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tags:
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- 高数复习模拟
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---
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### 一、选择题（共3小题，每小题6分，共18分）
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1. 下列级数发散的是（ ）。
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(A) $\sum_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$
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(B) $\sum_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$
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(C) $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$
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(D) $\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$
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2. 当 $x \to 0$时，与 $x - \sin x$同阶的无穷小是（ ）。
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(A) $x + \tan x$
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(B) $x \tan x$
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(C) $x^2 + \tan x$
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(D) $x^2 \tan x$
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3. 已知函数 $f(x)$在 $x = 1$的某邻域内三阶可导，且
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$$
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\lim_{x \to 1} \frac{f(x) - 2}{(x - 1)^2 \ln x} = \frac{1}{3},
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$$
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则（ ）。
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(A) $x = 1$为函数 $f(x)$的极大值点
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(B) 点 $(1, 2)$为曲线 $y = f(x)$的拐点
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(C) $x = 1$为函数 $f(x)$的极小值点
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(D) 点 $(1, 0)$为曲线 $y = f(x)$的拐点
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---
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### 二、填空题（共3小题，每小题6分，共18分）
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4. 极限 $$\lim_{n \to \infty} \frac{1}{n} \left( \tan \frac{\pi}{4n} + \tan \frac{2\pi}{4n} + \cdots + \tan \frac{n\pi}{4n} \right)$$ 的值为______。
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5. 已知 $\int f(x^2) dx = x \ln x + C，$则 $f'(2) = \underline{\quad }。$
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6. 极坐标曲线 $\rho = \theta$ 在点 $(\rho, \theta) = (\pi, \pi)$ 处的切线的直角坐标方程为______
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---
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### 三、解答题（共4小题，共64分）
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7. 计算数列极限
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$$
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\lim_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}。
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$$
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（12分）
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```text
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```
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7. (16分)已知函数 $f(x)$ 在 $[0,1]$上连续，在 $(0,1)$内可导，且 $f(0)=0$ ，$f(1)=1$ 。试证明：
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（1）（6分）存在 $\xi \in (0,1)$，使得
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$$
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f'(\xi) = 2\xi。
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$$
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（2）（10分）对任意正数 $a,b$，在 $(0,1)$内存在相异的两点 $x_1, x_2$，使得
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$$
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\frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = a+b。
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$$
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```text
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```
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7. （16分）已知当 $x \to 0$ 时，函数 $f(x) = \sqrt{a + bx^2} - \cos x$ 与 $x^2$ 是等价无穷小。
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（1）求参数 $a, b$ 的值；（6分）
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（2）计算极限 $\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分）
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```text
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```
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8. （20分）设函数 $f(x)$在 $[0,1]$上可导，$\int_{0}^{\frac{1}{2}} f(x) dx = 0$。证明存在 $\xi \in (0,1)$，使得
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$$f(\xi) = (1 - \xi) f'(\xi)。$$
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```text
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```
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# 高等数学（Ⅰ）考试试卷（A）卷 大题解答
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## 7. 计算极限
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$$
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\lim_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}
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$$
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**解：**
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令$L = \lim\limits_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}$，取对数得：
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$$
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\ln L = \lim_{x \to +\infty} \frac{\ln(x^2 + 2^x)}{x}
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$$
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当$x \to +\infty$时，$2^x \gg x^2$，故：
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$$
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x^2 + 2^x \sim 2^x
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$$
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$$
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\ln(x^2 + 2^x) = x\ln 2 + \ln\left(1 + \frac{x^2}{2^x}\right) \sim x\ln 2
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$$
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因此：
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$$
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\ln L = \lim_{x \to +\infty} \frac{x\ln 2}{x} = \ln 2
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$$
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$$
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L = e^{\ln 2} = 2
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$$
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**答案：**$\boxed{2}$
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---
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## 8. 中值定理证明题
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已知$f(x)$在$[0,1]$上连续，$(0,1)$内可导，$f(0)=0$，$f(1)=1$。
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**(1)** 证明存在$\xi \in (0,1)$使$f'(\xi) = 2\xi$
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**证：**
