《2025年绿色通道45分钟课时作业与单元测评高中数学选择性必修第二册人教版》

1. 函数$ f(x)=x^2 + 1 $，当自变量$ x $由$ 1 $增加到$ 1 + \Delta x $时，函数的平均变化率为 (C)

A.$ 2 $
B.$ \Delta x + (\Delta x)^2 $
C.$ \Delta x + 2 $
D.$ -\Delta x - 2 $

答案： 1.C $\because f(x) = x^2 + 1, \therefore \Delta y = f(1 + \Delta x) - f(1) =$
$(\Delta x)^2 + 2\Delta x$,$\therefore \frac{\Delta y}{\Delta x} = \frac{(\Delta x)^2 + 2\Delta x}{\Delta x} = \Delta x + 2$,故选 C.

2. (2024·河北保定一中月考)设函数$ f(x) $可导，则$ \lim\limits_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{\Delta x} $等于 (A)

A.$ f'(1) $
B.$ 3f'(1) $
C.$ \frac{1}{3}f'(1) $
D.$ f'(3) $

答案： 2.A 根据导数的定义，有$\lim_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{\Delta x} = f'(1)$.

3. 已知函数$ y = f(x) = x^3 $，则用平均变化率估计$ f(x) $在$ x = 1 $处的瞬时变化率为 (C)

A.$ 1 $
B.$ 2 $
C.$ 3 $
D.$ 4 $

答案： 3.C 函数$y = f(x) = x^3$在$[1,1 + \Delta x]$上的平均变化率为
$\frac{\Delta y}{\Delta x} = \frac{f(1 + \Delta x) - f(1)}{1 + \Delta x - 1} = \frac{(1 + \Delta x)^3 - 1}{\Delta x} = (\Delta x)^2 + 3\Delta x + 3$,
则$f'(1) = \lim_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{1 + \Delta x - 1} = \lim_{\Delta x \to 0} [(\Delta x)^2 + 3\Delta x + 3] = 3$,
故估计$f(x)$在$x = 1$处的瞬时变化率为3.

4. 已知$ f(x) $在$ \mathbf{R} $上的导数为$ f'(x) $，若$ \lim\limits_{\Delta x \to 0} \frac{f(3 - \Delta x) - f(3)}{3\Delta x} = 2 $，则$ f'(3) = $(C)

A.$ -2 $
B.$ 2 $
C.$ -6 $
D.$ 6 $

答案： 4.C 由于$\lim_{\Delta x \to 0} \frac{f(3 - \Delta x) - f(3)}{3\Delta x} =$
$\frac{1}{3} \lim_{\Delta x \to 0} \frac{f(3 - \Delta x) - f(3)}{- \Delta x} = \frac{1}{3} f'(3) = 2$,
则$f'(3) = -6$.

5. (多选)设$ f(x) $在$ x_0 $处可导，则下列式子中与$ f'(x_0) $相等的是 (AC)

A.$ \lim\limits_{\Delta x \to 0} \frac{f(x_0) - f(x_0 - 2\Delta x)}{2\Delta x} $
B.$ \lim\limits_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0 - \Delta x)}{\Delta x} $
C.$ \lim\limits_{\Delta x \to 0} \frac{f(x_0 + 2\Delta x) - f(x_0 + \Delta x)}{\Delta x} $
D.$ \lim\limits_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0 - 2\Delta x)}{\Delta x} $

答案： 5.AC A选项，$\lim_{\Delta x \to 0} \frac{f(x_0) - f(x_0 - 2\Delta x)}{2\Delta x} = f'(x_0)$.
B选项，$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0 - \Delta x)}{\Delta x} =$
$2\lim_{\Delta x \to 0} \frac{f(x_0 - \Delta x + 2\Delta x) - f(x_0 - \Delta x)}{2\Delta x} = 2f'(x_0)$.
C选项，$\lim_{\Delta x \to 0} \frac{f(x_0 + 2\Delta x) - f(x_0 + \Delta x)}{\Delta x} = f'(x_0)$.
D选项，$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0 - 2\Delta x)}{\Delta x} =$
$3\lim_{\Delta x \to 0} \frac{f(x_0 - 2\Delta x + 3\Delta x) - f(x_0 - 2\Delta x)}{3\Delta x} = 3f'(x_0)$.
故选 AC.

