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UPSC CDS 1 2021 Math Paper

Question : 72 of 100

Marks: +1, -0

x + \frac{1}{x} = 8

x^{2} + \frac{1}{x^{2}}

equal to?

Solution:

Concept:

Use the algebraic identity

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

to relate the given sum to the required sum of squares.

Explanation:

Given:

x + \frac{1}{x} = 8

Square both sides:

\left(x + \frac{1}{x}\right)^2 = 8^2

Expand the left side using the identity:

x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 64

Simplify the middle term:

x \cdot \frac{1}{x} = 1

, so

2 \times 1 = 2

Thus,

x^2 + 2 + \frac{1}{x^2} = 64

Subtract 2 from both sides:

x^2 + \frac{1}{x^2} = 64 - 2 = 62

Therefore, the required value is

62

Answer:

62

, which corresponds to option C.

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