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UPSC CDS 1 2022 Math Paper

Question : 3 of 100

Marks: +1, -0

x = 7 + 4\sqrt{3}

\sqrt{x} + \frac{1}{\sqrt{x}}

Solution:

Concept:

Express the given expression as a perfect square and rationalize the reciprocal.

Explanation:

We are given

x = 7 + 4\sqrt{3}

Notice that

7 + 4\sqrt{3} = 4 + 3 + 2 \cdot 2 \cdot \sqrt{3}

This can be written as

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

Therefore,

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

(positive root).

Now,

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

Rationalize by multiplying numerator and denominator by

2 - \sqrt{3}

\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}

Adding the two expressions:

\sqrt{x} + \frac{1}{\sqrt{x}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4

Answer:

4

(Option D)

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