新高考数学一轮复习精讲精练4.2 利用导数求单调性（提升版）（原卷版）

4.2利用导数求单调性（精讲）（提升版）思维导图思维导图考点呈现考点呈现例题剖析例题剖析考点一单调区间（无参）【例1-1】（2022·新疆）函数SKIPIF1<0的减区间是____________.【例1-2】（2022·广东·顺德一中）设曲线SKIPIF1<0在SKIPIF1<0上的单调递减区间是______.【例1-3】（江苏省苏州实验中学）已知函数f(x)满足SKIPIF1<0，则f(x)的单调递减区间为（

）A．(-∞,0) B．(1,+∞) C．(-∞,1) D．(0,+∞)【一隅三反】1．函数f(x)＝x＋2eq\r(1－x)的单调递增区间是()A．(0,1) B．(－∞，1)C．(－∞，0) D．(0，＋∞)2．（皖豫名校联盟体2022届）函数SKIPIF1<0的单调递减区间为__________．3．已知定义在区间(0，π)上的函数f(x)＝x＋2cosx，则f(x)的单调递增区间为．考点二已知单调性求参数【例2-1】（2022安徽省“皖东县中联盟）若函数SKIPIF1<0在区间SKIPIF1<0上单调递减，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【例2-2】（2022.广东）已知函数SKIPIF1<0在区间SKIPIF1<0上不是单调函数，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【一隅三反】1．（2022福建省）已知函数SKIPIF1<0在SKIPIF1<0上为单调递增函数，则实数m的取值范围为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（湖南省三湘名校教育联盟2022届）若SKIPIF1<0是R上的减函数，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（江西省宜春市八校2022届）已知函数SKIPIF1<0在区间SKIPIF1<0上存在单调减区间，则实数SKIPIF1<0的取值范围为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·宁夏吴忠）已知函数SKIPIF1<0存在三个单调区间，则实数SKIPIF1<0的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0考点三单调性的应用之解不等式【例3】（湖南省多所学校2022届）已知SKIPIF1<0，则SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【一隅三反】1．（陕西省西安地区八校2022届）已知函数SKIPIF1<0，则不等式SKIPIF1<0的解集为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（湖北省2022届）已知函数SKIPIF1<0，不等式SKIPIF1<0的解集为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．若函数f(x)＝lnx＋ex－sinx，则不等式f(x－1)≤f(1)的解集为．4．已知函数f(x)＝xsinx＋cosx＋x2，则不等式f(lnx)＋f

eq\b\lc\(\rc\)(\a\vs4\al\co1(ln\f(1,x)))<2f(1)的解集为．考点四单调性应用之比较大小【例4-1】（华大新高考联盟名校2022届）已知实数a，b，SKIPIF1<0，e为自然对数的底数，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【例4-2】（湖南师范大学附中2022届）下列两数的大小关系中正确的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【一隅三反】1．（2022年全国新高考I卷数学试题）设SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（山东省青州市2022届）设SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（江西省萍乡市2022届）设SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（湖北省二十一所重点中学2022届）已知SKIPIF1<0是自然对数的底数，设SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，下列说法正确的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0考点五含参函数的单调性讨论【例5-1】（2022广西节选）已知函数SKIPIF1<0，讨论SKIPIF1<0的单调性；【例5-2】（2022安徽）已知函数SKIPIF1<0，讨论f（x）的单调性；【例5-3】（安徽省江淮名校2022届）已知函数SKIPIF1<0，讨论SKIPIF1<0的单调性；【例5-4】（2022辽宁省沈阳市第二中学）已知函数SKIPIF1<0，讨论SKIPIF1<0的单调性；【一隅三反】1．（2022贵州省贵阳市五校）已知SKIPIF1<0，函数SKIPIF1<0，讨论SKIPIF1<0的单调性；2．（2022陕西省）已知函数SKIPIF1<0．讨论函数SKIPIF1<0的单调性；3．（重庆市第八中学校2022届高三下学期适应性月考（七）数学试题）已知SKIPIF1<0，讨论SKIPIF1<0的单调性；4．（2022江苏省）已知函数SKIPIF1<0，函数SKIPIF1<0的导函数为SKIPIF1<0．讨论函数SKIPIF1<0的单调性；4.2利用导数求单调性（精练）（提升版）题组一题组一单调区间1．（2022·天津·崇化中学）函数SKIPIF1<0的递增区间是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<02．（2022·四川省成都市新都一中）已知函数SKIPIF1<0的导函数为SKIPIF1<0，SKIPIF1<0，则函数SKIPIF1<0的单调递增区间为（

