当前位置：首页>文档>2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练

2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练

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2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练
2025版新教材高考数学第二轮复习专题练--3.1　导数的概念及运算（含答案）_2.2025数学总复习_2025年新高考资料_二轮复习_备考20252025版新教材高考数学第二轮复习专题练

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2025版新教材高考数学第二轮复习 专题三 导数及其应用 3.1 导数的概念及运算 五年高考 高考新风向 ex+2sinx 1.(2024全国甲理,6,5分,易)设函数f(x)= ,则曲线y=f(x)在点(0,1)处的切线与两坐 1+x2 标轴所围成的三角形的面积为 ( ) 1 1 1 2 A. B. C. D. 6 3 2 3 2.(2024新课标Ⅰ,13,5分,中)若曲线y=ex+x在点(0,1)处的切线也是曲线y=ln(x+1)+a的切 线,则a= . 考点 导数的运算及几何意义 1.(2020课标Ⅰ理,6,5分,易)函数f(x)=x4-2x3的图象在点(1, f(1))处的切线方程为 ( ) A.y=-2x-1 B.y=-2x+1 C.y=2x-3 D.y=2x+1 2.(2023全国甲文,8,5分,易)曲线y= ex 在点 ( e) 处的切线方程为 ( ) 1, x+1 2 e e A.y= x B.y= x 4 2 e e e 3e C.y= x+ D.y= x+ 4 4 2 4 3.(2021新高考Ⅰ,7,5分,中)若过点(a,b)可以作曲线y=ex的两条切线,则( ) A.eb0,函数f(x)的图象在点A(x , f(x )) 1 2 1 1 |AM| 和点B(x , f(x ))处的两条切线互相垂直,且分别交y轴于M,N两点,则 的取值范围是 2 2 |BN| . 9.(2022全国甲文,20,12分,中)已知函数f(x)=x3-x,g(x)=x2+a,曲线y=f(x)在点(x , f(x ))处的切 1 1 线也是曲线y=g(x)的切线. (1)若x =-1,求a; 1 (2)求a的取值范围. 10.(2020北京,19,15分,中)已知函数f(x)=12-x2. (1)求曲线y=f(x)的斜率等于-2的切线方程; (2)设曲线y=f(x)在点(t, f(t))处的切线与坐标轴围成的三角形的面积为S(t),求S(t)的最小值.11.(2020新高考Ⅰ,21,12分,中)已知函数f(x)=aex-1-ln x+ln a. (1)当a=e时,求曲线y=f(x)在点(1,f(1))处的切线与两坐标轴围成的三角形的面积; (2)若f(x)≥1,求a的取值范围. 三年模拟 练速度 1.(2024福建厦门一模,3)已知直线l与曲线y=x3-x在原点处相切,则l的倾斜角为 ( ) π π 3π 5π A. B. C. D. 6 4 4 6 2.(2024 湖北八市联考,6)已知函数 f(x)为偶函数,其图象在点(1, f(1))处的切线方程为 x-2y+1=0,记f(x)的导函数为f '(x),则f '(-1)= ( )1 1 A.- B. C.-2 D.2 2 2 3.(2024广东茂名一模,4)曲线f(x)=ex+ax在点(0,1)处的切线与直线y=2x平行,则a= ( ) A.-2 B.-1 C.1 D.2 4.(2024山东名校考试联盟联考,6)若曲线f(x)=ex在x=1处的切线与曲线g(x)=ln x+a也相 切,则a= ( ) 1 3 A. B.1 C. D.2 2 2 5.(2024湖南衡阳一模,7)若函数f(x)=x3+4与g(x)=x2-2x图象的交点为A,则曲线y=f(x)在点 A处的切线与坐标轴围成的三角形的面积为 ( ) 2 8 A.4 B.6 C. D. 3 3 lnx 6.