If the average of x and `1/x(x ne 0)` is M, then the average of `x^2` and `1/x^(2)` is:

Options

A. a) `1- M^(2)`
B. b) `1- 2M`
C. c) `2M^(2)-1`
D. d) `2M^(2) + 1`

Answer

To solve the problem step by step, we need to find the average of \( x^2 \) and \( \frac{1}{x^2} \) given that the average of \( x \) and \( \frac{1}{x} \) is \( M \).

### Step 1: Write the expression for the average of \( x \) and \( \frac{1}{x} \)
The average of \( x \) and \( \frac{1}{x} \) is given by:
\[ \text{Average} = \frac{x + \frac{1}{x}}{2} = M \]

### Step 2: Rearrange the equation to find \( x + \frac{1}{x} \)
From the average expression, we can rearrange it to find:
\[ x + \frac{1}{x} = 2M \]

### Step 3: Square both sides to find \( x^2 + \frac{1}{x^2} \)
Now, we will square both sides of the equation:
\[ \left(x + \frac{1}{x}\right)^2 = (2M)^2 \]
This expands to:
\[ x^2 + 2 + \frac{1}{x^2} = 4M^2 \]

### Step 4: Isolate \( x^2 + \frac{1}{x^2} \)
From the squared equation, we can isolate \( x^2 + \frac{1}{x^2} \):
\[ x^2 + \frac{1}{x^2} = 4M^2 - 2 \]

### Step 5: Find the average of \( x^2 \) and \( \frac{1}{x^2} \)
Now, we need to find the average of \( x^2 \) and \( \frac{1}{x^2} \):
\[ \text{Average} = \frac{x^2 + \frac{1}{x^2}}{2} \]
Substituting the value we found in Step 4:
\[ \text{Average} = \frac{4M^2 - 2}{2} \]
This simplifies to:
\[ \text{Average} = 2M^2 - 1 \]

### Final Answer
Thus, the average of \( x^2 \) and \( \frac{1}{x^2} \) is:
\[ \boxed{2M^2 - 1} \]