Untitled 21

6.

(1). \(f'(x) = \frac{(x-1)e^x(x+2) - (x-2)e^x}{(x+2)^2} = \frac{x^2e^x}{(x+2)^2} \ge 0. f(x)\) 在 \((-\infty, -2), (-2, +\infty) \uparrow\). \(f(0) = -1\). 已证 \(\frac{x-2}{x+2}e^x + 1 > 0\) 显然成立.

(2). \(g'(x) = \frac{(e^x-a)x^2 - 2x(e^x-ax-a)}{x^4} = \frac{xe^x - ax - 2e^x + 2ax + 2a}{x^3} = \frac{(x-2)e^x + a(x+2)}{x^3} \cdot (x+2)\). 令 \(q(x) = \frac{x-2}{x+2}e^x + a. q(x) \uparrow, q(0) = a-1 < 0. q(2) = a > 0. \exists x_0 \in (0, 2], q(x_0) = 0\). 故 \(a = -\frac{x_0-2}{x_0+2}e^{x_0}, x \in (0, 2]. h(a) = \varphi(x_0) = \frac{e^{x_0} \cdot (x_0+1)(x_0-2)}{x_0^2} \cdot \frac{e^{x_0}}{x_0+2} \dots = \frac{x_0}{x_0+2}e^{x_0}\). 在 \((0, 2] \uparrow\). 故 \(h(a) \in (\frac{1}{2}, \frac{e^2}{4}]\).

---

7.

(1). \(f'(x) = 3x^2 + a, f(x_0) = x_0^3 + ax_0 + \frac{1}{4}\). 切线 \(y = f'(x_0)(x-x_0) + f(x_0)\). 由定义知与原点判定 \(\begin{cases} f'(1) = 0 \\ f(x_0) - x_0 f'(x_0) = 0 \end{cases} \implies \begin{cases} x_0 = \frac{1}{2} \\ a = -\frac{3}{4} \end{cases}\).

(2). 由定义知 \(h(x)\) 在 \((1, +\infty)\) 恒小于 \(0\). 无零点. \(f'(x) = 0\). ① 当 \(a \in [0, +\infty)\) 时, \(f(x) \uparrow\). 且 \(f(0) = \frac{1}{4} > 0, f(1) > 0\). 故 \(h(x)\) 仅一个零点. ② \(a \in (-\infty, 0)\) 时. \(f'(-\sqrt{-\frac{a}{3}}) = f'(\sqrt{-\frac{a}{3}}) = 0. f(x)\) 在 \((0, \sqrt{-\frac{a}{3}}) \downarrow, (\sqrt{-\frac{a}{3}}, +\infty) \uparrow\). \(f(\sqrt{-\frac{a}{3}}) = \frac{1}{3}a\sqrt{-\frac{a}{3}} + \frac{2}{3}a\sqrt{-\frac{a}{3}} + \frac{1}{4} = \frac{2}{3}a\sqrt{-\frac{a}{3}} + \frac{1}{4} = 0 \implies a = -\frac{3}{4}\). \(f(\sqrt{-\frac{a}{3}}) > 0 \implies a > -\frac{3}{4}. f(\sqrt{-\frac{a}{3}}) < 0 \implies a < -\frac{3}{4}. f(1) = a + \frac{5}{4}\). (I). 当 \(a < -\frac{5}{4}\) 时, 仅一个零点. (II). \(a = -\frac{5}{4}\) 时, 两个零点. (III). \(-\frac{5}{4} < a < -\frac{3}{4}\) 时, 三个零点. (IV). \(a = -\frac{3}{4}\), 两个零点. \(a \in (-\frac{3}{4}, 0)\), 一个零点. 综上, \(a \in (-\infty, -\frac{5}{4}) \cup (-\frac{3}{4}, +\infty)\) 一个零点. \(a \in (-\frac{5}{4}, -\frac{3}{4})\) 三个零点. \(a \in \{-\frac{5}{4}, -\frac{3}{4}\}\) 两个.