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UPSC CDS 2 2023 Math Paper

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Question : 12 of 100

Marks: +1, -0

If x = 97 + 56√3, then what is the value of

\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}}

?

7
6
5
4

Solution:

Concept:

Use the identity

(a+b)^2 = a^2 + 2ab + b^2

to express

x

and its square root as perfect squares of binomials involving surds.

Explanation:

Given

x = 97 + 56\sqrt{3}

.

First,

x = (7 + 4\sqrt{3})^2

because

(7)^2 + 2 \cdot 7 \cdot 4\sqrt{3} + (4\sqrt{3})^2 = 49 + 56\sqrt{3} + 48 = 97 + 56\sqrt{3}

.

Therefore

\sqrt{x} = 7 + 4\sqrt{3}

(positive root).

Next,

\sqrt{x} = 7 + 4\sqrt{3} = (2 + \sqrt{3})^2

because

(2)^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

.

Hence

\sqrt[4]{x} = \sqrt{\sqrt{x}} = 2 + \sqrt{3}

.

Also

\frac{1}{\sqrt[4]{x}} = \frac{1}{2 + \sqrt{3}}

. Rationalize: multiply numerator and denominator by

2 - \sqrt{3}

to get

\frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}

.

Finally,

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

.

Answer:

4

(Option D).

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