200道高中数学题 跪求高二数学选择题200道

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?pwd=1234
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A. \u20133 B. \u20131 C. 1 D . 3
2.\u5df2\u77e5 \u5219 ( )
A. B. C. D.
3.\u8bbe\u51fd\u6570 \u5728x=2\u5904\u8fde\u7eed,\u5219a= ( )
A. B. C. D.
4. \u7684\u503c\u4e3a ( )
A. \u20131 B.0 C. D.1
5.\u51fd\u6570 \u662f ( )
A.\u5468\u671f\u4e3a \u7684\u5076\u51fd\u6570 B.\u5468\u671f\u4e3a \u7684\u5947\u51fd\u6570
C. \u5468\u671f\u4e3a2 \u7684\u5076\u51fd\u6570 D..\u5468\u671f\u4e3a2 \u7684\u5947\u51fd\u6570
6.\u4e00\u53f0X\u578b\u53f7\u81ea\u52a8\u673a\u5e8a\u5728\u4e00\u5c0f\u65f6\u5185\u4e0d\u9700\u8981\u5de5\u4eba\u7167\u770b\u7684\u6982\u7387\u4e3a0.8000,\u6709\u56db\u53f0\u8fd9\u79cd\u578b\u53f7\u7684\u81ea\u52a8\u673a\u5e8a\u5404\u81ea\u72ec\u7acb\u5de5\u4f5c,\u5219\u5728\u4e00\u5c0f\u65f6\u5185\u81f3\u591a2\u53f0\u673a\u5e8a\u9700\u8981\u5de5\u4eba\u7167\u770b\u7684\u6982\u7387\u662f ( )
A.0.1536 B. 0.1808 C. 0.5632 D. 0.9728
7.\u5728\u68f1\u957f\u4e3a1\u7684\u6b63\u65b9\u4f53\u4e0a\uff0c\u5206\u522b\u7528\u8fc7\u5171\u9876\u70b9\u7684\u4e09\u6761\u68f1\u4e2d\u70b9\u7684\u5e73\u9762\u622a\u8be5\u6b63\u65b9\u4f53,\u5219\u622a\u53bb8\u4e2a\u4e09\u68f1\u9525\u540e,\u5269\u4e0b\u7684\u51f8\u591a\u9762\u4f53\u7684\u4f53\u79ef\u662f ( )
A. B. C. D.
8. \u82e5\u53cc\u66f2\u7ebf \u7684\u7126\u70b9\u5230\u5b83\u76f8\u5bf9\u5e94\u7684\u51c6\u7ebf\u7684\u8ddd\u79bb\u662f2\uff0c\u5219k= \uff08 \uff09
A. 6 B. 8 C. 1 D. 4
9.\u5f53 \u65f6,\u51fd\u6570 \u7684\u6700\u5c0f\u503c\u662f ( )
A. 4 B. C.2 D.
10. \u53d8\u91cfx\u3001y\u6ee1\u8db3\u4e0b\u5217\u6761\u4ef6\uff1a
\u5219\u4f7fz=3x+2y\u7684\u503c\u6700\u5c0f\u7684\uff08x\uff0cy\uff09\u662f
A. ( 4.5 ,3 ) B. ( 3,6 ) C. ( 9, 2 ) D. ( 6, 4 )
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11. \u5982\u53f3\u4e0b\u56fe,\u5b9a\u5706\u534a\u5f84\u4e3aa\uff0c\u5706\u5fc3\u4e3a ( b ,c ), \u5219\u76f4\u7ebfax+by+c=0\u4e0e\u76f4\u7ebf x\u2013y+1=0\u7684\u4ea4\u70b9\u5728\u7b2c______\u8c61\u9650.
12. \u67d0\u73ed\u59d4\u4f1a\u75314\u540d\u7537\u751f\u4e0e3\u540d\u5973\u751f\u7ec4\u6210,\u73b0\u4ece\u4e2d\u9009\u51fa2\u4eba\u62c5\u4efb\u6b63\u526f\u73ed\u957f,\u5176\u4e2d\u81f3\u5c11\u67091\u540d\u5973\u751f\u5f53\u9009\u7684\u6982\u7387\u662f (\u7528\u5206\u6570\u4f5c\u7b54)____________.
13. \u5df2\u77e5\u590d\u6570z\u4e0e (z +2)2-8i \u5747\u662f\u7eaf\u865a\u6570,\u5219 z = .
14. \u7531\u56fe(1)\u6709\u9762\u79ef\u5173\u7cfb: \u5219\u7531(2) \u6709\u4f53\u79ef\u5173\u7cfb:






15. \u51fd\u6570 \u7684\u53cd\u51fd\u6570
16\u3001\u4e0d\u7b49\u5f0f \u5bf9\u4efb\u610f \u90fd\u6210\u7acb\uff0c\u5219 \u7684\u53d6\u503c\u8303\u56f4\u4e3a .


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\u9898\u53f712345678910
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(11) \u4e09 \uff0812\uff09 \uff0813\uff09\uff0d2i (14) (15) (16)


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1\uff0e\u4e0e\u76f4\u7ebf \u7684\u5e73\u884c\u7684\u629b\u7269\u7ebf \u7684\u5207\u7ebf\u65b9\u7a0b\u662f\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
2\uff0e\u590d\u6570 \u7684\u503c\u662f \uff08 \uff09
A\uff0e\uff0d16B\uff0e16C\uff0e D\uff0e
3\uff0e\u5df2\u77e5 \u7684\u89e3\u6790\u5f0f\u53ef\u53d6\u4e3a \uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
4\uff0e\u5df2\u77e5 \u4e3a\u975e\u96f6\u7684\u5e73\u9762\u5411\u91cf. \u7532\uff1a \uff08 \uff09
A\uff0e\u7532\u662f\u4e59\u7684\u5145\u5206\u6761\u4ef6\u4f46\u4e0d\u662f\u5fc5\u8981\u6761\u4ef6B\uff0e\u7532\u662f\u4e59\u7684\u5fc5\u8981\u6761\u4ef6\u4f46\u4e0d\u662f\u5145\u5206\u6761\u4ef6
C\uff0e\u7532\u662f\u4e59\u7684\u5145\u8981\u6761\u4ef6D\uff0e\u7532\u65e2\u4e0d\u662f\u4e59\u7684\u5145\u5206\u6761\u4ef6\u4e5f\u4e0d\u662f\u4e59\u7684\u5fc5\u8981\u6761\u4ef6
5\uff0e\u82e5 \uff0c\u5219\u4e0b\u5217\u4e0d\u7b49\u5f0f\u2460 \uff1b\u2461 \u2462 \uff1b\u2463 \u4e2d\uff0c\u6b63\u786e\u7684\u4e0d\u7b49\u5f0f\u6709 \uff08 \uff09
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6\uff0e\u5df2\u77e5\u692d\u5706 \u7684\u5de6\u3001\u53f3\u7126\u70b9\u5206\u522b\u4e3aF1\u3001F2\uff0c\u70b9P\u5728\u692d\u5706\u4e0a\uff0c\u82e5P\u3001F1\u3001F2\u662f\u4e00\u4e2a\u76f4\u89d2\u4e09\u89d2\u5f62\u7684\u4e09\u4e2a\u9876\u70b9\uff0c\u5219\u70b9P\u5230x\u8f74\u7684\u8ddd\u79bb\u4e3a\uff08 \uff09
A\uff0e B\uff0e3C\uff0e D\uff0e
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A\uff0e B\uff0e C\uff0e2D\uff0e4
8\uff0e\u5df2\u77e5\u6570\u5217{ }\u7684\u524dn\u9879\u548c \u5176\u4e2da\u3001b\u662f\u975e\u96f6\u5e38\u6570\uff0c\u5219\u5b58\u5728\u6570\u5217{ }\u3001{ }\u4f7f\u5f97\uff08 \uff09
A\uff0e \u4e3a\u7b49\u5dee\u6570\u5217\uff0c{ }\u4e3a\u7b49\u6bd4\u6570\u5217
B\uff0e \u548c{ }\u90fd\u4e3a\u7b49\u5dee\u6570\u5217
C\uff0e \u4e3a\u7b49\u5dee\u6570\u5217\uff0c{ }\u90fd\u4e3a\u7b49\u6bd4\u6570\u5217
D\uff0e \u548c{ }\u90fd\u4e3a\u7b49\u6bd4\u6570\u5217
9\uff0e\u51fd\u6570 \u6709\u6781\u503c\u7684\u5145\u8981\u6761\u4ef6\u662f \uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
10\uff0e\u8bbe\u96c6\u5408 \u5bf9\u4efb\u610f\u5b9e\u6570x\u6052\u6210\u7acb}\uff0c\u5219\u4e0b\u5217\u5173\u7cfb\u4e2d\u6210\u7acb\u7684\u662f\uff08 \uff09
A\uff0eP QB\uff0eQ PC\uff0eP=QD\uff0eP Q=
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11\uff0e\u5df2\u77e5\u5e73\u9762 \u6240\u6210\u7684\u4e8c\u9762\u89d2\u4e3a80\u00b0\uff0cP\u4e3a \u3001 \u5916\u4e00\u5b9a\u70b9\uff0c\u8fc7\u70b9P\u7684\u4e00\u6761\u76f4\u7ebf\u4e0e \u3001 \u6240\u6210\u7684\u89d2\u90fd\u662f30\u00b0\uff0c\u5219\u8fd9\u6837\u7684\u76f4\u7ebf\u6709\u4e14\u4ec5\u6709____________\u6761.
12\u8bbe\u968f\u673a\u53d8\u91cf \u7684\u6982\u7387\u5206\u5e03\u4e3a .
13\uff0e\u5c06\u6807\u53f7\u4e3a1\uff0c2\uff0c\u2026\uff0c10\u768410\u4e2a\u7403\u653e\u5165\u6807\u53f7\u4e3a1\uff0c2\uff0c\u2026\uff0c10\u768410\u4e2a\u76d2\u5b50\u5185\uff0c\u6bcf\u4e2a\u76d2\u5185\u653e\u4e00\u4e2a\u7403\uff0c\u5219\u6070\u597d\u67093\u4e2a\u7403\u7684\u6807\u53f7\u4e0e\u5176\u6240\u5728\u76d2\u5b50\u7684\u6807\u53f7\u4e0d\u4e00\u81f4\u7684\u653e\u5165\u65b9\u6cd5\u5171\u6709
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14\uff0e\u8bbeA\u3001B\u4e3a\u4e24\u4e2a\u96c6\u5408\uff0c\u4e0b\u5217\u56db\u4e2a\u547d\u9898\uff1a\u2460A B \u5bf9\u4efb\u610f \u2461A B \u2462A B A B \u2463A B \u5b58\u5728
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15\uff0e\u67d0\u65e5\u4e2d\u534812\u65f6\u6574\uff0c\u7532\u8239\u81eaA\u5904\u4ee516km/h\u7684\u901f\u5ea6\u5411\u6b63\u4e1c\u884c\u9a76\uff0c\u4e59\u8239\u81eaA\u7684\u6b63\u531718km\u5904\u4ee524km/h\u7684\u901f\u5ea6\u5411\u6b63\u5357\u884c\u9a76\uff0c\u5219\u5f53\u65e512\u65f630\u5206\u65f6\u4e24\u8239\u4e4b\u95f4\u8ddd\u79bb\u5bf9\u65f6\u95f4\u7684\u53d8\u5316\u7387\u662f _________________km/h.
16\uff0e\u82e5\u51fd\u6570f(x)=2cos( )\u7684\u5468\u671f\u4e3aT\uff0c\u4e14T\u2208( , )\uff0c\u5219\u6b63\u6574\u6570k\u7684\u503c\u4e3a .