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构造$g(x) = f(x) - x^2$
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-$g(0) = f(0) - 0 = 0$
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-$g(1) = f(1) - 1 = 0$
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由罗尔定理，存在$\xi \in (0,1)$使$g'(\xi) = 0$，即：
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$$
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f'(\xi) - 2\xi = 0 \quad \Rightarrow \quad f'(\xi) = 2\xi
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$$
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---
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**(2)** 对任意$a,b > 0$，存在相异$x_1, x_2 \in (0,1)$使：
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$$
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\frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = a+b
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$$
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**证：**
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由介值定理，存在$c \in (0,1)$使$f(c) = \dfrac{a}{a+b}$（因$0 < \frac{a}{a+b} < 1$）
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1. 在$[0,c]$上用拉格朗日中值定理：
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$$
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f'(x_1) = \frac{f(c)-f(0)}{c-0} = \frac{\frac{a}{a+b}}{c}, \quad x_1 \in (0,c)
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$$
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$$
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\frac{a}{f'(x_1)} = a \cdot \frac{c}{\frac{a}{a+b}} = c(a+b)
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$$
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2. 在$[c,1]$上用拉格朗日中值定理：
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$$
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f'(x_2) = \frac{f(1)-f(c)}{1-c} = \frac{1 - \frac{a}{a+b}}{1-c} = \frac{\frac{b}{a+b}}{1-c}, \quad x_2 \in (c,1)
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$$
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$$
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\frac{b}{f'(x_2)} = b \cdot \frac{1-c}{\frac{b}{a+b}} = (1-c)(a+b)
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$$
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相加得：
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$$
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\frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = c(a+b) + (1-c)(a+b) = a+b
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$$
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且$x_1 < c < x_2$，故$x_1 \neq x_2$。
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---
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## 9. 等价无穷小与参数确定
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已知$f(x) = \sqrt{a + bx^2} - \cos x \sim x^2 \ (x\to 0)$
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**(1)** 求$a, b$
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**解：**
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展开至$x^4$：
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$$
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\sqrt{a+bx^2} = \sqrt{a} + \frac{b\sqrt{a}}{2a}x^2 - \frac{b^2\sqrt{a}}{8a^2}x^4 + o(x^4)
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$$
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$$
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\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)
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$$
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代入$f(x)$：
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$$
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f(x) = (\sqrt{a}-1) + \left(\frac{b}{2\sqrt{a}} + \frac{1}{2}\right)x^2 + \left(-\frac{b^2}{8a^{3/2}} - \frac{1}{24}\right)x^4 + o(x^4)
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$$
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由$f(x) \sim x^2$得：
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1.$\sqrt{a}-1 = 0 \Rightarrow a = 1$
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2.$\frac{b}{2} + \frac{1}{2} = 1 \Rightarrow b = 1$
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**(2)** 计算$\lim\limits_{x\to 0} \dfrac{f(x) - x^2}{x^4}$
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**解：**
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$a=1, b=1$时：
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$$
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f(x) = \sqrt{1+x^2} - \cos x = x^2 - \frac{1}{6}x^4 + o(x^4)
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$$
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$$
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f(x) - x^2 = -\frac{1}{6}x^4 + o(x^4)
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$$
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$$
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\lim_{x\to 0} \frac{f(x) - x^2}{x^4} = -\frac{1}{6}
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$$
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**答案：**
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(1)$a=1, b=1$
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(2)$-\dfrac{1}{6}$
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---
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## 10. 积分条件与微分方程形式的中值定理
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已知$f(x)$在$[0,1]$上可导，$\int_0^{1/2} f(x) dx = 0$
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证明存在$\xi \in (0,1)$使$f(\xi) = (1-\xi)f'(\xi)$
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**证：**
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令$g(x) = (1-x)f(x)$，则：
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$$
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g'(x) = -f(x) + (1-x)f'(x)
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$$
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要证等式等价于$g'(\xi) = 0$
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设$h(x) = \int_0^x f(t)dt$，由已知$h(1/2) = 0$，且$h(0)=0$
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由罗尔定理，存在$c \in (0,1/2)$使$h'(c) = f(c) = 0$
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于是：
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$$
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g(c) = (1-c)f(c) = 0, \quad g(1) = (1-1)f(1) = 0
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$$
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在$[c,1]$上对$g(x)$应用罗尔定理，存在$\xi \in (c,1) \subset (0,1)$使：
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$$
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g'(\xi) = 0 \quad \Rightarrow \quad f(\xi) = (1-\xi)f'(\xi)
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$$
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证毕。
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