《2025年金版新学案高三总复习数学北师大版》

典例 1 (1)(2023·全国甲卷)曲线 $ y = \dfrac{e^x}{x + 1} $ 在点 $\left(1, \dfrac{e}{2}\right)$ 处的切线方程为(C)

A.$ y = \dfrac{e}{4}x $
B.$ y = \dfrac{e}{2}x $
C.$ y = \dfrac{e}{4}x + \dfrac{e}{4} $
D.$ y = \dfrac{e}{2}x + \dfrac{3e}{4} $

答案： 典例1
(1)C
(1)设曲线$y = \frac{e^{x}}{x + 1}$在点$(1,\frac{e}{2})$处的切线方程为$y - \frac{e}{2} = k(x - 1)$,因为$y^{\prime} = \frac{e^{x}(x + 1) - e^{x}}{(x + 1)^{2}} = \frac{xe^{x}}{(x + 1)^{2}}$,所以$k = y^{\prime}|_{x = 1} = \frac{e}{4}$,所以$y - \frac{e}{2} = \frac{e}{4}(x - 1)$,所以曲线$y = \frac{e^{x}}{x + 1}$在点$(1,\frac{e}{2})$处的切线方程为$y = \frac{e}{4}x + \frac{e}{4}$.故选C.

(2)(双空题)(2022·新高考Ⅱ卷)曲线 $ y = \ln |x| $ 过坐标原点的两条切线的方程为$y = \frac{1}{e}x$，$y = -\frac{1}{e}x$.

答案：
(2)$y = \frac{1}{e}x$ $y = -\frac{1}{e}x$(不分先后)
(2)由题意可知,函数的定义域为$\{x|x\neq0\}$.易证函数$y = \ln|x|$为偶函数,当$x ＞ 0$时,$y = \ln x$,设切点坐标为$(x_{0},\ln x_{0})$,因为$y^{\prime} = \frac{1}{x}$,所以切线斜率$k = y^{\prime}|_{x = x_{0}} = \frac{1}{x_{0}}$,故切线方程为$y - \ln x_{0} = \frac{1}{x_{0}}(x - x_{0})$,又知切线过原点$(0,0)$,所以$-\ln x_{0} = -1$,所以$x_{0} = e$,故切线方程为$y - 1 = \frac{1}{e}(x - e)$,即$y = \frac{1}{e}x$.由偶函数图象的对称性可知,另一条切线方程为$y = -\frac{1}{e}x$.故过坐标原点的两条切线方程为$y = \frac{1}{e}x$和$y = -\frac{1}{e}x$.

典例 2 (1) 已知曲线 $ y = x^2 - \ln x $ 在点 $ A $ 处的切线与直线 $ x + y - 2 = 0 $ 垂直，则点 $ A $ 的横坐标为(D)

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

答案： 典例2
(1)D
(1)设$f(x) = x^{2} - \ln x$,点$A(x_{0},y_{0})$,则$f^{\prime}(x) = 2x - \frac{1}{x}$.由在点A处的切线与直线$x + y - 2 = 0$垂直可得$f^{\prime}(x_{0}) = 1$,即$2x_{0} - \frac{1}{x_{0}} = 1$,又$x_{0} ＞ 0$,所以$x_{0} = 1$.故选D.

(2)(2025·北京西城模拟)已知直线 $ y = ex - 2 $ 是曲线 $ y = \ln x $ 的切线，则切点坐标为(A)

A.$\left(\dfrac{1}{e}, -1\right)$
B.$(e, 1)$
C.$\left(\dfrac{1}{e}, 1\right)$
D.$(0, 1)$

答案：
(2)A
(2)设切点坐标为$(t,\ln t)$,因为$(\ln x)^{\prime} = \frac{1}{x}$,所以在点$(t,\ln t)$处切线的斜率为$\frac{1}{t}$,所以曲线$y = \ln x$在点$(t,\ln t)$处的切线方程为$y - \ln t = \frac{1}{t}(x - t)$,即$y = \frac{1}{t}x + \ln t - 1$,所以$\begin{cases}\frac{1}{t} = e\\\ln t - 1 = - 1\end{cases}$,解得$t = \frac{1}{e}$,所以切点为$(\frac{1}{e}, - 1)$.故选A.

典例 3 (1)(2025·福建福州模拟)若曲线 $ y = \dfrac{ax + 1}{e^x} $ 有且仅有一条过坐标原点的切线，则正数 $ a $ 的值为(A)

A.$\dfrac{1}{4}$
B.$\dfrac{\sqrt{2}}{4}$
C.$\dfrac{1}{3}$
D.$\dfrac{\sqrt{3}}{3}$

答案： 典例3
(1)A
(1)设$y = f(x) = \frac{ax + 1}{e^{x}}$,则$f^{\prime}(x) = \frac{-ax + a - 1}{e^{x}}$,设切点为$(x_{0},\frac{ax_{0} + 1}{e^{x_{0}}})$,则$f^{\prime}(x_{0}) = \frac{-ax_{0} + a - 1}{e^{x_{0}}}$,所以切线方程为$y - \frac{ax_{0} + 1}{e^{x_{0}}} = \frac{-ax_{0} + a - 1}{e^{x_{0}}}(x - x_{0})$,又该切线过原点,所以$0 - \frac{ax_{0} + 1}{e^{x_{0}}} = \frac{-ax_{0} + a - 1}{e^{x_{0}}}(0 - x_{0})$,整理得$ax_{0}^{2} + x_{0} + 1 = 0$①,因为曲线$y = f(x)$只有一条过坐标原点的切线,所以方程①只有一个解,故$\Delta = 1 - 4a = 0$,解得$a = \frac{1}{4}$.故选A.

