If `A = x^2+1/x^2 , B = x-1/x` then minimum value of `A/B` is

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `sqrt(2)+2`
D. none of these

Answer

To find the minimum value of \( \frac{A}{B} \) where \( A = x^2 + \frac{1}{x^2} \) and \( B = x - \frac{1}{x} \), we can follow these steps:

### Step 1: Write the expression for \( \frac{A}{B} \)

We start with the definitions of \( A \) and \( B \):
\[ A = x^2 + \frac{1}{x^2}, \quad B = x - \frac{1}{x} \]
Thus,
\[ \frac{A}{B} = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}} \]

### Step 2: Simplify \( \frac{A}{B} \)

To simplify this expression, we can rewrite \( A \) in a different form. We can express \( A \) as:
\[ A = \left( x - \frac{1}{x} \right)^2 + 2 \]
This is because:
\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2} \Rightarrow x^2 + \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 + 2 \]

Now substituting this into our expression for \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{\left( x - \frac{1}{x} \right)^2 + 2}{x - \frac{1}{x}} \]

### Step 3: Let \( t = x - \frac{1}{x} \)

Let \( t = x - \frac{1}{x} \). Then we can rewrite the expression as:
\[ \frac{A}{B} = \frac{t^2 + 2}{t} \]
This simplifies to:
\[ \frac{A}{B} = t + \frac{2}{t} \]

### Step 4: Find the minimum value of \( t + \frac{2}{t} \)

To find the minimum value of \( t + \frac{2}{t} \), we can use the AM-GM inequality:
\[ t + \frac{2}{t} \geq 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2} \]
Equality holds when \( t = \sqrt{2} \).

### Step 5: Conclusion

Thus, the minimum value of \( \frac{A}{B} \) is:
\[ \frac{A}{B} \geq 2\sqrt{2} \]
Therefore, the minimum value of \( \frac{A}{B} \) is \( 2\sqrt{2} \).

### Final Answer:
The minimum value of \( \frac{A}{B} \) is \( 2\sqrt{2} \).

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