Least positive integral value of 'a' for which `log_((x+(1)/(x)))(a^(2)-3a+3) gt 0, (x gt 0)` :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the least positive integral value of \( a \) for which

\[ \log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0 \quad (x > 0). \]

### Step 1: Understand the logarithmic inequality

The inequality \( \log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0 \) implies that the argument of the logarithm must be greater than 1, since the base \( x + \frac{1}{x} > 1 \) for \( x > 0 \).

### Step 2: Set up the inequality

Thus, we can rewrite the inequality as:

\[ a^2 - 3a + 3 > 1. \]

### Step 3: Simplify the inequality

Subtracting 1 from both sides gives us:

\[ a^2 - 3a + 2 > 0. \]

### Step 4: Factor the quadratic expression

Now, we can factor the quadratic:

\[ (a - 1)(a - 2) > 0. \]

### Step 5: Analyze the factors

To find the solution to the inequality \( (a - 1)(a - 2) > 0 \), we need to determine the intervals where this product is positive. This occurs when both factors are either positive or negative.

### Step 6: Identify critical points

The critical points are \( a = 1 \) and \( a = 2 \). We can test the intervals:

1. For \( a < 1 \): Both \( (a - 1) < 0 \) and \( (a - 2) < 0 \) → Product is positive.
2. For \( 1 < a < 2 \): \( (a - 1) > 0 \) and \( (a - 2) < 0 \) → Product is negative.
3. For \( a > 2 \): Both \( (a - 1) > 0 \) and \( (a - 2) > 0 \) → Product is positive.

### Step 7: Write the solution in interval notation

Thus, the solution to the inequality is:

\[ a < 1 \quad \text{or} \quad a > 2. \]

### Step 8: Find the least positive integral value of \( a \)

Since we are looking for the least positive integral value of \( a \), we consider the interval \( a > 2 \). The smallest integer greater than 2 is 3.

### Conclusion

Therefore, the least positive integral value of \( a \) for which the inequality holds is:

\[ \boxed{3}. \]