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UPSC CDS II 2025 Mathematics Paper

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Question : 28 of 100

Marks: +1, -0

If

p=\frac{\sqrt{5}-2}{\sqrt{5}+2}

q=\frac{\sqrt{5}+2}{\sqrt{5}-2}

(\frac{p}{q}+\frac{q}{p})

equal to?

18
8√5
322
72√5

Solution:

Concept:

Recognize that

p

and

q

are reciprocals because

pq = 1

. Thus,

\frac{p}{q} + \frac{q}{p} = p^2 + q^2

.

Explanation:

First, square

p

:

p^2 = \left(\frac{\sqrt{5}-2}{\sqrt{5}+2}\right)^2 = \frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)^2} = \frac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}}

.

Similarly,

q^2 = \frac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}}

.

Add them:

\frac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}} + \frac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}} = \frac{(9 - 4\sqrt{5})^2 + (9 + 4\sqrt{5})^2}{(9 + 4\sqrt{5})(9 - 4\sqrt{5})}

.

The denominator simplifies:

9^2 - (4\sqrt{5})^2 = 81 - 80 = 1

.

The numerator:

(81 + 80) + (81 + 80) = 161 + 161 = 322

.

Hence, the sum equals

322

.

Answer:

322

(option C).

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