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 inequality \( \log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0 \) for \( x > 0 \), we will follow these steps:

### Step 1: Rewrite the logarithmic inequality
The given inequality can be rewritten as:
\[ \log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0 \]
This implies that:
\[ a^2 - 3a + 3 > (x + \frac{1}{x})^0 \]
Since any number raised to the power of 0 is 1, we have:
\[ a^2 - 3a + 3 > 1 \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ a^2 - 3a + 2 > 0 \]

### Step 3: Factor the quadratic expression
Next, we factor the quadratic expression:
\[ a^2 - 3a + 2 = (a - 1)(a - 2) \]
Thus, we need to solve:
\[ (a - 1)(a - 2) > 0 \]

### Step 4: Determine the intervals
To find the intervals where this product is positive, we analyze the critical points:
- The critical points are \( a = 1 \) and \( a = 2 \).
- The sign of the product \( (a - 1)(a - 2) \) changes at these points.

Testing intervals:
1. For \( a < 1 \): Choose \( a = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \) (Positive)
2. For \( 1 < a < 2 \): Choose \( a = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = -0.25 < 0 \) (Negative)
3. For \( a > 2 \): Choose \( a = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \) (Positive)

### Step 5: Identify the valid intervals
From the analysis, we find:
- The product \( (a - 1)(a - 2) > 0 \) for \( a < 1 \) or \( a > 2 \).

### Step 6: Find the least positive integer value of \( a \)
The positive intervals are \( a > 2 \). The least positive integer satisfying this condition is:
\[ a = 3 \]

### Final Answer
Thus, the least positive integral value of \( a \) for which the inequality holds is:
\[ \boxed{3} \]