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1、高考数学二轮复习：专题十 选择题的解题方法与技巧高考数学二轮复习:专题十 选择题的解题方法与技巧 【重点知识回顾】 2 高考数学选择题占总分值的( 5其解答特点是“四选一”，快速、准确、无误地选择好这个“一”是十分重要的( 选择题和其它题型相比，解题思路和方法有着一定的区别，产生这种现象的原因在于选择题有着与其它题型明显不同的特点:?立意新颖、构思精巧、迷惑性强、题材内容相关相近，真假难分;?技巧性高、灵活性大、概念性强、题材内容储蓄多变、解法奇特;?知识面广、跨度较大、切入点多、综合性强( 正因为这些特点，使得选择题还具有区别与其它题型的考查功能:?能在较大的知识范围内，实现对基础知识、基本
2、技能和基本思想方法的考查;?能比较确切地考查考生对概念、原理、性质、法则、定理和公式的掌握和理解情况;?在一定程度上，能有效地考查逻辑思维能力，运算能力、空间想象能力及灵活和综合地运用数学知识解决问题的能力( 【典型例题】 (一)直接法 直接从题目条件出发，运用有关概念、性质、定理、法则和公式等知识，通过严密的推理和准确的运算，从而得出正确的结论，然后对照题目所给出的选择支“对号入座”作出相应的选择、涉及概念、性质的辨析或运算较简单的题目常用直接法( 212|x|()sin()fx,x,，例1、关于函数，看下面四个结论: 3213x,2007f(x), ?是奇函数;?当时，恒成立;?的最大值是
3、;f(x)f(x)221,?的最小值是(其中正确结论的个数为: f(x)2A(1个 B(2个 C(3个 D(4个 211,cos2x21122|x|x|x|f(x),sinx,()，,()，,1,cos2x,()【解析】， 3223223x,1000,f(x)?为偶函数，结论?错;对于结论?，当时，2， x,2007,sin1000,01211000,(1000)()f,?，结论?错( 232113123|x|,1,cos2x,1,1,cos2x,1cos2(),x,又?，?，从而，结论?错( 222232standards and accompanied by the manufacture
4、rs quality certificate. Each batch of material storage test should be required before acceptance, we will test results submitted to the supervisor in a timely manner. 11.2.2 cement (1) shall be under the supervision of construction drawings or instructions, use grout of cement. Used for backfill gro
5、uting and the Grouting cement strength should not be lower than PO42.5. (2) Grouting cement must meet the required quality standards, lumped the cement shall be used. Cement should not be stored for too long, the factory of cement should not be used for more than three months. 11.2.3 water filling w
6、ith water, the water used in mixing concrete requirements, mixing water temperature must not exceed 40. C。 11.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of various admixtures shall conform to DL/T5148-2001 5th. section 1.6, the
7、 incorporation should be determined through experiments, test results should be submitted to the supervisor. 11.2.5 additives approved by the supervisor, can be mixed with quick-setting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or . Then the next cycle 21122|x
8、|2|x|()sin()中，?f(x),， fx,x,，sinx,0,(),13223等号当且仅当x=0时成立，可知结论?正确( 【题后反思】 直接法是解答选择题最常用的基本方法，低档选择题可用此法迅速求解，直接法运用的范围很广，只要运算正确必能得到正确的答案，提高直接法解选择题的能力，准确地把握中档题的“个性”，用简便方法巧解选择题，是建在扎实掌握“三基”的基础上的，否则一味求快则会快中出错( (二)排除法 排除法也叫筛选法或淘汰法，使用排除法的前提条件是答案唯一，具体的做法是采用简捷有效的手段对各个备选答案进行“筛选”，将其中与题干相矛盾的干扰支逐一排除，从而获得正确结论( 22例2、直线
9、与圆的图象可能是: ax,y，b,0x，y,2ax，2by,0y y y y O x x x O x O O A( B( C( D( 【解析】由圆的方程知圆必过原点，?排除A、C选项，圆心(a，-b), 由B、D两图知(直线方程可化为，可知应选B( a,0,b,0y,ax，b【题后反思】 用排除法解选择题的一般规律是: (1)对于干扰支易于淘汰的选择题，可采用筛选法，能剔除几个就先剔除几个; (2)允许使用题干中的部分条件淘汰选择支; (3)如果选择支中存在等效命题，那么根据规定-答案唯一，等效命题应该同时排除; (4)如果选择支存在两个相反的，或互不相容的判断，那么其中至少有一个是假的; (
