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Telangana TET Paper 1 Exam Model Paper 2

Section: Mathematics

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Question : 98 of 150

Marks: +1, -0

If

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

, then find the value of

x^{3}+\frac{1}{x^{3}}

.

$-2$
$-1$
0
2

Solution:

Concept:

Use algebraic identities for

(x + \frac{1}{x})

to find

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

and then

x^3 + \frac{1}{x^3}

.

Explanation:

We are given

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

.

First compute

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

using the identity:

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

.

Substitute the value:

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

.

Now use the identity for cubes:

x^3 + \frac{1}{x^3} = (x + \frac{1}{x}) \times (x^2 + \frac{1}{x^2}) - (x + \frac{1}{x})

.

Plug in the known values:

x^3 + \frac{1}{x^3} = 1 \times (-1) - 1 = -1 - 1 = -2

.

Answer:

The value of

x^{3}+\frac{1}{x^{3}}

is

-2

. Option A.

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