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1\uff0eD 2\uff0eA 3\uff0eC 4\uff0eB 5\uff0eB 6\uff0eD 7\uff0eB 8\uff0eC 9\uff0eC 10\uff0eA
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11. 4 12\uff0e4 13\uff0e240 14\uff0e\uff084\uff09 15\uff0e\uff0d1.6 16.26\uff0c27\uff0c28


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1\uff0e\u590d\u6570 \u7684\u503c\u662f\uff08 \uff09
A\uff0e B\uff0e\uff0d C\uff0e4D\uff0e\uff0d4
2\uff0e\u5982\u679c\u53cc\u66f2\u7ebf \u4e0a\u4e00\u70b9P\u5230\u53f3\u7126\u70b9\u7684\u8ddd\u79bb\u7b49\u4e8e \uff0c\u90a3\u4e48\u70b9P\u5230\u53f3\u51c6\u7ebf\u7684\u8ddd\u79bb\u662f\uff08 \uff09
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3\uff0e\u8bbe \u662f\u51fd\u6570 \u7684\u53cd\u51fd\u6570\uff0c\u82e5 \uff0c\u5219 \u7684\u503c\u4e3a\uff08 \uff09
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4\uff0e\u628a\u6b63\u65b9\u5f62ABCD\u6cbf\u5bf9\u89d2\u7ebfAC\u6298\u8d77\uff0c\u5f53A\u3001B C\u3001D\u56db\u70b9\u4e3a\u9876\u70b9\u7684\u4e09\u68f1\u9525\u4f53\u79ef\u6700\u5927\u65f6\uff0c\u76f4\u7ebfBD\u4e0e\u5e73\u9762ABC\u6240\u6210\u7684\u89d2\u7684\u5927\u5c0f\u4e3a\uff08 \uff09
A\uff0e90\u00b0B\uff0e60\u00b0C\uff0e45\u00b0D\uff0e30\u00b0
5\uff0e\u67d0\u516c\u53f8\u7532\u3001\u4e59\u3001\u4e19\u3001\u4e01\u56db\u4e2a\u5730\u533a\u5206\u522b\u6709150 \u4e2a\u3001120\u4e2a\u3001180\u4e2a\u3001150\u4e2a\u9500\u552e\u70b9 \u516c\u53f8\u4e3a\u4e86\u8c03\u67e5\u4ea7\u54c1\u9500\u552e\u7684\u60c5\u51b5\uff0c\u9700\u4ece\u8fd9600\u4e2a\u9500\u552e\u70b9\u4e2d\u62bd\u53d6\u4e00\u4e2a\u5bb9\u91cf\u4e3a100\u7684\u6837\u672c\uff0c\u8bb0\u8fd9\u9879\u8c03\u67e5\u4e3a\u2460\uff1b\u5728\u4e19\u5730\u533a\u4e2d\u670920\u4e2a\u7279\u5927\u578b\u9500\u552e\u70b9\uff0c\u8981\u4ece\u4e2d\u62bd\u53d67\u4e2a\u8c03\u67e5\u5176\u6536\u5165\u548c\u552e\u540e\u670d\u52a1\u7b49\u60c5\u51b5\uff0c\u8bb0\u8fd9\u9879\u8c03\u67e5\u4e3a\u2461 \u5219\u5b8c\u6210\u2460\u3001\u2461\u8fd9\u4e24\u9879\u8c03\u67e5\u5b9c\u91c7\u7528\u7684\u62bd\u6837\u65b9\u6cd5\u4f9d\u6b21\u662f \uff08 \uff09
A\uff0e\u5206\u5c42\u62bd\u6837\u6cd5\uff0c\u7cfb\u7edf\u62bd\u6837\u6cd5B\uff0e\u5206\u5c42\u62bd\u6837\u6cd5\uff0c\u7b80\u5355\u968f\u673a\u62bd\u6837\u6cd5
C\uff0e\u7cfb\u7edf\u62bd\u6837\u6cd5\uff0c\u5206\u5c42\u62bd\u6837\u6cd5D\uff0e\u7b80\u5355\u968f\u673a\u62bd\u6837\u6cd5\uff0c\u5206\u5c42\u62bd\u6837\u6cd5
6\uff0e\u8bbe\u51fd\u6570 \u5219\u5173\u4e8ex\u7684\u65b9\u7a0b \u89e3\u7684\u4e2a\u6570\u4e3a\uff08 \uff09
A\uff0e1B\uff0e2C\uff0e3D\uff0e4
7\uff0e\u8bbe \u5219\u4ee5\u4e0b\u4e0d\u7b49\u5f0f\u4e2d\u4e0d\u6052\u6210\u7acb\u7684\u662f\uff08 \uff09
A\uff0e B\uff0e
C\uff0e D\uff0e
8\uff0e\u6570\u5217 \uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
9\uff0e\u8bbe\u96c6\u5408 \uff0c\u90a3\u4e48
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A\uff0e B\uff0e
C\uff0e D\uff0e
10\uff0e\u4ece\u6b63\u65b9\u4f53\u7684\u516b\u4e2a\u9876\u70b9\u4e2d\u4efb\u53d6\u4e09\u4e2a\u70b9\u4e3a\u9876\u70b9\u4f5c\u4e09\u89d2\u5f62\uff0c\u5176\u4e2d\u76f4\u89d2\u4e09\u89d2\u5f62\u7684\u4e2a\u6570\u4e3a\uff08 \uff09
A\uff0e56B\uff0e52C\uff0e48D\uff0e40
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11\uff0e\u8bbe \u5206\u522b\u662f\u5b9a\u4e49\u5728R\u4e0a\u7684\u5947\u51fd\u6570\u548c\u5076\u51fd\u6570\uff0c\u5f53 \u65f6\uff0c \u4e14 \u5219\u4e0d\u7b49\u5f0f \u7684\u89e3\u96c6\u662f________________________.
12\uff0e\u5df2\u77e5\u5411\u91cfa= \uff0c\u5411\u91cfb= \uff0c\u5219|2a\uff0db|\u7684\u6700\u5927\u503c\u662f .
13\uff0e\u540c\u65f6\u629b\u4e24\u679a\u76f8\u540c\u7684\u5747\u5300\u786c\u5e01\uff0c\u968f\u673a\u53d8\u91cf\u03be=1\u8868\u793a\u7ed3\u679c\u4e2d\u6709\u6b63\u9762\u5411\u4e0a\uff0c\u03be=0\u8868\u793a\u7ed3\u679c\u4e2d\u6ca1\u6709\u6b63\u9762\u5411\u4e0a\uff0c\u5219E\u03be= .
14\uff0e\u82e5 \u7684\u5c55\u5f00\u5f0f\u4e2d\u7684\u5e38\u6570\u9879\u4e3a84\uff0c\u5219n= .
15\uff0e\u8bbeF\u662f\u692d\u5706 \u7684\u53f3\u7126\u70b9\uff0c\u4e14\u692d\u5706\u4e0a\u81f3\u5c11\u670921\u4e2a\u4e0d\u540c\u7684\u70b9Pi\uff08i=1\uff0c2\uff0c3\uff0c\u2026\uff09\uff0c\u4f7f|FP1|\uff0c|FP2|\uff0c|FP3|\uff0c\u2026\u7ec4\u6210\u516c\u5dee\u4e3ad\u7684\u7b49\u5dee\u6570\u5217\uff0c\u5219d\u7684\u53d6\u503c\u8303\u56f4\u4e3a .
16.\u5c06\u6b63\u65b9\u5f62 \u6cbf\u5bf9\u89d2\u7ebf \u6298\u6210\u76f4\u4e8c\u9762\u89d2 ,\u6709\u5982\u4e0b\u56db\u4e2a\u7ed3\u8bba:
\u2460 \u2461 \u662f\u7b49\u8fb9\u4e09\u89d2\u5f62
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1.D 2.A 3.B 4.C 5.B 6.C 7.B 8.C 9.A 10.C
11. 12\uff0e4 13\uff0e0.75 14\uff0e9 15\uff0e
16.\u2460\u2461\u2463