10、5)如果选择支之间存在包含关系，必须根据题意才能判定( (三)特例法 特例法也称特值法、特形法( 就是运用满足题设条件的某些特殊值、特殊关系或特殊图形对选项进行检验或推理，从而得到正确选项的方法，常用的特例有特殊数值、特殊数列、特殊函数、特殊图形、特殊角、特殊位置等( . Then the next cyclesetting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or .-supervisor, can be mixed with quickould be determ
11、ined through experiments, test results should be submitted to the supervisor. 11.2.5 additives approved by the 2001 5th. section 1.6, the incorporation sh-ious admixtures shall conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water
12、glass. Quality of var 。xing water temperature must not exceed 40. Cs, mishould not be used for more than three months. 11.2.3 water filling with water, the water used in mixing concrete requirementrequired quality standards, lumped the cement shall be used. Cement should not be stored for too long,
13、the factory of cement t thent. Used for backfill grouting and the Grouting cement strength should not be lower than PO42.5. (2) Grouting cement must meea timely manner. 11.2.2 cement (1) shall be under the supervision of construction drawings or instructions, use grout of ceme efore acceptance, we w
14、ill test results submitted to the supervisor instandards and accompanied by the manufacturers quality certificate. Each batch of material storage test should be required b2 ,x,2,1,x,0,1f(x),例3、设函数，若，则的取值范围为: f(x),1x,002,x,x,0,A(-1，1) B() C( D( ,1,，,(,2):(0,，,)(,1):(1,，,)121【解析】?，?不符合题意，?排除选项A、B、C，故应
15、选D( f(),122232例4、已知函数的图像如图所示，则b的取值范围是: f(x),ax，bx，cx，dy A( B( (,0)(0,1)C(1，2) D( (2,，,)O 1 x 2 32【解析】设函数， f(x),x(x,1)(x,2),x,3x，2x此时( a,1,b,3,c,2,d,0【题后反思】 这类题目若是脚踏实地地求解，不仅运算量大，而且极易出错，而通过选择特殊点进行运算，既快又准，但要特别注意，所选的特殊值必须满足已知条件( (四)验证法 又叫代入法，就是将各个选择项逐一代入题设进行检验，从而获得正确的判断，即将各个选择支分别作为条件，去验证命题，能使命题成立的选择支就是应
16、选的答案( 例5、在下列四个函数中，满足性质:“对于区间(1，2)上的任意，x,x(x,x)1212恒成立”的只有: |f(x),f(x)|,|x,x|12121x2f(x),A( B( C( D( f(x),|x|f(x),2f(x),xx|f(x),f(x)|1121f(x),1【解析】当时， ，所以|f(x),f(x)|,|x,x|1212x|x,x|xx|1212恒成立，故选A( 222例6、若圆上恰有相异两点到直线的距离等于1，4x,3y，25,0x，y,r(r,0)则r的取值范围是: A(4，6 B(4,6) C(4,6 D(4,6) r,4【解析】圆心到直线4x,3y，25,0的
17、距离为5，则当时，圆上只有一个点到直r,6线的距离为1，当时，圆上有三个点到直线的距离等于1，故应选D( setting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or . Then the next cycle-quick periments, test results should be submitted to the supervisor. 11.2.5 additives approved by the supervisor, can be mixed with2
18、001 5th. section 1.6, the incorporation should be determined through ex-ious admixtures shall conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of var 。ot exceed 40. Cmust nthan three months. 11.2.3 water filling
19、 with water, the water used in mixing concrete requirements, mixing water temperature lumped the cement shall be used. Cement should not be stored for too long, the factory of cement should not be used for more ards, ng and the Grouting cement strength should not be lower than PO42.5. (2) Grouting c