(2)(2022·新高考Ⅰ卷)若曲线 $ y = (x + a)e^x $ 有两条过坐标原点的切线，则实数 $ a $ 的取值范围是$(-\infty, - 4)\cup(0, +\infty)$.

答案：
(2)$(-\infty, - 4)\cup(0, +\infty)$
(2)因为$y = (x + a)e^{x}$,所以$y^{\prime} = (x + a + 1)e^{x}$.设切点为$A(x_{0},(x_{0} + a)e^{x_{0}})$,O为坐标原点,依题意,当曲线的切线过原点时,切线斜率$k_{OA} = y^{\prime}|_{x = x_{0}} = (x_{0} + a + 1)e^{x_{0}} = \frac{(x_{0} + a)e^{x_{0}}}{x_{0}}$,化简得$x_{0}^{2} + ax_{0} - a = 0$.因为曲线$y = (x + a)e^{x}$有两条过坐标原点的切线,所以关于$x_{0}$的方程$x_{0}^{2} + ax_{0} - a = 0$有两个不同的根,所以$\Delta = a^{2} + 4a ＞ 0$,解得$a ＜ - 4$或$a ＞ 0$,所以实数$a$的取值范围是$(-\infty, - 4)\cup(0, +\infty)$.

(1) 如图，函数 $ y = f(x) $ 的图象在点 $ P(1, y_0) $ 处的切线是 $ l $，则 $ f(1) + f'(1) =$(C)
(第1题图)
A. $ 1 $
B. $ 2 $
C. $ 0 $
D. $ -1 $
(2)(2025·广西桂林模拟)过点 $(1, 4)$ 且与曲线 $ f(x) = x^3 + x + 2 $ 相切的直线方程为(C)
A. $ 4x - y = 0 $
B. $ 7x - 4y + 9 = 0 $
C. $ 4x - y = 0 $ 或 $ 7x - 4y + 9 = 0 $
D. $ 4x - y = 0 $ 或 $ 4x - 7y + 24 = 0 $
(3)(2025·四川德阳模拟)已知直线 $ y = ax - 1 $ 与曲线 $ f(x) = \ln (ex) $ 相切，则 $ a $ 的值为(D)
A. $\dfrac{1}{e}$
B. $ 1 $
C. $\sqrt{e}$
D. $ e $

答案： 对点训练.
(1)C
(2)C
(3)D
(1)由图象可得切线过点$(2,0)$,$(0,2)$,所以切线$l$的方程为$\frac{x}{2} + \frac{y}{2} = 1$,即$y = 2 - x$,所以切线的斜率为$-1$,所以$f^{\prime}(1) = -1$.因为点$P(1,y_{0})$在切线上,所以$y_{0} = 2 - 1 = 1$,所以$f(1) = 1$,所以$f(1) + f^{\prime}(1) = 1 - 1 = 0$.故选C.
(2)设切点为$(x_{0},y_{0})$,$f^{\prime}(x) = 3x^{2} + 1$,则过点$(1,4)$的曲线$y = f(x)$切线为$l:y - y_{0} = (3x_{0}^{2} + 1)(x - x_{0})$,有$\begin{cases}(3x_{0}^{2} + 1)(1 - x_{0}) = 4 - y_{0}\\y_{0} = x_{0}^{3} + x_{0} + 2\end{cases}$,解得$\begin{cases}x_{0} = 1\\y_{0} = 4\end{cases}$或$\begin{cases}x_{0} = -\frac{1}{2}\\y_{0} = \frac{11}{8}\end{cases}$代入$l$可得$4x - y = 0$或$7x - 4y + 9 = 0$.故选C.
(3)由$f(x) = \ln(ex)$,可得$f^{\prime}(x) = \frac{e}{ex} = \frac{1}{x}$,设切点为$(x_{0},\ln(ex_{0}))$,则$f^{\prime}(x_{0}) = \frac{1}{x_{0}}$,则切线方程为$y - \ln(ex_{0}) = \frac{1}{x_{0}}(x - x_{0})$,即$y = \frac{1}{x_{0}}x + \ln(ex_{0}) - 1$,又直线$y = ax - 1$与曲线$f(x) = \ln(ex)$相切,所以$\begin{cases}\frac{1}{x_{0}} = a\\\ln(ex_{0}) - 1 = - 1\end{cases}$,解得$\begin{cases}x_{0} = \frac{1}{e}\\a = e\end{cases}$故选D.