2010\u5c4a\u9ad8\u8003\u9009\u62e9\u9898\u548c\u586b\u7a7a\u9898\u4e13\u9879\u8bad\u7ec3\uff086\uff09
\u4e00\u3001\u9009\u62e9\u9898:
1.\u8bbe\u96c6\u5408P={1\uff0c2\uff0c3\uff0c4}\uff0cQ={ }\uff0c\u5219P\u2229Q\u7b49\u4e8e ( )
(A){1\uff0c2} (B) {3\uff0c4} (C) {1} (D) {-2\uff0c-1\uff0c0\uff0c1\uff0c2}
2.\u51fd\u6570y=2cos2x+1(x\u2208R)\u7684\u6700\u5c0f\u6b63\u5468\u671f\u4e3a ( )
(A) (B) (C) (D)
3.\u4ece4\u540d\u7537\u751f\u548c3\u540d\u5973\u751f\u4e2d\u9009\u51fa4\u4eba\u53c2\u52a0\u67d0\u4e2a\u5ea7\u8c08\u4f1a\uff0c\u82e5\u8fd94\u4eba\u4e2d\u5fc5\u987b\u65e2\u6709\u7537\u751f\u53c8\u6709\u5973\u751f\uff0c\u5219\u4e0d\u540c\u7684\u9009\u6cd5\u5171\u6709 ( )
(A)140\u79cd (B)120\u79cd (C)35\u79cd (D)34\u79cd
4.\u4e00\u5e73\u9762\u622a\u4e00\u7403\u5f97\u5230\u76f4\u5f84\u662f6cm\u7684\u5706\u9762\uff0c\u7403\u5fc3\u5230\u8fd9\u4e2a\u5e73\u9762\u7684\u8ddd\u79bb\u662f4cm\uff0c\u5219\u8be5\u7403\u7684\u4f53\u79ef\u662f ( )
(A) (B) (C) (D)
5.\u82e5\u53cc\u66f2\u7ebf \u7684\u4e00\u6761\u51c6\u7ebf\u4e0e\u629b\u7269\u7ebf \u7684\u51c6\u7ebf\u91cd\u5408\uff0c\u5219\u53cc\u66f2\u7ebf\u79bb\u5fc3\u7387\u4e3a ( )
(A) (B) (C) 4 (D)
6.\u67d0\u6821\u4e3a\u4e86\u4e86\u89e3\u5b66\u751f\u7684\u8bfe\u5916\u9605\u8bfb\u60c5\u51b5\uff0c\u968f\u673a\u8c03\u67e5\u4e8650\u540d\u5b66\u751f\uff0c\u5f97\u5230\u4ed6\u4eec\u5728\u67d0\u4e00\u5929\u5404\u81ea\u8bfe\u5916\u9605\u8bfb\u6240\u7528\u65f6\u95f4\u7684\u6570\u636e\uff0c\u7ed3\u679c\u7528\u53f3\u4fa7\u7684\u6761\u5f62\u56fe\u8868\u793a. \u6839\u636e\u6761\u5f62\u56fe\u53ef\u5f97\u8fd950\u540d\u5b66\u751f\u8fd9\u4e00\u5929\u5e73\u5747\u6bcf\u4eba\u7684\u8bfe\u5916\u9605\u8bfb\u65f6\u95f4\u4e3a ( )
(A)0.6\u5c0f\u65f6 (B)0.9\u5c0f\u65f6
(C)1.0\u5c0f\u65f6 (D)1.5\u5c0f\u65f6

7. \u7684\u5c55\u5f00\u5f0f\u4e2dx3\u7684\u7cfb\u6570\u662f ( )
(A)6 (B)12 (C)24 (D)48
8.\u82e5\u51fd\u6570 \u7684\u56fe\u8c61\u8fc7\u4e24\u70b9(-1\uff0c0)\u548c(0\uff0c1)\uff0c\u5219 ( )
(A)a=2,b=2 (B)a= ,b=2 (C)a=2,b=1 (D)a= ,b=
9.\u5c06\u4e00\u9897\u8d28\u5730\u5747\u5300\u7684\u9ab0\u5b50\uff08\u5b83\u662f\u4e00\u79cd\u5404\u9762\u4e0a\u5206\u522b\u6807\u6709\u70b9\u65701\uff0c2\uff0c3\uff0c4\uff0c5\uff0c6\u7684\u6b63\u65b9\u4f53\u73a9\u5177\uff09\u5148\u540e\u629b\u63b73\u6b21\uff0c\u81f3\u5c11\u51fa\u73b0\u4e00\u6b216\u70b9\u5411\u4e0a\u7684\u6982\u7387\u662f ( )
(A) (B) (C) (D)
10.\u51fd\u6570 \u5728\u95ed\u533a\u95f4[-3\uff0c0]\u4e0a\u7684\u6700\u5927\u503c\u3001\u6700\u5c0f\u503c\u5206\u522b\u662f ( )
(A)1\uff0c-1 (B)1\uff0c-17 (C)3\uff0c-17 (D)9\uff0c-19
\u4e8c\u3001\u586b\u7a7a\u9898:
11.\u8bbek>1\uff0cf(x)=k(x-1)(x\u2208R) . \u5728\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfbxOy\u4e2d\uff0c\u51fd\u6570y=f(x)\u7684\u56fe\u8c61\u4e0ex\u8f74\u4ea4\u4e8eA\u70b9\uff0c\u5b83\u7684\u53cd\u51fd\u6570y=f -1(x)\u7684\u56fe\u8c61\u4e0ey\u8f74\u4ea4\u4e8eB\u70b9\uff0c\u5e76\u4e14\u8fd9\u4e24\u4e2a\u51fd\u6570\u7684\u56fe\u8c61\u4ea4\u4e8eP\u70b9. \u5df2\u77e5\u56db\u8fb9\u5f62OAPB\u7684\u9762\u79ef\u662f3\uff0c\u5219k\u7b49\u4e8e____________________.
12.\u4e8c\u6b21\u51fd\u6570y=ax2+bx+c(x\u2208R)\u7684\u90e8\u5206\u5bf9\u5e94\u503c\u5982\u4e0b\u8868\uff1a

