{2-\sqrt{3}}} - \frac{2\sqrt{2}}{2+\sqrt{2}}

Multiplying and dividing by

\sqrt{2}

in first two terms and rationalising the third term, we get,

= \frac{\sqrt{2} \times \sqrt{6} + 2\sqrt{2}}{\sqrt{2} \times \sqrt{2} + \sqrt{4 + 2\sqrt{3}}}

- \frac{\sqrt{2} \times \sqrt{6} - 2\sqrt{2}}{\sqrt{2} \times \sqrt{2} - \sqrt{4 - 2\sqrt{3}}}

- \frac{2\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}

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

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

= \frac{2\sqrt{3}+2\sqrt{2}}{3+\sqrt{3}}

- \frac{2\sqrt{3}-2\sqrt{2}}{3-\sqrt{3}} - 2\sqrt{2} + 2

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

-2\sqrt{2}+2

= \frac{12\sqrt{2}-12}{6} - 2\sqrt{2} + 2

= 2\sqrt{2} - 2 - 2\sqrt{2} + 2 = 0

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