已知函数 f(x) = (1)/(2)x^2 - aln x (a ∈ R)。 求函数 f(x)的单调区间； 若 a=2, 求函数 f(x)在区间[1, e]上的最值。---**Extraction Content:** 11. (12 分) 已知函数 f(x) = (1)/(2)x^2 - aln x (a ∈ R)。 求函数 f(x)的单调区间； 若 a=2, 求函数 f(x)在区间[1, e]上的最值。

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Let's analyze the function f(x) = (1/2)x² - a ln x. First, we determine the domain. Since ln x requires x > 0, our domain is (0, +∞). Next, we find the derivative: f'(x) = x - a/x = (x² - a)/x. The sign of f'(x) determines the monotonic intervals. Now we analyze the sign of f'(x) = (x² - a)/x. For Case 1, when a ≤ 0, we have x² - a ≥ x² > 0 for all x > 0, so f'(x) > 0 and f(x) is increasing on (0, +∞). For Case 2, when a > 0, setting x² - a = 0 gives x = √a. When x ∈ (0, √a), f'(x) < 0 so f(x) is decreasing. When x ∈ (√a, +∞), f'(x) > 0 so f(x) is increasing. For the special case a = 2, we have f(x) = (1/2)x² - 2ln x. The derivative is f'(x) = x - 2/x = (x² - 2)/x. Setting f'(x) = 0 gives x = √2. Since 1 < √2 ≈ 1.414 < e ≈ 2.718, the critical point √2 lies within our interval [1, e]. This critical point will be key to finding the extrema. Now we calculate the function values at the critical point and endpoints. At x = 1: f(1) = 1/2. At x = √2: f(√2) = 1 - ln 2, which is approximately -0.193. At x = e: f(e) = (1/2)e² - 2, which is approximately 1.695. Since the function decreases from x = 1 to x = √2, then increases from x = √2 to x = e, the minimum occurs at √2 and the maximum at x = e. To summarize our complete analysis: For the monotonic intervals of f(x) = (1/2)x² - a ln x, if a ≤ 0, the function is increasing on (0, +∞). If a > 0, the function is decreasing on (0, √a) and increasing on (√a, +∞). For the special case when a = 2 on the interval [1, e], the minimum value is 1 - ln 2 occurring at x = √2, and the maximum value is (1/2)e² - 2 occurring at x = e. This completes our function analysis.