）A．SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·北京·首都师范大学附属中学三模）下列函数中，既是偶函数又在SKIPIF1<0上单调递减的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·黑龙江·哈师大附中高二期中）函数SKIPIF1<0，SKIPIF1<0的增区间为___________.5．（2022·四川·射洪中学）函数SKIPIF1<0的单调增区间为______．题组二题组二已知单调性求参数1．（2022·浙江宁波）若函数SKIPIF1<0在区间SKIPIF1<0上单调递增，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·广东东莞）若函数SKIPIF1<0在SKIPIF1<0上单调递增，则实数a的取值范围是（）A．（-1，1） B．SKIPIF1<0 C．（-1，+∞） D．（-1，0）3．（2022·天津一中）已知函数SKIPIF1<0的单调递减区间是SKIPIF1<0，则SKIPIF1<0（

）A．3 B．SKIPIF1<0 C．2 D．SKIPIF1<04．（2022·山东聊城）若函数SKIPIF1<0在区间SKIPIF1<0上单调递减，则实数m的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05（2022·福建宁德）若函数SKIPIF1<0在SKIPIF1<0上单调递增，则实数SKIPIF1<0的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2022·黑龙江·齐齐哈尔市第八中学校）若函数SKIPIF1<0在区间SKIPIF1<0内存在单调递增区间，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2022·河北唐山）已知函数SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0单调递增，a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2022·河南·南阳中学）若函数SKIPIF1<0在区间SKIPIF1<0上单调递增，则实数SKIPIF1<0的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2022·福建泉州·高二期中）已知函数SKIPIF1<0为减函数，则a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2022·山东潍坊·高二阶段练习）已知函数SKIPIF1<0在R上单调递增，则实数a的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0题组三题组三单调性的应用之解不等式1．（2022·陕西·西北工业大学附属中学模拟预测）已知函数SKIPIF1<0，则不等式SKIPIF1<0的解集为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·河北唐山·三模）已知函数SKIPIF1<0则使不等式SKIPIF1<0成立的实数x的取值范围为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·湖北·房县第一中学模拟预测）已知函数SKIPIF1<0，不等式SKIPIF1<0的解集为（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·甘肃·兰州一中）已知SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0成立，则实数SKIPIF1<0的取值范围是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·河南）已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·湖南·邵阳市第二中学模拟预测）已知函数SKIPIF1<0，若不等式SKIPIF1<0对SKIPIF1<0恒成立，则实数SKIPIF1<0的取值范围______．题组四题组四单调性应用之比较大小1．（贵州省毕节市2022届）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0（SKIPIF1<0为自然对数的底数），则SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小关系为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（广西贵港市高级中学2022届）已知SKIPIF1<0，则下列结论正确的是（

）A．b＞c＞a B．a＞b＞cC．b＞a＞c D．c＞b＞a3．（河北省邯郸市2022届）已知函数SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）.A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（江西师范大学附属中学2022届）设SKIPIF1<0．则a，b，c大小关系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022届高三下学期临考冲刺原创卷（三）数学试题）已知SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（江苏省苏州市2022届）已知SKIPIF1<0，则SKIPIF1<0的大小关系为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（新疆乌鲁木齐地区2022届）设SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（新疆乌鲁木齐地区2022届）设SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．（河南省郑州市2022届）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则它们的大小关系正确的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（陕西省西安中学2022届）已知SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，则（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<011．（湖北省省级示范高中2022届）已知：SKIPIF1<0，SKIPI