x-3-2-101234
y60-4-6-6-406



\u5219\u4e0d\u7b49\u5f0fax2+bx+c>0\u7684\u89e3\u96c6\u662f_______________________.
13.\u4ee5\u70b9(1\uff0c2)\u4e3a\u5706\u5fc3\uff0c\u4e0e\u76f4\u7ebf4x+3y-35=0\u76f8\u5207\u7684\u5706\u7684\u65b9\u7a0b\u662f________________.
14.\u8bbe\u6570\u5217{an}\u7684\u524dn\u9879\u548c\u4e3aSn\uff0cSn= (\u5bf9\u4e8e\u6240\u6709n\u22651)\uff0c\u4e14a4=54\uff0c\u5219a1\u7684\u6570\u503c\u662f_______.
15.\u5e73\u9762\u5411\u91cf \u4e2d\uff0c\u5df2\u77e5 =(4\uff0c-3)\uff0c =1\uff0c\u4e14 =5\uff0c\u5219\u5411\u91cf =__________.
16\uff0e\u6709\u4e0b\u5217\u547d\u9898\uff1a
\u2460G= \uff08G\u22600\uff09\u662fa\uff0cG\uff0cb\u6210\u7b49\u6bd4\u6570\u5217\u7684\u5145\u5206\u975e\u5fc5\u8981\u6761\u4ef6\uff1b\u2461\u82e5\u89d2\u03b1\uff0c\u03b2\u6ee1\u8db3
cos\u03b1cos\u03b2=1\uff0c\u5219sin\uff08\u03b1+\u03b2\uff09=0;\u2462\u82e5\u4e0d\u7b49\u5f0f|x\uff0d4|+\uff5cx\uff0d3|<a\u7684\u89e3\u96c6\u975e\u7a7a\uff0c\u5219\u5fc5\u6709a\u22651;\u2463\u51fd\u6570y=sinx+sin\uff5cx\uff5c\u7684\u503c\u57df\u662f\uff3b\uff0d2\uff0c2\uff3d.
\u5176\u4e2d\u9519\u8bef\u547d\u9898\u7684\u5e8f\u53f7\u662f .\uff08\u628a\u4f60\u8ba4\u4e3a\u9519\u8bef\u7684\u547d\u9898\u7684\u5e8f\u53f7\u90fd\u586b\u4e0a\uff09


\u53c2\u8003\u7b54\u6848
\u4e00\u3001\u9009\u62e9\u9898
ABDCA BCADC
\u4e8c\u3001\u586b\u7a7a\u9898
11. 12\u3001 \u6216 13\u3001 14\u30012 15\u3001 16\u3001\u2462

2010\u5c4a\u9ad8\u8003\u9009\u62e9\u9898\u548c\u586b\u7a7a\u9898\u4e13\u9879\u8bad\u7ec3\uff087\uff09
\u4e00\u3001\u9009\u62e9\u9898\uff1a
1\uff0e\u82e5 \u7684\u7ec8\u8fb9\u6240\u5728\u8c61\u9650\u662f\uff08 \uff09
A\uff0e\u7b2c\u4e00\u8c61\u9650B\uff0e\u7b2c\u4e8c\u8c61\u9650C\uff0e\u7b2c\u4e09\u8c61\u9650D\uff0e\u7b2c\u56db\u8c61\u9650
2\uff0e\u5bf9\u4e8e \uff0c\u7ed9\u51fa\u4e0b\u5217\u56db\u4e2a\u4e0d\u7b49\u5f0f
\u2460 \uff1b\u2461 \uff1b\u2462 \uff1b\u2463
\u5176\u4e2d\u6210\u7acb\u7684\u662f\uff08 \uff09
A\uff0e\u2460\u4e0e\u2462B\uff0e\u2460\u4e0e\u2463 C\uff0e\u2461\u4e0e\u2462D\uff0e\u2461\u4e0e\u2463
3\uff0e\u5df2\u77e5\u03b1\u3001\u03b2\u662f\u4e0d\u540c\u7684\u4e24\u4e2a\u5e73\u9762\uff0c\u76f4\u7ebf \uff0c\u547d\u9898 \u65e0\u516c\u5171\u70b9\uff1b\u547d\u9898 . \u5219 \u7684\uff08 \uff09
A\uff0e\u5145\u5206\u800c\u4e0d\u5fc5\u8981\u7684\u6761\u4ef6 B\uff0e\u5fc5\u8981\u800c\u4e0d\u5145\u5206\u7684\u6761\u4ef6
C\uff0e\u5145\u8981\u6761\u4ef6D\uff0e\u65e2\u4e0d\u5145\u5206\u4e5f\u4e0d\u5fc5\u8981\u7684\u6761\u4ef6
4\uff0e\u8bbe\u590d\u6570z\u6ee1\u8db3 \uff08 \uff09
A\uff0e0 B\uff0e1 C\uff0e D\uff0e2
5\uff0e\u7532\u3001\u4e59\u4e24\u4eba\u72ec\u7acb\u5730\u89e3\u540c\u4e00\u95ee\u9898\uff0c\u7532\u89e3\u51b3\u8fd9\u4e2a\u95ee\u9898\u7684\u6982\u7387\u662fp1\uff0c\u4e59\u89e3\u51b3\u8fd9\u4e2a\u95ee\u9898\u7684\u6982\u7387\u662fp2\uff0c\u90a3\u4e48\u6070\u597d\u67091\u4eba\u89e3\u51b3\u8fd9\u4e2a\u95ee\u9898\u7684\u6982\u7387\u662f\uff08 \uff09
A\uff0e B.
C\uff0e D\uff0e
6\uff0e\u5df2\u77e5\u70b9 \u3001 \uff0c\u52a8\u70b9 \uff0c\u5219\u70b9P\u7684\u8f68\u8ff9\u662f\uff08 \uff09
A\uff0e\u5706B\uff0e\u692d\u5706C\uff0e\u53cc\u66f2\u7ebfD\uff0e\u629b\u7269\u7ebf
7\uff0e\u5df2\u77e5\u51fd\u6570 \uff0c\u5219\u4e0b\u5217\u547d\u9898\u6b63\u786e\u7684\u662f\uff08 \uff09
A\uff0e \u662f\u5468\u671f\u4e3a1\u7684\u5947\u51fd\u6570B\uff0e \u662f\u5468\u671f\u4e3a2\u7684\u5076\u51fd\u6570
C\uff0e \u662f\u5468\u671f\u4e3a1\u7684\u975e\u5947\u975e\u5076\u51fd\u6570D\uff0e \u662f\u5468\u671f\u4e3a2\u7684\u975e\u5947\u975e\u5076\u51fd\u6570
8\uff0e\u5df2\u77e5\u968f\u673a\u53d8\u91cf \u7684\u6982\u7387\u5206\u5e03\u5982\u4e0b\uff1a

12345678910










m
\u5219 \uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
9\uff0e\u5df2\u77e5\u70b9 \u3001 \uff0c\u52a8\u70b9P\u6ee1\u8db3 . \u5f53\u70b9P\u7684\u7eb5\u5750\u6807\u662f \u65f6\uff0c\u70b9P\u5230\u5750\u6807\u539f\u70b9\u7684\u8ddd\u79bb\u662f\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e2
10\uff0e\u8bbeA\u3001B\u3001C\u3001D\u662f\u7403\u9762\u4e0a\u7684\u56db\u4e2a\u70b9\uff0c\u4e14\u5728\u540c\u4e00\u5e73\u9762\u5185\uff0cAB=BC=CD=DA=3\uff0c\u7403\u5fc3\u5230\u8be5\u5e73\u9762\u7684\u8ddd\u79bb\u662f\u7403\u534a\u5f84\u7684\u4e00\u534a\uff0c\u5219\u7403\u7684\u4f53\u79ef\u662f\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
\u4e8c\u3001\u586b\u7a7a\u9898\uff1a
11\uff0e\u6709\u4e24\u6392\u5ea7\u4f4d\uff0c\u524d\u639211\u4e2a\u5ea7\u4f4d\uff0c\u540e\u639212\u4e2a\u5ea7\u4f4d\uff0c\u73b0\u5b89\u63922\u4eba\u5c31\u5ea7\uff0c\u89c4\u5b9a\u524d\u6392\u4e2d\u95f4\u76843\u4e2a\u5ea7\u4f4d\u4e0d\u80fd\u5750\uff0c\u5e76\u4e14\u8fd92\u4eba\u4e0d\u5de6\u53f3\u76f8\u90bb\uff0c\u90a3\u4e48\u4e0d\u540c\u6392\u6cd5\u7684\u79cd\u6570\u662f_____________________.
12\uff0e\u82e5\u7ecf\u8fc7\u70b9P\uff08\uff0d1\uff0c0\uff09\u7684\u76f4\u7ebf\u4e0e\u5706 \u76f8\u5207\uff0c\u5219\u6b64\u76f4\u7ebf\u5728y\u8f74\u4e0a\u7684\u622a\u8ddd\u662f .
13\uff0e = .
14\uff0e\u5982\u56fe\uff0c\u56db\u68f1\u67f1ABCD\u2014A1B1C1D1\u7684\u5e95\u9762ABCD\u4e3a\u6b63\u65b9\u5f62\uff0c\u4fa7\u68f1\u4e0e\u5e95\u9762\u8fb9\u957f\u5747\u4e3a2a\uff0c\u4e14 \uff0c\u5219\u4fa7\u68f1AA1\u548c\u622a\u9762B1D1DB\u7684\u8ddd\u79bb\u662f .
15\uff0e\u53e3\u888b\u5185\u88c5\u670910\u4e2a\u76f8\u540c\u7684\u7403\uff0c\u5176\u4e2d5\u4e2a\u7403\u6807\u6709\u6570\u5b570\uff0c5\u4e2a\u7403\u6807\u6709\u6570\u5b571\uff0c\u82e5\u4ece\u888b\u4e2d\u6478\u51fa5\u4e2a\u7403\uff0c\u90a3\u4e48\u6478\u51fa\u76845\u4e2a\u7403\u6240\u6807\u6570\u5b57\u4e4b\u548c\u5c0f\u4e8e2\u6216\u5927\u4e8e3\u7684\u6982\u7387\u662f .\uff08\u4ee5\u6570\u503c\u4f5c\u7b54\uff09
16\uff0e\u5b9a\u4e49\u8fd0\u7b97 \u4e3a: \u4f8b\u5982\uff0c ,\u5219\u51fd\u6570f(x)= \u7684\u503c\u57df\u4e3a\uff0e




