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CTET Paper II (Mathematics and Science) 11 Jan 2023 Solved Paper

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Question : 40 of 150

Marks: +1, -0

If

x^{4}+\frac{1}{x^{4}}=727

, x > 0, then the value of

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

is:

3√3
$\sqrt{21}$
5
7

Solution:

Concept:

Given

x^{4} + \frac{1}{x^{4}}

, we use the relationship:

x^{2} + \frac{1}{x^{2}} = \sqrt{x^{4} + \frac{1}{x^{4}} + 2}

, and then

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

. Finally,

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

.

Explanation:

First, compute

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

:

x^{2} + \frac{1}{x^{2}} = \sqrt{727 + 2} = \sqrt{729} = 27

.

Next, find

x + \frac{1}{x}

:

x + \frac{1}{x} = \sqrt{27 + 2} = \sqrt{29}

.

Finally, use

x - \frac{1}{x} = \sqrt{(x + \frac{1}{x})^{2} - 4} = \sqrt{29 - 4} = \sqrt{25} = 5

.

Since

x > 0

, the value is positive.

Answer:

5 (Option C)

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