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

Options

A. `1-M^(2)`
B. `1-2M`
C. `2M^(2)-1`
D. `2M^(2)+1`

Answer

To solve the problem step by step, let's follow the reasoning provided in the video transcript:

### Step 1: Understand the Average of x and 1/x
The average of \( x \) and \( \frac{1}{x} \) is given by:
\[ \text{Average} = \frac{x + \frac{1}{x}}{2} \]
According to the problem, this average is equal to \( M \). Therefore, we can write:
\[ \frac{x + \frac{1}{x}}{2} = M \]

### Step 2: Rearranging the Equation
To eliminate the fraction, we multiply both sides by 2:
\[ x + \frac{1}{x} = 2M \]

### Step 3: Square Both Sides
Next, we square both sides of the equation to find \( x^2 + \frac{1}{x^2} \):
\[ \left( x + \frac{1}{x} \right)^2 = (2M)^2 \]
Expanding the left side using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ x^2 + 2 + \frac{1}{x^2} = 4M^2 \]

### Step 4: Isolate x^2 + 1/x^2
Now, 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 1/x^2
The average of \( x^2 \) and \( \frac{1}{x^2} \) is given by:
\[ \text{Average} = \frac{x^2 + \frac{1}{x^2}}{2} \]
Substituting the expression we found in Step 4:
\[ \text{Average} = \frac{4M^2 - 2}{2} \]

### Step 6: Simplify the Expression
Now, simplify the expression:
\[ \text{Average} = 2M^2 - 1 \]

### Conclusion
Thus, the average of \( x^2 \) and \( \frac{1}{x^2} \) is:
\[ \text{Average} = 2M^2 - 1 \]