If the average of x and \(\frac{1 }{x}\) \((x\neq 0)\) is M, then the average of \(x^{2} and \frac{1 }{x^{2}}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: \(2m^{2}-1\)
Given is the average of x and \(\frac{1 }{x}\) \((x\neq 0)\) is M i.e.
\(\frac{x + \frac{1}{×}}{2}\) = m
⇒ x + \(\frac{1}{×}\) = 2m
x2 + \(\frac{1}{×^2}\) = 4m2 - 2
Avg. = \(\frac{4m^2 - 2}{2}\) = 2m2 - 1