(2024辽宁葫芦岛学业质量监测,8)已知直线y=ax-1与曲线y= 相切,则a的值为 ( x ) 1 1−ln2 A.1 B. C. D.2e2 e 4 7.(多选)(2024河北质量监测,10)过点A(1,2)与曲线f(x)=x3+x相切的直线为( ) A.2x+y-4=0 B.3x-y-1=0 C.4x-y-2=0 D.7x-4y+1=0 练思维 1.(2024湖南长沙适应性考试,7)已知直线y=a与函数f(x)=ex,g(x)=ln x的图象分别相交于 A,B两点.设k 为曲线y=f(x)在点A处切线的斜率,k 为曲线y=g(x)在点B处切线的斜率,则 1 2 k k 的最大值为 ( ) 1 2 1 A. B.1 C.e D.ee e 2.(2024河北唐山期末,7)已知函数f(x)=sin πx,x∈(0,2)的图象与直线y=a(x-1)有3个交点,则 实数a的取值范围为 ( ) A.(-∞,0) B.(-1,0) C.(-∞,-π) D.(-π,0) 3.(2024福建部分地市质量检测,7)若直线y=ax+b与曲线y=ex相切,则a+b的取值范围为 ( ) A.(-∞,e] B.[2,e] C.[e,+∞) D.[2,+∞)4.(2024湘豫名校联考一模,15)已知曲线y=ex-1与曲线y=f(x)关于直线x-y=0对称,则与两 曲线均相切的直线的方程为 . 5.(2024山东日照联考)已知函数f(x)=x+sin x的图象上存在三个不同的点A,B,C,使得曲线 y=f(x)在A,B,C三点处的切线重合,则此切线的方程为 .(写出符合要求的一条切线 即可) 6.(2024河北石家庄模拟,15)已知函数f(x)=eax-ex-b在x=0处的切线为x轴. (1)求a,b的值; (2)求f(x)的单调区间. 7.(2024湖北武汉四调,16)已知函数f(x)=ln x-ax+x2. (1)若a=-1,求曲线y=f(x)在点(1, f(1))处的切线方程; (2)讨论f(x)的单调性.8.(2024河南新乡三模,15)已知函数f(x)=xln x. (1)求f(x)的极值; (2)若过点(a,b)可以作两条直线与曲线y=f(x)相切,证明:b0,函数f(x)的图象在点A(x , f(x )) 1 2 1 1 |AM| 和点B(x , f(x ))处的两条切线互相垂直,且分别交y轴于M,N两点,则 的取值范围是 2 2 |BN| (0,1) . 9.(2022全国甲文,20,12分,中)已知函数f(x)=x3-x,g(x)=x2+a,曲线y=f(x)在点(x , f(x ))处的切 1 1 线也是曲线y=g(x)的切线. (1)若x =-1,求a; 1 (2)求a的取值范围. 解析 解法一:由题意可知f '(x)=3x2-1, f(x )= -x ,则曲线y=f(x)在点(x , f(x ))处的切线方 1 x3 1 1 1 1 程为y-( -x )=(3 -1)(x-x ),即y=(3 -1)x-2 . x3 1 x2 1 x2 x3 1 1 1 1 因为曲线y=f(x)在点(x , f(x ))处的切线也是曲线y=g(x)的切线,所以{y=(3x2−1)x−2x3, 1 1 1 1 y=x2+a 有且仅有一组解, 即方程x2-(3 -1)x+2 +a=0有两个相等的实数根, x2 x3 1 1 从而Δ=(3 -1)2-4(2 +a)=0⇔4a=9 -8 -6 +1. x2 x3 x4 x3 x2 1 1 1 1 1 (1)若x =-1,则4a=12⇔a=3. 1 (2)4a=9 -8 -6 +1, x4 x3 x2 1 1 1 令h(x)=9x4-8x3-6x2+1, 则h'(x)=36x3-24x2-12x=12x(x-1)(3x+1), 1 令h'(x)>0,得- 1, 3 1 令h'(x)<0,得x<- 或00时,S(t)= · ·(t2+12)= , 2 2t 4t 则S'(t)=3(t2−4)(t2+12), 4t2 当02时,S'(t)>0,此时S(t)在(2,+∞)上单调递增, 所以S(t) =S(2)=32. min ②当t<0时,S(t)=-(t2+12) 2,则S'(t)=-3(t2−4)(t2+12), 4t 4t2 当t<-2时,S'(t)<0,此时S(t)在(-∞,-2)上单调递减; 当-20,此时S(t)在(-2,0)上单调递增, 所以S(t) =S(-2)=32. min 综上所述,当t=±2时,S(t)取最小值,为32. 