Question

if the mean of x and \[\dfrac{1}{x}\]is M, then the mean of $ {x^2}and\dfrac{1}{{{x^2}}} $ is
A) $ {M^2} $
B) $ \dfrac{{{M^2}}}{4} $
C) $ 2{M^2} - 1 $
D) $ 2{M^2} + 1 $

Answer 1

Hint:
Mean is equal to the sum of all the given observations divided by the total number of observations for a given data set.
So, Mean $ = \dfrac{\text{sum of observations}}{\text{Total number of observations}} $ . After applying the formula of the mean written above on x and \[\dfrac{1}{x}\], we need to take the square on both sides.

Complete step by step solution:
 it is given that the mean of x and \[\dfrac{1}{x}\] is M.
I.e. $ \dfrac{{x + \dfrac{1}{x}}}{2} = M $ , multiplying 2 on both sides , we get $ x + \dfrac{1}{x} = 2M $ .
Step2
On taking squares on both the sides of the above equation; we get
 $ {\left( {x + \dfrac{1}{x}} \right)^2} = {\left( {2M} \right)^2} $
Step 3
By using the identity, we will open the brackets in the above equation .
 $ {x^2} + \dfrac{1}{{{x^2}}} + 2.x.\dfrac{1}{x} = 4{M^2} $
It will become, $ {x^2} + \dfrac{1}{{{x^2}}} + 2 = 4{M^2} $
Step 4
On subtracting 2 form both the sides, $ {x^2} + \dfrac{1}{{{x^2}}} = 4{M^2} - 2..............(i) $
Step 5
We know that the mean of $ {x^2} $ and $ \dfrac{1}{{{x^2}}} $ can be calculated by using the formula of mean as :
Mean = $ \dfrac{{{x^2} + \dfrac{1}{{{x^2}}}}}{2} $ ; from the equation (i) we know that $ {x^2} + \dfrac{1}{{{x^2}}} = 4{M^2} - 2 $ .
So, on dividing both sides by 2, we get $ \dfrac{{{x^2} + \dfrac{1}{{{x^2}}}}}{2} = \dfrac{{4{M^2} - 2}}{2} $ .
Step 6
On solving the right hand side; we get $ \dfrac{{{x^2} + \dfrac{1}{{{x^2}}}}}{2} = \dfrac{{4{M^2}}}{2} - \dfrac{2}{2} $
 $ \Rightarrow \dfrac{{{x^2} + \dfrac{1}{{{x^2}}}}}{2} = 2{M^2} - 1 $ .

Hence, the mean of $ {x^2} \text{and } \dfrac{1}{{{x^2}}} $ is $ 2{M^2} - 1 $ .

Note:
Mean or the Arithmetic mean is also known as the expected value average. In general, mean or the average can be defined as the fraction having sum of all the given observations as the numerators and the total number of observations as denominators.
For example: mean of 11, 12, 13, 14 can be calculated as Mean $ = \dfrac{{11 + 12 + 13 + 14}}{4} = 12.5 $ .
As the given observations are 4 in number. By taking squares of the expressions on the both sides, the calculations of the new required mean become easier.