4.解:(1)当a?0时,f(x)?x?|x|?1为偶函数,
21 当a?0时,f(x)?x?|x?a|?为非奇非偶函数;
(2)当x?a时,f(x)?x?x?a?1?(x?)?a? 当a?221223, 4113时,f(x)min?f()?a?, 2241 当a?时,f(x)min不存在;
21232当x?a时,f(x)?x?x?a?1?(x?)?a?,
241f(a)?2a?,1 当a??时,f(x)min?2113 当a??时,f(x)min?f(?)??a?。
224(数学1必修)第一章(下) [提高训练C组] 一、选择题
1. D f??x???x?a??x?a?x?a?x?a??f(x), 画出h(x)的图象可观察到它关于原点对称
或当x?0时,?x?0,则h(?x)?x?x??(?x?x)??h(x); 当x?0时,?x?0,则h(?x)??x?x??(x?x)??h(x);
2222?h(?x)??h(x)
2. C a?2a?2533335?(a?1)2??,f(?)?f()?f(a2?2a?) 2222223. B 对称轴x?2?a,2?a?4,a??2
4. D 由x?f(x)?0得??x?0?x?0或?而f(?3)?0,f(3)?0
?f(x)?0?f(x)?0 即??x?0?x?0或?
?f(x)?f(?3)?f(x)?f(3)335. D 令F(x)?f(x)?4?ax?bx,则F(x)?ax?bx为奇函数 F(?2)?f(?2)?4?6,F(2)?f(2)?4??6,f(2)??10
46
6. B f(?x)??x3?1??x3?1?x3?1?x3?1?f(x)为偶函数
(a,f(a)一定在图象上,而 )f(a)?f?(a,∴)(a,f?(a)一定在图象上)二、填空题
1. x(1?3x) 设x?0,则?x?0,f(?x)??x(1?3?x)??x(1?3x)
∵f(?x)??f(x)∴?f(x)??x(1?3x)
2. a?0且b?0 画出图象,考虑开口向上向下和左右平移
x271113. f(x)?,f()?,f(x)?f()?1 222x1?xx1?x1111f(1)?,f(2)?f()?1,f(3)?f()?1,f(4)?f()?1
22344. (,??) 设x1?x2??2,则f(x1)?f(x2),而f(x1)?f(x2)
12?ax1?1ax2?12ax1?x2?2ax2?x1(x1?x2)(2a?1)????0,则2a?1?0 x1?2x2?2(x1?2)(x2?2)(x1?2)(x2?2)5. ?1,4? 区间[3,6]是函数f(x)?三、解答题
4的递减区间,把3,6分别代入得最大、小值 x?21. 解:(1)令x?y?1,则f(1)?f(1)?f(1),f(1)?0
(2)f(?x)?f(3?x)??2f()
1211f(?x)?f()?f(3?x)?f()?0?f(1)
22x3?xx3?xf(?)?f()?f(1),f(??)?f(1)
2222?x??2?0??3?x则??0,?1?x?0。 ?2?x3?x??2?2?1?2. 解:对称轴x?3a?1, 当3a?1?0,即a?12时,?0,1?是f(x)的递增区间,f(x)min?f(0)?3a; 322当3a?1?1,即a?时,?0,1?是f(x)的递减区间,f(x)min?f(1)?3a?6a?3;
347
12?a?时,f(x)min?f(3a?1)??6a2?6a?1。 33aa3.解:对称轴x?,当?0,即a?0时,?0,1?是f(x)的递减区间,
22当0?3a?1?1,即
2则f(x)max?f(0)??4a?a??5,得a?1或a??5,而a?0,即a??5;
a?1,即a?2时,?0,1?是f(x)的递增区间,则f(x)max?f(1)??4?a2??5, 2a得a?1或a??1,而a?2,即a不存在;当0??1,即0?a?2时,
2a555则f(x)max?f()??4a??5,a?,即a?;∴a??5或 。
24443a2121214.解:f(x)??(x?)?a,f(x)?a?,得?1?a?1,
23666当
对称轴x?a31?11?,当?1?a?时,?,?是f(x)的递减区间,而f(x)?, 348?42?12a313??,a?1与?1?a?矛盾,即不存在; 288411?3a1a11423当?a?1时,对称轴x?,而??,且?? 434333281a313即f(x)min?f()???,a?1,而?a?1,即a?1