名师点拨 本题主要考查导数在研究函数中的应用和导数的概念及几何意义.本题第(2)问先求 出切线与x轴和y轴的交点,再求出三角形的面积表达式,分t>0和t<0两种情况,也可以只 研究t>0时S(t)的最小值,由上面解析知当t=2时,S(t)取最小值,S(t) =S(2)=32,利用f(x)=12- min x2是偶函数,图象关于y轴对称,得出t<0时,S(t) =S(-2)=32. min 11.(2020新高考Ⅰ,21,12分,中)已知函数f(x)=aex-1-ln x+ln a. (1)当a=e时,求曲线y=f(x)在点(1,f(1))处的切线与两坐标轴围成的三角形的面积; (2)若f(x)≥1,求a的取值范围. 1 解析 f(x)的定义域为(0,+∞), f '(x)=aex-1- . x (1)当 a=e 时, f(x)=ex-ln x+1, f '(1)=e-1,曲线 y=f(x)在点(1,f(1))处的切线方程为 y- (e+1)=(e-1)(x-1),即y=(e-1)x+2. −2 直线y=(e-1)x+2在x轴,y轴上的截距分别为 ,2. e−1 2 −2 因此所求三角形的面积为 易错:容易忽略三角形的面积应大于0而把结果写成 e−1 e−1 . (2)解法一:当00.所以当 x x=1时, f(x)取得最小值,最小值为f(1)=1,从而f(x)≥1. 当a>1时, f(x)=aex-1-ln x+ln a>ex-1-ln x≥1. 综上,a的取值范围是[1,+∞). 解法二:由f(x)≥1,可得aex-1-ln x+ln a≥1, 即ex−1+lna-ln x+ln a≥1, 即ex−1+lna+ln a+x-1≥ln x+x=eln x+ln x. 令g(t)=et+t,则g'(t)=et+1>0, ∴g(t)在R上单调递增. ∵g(ln a+x-1)≥g(ln x), ∴ln a+x-1≥ln x, 即ln a≥ln x-x+1. 1 1−x 令h(x)=ln x-x+1,∴h'(x)= -1= , x x 当00,函数h(x)单调递增, 当x>1时,h'(x)<0,函数h(x)单调递减, ∴h(x)≤h(1)=0, ∴ln a≥0,∴a≥1, 故a的取值范围为[1,+∞). 思路导引 (1)根据导数的几何意义即可求出切线方程,可得三角形的面积; (2)解法一:对a进行分类讨论,看哪种情况下可使f(x)≥1; 解法二:不等式等价于ex−1+lna+ln a+x-1≥ln x+x=eln x+ln x,令g(t)=et+t,根据函数单调性可得 ln a≥ln x-x+1,再构造函数h(x)=ln x-x+1,利用导数求出函数的最值,即可求出a的取值范 围.三年模拟 练速度 1.(2024福建厦门一模,3)已知直线l与曲线y=x3-x在原点处相切,则l的倾斜角为 ( C ) π π 3π 5π A. B. C. D. 6 4 4 6 2.(2024 湖北八市联考,6)已知函数 f(x)为偶函数,其图象在点(1, f(1))处的切线方程为 x-2y+1=0,记f(x)的导函数为f '(x),则f '(-1)= ( A ) 1 1 A.- B. C.-2 D.2 2 2 3.(2024广东茂名一模,4)曲线f(x)=ex+ax在点(0,1)处的切线与直线y=2x平行,则a= ( C ) A.-2 B.-1 C.1 D.2 4.(2024山东名校考试联盟联考,6)若曲线f(x)=ex在x=1处的切线与曲线g(x)=ln x+a也相 切,则a= ( D ) 1 3 A. B.1 C. D.2 2 2 5.