20、ement must meet the required quality standnt (1) shall be under the supervision of construction drawings or instructions, use grout of cement. Used for backfill groutiefore acceptance, we will test results submitted to the supervisor in a timely manner. 11.2.2 cemestandards and accompanied by the ma
21、nufacturers quality certificate. Each batch of material storage test should be required b3 【题后反思】 代入验证法适用于题设复杂、结论简单的选择题，这里选择把选项代入验证，若第一个恰好满足题意就没有必要继续验证了，大大提高了解题速度( (五)数形结合法 “数缺形时少直观，形少数时难入微”，对于一些具体几何背景的数学题，如能构造出与之相应的图形进行分析，则能在数形结合，以形助数中获得形象直观的解法( 例7、若函数满足，且时，则y,f(x)(x,R)f(x，2),f(x)x,1,1f(x),|x|函数的图像
22、与函数的图像的交点个数为: y,log|x|y,f(x)(x,R)3y A(2 B(3 C(4 D(无数个 y,log|x|3Y=f(x) 【解析】由已知条件可做出函数及 y,log|x|f(x)3x -3 -1 1 2 -2 3 的图像，如下图，由图像可得其交点的个数为4个， 故应选C( ,x,2x,1,x,0,1f(x),例8、设函数，若若，则的取值范围为: f(x),1f(x),1x,0002,x,x,0,A(-1，1) B( (,2):(0,，,)y C() D( ,1,，,(,1):(1,，,)1 x 【解析】在同一直角坐标系中，做出函数 f(x)O 1 -1 和直线x=1的图像，它
23、们相交于(-1，1)和 (1，1)两点，则f(x),1，得x,1或x,1，故选D( 000【题后反思】 严格地说，图解法并非属于选择题解题思路范畴，而是一种数形结合的解题策略，但它在解有关选择题时非常简便有效，不过运用图解法解题一定要对有关函数图象、方程曲线、几何图形较熟悉，否则错误的图像反会导致错误的选择( (六)逻辑分析法 分析法就是根据结论的要求，通过对题干和选择支的关系进行观察分析、寻求充分条件，发现规律，从而做出正确判断的一种方法，分析法可分为定性分析法和定量分析法( f(x),log(x，1)例9、若定义在区间(-1，0)内的函数满足f(x),0，则a的取值2a范围是: 111(0
24、,)(0,(,，,)(0,，,)A( B( C( D( 222. Then the next cyclesetting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or .-supervisor, can be mixed with quickould be determined through experiments, test results should be submitted to the supervisor. 11.2.5 additives approved
25、by the 2001 5th. section 1.6, the incorporation sh-ious admixtures shall conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of var 。xing water temperature must not exceed 40. Cs, mishould not be used for more than
26、 three months. 11.2.3 water filling with water, the water used in mixing concrete requirementrequired quality standards, lumped the cement shall be used. Cement should not be stored for too long, the factory of cement t thent. Used for backfill grouting and the Grouting cement strength should not be
27、 lower than PO42.5. (2) Grouting cement must meea timely manner. 11.2.2 cement (1) shall be under the supervision of construction drawings or instructions, use grout of ceme efore acceptance, we will test results submitted to the supervisor instandards and accompanied by the manufacturers quality ce
28、rtificate. Each batch of material storage test should be required b4 【解析】要使成立，只要2a和x+1同时大于1或同时小于1成立，当f(x),0x,(,1,0)时，则，故选A( x，1,(0,1)2a,(0,1)n!例10、用n个不同的实数可得个不同的排列，每个排列为一行写成一a,a,a？,a123nnn!个行的矩阵，对第i行，记， a,a,a？,ab,a，2a,3a，？，(,1)ai1i2i3inii1i2i3in1 2 3 ()例如用1、2、3排数阵如图所示，由于此数阵中每一列各 i,1,2,3,？,n1 3 2 2 1
