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#93b1108f-7cd7-4f1a-9dfe-a57f2c51f3f9基础填空题导数与切线导数
考点来源：一数《【导数热门】零点与找点！姥姥级串讲！》一数《「导数含参单调性讨论」一小时大串讲！》一数《【高二】导数公式+切线+单调性极值！一课搞定！》一数《【高考最后十课】导数大题篇！2025高考冲刺！》一数《【导数必学】求导后无从下手？三个思路轻松搞定！》一数《BV1mcPMzbEvK》

3．（2023•徐汇区校级一模）若直线y=kx+by=kx+b是曲线f(x)=ex−2f(x)=e^{x-2}与g(x)=ex+2022−2022g(x)=e^{x+2022}-2022的公切线，则k=(k=(　　))

解析
【解答】解：设直线y=kx+by=kx+b与f(x)f(x)的图象相切于点P1(x1P_{1}(x_{1}，y1)y_{1})，与g(x)g(x)的图象相切于点P2(x2P_{2}(x_{2}，y2)y_{2})， 又f′(x)=ex−2f\prime (x)=e^{x-2}，g′(x)=ex+2022g\prime (x)=e^{x+2022}，且y1=ex1−2{y}_{1}={e}^{{x}_{1}-2}，y2=ex2+2022−2022{y}_{2}={e}^{{x}_{2}+2022}-2022． 曲线y=f(x)y=f(x)在点P1(x1P_{1}(x_{1}，y1)y_{1})处的切线方程为y−ex1−2=ex1−2(x−x1)y-{e}^{{x}_{1}-2}={e}^{{x}_{1}-2}(x-{x}_{1})， 曲线y=g(x)y=g(x)在点P2(x2P_{2}(x_{2}，y2)y_{2})处的切线方程为y−ex2+2022+2022=ex2+2022(x−x2)y-{e}^{{x}_{2}+2022}+2022={e}^{{x}_{2}+2022}(x-{x}_{2})． 故{ex1−2=ex2+2022ex1−2(1−x1)=ex2+2022(1−x2)−2022\left\{\begin{array}{l}{{e}^{{x}_{1}-2}={e}^{{x}_{2}+2022}}\\ {{e}^{{x}_{1}-2}(1-{x}_{1})={e}^{{x}_{2}+2022}(1-{x}_{2})-2022}\end{array}\right.，解得x1−x2=2024x_{1}-x_{2}=2024， 故k=y1−y2x1−x2=ex1−2−ex2+2022+2022x1−x2=20222024=10111012k=\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}=\frac{{e}^{{x}_{1}-2}-{e}^{{x}_{2}+2022}+2022}{{x}_{1}-{x}_{2}}=\frac{2022}{2024}=\frac{1011}{1012}． 故选：AA．

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