If \(K + \frac{1}{K} = 3 \), then what is the value of \({k^2} + \frac{1}{{{k^2}}}\)?

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SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 4)
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  1. 9
  2. 6
  3. 7
  4. 5

Answer (Detailed Solution Below)

Option 3 : 7
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Detailed Solution

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Given:

\(K + \frac{1}{K} = 3 \)

Formula used:

(a + b)2 = a+ 2ab + b2

Calculation:

Let us square both sides:

\((K + \frac{1}{K} )^2\) = 32

k2 + (2 × k × \(1 \over k\)) + \(1 \over k^2\) = 9

k2 + 2\(1 \over k^2\) = 9

k2 + \(1 \over k^2\) = 9 - 2

k2 + \(1 \over k^2\) = 7

∴ The value of k2 + \(\bf 1 \over k^2\)  is 7.

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