29、 3 数之和都是12，所以，那么用1， b，b，？，b,12，2，12,3，12,241262 3 1 3 2 1 2，3，4，5形成的数阵中， b，b，？，b,121203 1 2 A(-3600 B(1800 C(-1080 D(-720 n,3【解析】时，每一列之和为， 3!,63!,2!,12b，b，？，b,12，(,1，2,3),24126n,55!,65!,4!,360时，每一列之和为，b，b，？，b,360，(,1，2,3，4,5),108012120故选C( 【题后反思】 分析法实际是一种综合法，它要求在解题的过程中必须保持和平的心态、仔细、认真的去分析、学习、掌握、验证学习的
30、结果，再运用所学的知识解题，对考察学生的学习能力要求较高( (七)极端值法 从有限到无限，从近似到精确，从量变到质变，应用极端值法解决某些问题，可以避开抽象、复杂的运算，隆低难度，优化解题过程( ,(0,)例11、对任意都有: ,2A( B( sin(sin,),cos,cos(cos,)sin(sin,),cos,cos(cos,)C( D( sin(cos,),cos(sin,),cos,sin(cos,),cos,cos(sin,),0【解析】当时，故排除A、B， sin(sin,),0cos,1,cos(cos,),cos1,cos,0,当,时，cos(sin,),cos1，故排除C，
31、因此选D( 2,0,例12、设a,sin,，cos,b,sin,，cos,，且,，则 422222222a，ba，ba，ba，ba,b,a,b,A( B( 2222periments, test results should be submitted to the supervisor. 11.2.5 additives approved by the supervisor, can be mixed with2001 5th. section 1.6, the incorporation should be determined through ex-ious admixtures shal
32、l conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of var 。ot exceed 40. Cmust nthan three months. 11.2.3 water filling with water, the water used in mixing concrete requirements, mixing water temperature lumped
33、 the cement shall be used. Cement should not be stored for too long, the factory of cement should not be used for more ards, ng and the Grouting cement strength should not be lower than PO42.5. (2) Grouting cement must meet the required quality standnt (1) shall be under the supervision of construct
34、ion drawings or instructions, use grout of cement. Used for backfill groutiefore acceptance, we will test results submitted to the supervisor in a timely manner. 11.2.2 cemestandards and accompanied by the manufacturers quality certificate. Each batch of material storage test should be required bset
35、ting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or . Then the next cycle-quick 5 22222222a，ba，ba，ba，bC( D( a,b,a,b,2222223a，b,【解析】?，?令，则， 00,1,2,a,b,4422易知:，故应选A( 1,1.5,2,1.5【题后反思】 有一类比较大小的问题，使用常规方法难以奏效(或过于繁杂)，又无特殊值可取，在这种情况下，取极限往往会收到意想不到的效果( (八)估值法 由于选择题提供了唯一正确
36、的选择支，解答又无需过程，因此可通过猜测、合情推理、估算而获得答案，这样往往可以减少运算量，避免“小题大做”( 例13、如图，在多面体ABCDEF中，已知面ABCD是边长为3的正方形，EF/AB，3E EF,，EF与面AC的距离为2，则该多面体的体积为: F 2915D C A( B(5 C(6 D( 22【解析】由已知条件可知，EF/面ABCD，则F到平面ABCD A B 12V,，3，2,6的距离为2，?，而该多面体的体积必大于6，故选D( F,ABCD3例14、已知过球面上A、B、C三点的截面和球心的距离等于球半径的一半，且AB=BC=CA=2，则球面面积是: ,168644,A( B(
37、 C( D( 93923,ABC【解析】设球的半径为R，的外接圆半径，则r,31622S,4R,4,r,5,，故选D( 球3【题后反思】 有些问题，由于受条件限制，无法(有时也没有必要)进行精确的运算和判断，而又能依赖于估算，估算实质上是一种数字意义，它以正确的算理为基础，通过合理的观察、比较、判断、推理，从而做出正确的判断、估算、省去了很多推导过程和比较复杂的计算，节省了时间，从而显得快捷(其应用广泛，它是人们发现问题、研究问题、解决问题的一种重要的运算方法( (九)割补法 “级割善补”是解决几何问题常用的方法，巧妙地利用割补法，可以将不规则的图形转化2001 5th. section 1.
