Test Index

UPSC NDA I 2025 Mathematics Paper

© examsnet.com

Question : 14 of 120

Marks: +1, -0

If

x^{2}− x +1 = 0

(x-\frac{1}{x})^{2}+(x-\frac{1}{x})^{4}+(x-\frac{1}{x})^{8}

81
85
87
90

Solution:

Concept:

Given

x^2 - x + 1 = 0

, divide by

x

to get

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

. Then use

\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4

to find its value.

Explanation:

Divide

x^2 - x + 1 = 0

by

x

(since

x \neq 0

):

x - 1 + \frac{1}{x} = 0

, so

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

.

Compute

\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 = 1^2 - 4 = -3

.

Now evaluate each term:

\left(x - \frac{1}{x}\right)^2 = -3

.

\left(x - \frac{1}{x}\right)^4 = \left[\left(x - \frac{1}{x}\right)^2\right]^2 = (-3)^2 = 9

.

\left(x - \frac{1}{x}\right)^8 = \left[\left(x - \frac{1}{x}\right)^4\right]^2 = 9^2 = 81

.

Sum them:

-3 + 9 + 81 = 87

.

Answer:

87 (Option C)

© examsnet.com

Go to Question:

Prev Question

Next Question