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UPTET Paper 1 Model Test 3

Section: Mathematics

Question : 105 of 150

Marks: +1, -0

x=5+2\sqrt{6}

\sqrt{x}-\frac{1}{\sqrt{x}}

is:

Solution:

Concept:

We simplify the given expression using surds. Recognize that

5+2\sqrt{6}

is a perfect square of

(\sqrt{3}+\sqrt{2})

and then rationalize the reciprocal.

Explanation:

First, write

x = 5+2\sqrt{6} = (\sqrt{3})^2 + (\sqrt{2})^2 + 2\sqrt{3}\sqrt{2} = (\sqrt{3}+\sqrt{2})^2

\sqrt{x} = \sqrt{(\sqrt{3}+\sqrt{2})^2} = \sqrt{3}+\sqrt{2}

(positive value).

Now find

\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}+\sqrt{2}}

. Rationalize by multiplying numerator and denominator by

\sqrt{3}-\sqrt{2}

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

Finally, compute

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

Answer:

2\sqrt{2}

which is option A.

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