(2024湖南衡阳一模,7)若函数f(x)=x3+4与g(x)=x2-2x图象的交点为A,则曲线y=f(x)在点 A处的切线与坐标轴围成的三角形的面积为 ( B ) 2 8 A.4 B.6 C. D. 3 3 lnx 6.(2024辽宁葫芦岛学业质量监测,8)已知直线y=ax-1与曲线y= 相切,则a的值为 ( x A ) 1 1−ln2 A.1 B. C. D.2e2 e 4 7.(多选)(2024河北质量监测,10)过点A(1,2)与曲线f(x)=x3+x相切的直线为( CD ) A.2x+y-4=0 B.3x-y-1=0 C.4x-y-2=0 D.7x-4y+1=0 练思维 1.(2024湖南长沙适应性考试,7)已知直线y=a与函数f(x)=ex,g(x)=ln x的图象分别相交于 A,B两点.设k 为曲线y=f(x)在点A处切线的斜率,k 为曲线y=g(x)在点B处切线的斜率,则 1 2 k k 的最大值为 ( A ) 1 21 A. B.1 C.e D.ee e 2.(2024河北唐山期末,7)已知函数f(x)=sin πx,x∈(0,2)的图象与直线y=a(x-1)有3个交点,则 实数a的取值范围为 ( D ) A.(-∞,0) B.(-1,0) C.(-∞,-π) D.(-π,0) 3.(2024福建部分地市质量检测,7)若直线y=ax+b与曲线y=ex相切,则a+b的取值范围为 ( A ) A.(-∞,e] B.[2,e] C.[e,+∞) D.[2,+∞) 4.(2024湘豫名校联考一模,15)已知曲线y=ex-1与曲线y=f(x)关于直线x-y=0对称,则与两 曲线均相切的直线的方程为 x - y = 0 . 5.(2024山东日照联考)已知函数f(x)=x+sin x的图象上存在三个不同的点A,B,C,使得曲线 y=f(x)在A,B,C三点处的切线重合,则此切线的方程为 y = x +1 ( 或 y = x -1 ) .(写出符合 要求的一条切线即可) 6.(2024河北石家庄模拟,15)已知函数f(x)=eax-ex-b在x=0处的切线为x轴. (1)求a,b的值; (2)求f(x)的单调区间. 解析 (1)因为f(x)=eax-ex-b,所以f '(x)=aeax-e, 依题意f(0)=0且f '(0)=0, 所以{e0−b=0, 解得{a=e, ae0−e=0, b=1. (2)由(1)可得f(x)=eex-ex-1,定义域为R, 又f '(x)=eex+1-e=e(eex-1), 令g(x)=f '(x)=eex+1-e,则g'(x)=eex+2>0, 所以 g(x)在定义域 R 上单调递增,即 f '(x)在 R 上单调递增.又 f '(0)=0,所以当 x<0 时 f '(x)<0,当x>0时f '(x)>0, 所以f(x)的单调递减区间为(-∞,0),单调递增区间为(0,+∞). 7.(2024湖北武汉四调,16)已知函数f(x)=ln x-ax+x2. (1)若a=-1,求曲线y=f(x)在点(1, f(1))处的切线方程; (2)讨论f(x)的单调性.1 解析 (1)a=-1时, f(x)=ln x+x+x2, f '(x)= +1+2x, x 则f '(1)=4, f(1)=2, 所以所求切线方程为y=4(x-1)+2,整理得y=4x-2. 1 2x2−ax+1 (2)f '(x)= -a+2x= , x x 因为x>0,所以a≤0时, f '(x)>0, f(x)在(0,+∞)上单调递增, 当a>0时,对于y=2x2-ax+1,Δ=a2-8, 若02 ,令2x2-ax+1=0,得x=a±√a2−8>0,00, f(x)单调递增; √2 4 4 x>a+√a2−8时, f '(x)>0, f(x)单调递增; 4 a−√a2−82 时, f(x)在( a−√a2−8),(a+√a2−8 )上单调递增,在(a−√a2−8 a+√a2−8) √2 0, ,+∞ , 4 4 4 4 上单调递减. 8.