\u53c2\u8003\u7b54\u6848:
\u4e00\u3001\u9009\u62e9\u9898\uff1a
1.D 2.D 3.B 4.C 5.B 6.D 7.B 8.C 9.A 10.A
\u4e8c\u3001\u586b\u7a7a\u9898\uff1a
11.346 12\uff0e1 13\uff0e 14\uff0ea 15\uff0e
16\uff0e\u7531\u9898\u610f\u53ef\u5f97\u51fd\u6570\u5728\u4e00\u4e2a\u5468\u671f\u5185\u7684\u8868\u8fbe\u5f0f\uff0e\u5373\uff1a
\uff0c\u4f5c\u51fa\u56fe\u8c61\u6613\u5f97\u51fd\u6570\u7684\u503c\u57df\u4e3a \uff0e


2010\u5c4a\u9ad8\u8003\u9009\u62e9\u9898\u548c\u586b\u7a7a\u9898\u4e13\u9879\u8bad\u7ec3\uff088\uff09
\u4e00\u3001\u9009\u62e9\u9898 \uff1a
1\uff0e(1\uff0di)2•i= \uff08 \uff09 A\uff0e2\uff0d2iB\uff0e2+2i C\uff0e\uff0d2D\uff0e2
2\uff0e\u5df2\u77e5\u51fd\u6570 \uff08 \uff09
A\uff0ebB\uff0e\uff0dbC\uff0e D\uff0e\uff0d
3\uff0e\u5df2\u77e5 \u3001 \u5747\u4e3a\u5355\u4f4d\u5411\u91cf\uff0c\u5b83\u4eec\u7684\u5939\u89d2\u4e3a60\u00b0\uff0c\u90a3\u4e48| +3 |=\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e4
4\uff0e\u51fd\u6570 \u7684\u53cd\u51fd\u6570\u662f\uff08 \uff09
A\uff0ey=x2\uff0d2x+2(x<1)B\uff0ey=x2\uff0d2x+2(x\u22651)
C\uff0ey=x2\uff0d2x (x<1)D\uff0ey=x2\uff0d2x (x\u22651)
5\uff0e \u7684\u5c55\u5f00\u5f0f\u4e2d\u5e38\u6570\u9879\u662f\uff08 \uff09
A\uff0e14B\uff0e\uff0d14C\uff0e42D\uff0e\uff0d42
6\uff0e\u8bbeA\u3001B\u3001I\u5747\u4e3a\u975e\u7a7a\u96c6\u5408\uff0c\u4e14\u6ee1\u8db3A B I\uff0c\u5219\u4e0b\u5217\u5404\u5f0f\u4e2d\u9519\u8bef\u7684\u662f\uff08 \uff09
A\uff0e( A)\u222aB=IB\uff0e( A)\u222a( B)=I
C\uff0eA\u2229( B)= D\uff0e( A) ( B)= B
7\uff0e\u692d\u5706 \u7684\u4e24\u4e2a\u7126\u70b9\u4e3aF1\u3001F2\uff0c\u8fc7F1\u4f5c\u5782\u76f4\u4e8ex\u8f74\u7684\u76f4\u7ebf\u4e0e\u692d\u5706\u76f8\u4ea4\uff0c\u4e00\u4e2a\u4ea4\u70b9\u4e3aP\uff0c\u5219 =\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e4
8\uff0e\u8bbe\u629b\u7269\u7ebfy2=8x\u7684\u51c6\u7ebf\u4e0ex\u8f74\u4ea4\u4e8e\u70b9Q\uff0c\u82e5\u8fc7\u70b9Q\u7684\u76f4\u7ebfl\u4e0e\u629b\u7269\u7ebf\u6709\u516c\u5171\u70b9\uff0c\u5219\u76f4\u7ebfl \u7684\u659c\u7387\u7684\u53d6\u503c\u8303\u56f4\u662f\uff08 \uff09
A\uff0e[\uff0d \uff0c ]B\uff0e[\uff0d2\uff0c2]C\uff0e[\uff0d1\uff0c1]D\uff0e[\uff0d4\uff0c4]
9\uff0e\u4e3a\u4e86\u5f97\u5230\u51fd\u6570 \u7684\u56fe\u8c61\uff0c\u53ef\u4ee5\u5c06\u51fd\u6570 \u7684\u56fe\u8c61\uff08 \uff09
A\uff0e\u5411\u53f3\u5e73\u79fb \u4e2a\u5355\u4f4d\u957f\u5ea6B\uff0e\u5411\u53f3\u5e73\u79fb \u4e2a\u5355\u4f4d\u957f\u5ea6
C\uff0e\u5411\u5de6\u5e73\u79fb \u4e2a\u5355\u4f4d\u957f\u5ea6D\uff0e\u5411\u5de6\u5e73\u79fb \u4e2a\u5355\u4f4d\u957f\u5ea6
10\uff0e\u5df2\u77e5\u6b63\u56db\u9762\u4f53ABCD\u7684\u8868\u9762\u79ef\u4e3aS\uff0c\u5176\u56db\u4e2a\u9762\u7684\u4e2d\u5fc3\u5206\u522b\u4e3aE\u3001F\u3001G\u3001H.\u8bbe\u56db\u9762\u4f53EFGH\u7684\u8868\u9762\u79ef\u4e3aT\uff0c\u5219 \u7b49\u4e8e\uff08 \uff09
A\uff0e B\uff0e C\uff0e D\uff0e
\u4e8c\u3001\u586b\u7a7a\u9898\uff1a
11\uff0e\u4ece\u6570\u5b571\uff0c2\uff0c3\uff0c4\uff0c5\uff0c\u4e2d\uff0c\u968f\u673a\u62bd\u53d63\u4e2a\u6570\u5b57\uff08\u5141\u8bb8\u91cd\u590d\uff09\u7ec4\u6210\u4e00\u4e2a\u4e09\u4f4d\u6570\uff0c\u5176\u5404\u4f4d\u6570\u5b57\u4e4b\u548c\u7b49\u4e8e9\u7684\u6982\u7387\u4e3a________________.
12\uff0e\u4e0d\u7b49\u5f0f|x+2|\u2265|x|\u7684\u89e3\u96c6\u662f .
13\uff0e\u7531\u52a8\u70b9P\u5411\u5706x2+y2=1\u5f15\u4e24\u6761\u5207\u7ebfPA\u3001PB\uff0c\u5207\u70b9\u5206\u522b\u4e3aA\u3001B\uff0c\u2220APB=60\u00b0\uff0c\u5219\u52a8\u70b9P\u7684\u8f68\u8ff9\u65b9\u7a0b\u4e3a .
14\uff0e\u5df2\u77e5\u6570\u5217{an},\u6ee1\u8db3a1=1\uff0can=a1+2a2+3a3+\u2026+(n\uff0d1)an\uff0d1(n\u22652),\u5219{an}\u7684\u901a\u9879
15\uff0e\u5df2\u77e5a\u3001b\u4e3a\u4e0d\u5782\u76f4\u7684\u5f02\u9762\u76f4\u7ebf\uff0c\u03b1\u662f\u4e00\u4e2a\u5e73\u9762\uff0c\u5219a\u3001b\u5728\u03b1\u4e0a\u7684\u5c04\u5f71\u6709\u53ef\u80fd\u662f .
\u2460\u4e24\u6761\u5e73\u884c\u76f4\u7ebf \u2461\u4e24\u6761\u4e92\u76f8\u5782\u76f4\u7684\u76f4\u7ebf \u2462\u540c\u4e00\u6761\u76f4\u7ebf \u2463\u4e00\u6761\u76f4\u7ebf\u53ca\u5176\u5916\u4e00\u70b9
\u5728\u4e0a\u9762\u7ed3\u8bba\u4e2d\uff0c\u6b63\u786e\u7ed3\u8bba\u7684\u7f16\u53f7\u662f \uff08\u5199\u51fa\u6240\u6709\u6b63\u786e\u7ed3\u8bba\u7684\u7f16\u53f7\uff09.
16\u3001\u82e5\u51fd\u6570 \u5728\u533a\u95f4 \u4e0a\u662f\u51cf\u51fd\u6570\uff0c\u5219\u5b9e\u6570 \u7684\u53d6\u503c\u8303\u56f4\u662f \u3002


