<p>Least positive integral value of \(a\) for which \(\log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0\) (where \(x > 0\)):</p>
Step-by-Step Solution
Key Concept: For a logarithm to be positive, the argument must be greater than the base when base > 1, or between 0 and 1 when 0 < base < 1. We must first determine the range of the base (x + 1/x) for x > 0, then apply the appropriate inequality.
**Step 1: Determine the range of the base.**
For $x > 0$, by the AM-GM inequality, we have:
$$x + \frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2$$
Equality holds when $x = 1$.
Thus, the base of the logarithm, $x + \frac{1}{x}$, is always greater than or equal to 2. This implies that the base is always greater than 1.
**Step 2: Apply logarithm inequality rule.**
For a logarithm $\log_b(y) > 0$, if the base $b > 1$, then the argument $y$ must be greater than 1.
From Step 1, we established that the base $x + \frac{1}{x} \ge 2$, which means $x + \frac{1}{x} > 1$.
Therefore, for $\log_{(x + \frac{1}{x})}(a^2 - 3a + 3) > 0$ to hold, we must have:
$$a^2 - 3a + 3 > 1$$
**Step 3: Solve the inequality.**
Subtracting 1 from both sides of the inequality:
$$a^2 - 3a + 2 > 0$$
Factoring the quadratic expression:
$$(a - 1)(a - 2) > 0$$
This inequality holds when $a < 1$ or $a > 2$.
**Step 4: Find the least positive integral value.**
We seek the least positive integral value of $a$ that satisfies $a < 1$ or $a > 2$.
\begin{itemize}
\item If $a < 1$, there are no positive integers that satisfy this condition.
\item If $a > 2$, the positive integers satisfying this condition are $3, 4, 5, \dots$. The least among these is $a = 3$.
\end{itemize}
Thus, the least positive integral value of $a$ for which the inequality holds is $3$.
Correct Answer: B