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Telangana TET Paper 1 Exam Model Paper 1

Section: Mathematics

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Question : 102 of 150

Marks: +1, -0

If x = 5 +

2\sqrt{6}

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

is equal to

$\sqrt{2}$
$2\sqrt{2}$
$\sqrt{3}$
$2\sqrt{3}$

Solution:

Concept:

Recognize that

x = 5 + 2\sqrt{6}

can be written as a perfect square

(\sqrt{a} + \sqrt{b})^2

.

Explanation:

First, express

x

as

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

.

So

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

.

Now compute

x - 1 = 5 + 2\sqrt{6} - 1 = 4 + 2\sqrt{6}

.

Factor

2\sqrt{2}

:

4 + 2\sqrt{6} = 2\sqrt{2}(\sqrt{2} + \sqrt{3})

.

Thus

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

.

Answer:

B.

2\sqrt{2}

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