\u53c2\u8003\u7b54\u6848
\u4e00\u3001\u9009\u62e9\u9898
DBCBABCCBA
\u4e8c\u3001\u586b\u7a7a\u9898\uff1a
11. 12\uff0e{x|x\u2265\uff0d1} 13\uff0ex2+y2=4
14\uff0e 15\uff0e\u2460\u2461\u2463 16\u3001


\u62b1\u6b49 \u5e76\u975e\u5168\u90e8\u90fd\u662f\u9ad8\u4e8c\u7684 \u770b\u7740\u6284\u6284\u5427 \u6216\u8005\u6284\u4e00\u4e9b\u77e5\u9053\u4e0a\u522b\u4eba\u95ee\u51fa\u6765\u7684\u9009\u62e9

、选择题

(1)若x∈R,下列不等式中解法正确的是 ( )

(A)x2>2x>±

(B)(x-1)2<21-<x<1+

(C)ax+b<0x<-

(D)<1-2xx2-1<(1-2x)23x2-4x+2>0

∵△=16-24<0 ∴无解.

(2)下列各对不等式中同解的是 ( )

(A)(2a+7)x>a+3与x>

(B)lg(x-a)2<0与(x-a)2<1

(C)<1与≤1

(D)(x-a)(x-b)>0与>0

(3)不等式4x>的解集是 ( )

(A){x|x<-或x>} (B){x|x>-且x≠}

(C){x|-<x<0或x>} (D){x|-<x<}

(4)不等式ax2+bx+2>0的解集是{x|-<x<},则a+b的值为 ( )

(A)10 (B)-10 (C)14 (D)-14

(5)不等式(x-1)≥0的解集是 ( )

(A){x|x>1} (B){x|x≥1}

(C){x|x≥1或x-2} (D){x|x<-2或x≥2

(6)不等式≥0的解集是 ( )

(A){x|-2≤x≤2} (B){x|-≤x<0或0<x≤2}

(C){x|-2<x≤0或0>x≤2} (D){x|-≤x<0或0<x≤}

(7)不等式|-3|<1的解集是 ( )

(A){x|5<x<16} (B){x|6<x<18}

(C){x|7<x<20 (D){x|8<x<22

(8)已知集合A=,B=,则A∩B用区间表示为 ( )

(A) (B)(-∞,0)∪

(C)(1,+ ∞) (D) (-∞,0)∪

(9)不等式>4的解集是 ( )

(A){x|x<100} (B){x|0<x<100}

(C){x|x<} (D)

(10)若集合M={x|x2-5x-6<0,N={x|lg(x+1)2<2},全集I=R,则为 ( )

(A){x|x≤1}∪{x|6≤x<9} (B){x|-1<x<6}

(C){x|-11<x≤-1或6≤x<9} (D){x|-11<x<9}

(11)不等式log(3x2+2x-1) <1的解集是 ( )

(A){x|-2<x<0} (B){x|0<x<1或-2<x<-1}

(C){x|-2<x<-1 (D){x|-2<x<-1或<x<1

(12)不等式(a-2)x2+2(a-2)x-4<0,对任意实数x恒成立,则a的取值范围是 ( )

(A)(-2,2) (B)(-2,2]

(C)(-∞,-2)∪(2,+∞) (D)(-∞,-2)∪[2,+∞)

(13)如果loga<1,则a的取值范围是 ( )

(A) (B)

(C) (D)∪(1,+∞)

(14)不等式<2对一切实数x都成立,则a的取值范围是 ( )

(A)a> (B)a<

(C) 0<a< (D) <a<1

(15)若关于x的方程x2-x-(m+1)=0在[-1,1]上有解,则m的取值范围是 ( )

(A)m≥- (B)-≤m≤-1

(C)-≤m≤1 (D)m≤1

二、填空题

(1)不等式≥1的解集是__________.

(2)不等式(x2-4x-5)(x2-4)≤0的解集是__________.

(3)使不等式>x+1成立的x的取值范围是_______.

(4)不等式|2x2-5|>3x的解集是________.

(5)不等式lg<0的解集是__________.

(6)不等式5≥0.2的解集是________.

三、解答题

(1)解不等式≥x.

(2)解不等式log3x+logx27<4.

(3)解不等式|-2x|≥1.

(4)已知:a>0,a≠1,解不等式

loga(4+3x-x2)-loga(2x-1)>loga2.

(5)若(a-2)x2+1≤(a-2)x对任意实数x都成立,求a的取值范围.

(6)如果偶函数f(x)在x∈[0,+∞)上是增函数,且f(log427·log272)=0,求不等式f(logax)>0 (a>0且a≠1)的解集.

例1.求函数的解析式

(1) f9[(x 1)= , 求f (x); 答案:f (x)=x2-x+1(x≠1)

练习1:已知f( 1)= x 2 ,求f(x) 答案:f (x)=x2-1(x≥1)

(2) f (x) = 3x2 1, g (x) = 2x -1 , 求f[g(x)];答案:f[g(x)]=12x2-12x+4

练习2:已知:g(x)=x 1,f[g (x)]=2x2 1,求f(x-1) 答案:f(x-1)=2x2-8x 9

(3)如果函数f (x)满足af (x) f()=ax,x∈R且x≠0,a为常数,且a≠±1,求f (x)的表达式。答案:f (x)= (x∈R且x≠0)

练习3: 2f (x) - f (-x) = lg (x 1), 求 f (x).

答案:f(x)= lg(x 1) lg(1-x) (-1<x<1)

例2.已知f (x)是一次函数,并且满足3f (x 1) - 2f (x-1)=2x 17,求f (x).

答案:f (x)=2x 7.

练习4:已知f (x)是二次函数,满足f(0)=1且f (x 1) - f (x)=2x,求f (x)

答案:f (x) = x2- x 1

例3.设f(x)是R上的函数,且满足f(0)=1,并且对任意实数x,y

有f(x-y)=f(x)-y(2x-y 1),求f(x) 答案:f (x) =x2 x 1

练习5:函数f(x)对任何x∈R恒有f(xx)=f(x1) f(x2),已知f(8)=3,则f()=

例4.已知函数y=f(x)的图像如图所示,求f(x)

练习6:已知函数f(x)的图像是由两条射线和开口向下的抛物线组成,

求f(x)解析式

例5.已知定义在R上的函数y=f(x)关于直线x=2对称并且x∈[0,2]上的解析式为y=2x-1,则f(x)在x∈[2,4]上的解析式为 y=7-2x

练习7:设函数y=f(x)关于直线x=1对称,若当x≤1时,y=x2 1,

则当x>1 时,f(x)= x2-4x 5

课堂小结:求函数的解析式的方法较多,应根椐题意灵活选择,但不论是哪种方法都应注意自变量的取值范围,对于实际问题材,同样需注意这一点,应保证各种有关量均有意义。

布置作业:

1、若g(x)=1-2x , f[g(x)] = (x≠0),求f()的值。

2、已知f(x - )=x , 求f(x-1)的表达式.

3、已知f(x)=9x 1,g(x)=x,则满足f[g(x)]= g[f(x)] 的x的值为多少?