6. 若可导函数$ f(x) $的图象过原点，且满足$ \lim\limits_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x} = -1 $，则$ f'(0) $等于 (C)

A.$ -2 $
B.$ 2 $
C.$ -1 $
D.$ 1 $

答案： 6.C $\because f(x)$的图象过原点,$\therefore f(0) = 0$,
$\therefore f'(0) = \lim_{\Delta x \to 0} \frac{f(0 + \Delta x) - f(0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x} = -1$.

7. 已知函数$ y = f(x) $在$ x = x_0 $处的导数为$ 1 $，则$ \lim\limits_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = $1$ $。

答案： 7.1 解析 根据题意，由导数的概念可得
$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = f'(x_0)$，又由函数$f(x)$在
$x = x_0$处的导数为1，即$f'(x_0) = 1$，得
$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = 1$.

8. 已知函数$ y = f(x) = 2x^2 + 1 $在$ x = x_0 $处的瞬时变化率为$ -8 $，则$ f(x_0) = $9$ $。

答案： 8.9 解析 由题知$-8 = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} (2\Delta x + 4x_0) = 4x_0$，解
得$x_0 = -2$，所以$f(x_0) = f(-2) = 2 × (-2)^2 + 1 = 9$.

9. 在函数$ y = f(x) = x^2 + 3 $的图象上取一点$ P(1, 4) $及附近一点$ (1 + \Delta x, 4 + \Delta y) $。求：
(1) $ \frac{\Delta y}{\Delta x} $；
(2) $ f'(1) $。

答案： 9.解
(1)$\frac{\Delta y}{\Delta x} = \frac{f(1 + \Delta x) - f(1)}{\Delta x} =$
$\frac{(1 + \Delta x)^2 + 3 - (1^2 + 3)}{\Delta x} = 2 + \Delta x$.
(2)$f'(1) = \lim_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{\Delta x} = \lim_{\Delta x \to 0} (2 + \Delta x) = 2$.

10. (1) 利用导数的定义求函数$ y = x^2 + \frac{1}{x} + 5 $在$ x = 2 $处的导数。
(2) 利用导数的定义求函数$ f(x) = \frac{1}{x} - \sqrt{x} $在$ x = 4 $处的导数。

答案： 10.解
(1)当$x = 2$时，$\Delta y = (2 + \Delta x)^2 + \frac{1}{2 + \Delta x} + 5 -$
$(2^2 + \frac{1}{2} + 5) = 4\Delta x + (\Delta x)^2 + \frac{ - \Delta x}{2(2 + \Delta x)}$，
所以$\frac{\Delta y}{\Delta x} = 4 + \Delta x - \frac{1}{4 + 2\Delta x}$，
所以$y'\big|_{x=2} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} (4 + \Delta x - \frac{1}{4 + 2\Delta x}) =$
$4 - \frac{1}{4} = \frac{15}{4}$.
(2)当$x = 4$时，$\Delta y = f(4 + \Delta x) - f(4) = \frac{1}{4 + \Delta x} -$
$\sqrt{4 + \Delta x} - (\frac{1}{4} - \sqrt{4}) = (\frac{1}{4 + \Delta x} - \frac{1}{4}) - (\sqrt{4 + \Delta x} - 2)$
$2) = \frac{ - \Delta x}{4(4 + \Delta x)} - \frac{\Delta x}{\sqrt{4 + \Delta x} + 2}$
所以$\frac{\Delta y}{\Delta x} = \frac{ - 1}{4(4 + \Delta x)} - \frac{1}{\sqrt{4 + \Delta x} + 2}$
所以$f'(4) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} [\frac{ - 1}{4(4 + \Delta x)} - \frac{1}{\sqrt{4 + \Delta x} + 2}] =$
$- \frac{5}{16}$