38、6, the incorporation sh-ious admixtures shall conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of var 。xing water temperature must not exceed 40. Cs, mishould not be used for more than three months. 11.2.3 water
39、 filling with water, the water used in mixing concrete requirementrequired quality standards, lumped the cement shall be used. Cement should not be stored for too long, the factory of cement t thent. Used for backfill grouting and the Grouting cement strength should not be lower than PO42.5. (2) Gro
40、uting cement must meea timely manner. 11.2.2 cement (1) shall be under the supervision of construction drawings or instructions, use grout of ceme efore acceptance, we will test results submitted to the supervisor instandards and accompanied by the manufacturers quality certificate. Each batch of ma
41、terial storage test should be required b. Then the next cyclesetting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or .-supervisor, can be mixed with quickould be determined through experiments, test results should be submitted to the supervisor. 11.2.5 additives
42、approved by the 6 A 为规则的图形，这样可以使问题得到简化，从而缩短解题时间( D 例15、一个四面体的所有棱长都为，四个顶点在同一 2C 球面上，则此球的表面积为: B 3,4,6,A( B( C( D( 33,【解析】如图，将正四面体ABCD补成正方体，则正四面体、正方体的中心与其外接球3的球心共一面，因为正四面体棱长为，所以正方体棱长为1，从而外接球半径，2R,2故，选A( S,3,球【题后反思】 “割”即化整为零，各个击破，将不易求解的问题，转化为易于求解的问题;“补”即代分散不集中，着眼整体，补成一个“规则图形”来解决问题，当我们遇到不规则的几何图形或几何体时，自然
43、要想到“割补法”( 【模拟演练】 2,22，,(1)已知是锐角，且，则的取值范围是: ,cos,，cos,313131313，)，)A( B( C( D( 222422242,x (2)若，则A交B补中元素B,x|logx|,1,x,RA,x|2,2,8,x,Z2的个数为: A(0 B(1 C(2 D(3 1x，1M:N,N,x|,2,4,x,Z (3)已知集合，则 M,1,12A( B( C( D( ,1,1,10,1,022 (4)过原点的直线与圆相切，若切点在第三象限，则该直线的方程x，y，4x，3,0是: 33y,xy,xA( B( C( D( y,3xy,3x3302n (5)如果n
44、是正偶数，则C，C，？，C, nnnnn,1n，1n,1222A( B( C( D( (n,1)，2f(x),Msin(,x，,)(,0) (6)函数，则区间a，b上是增函数，且setting Cement Grout agents, reducing agents, stabilizers and the supervisor instructions or . Then the next cycle-quick periments, test results should be submitted to the supervisor. 11.2.5 additives approved b
45、y the supervisor, can be mixed with2001 5th. section 1.6, the incorporation should be determined through ex-ious admixtures shall conform to DL/T514811.2.4 additives approved by the supervisor, can be mixed with sand in cement slurry, fly ash and water glass. Quality of var 。ot exceed 40. Cmust ntha
46、n three months. 11.2.3 water filling with water, the water used in mixing concrete requirements, mixing water temperature lumped the cement shall be used. Cement should not be stored for too long, the factory of cement should not be used for more ards, ng and the Grouting cement strength should not
47、be lower than PO42.5. (2) Grouting cement must meet the required quality standnt (1) shall be under the supervision of construction drawings or instructions, use grout of cement. Used for backfill groutiefore acceptance, we will test results submitted to the supervisor in a timely manner. 11.2.2 cem
48、estandards and accompanied by the manufacturers quality certificate. Each batch of material storage test should be required b7 ，则函数在a，b上是: f(a),M,f(b),Mg(x),Mcos(,x，,)A(增函数 B(减函数 C(有最大值M D(有最小值M ,(7)函数的最小正周期是: f(x),sin(,2x)，sin2x3,A( B( C(2 D(4 ,2(8)过点A(1，-1)，B(-1，1)且圆心在直线上的圆的方程是: x，y,2,02222 A( B( (x,3)，(y，1),4(x，3)，(y,1),42222C( D( (x,1)，(y,1),4(x，1)，(y，1),4x,0 (9)定义在上的奇函数，在上为增函数，当时，(,0):(0,，,)f(x)(0,，,)f(x)y 的图像如下图