4、已知f(x)为一次函数且f[f(x)] = 9x 4,求f(x).
回答者:542839777 - 初入江湖 三级 8-2 16:13
分数好少
回答者:tm19880202 - 助理 二级 8-3 21:14
历届高考中的“不等式”试题汇编大全
一、选择题:

6.(2006江西理)若a0,b0,则不等式-ba等价于( )
A.x0或0x B.-x C.x-或x D.x或x

8.(2006陕西文)设x、y为正数,则有(x+y)()的最小值为
A.15 B.12 C.9 D.6
9.(2006陕西理)已知不等式(x+y)( + )≥9对任意正实数x,y恒成立,则正实数a的最小值为( )
A.2 B.4 C.6 D.8

10.(2006上海理)若关于的不等式≤+4的解集是M,则对任意实常数,总有[答]( )
(A)2∈M,0∈M; (B)2M,0M; (C)2∈M,0M; (D)2M,0∈M.

12.(2006重庆理)若a,b,c>0且a(a+b+c)+bc=4-2,则2a+b+c的最小值为
(A)-1 (B) +1
(C) 2+2 (D) 2-2
4、(2005湖南理)集合A={x|<0},B={x||x-b|<a},若“a=1”是“A∩B≠Φ”的充分条件,则b的取值范围是( )
A、-2≤b<0 B、0<b≤2 C、-3<b<-1 D、-1≤b<2
5.(2005湖南文)设集合A={x|<0,B={x || x -1|<a,若“a=1”是“A∩B≠”的( )
A.充分不必要条件 B.必要不充分条件
C.充要条件 D.既不充分又不必要条件

10.(2005全国卷Ⅱ理科)已知集合M={x∣-3x -28 ≤0},N = {x|-x-6>0},则M∩N 为( )
(A){x|- 4≤x< -2或3<x≤7} (B){x|- 4<x≤ -2或 3≤x<7 }
(C){x|x≤ - 2或 x> 3 } (D){x|x<- 2或x≥3}
11.(2005北京理科)设全集U=R,集合M={x| x>1,P={x| x2>1},则下列关系中正确的是
A.M=P B.P M C.M P( D)
12.(2005北京文科)设全集U=R,集合M={x| x>1,P={x| x2>1},则下列关系中正确的是
A.M=P B.P M C.M P( D)

(2004年)
1.(2004安徽春招文、理)不等式|2x2-1|≤1的解集为
A.{x|-1≤x≤1} B.{x|-2≤x≤2} C.{x|0≤x≤2} D.{x|-2≤x≤0}
2.(2004北京春招理) 已知三个不等式:(其中a,b,c,d均为实数),用其中两个不等式作为条件,余下的一个不等式作为结论组成一个命题,可组成的正确命题的个数是( )
A. 0 B. 1 C. 2 D. 3

6.(2004湖北理科)设集合P={m|-1<m<0}, Q={m∈R|mx2+4mx-4<0对任意实数恒成立},
则下列关系中成立的是( )
(A ) P Q (B) Q P (C)P=Q (D)P∩Q=

13.(2004全国卷Ⅱ文、理)已知集合M={x|x2<4,N={x|x2-2x-3<0,则集合M∩N=
(A){x|x<-2 (B){x|x>3} (C){x|-1<x<2 (D){x|2<x<3

5.(2003天津文)不等式的解集是( )
A.(0,2) B.(2,+∞)
C.(2,4) D.(-∞,0)∪(2,+∞)

8. (2002广东、江苏、河南、全国文理、天津文理)不等式(1+x)(1-|x|)>0的解集是
A.{x|0≤x<1} B.{x|x<0且x≠-1}
C.{x|-1<x<1} D.{x|x<1且x≠-1}9. (2002年广东、江苏、河南,全国文)已知0<x<y<a<1,则有
A.loga(xy)<0 B.0<loga(xy)<1
C.1<loga(xy)<2 D.loga(xy)>210 �

二.填空题:
.

2.(2006上海理)三个同学对问题“关于的不等式+25+|-5|≥在[1,12]上恒成立,求实数的取值范围”提出各自的解题思路.
甲说:“只须不等式左边的最小值不小于右边的最大值”.
乙说:“把不等式变形为左边含变量的函数,右边仅含常数,求函数的最值”.
丙说:“把不等式两边看成关于的函数,作出函数图像”.
参考上述解题思路,你认为他们所讨论的问题的正确结论,即的取值范围是 .

。.

= .

7.(2004江苏)二次函数y=ax2+bx+c(x∈R)的部分对应值如下表:

x -3 -2 -1 0 1 2 3 4
y 6 0 -4 -6 -6 -4 0 6

则不等式ax2+bx+c>0的解集是_______________________.

8.(2004全国1卷理)不等式|x+2|≥|x|的解集是 .

9.(2004全国1卷文)不等式x+x3≥0的解集是 . .

三、解答题:

2.(2005北京理)设f(x)是定义在[0, 1]上的函数,若存在x*∈(0,1),使得f(x)在[0, x*]上单调递增,在[x*,1]上单调递减,则称f(x)为[0, 1]上的单峰函数,x*为峰点,包含峰点的区间为含峰区间. 对任意的[0,l]上的单峰函数f(x),下面研究缩短其含峰区间长度的方法.
(I)证明:对任意的x1,x2∈(0,1),x1<x2,若f(x1)≥f(x2),则(0,x2)为含峰区间;若f(x1)≤f(x2),则(x*,1)为含峰区间;
(II)对给定的r(0<r<0.5),证明:存在x1,x2∈(0,1),满足x2-x1≥2r,使得由(I)所确定的含峰区间的长度不大于 0.5+r;
(III)选取x1,x2∈(0, 1),x1<x2,由(I)可确定含峰区间为(0,x2)或(x1,1),在所得的含峰区间内选取x3,由x3与x1或x3与x2类似地可确定一个新的含峰区间.在第一次确定的含峰区间为(0,x2)的情况下,试确定x1,x2,x3的值,满足两两之差的绝对值不小于0.02,且使得新的含峰区间的长度缩短到0.34.
(区间长度等于区间的右端点与左端点之差)

3.(2005湖北理)已知不等式为大于2的整数,表示不超过的最大整数. 设数列的各项为正,且满足
(Ⅰ)证明
(Ⅱ)猜测数列是否有极限?如果有,写出极限的值(不必证明);
(Ⅲ)试确定一个正整数N,使得当时,对任意b>0,都有

4.(2005江西理、文)
已知函数(a,b为常数)且方程f(x)-x+12=0有两个实根为x1=3, x2=4.
(1)求函数f(x)的解析式;
(2)设k>1,解关于x的不等式;

8.(2005天津文、理)某人在一山坡P处观看对面山项上的一座铁塔,如图所示,塔高BC=80(米),塔所在的山高OB=220(米),OA=200(米),图中所示的山坡可视为直线l且点P在直线l上,与水平地面的夹角为a ,tana=1/2试问此人距水平地面多高时,观看塔的视角∠BPC最大(不计此人的身高)

11.(2005浙江理)已知函数f(x)和g(x)的图象关于原点对称,且f(x)=x2=2x.
(Ⅰ)求函数g(x)的解析式;
(Ⅱ)解不等式g(x)≥f(x)-|x-1|.

(2004年)
1.(2004安徽春招理)解关于x的不等式:loga3x<3logax(a>0且a≠1)

5.(2004福建理)已知f(x)=(x∈R)在区间[-1,1]上是增函数。
(Ⅰ)求实数a的值组成的集合A;
(Ⅱ)设关于x的方程f(x)=的两个非零实根为x1、x2.试问:是否存在实数m,使得不等式m2+tm+1≥|x1-x2|对任意a∈A及t∈[-1,1]恒成立?若存在,求m的取值范围;若不存在,请说明理由。

6.(2004福建文)已知f(x)=在区间[-1,1]上是增函数.
(Ⅰ)求实数a的值组成的集合A;
(Ⅱ)设关于x的方程f(x)=的两个非零实根为x1、x2.试问:是否存在实数m,使得不等式m2+tm+1≥|x1-x2|对任意a∈A及t∈[-1,1]恒成立?若存在,求m的取值范围;若不存在,请说明理由.

11.(2004全国2卷文)若函数f(x)=x3-ax2+(a-1)x+1在区间(1,4) 内为减函数,在区间(6,+∞)上为增函数,试求实数a的取值范围。

12.(2004全国Ⅲ卷文、理)某村计划建造一个室内面积为800的矩形蔬菜温室。在温室内,沿左.右两侧与后侧内墙各保留1宽的通道,沿前侧内墙保留3宽的空地。当矩形温室的边长各为多少时?蔬菜的种植面积最大。最大种植面积是多少?

13、(2004上海文、理)
某单位用木料制作如图所示的框架, 框架的下部是边长分别为x、y(单位:m)的矩形.上部是等腰直角三角形. 要求框架围成的总面积8cm2. 问x、y分别为多少(精确到0.001m) 时用料最省?

