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UPSC CDS 2 2022 Math Paper

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Question : 1 of 100

Marks: +1, -0

x^{2} - 20 = \sqrt{20+\sqrt{20+\sqrt{20+\dots \infty}}}

, then what is x equal to ?

4
5
√5
2√5

Solution:

Concept:

For an infinite nested radical of the form

\sqrt{a+\sqrt{a+\sqrt{a+\dots \infty}}}

, the value is given by

\frac{\sqrt{4a+1}+1}{2}

.

Explanation:

Let

y = \sqrt{20+\sqrt{20+\sqrt{20+\dots \infty}}}

.

Then

y = \frac{\sqrt{4(20)+1}+1}{2}

.

Compute:

4 \times 20 = 80

, so

\sqrt{80+1} = \sqrt{81} = 9

.

Thus

y = \frac{9+1}{2} = \frac{10}{2} = 5

.

Given

x^2 - 20 = y

, so

x^2 - 20 = 5

.

Hence

x^2 = 25

, so

x = 5

(taking the positive root).

Answer:

x = 5

(Option B).

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