If the average of x and \(\rm \frac{1}{x}(x\ne0)\) is M, then the average of x² and \(\rm \frac{1}{x^2}\) is:

Question

Question

This question was previously asked in

MP Vyapam Group 4 (Assistant Grade-3/Stenographer) Official Paper (Held On: 16 July, 2023 Shift 1)

Answer 1

Given:

Average of x and 1/x = M

Formula used:

Average = Sum of numbers / Number of numbers

(a + b)² = a² + b² + 2ab

Calculation:

Given that the average of x and 1/x is M:

(x + 1/x) / 2 = M

⇒ x + 1/x = 2M

⇒ (x + 1/x)² = (2M)²

⇒ x² + 2(x)(1/x) + 1/x² = 4M²

⇒ x² + 2 + 1/x² = 4M²

⇒ x² + 1/x² = 4M² - 2

Now, the average of x2 and 1/x2, which is:

(x2 + 1/x2) / 2

Average = (4M² - 2) / 2

Average = 2M² - 1

∴ The average of x² and 1/x² is 2M² - 1.

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