(2024河南新乡三模,15)已知函数f(x)=xln x. (1)求f(x)的极值; (2)若过点(a,b)可以作两条直线与曲线y=f(x)相切,证明:b0, f(x)单 0, ,+∞ e e e 调递增,(3分) 故当x=1时, f(x)取得极小值,且极小值为f(1)=-1,无极大值. (5分) e e e (2)证明:设切点为(x ,x ln x ),则切线的方程为y-x ln x =(1+ln x )(x-x ), (6分) 0 0 0 0 0 0 0 则b-x ln x =(1+ln x )(a-x ),整理得b=aln x -x +a. (7分) 0 0 0 0 0 0 由过点(a,b)可以作两条直线与曲线y=f(x)相切,可得方程b=aln x-x+a有两个不相等的正根. (9分) a−x 令g(x)=aln x-x+a,则g'(x)= . (10分) x 当a≤0时,g'(x)<0,g(x)单调递减,则方程b=aln x-x+a最多只有一个正根,不符合题意. (11 分) 当 a>0 时,若 x∈(0,a),则 g'(x)>0,g(x)单调递增,若 x∈(a,+∞),则 g'(x)<0,g(x)单调递减,则 g(x) =g(a)=aln a. (12分) max 故要使得方程b=aln x-x+a有两个不相等的正根,则b
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  1. CONNECT:[ UseTime:0.000800s ] mysql:host=127.0.0.1;port=3306;dbname=wenku;charset=utf8mb4
  2. SHOW FULL COLUMNS FROM `fenlei` [ RunTime:0.001186s ]
  3. SELECT * FROM `fenlei` WHERE `fid` = 0 [ RunTime:0.000491s ]
  4. SELECT * FROM `fenlei` WHERE `fid` = 63 [ RunTime:0.000440s ]
  5. SHOW FULL COLUMNS FROM `set` [ RunTime:0.000873s ]
  6. SELECT * FROM `set` [ RunTime:0.000382s ]
  7. SHOW FULL COLUMNS FROM `wendang` [ RunTime:0.001035s ]
  8. SELECT * FROM `wendang` WHERE `id` = 370704 LIMIT 1 [ RunTime:0.000888s ]
  9. UPDATE `wendang` SET `liulancishu` = 1 WHERE `id` = 370704 [ RunTime:0.013879s ]
  10. UPDATE `wendang` SET `lasttime` = 1773095877 WHERE `id` = 370704 [ RunTime:0.000764s ]
  11. SELECT * FROM `wendang` WHERE `id` < 370704 ORDER BY `id` DESC LIMIT 1 [ RunTime:0.000806s ]
  12. SELECT * FROM `wendang` WHERE `id` > 370704 ORDER BY `id` ASC LIMIT 1 [ RunTime:0.000692s ]
  13. SELECT * FROM `wendang` WHERE `id` <> 370704 ORDER BY `id` DESC LIMIT 10 [ RunTime:0.001613s ]
  14. SELECT * FROM `wendang` WHERE `id` <> 370704 ORDER BY `id` DESC LIMIT 10,10 [ RunTime:0.002689s ]
  15. SELECT * FROM `wendang` WHERE `id` <> 370704 ORDER BY `id` DESC LIMIT 20,10 [ RunTime:0.002338s ]
0.111842s