22884∴a?1
即f(x)min?f()?新课程高中数学训练题组参考答案(咨询13976611338)
(数学1必修)第二章 基本初等函数(1)[基础训练A组] 一、选择题
x2,(x?0) 1. D y?x?x,对应法则不同;y?x2y?alogax?x,(x?0);y?logaax?x(x?R)
ax?1a?x?1ax?1,f(?x)??x???f(x),为奇函数; 2. D 对于y?xa?1a?11?axxlg(1?x2)lg(1?x2)y?对于y?,显然为奇函数;显然也为奇函数; ?xx?3?3x对于y?loga1?x1?x1?x??loga??f(x),为奇函数; ,f(?x)?loga1?x1?x1?x?x?x3. D 由y??3得?y?3,(x,y)?(?x,?y),即关于原点对称;
48
4. B x?x32?1?(x?x)?2?3,x?x?3212?1212?12212?12?5 x?x?(x?x)(x?1?x?1)?25
2?x?1 35. D log1(3x?2)?0?log11,0?3x?2?1,226060.7?60=1,log0.76?0 6. D 0.7?0.7=1,当a,b范围一致时,logab?0;当a,b范围不一致时,logab?0 注意比较的方法,先和0比较,再和1比较 7. D 由f(lnx)?3x?4?3e二、填空题 1.
3lnx?4得f(x)?3ex?4
2?88?54?916?2
123135258389492?2,2?2,4?2,8?2,16?2,
而
13241???? 385922. 16
810?410230?220220(1?210)?12?12?28?16 41122108?42?22(1?2)?13. ?2 原式?log25?2?log25?log25?2?log25??2
x2224. 0 (x?2)?(y?1)?0,x?2且y?1,logx(y)?log2(1)?0
3?x?3x?3?x?3?x?3,x??1 5. ?1 x1?3?6. ?x|x??11?12x?1?0,且y?1 ?,?y|y?0,且y?1? 2x?1?0,x?;y?82?27. 奇函数 f(?x)?x2lg(?x?x2?1)??x2lg(x?x2?1)??f(x) 三、解答题
x1.解:a?6?5,a?x?6?5,ax?a?x?26 a2x?a?2x?(ax?a?x)2?2?22
a3x?a?3x(ax?a?x)(a2x?1?a?2x)??23
ax?a?xax?a?x 49
2.解:原式?1?3?lg3?2?lg300
?2?2?lg3?lg3?2
?61?x?0,?1?x?1且x?0,即定义域为(?1,0)(0,1); 1?x11?x11?x f(?x)??log2???log2??f(x)为奇函数;
?x1?xx1?x12和)(0上为减函数。,1 f(x)??log2(1? )在(?1,01x?1x3.解:x?0且
?2x?1?022?4.解:(1)?2x?1?1,x?,且x?1,即定义域为(,1)(1,??);
33?3x?2?0?5?4(2)令u?x?4x,x?[0,5),则?4?u?5,()?y?(),
2131311?y?81,即值域为(,81]。 243243(数学1必修)第二章 基本初等函数(1)[综合训练B组] 一、选择题
1112321. A logaa?3loga(2a),loga(2a)?,a3?2a,a?8a,a?,a? 3842. A loga(b?1)?0,且logab?1,a?b?2 3. D 令x?8(x?0),x?8?6162,f(8)?f(x6)?log2x?log22 4. B 令f(x)?lgx,f(?x)?lg?x?lgx?f(x),即为偶函数 令u?x,x?0时,u是x的减函数,即y?lgx在区间(??,0)上单调递减 5. B f(?x)?lg1?x1?x??lg??f(x).则f(?a)??f(a)??b. 1?x1?x6. A 令u?x?1,(0,1)是u的递减区间,即a?1,(1,??)是u的 递增区间,即f(x)递增且无最大值。 二、填空题 1.
1x?x?xx f(x)?f(?x)?2?2lga?2?2lga 10 50
新课程高中数学测试题组(必修1)全套含答案
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