15.(2004北京文、理) 某段城铁线路上依次有A、B、C三站,AB=5km,BC=3km,在列车运行时刻表上,规定列车8时整从A站发车,8时07分到达B站并停车1分钟,8时12分到达C站,在实际运行中,假设列车从A站正点发车,在B站停留1分钟,并在行驶时以同一速度匀速行驶,列车从A站到达某站的时间与时刻表上相应时间之差的绝对值称为列车在该站的运行误差。
(I)分别写出列车在B、C两站的运行误差
(II)若要求列车在B,C两站的运行误差之和不超过2分钟,求的取值范围

16.(2004北京文、理)
给定有限个正数满足条件T:每个数都不大于50且总和L=1275。现将这些数按下列要求进行分组,每组数之和不大于150且分组的步骤是:
首先,从这些数中选择这样一些数构成第一组,使得150与这组数之和的差与所有可能的其他选择相比是最小的,称为第一组余差;
然后,在去掉已选入第一组的数后,对余下的数按第一组的选择方式构成第二组,这时的余差为;如此继续构成第三组(余差为)、第四组(余差为)、……,直至第N组(余差为)把这些数全部分完为止。
(I)判断的大小关系,并指出除第N组外的每组至少含有几个数

6.(2003全国理,广东)
已知c>0,设P:函数在R上单调递减Q:不等式x+|x-2c|>1的解集为R.如果P和Q有且仅有一个正确,求c的取值范围

7.(2003全国文、理,广东)
在某海滨城市附近海面有一台风,据监测,当前台风中心位于城市O(如图)的东偏南方向300km的海面P处,并以20km/h的速度向西偏北45°方向移动. 台风侵袭的范围为圆形区域,当前半径为60km,并以10km/h的速度不断增大. 问几小时后该城市开始受到台风的侵袭?

都是我经过筛选的过去的高考题,希望能令你满意

一、选择题

(1)若x∈R,下列不等式中解法正确的是 ( )

(A)x2>2x>±

(B)(x-1)2<21-<x<1+

(C)ax+b<0x<-

(D)<1-2xx2-1<(1-2x)23x2-4x+2>0

∵△=16-24<0 ∴无解.

(2)下列各对不等式中同解的是 ( )

(A)(2a+7)x>a+3与x>

(B)lg(x-a)2<0与(x-a)2<1

(C)<1与≤1

(D)(x-a)(x-b)>0与>0

(3)不等式4x>的解集是 ( )

(A){x|x<-或x>} (B){x|x>-且x≠}

(C){x|-<x<0或x>} (D){x|-<x<}

(4)不等式ax2+bx+2>0的解集是{x|-<x<},则a+b的值为 ( )

(A)10 (B)-10 (C)14 (D)-14

(5)不等式(x-1)≥0的解集是 ( )

(A){x|x>1} (B){x|x≥1}

(C){x|x≥1或x-2} (D){x|x<-2或x≥2

(6)不等式≥0的解集是 ( )

(A){x|-2≤x≤2} (B){x|-≤x<0或0<x≤2}

(C){x|-2<x≤0或0>x≤2} (D){x|-≤x<0或0<x≤}

(7)不等式|-3|<1的解集是 ( )

(A){x|5<x<16} (B){x|6<x<18}

(C){x|7<x<20 (D){x|8<x<22

(8)已知集合A=,B=,则A∩B用区间表示为 ( )

(A) (B)(-∞,0)∪

(C)(1,+ ∞) (D) (-∞,0)∪

(9)不等式>4的解集是 ( )

(A){x|x<100} (B){x|0<x<100}

(C){x|x<} (D)

(10)若集合M={x|x2-5x-6<0,N={x|lg(x+1)2<2},全集I=R,则为 ( )

(A){x|x≤1}∪{x|6≤x<9} (B){x|-1<x<6}

(C){x|-11<x≤-1或6≤x<9} (D){x|-11<x<9}

(11)不等式log(3x2+2x-1) <1的解集是 ( )

(A){x|-2<x<0} (B){x|0<x<1或-2<x<-1}

(C){x|-2<x<-1 (D){x|-2<x<-1或<x<1

(12)不等式(a-2)x2+2(a-2)x-4<0,对任意实数x恒成立,则a的取值范围是 ( )

(A)(-2,2) (B)(-2,2]

(C)(-∞,-2)∪(2,+∞) (D)(-∞,-2)∪[2,+∞)

(13)如果loga<1,则a的取值范围是 ( )

(A) (B)

(C) (D)∪(1,+∞)

(14)不等式<2对一切实数x都成立,则a的取值范围是 ( )

(A)a> (B)a<

(C) 0<a< (D) <a<1

(15)若关于x的方程x2-x-(m+1)=0在[-1,1]上有解,则m的取值范围是 ( )

(A)m≥- (B)-≤m≤-1

(C)-≤m≤1 (D)m≤1

二、填空题

(1)不等式≥1的解集是__________.

(2)不等式(x2-4x-5)(x2-4)≤0的解集是__________.

(3)使不等式>x+1成立的x的取值范围是_______.

(4)不等式|2x2-5|>3x的解集是________.

(5)不等式lg<0的解集是__________.

(6)不等式5≥0.2的解集是________.

三、解答题

(1)解不等式≥x.

(2)解不等式log3x+logx27<4.

(3)解不等式|-2x|≥1.

(4)已知:a>0,a≠1,解不等式

loga(4+3x-x2)-loga(2x-1)>loga2.

(5)若(a-2)x2+1≤(a-2)x对任意实数x都成立,求a的取值范围.

(6)如果偶函数f(x)在x∈[0,+∞)上是增函数,且f(log427·log272)=0,求不等式f(logax)>0 (a>0且a≠1)的解集.

例1.求函数的解析式

(1) f9[(x 1)= , 求f (x); 答案:f (x)=x2-x+1(x≠1)

练习1:已知f( 1)= x 2 ,求f(x) 答案:f (x)=x2-1(x≥1)

(2) f (x) = 3x2 1, g (x) = 2x -1 , 求f[g(x)];答案:f[g(x)]=12x2-12x+4

练习2:已知:g(x)=x 1,f[g (x)]=2x2 1,求f(x-1) 答案:f(x-1)=2x2-8x 9

(3)如果函数f (x)满足af (x) f()=ax,x∈R且x≠0,a为常数,且a≠±1,求f (x)的表达式。答案:f (x)= (x∈R且x≠0)

练习3: 2f (x) - f (-x) = lg (x 1), 求 f (x).

答案:f(x)= lg(x 1) lg(1-x) (-1<x<1)

例2.已知f (x)是一次函数,并且满足3f (x 1) - 2f (x-1)=2x 17,求f (x).

答案:f (x)=2x 7.

练习4:已知f (x)是二次函数,满足f(0)=1且f (x 1) - f (x)=2x,求f (x)

答案:f (x) = x2- x 1

例3.设f(x)是R上的函数,且满足f(0)=1,并且对任意实数x,y

有f(x-y)=f(x)-y(2x-y 1),求f(x) 答案:f (x) =x2 x 1

练习5:函数f(x)对任何x∈R恒有f(xx)=f(x1) f(x2),已知f(8)=3,则f()=

例4.已知函数y=f(x)的图像如图所示,求f(x)

练习6:已知函数f(x)的图像是由两条射线和开口向下的抛物线组成,

求f(x)解析式

例5.已知定义在R上的函数y=f(x)关于直线x=2对称并且x∈[0,2]上的解析式为y=2x-1,则f(x)在x∈[2,4]上的解析式为 y=7-2x

练习7:设函数y=f(x)关于直线x=1对称,若当x≤1时,y=x2 1,

则当x>1 时,f(x)= x2-4x 5

课堂小结:求函数的解析式的方法较多,应根椐题意灵活选择,但不论是哪种方法都应注意自变量的取值范围,对于实际问题材,同样需注意这一点,应保证各种有关量均有意义。

布置作业:

1、若g(x)=1-2x , f[g(x)] = (x≠0),求f()的值。

2、已知f(x - )=x , 求f(x-1)的表达式.

3、已知f(x)=9x 1,g(x)=x,则满足f[g(x)]= g[f(x)] 的x的值为多少?

4、已知f(x)为一次函数且f[f(x)] = 9x 4,求f(x).

同底数对数相加真数相乘
定义域x+1>0 1-x>0
1>x>-1
y=lg(x+1)+lg(1-x)=lg(1-x^2)
0<x^2<1 0<1-x^2<1
值域(-∞,0)

满足定义域即可1/x-3>0 0<x<1/3

http://www.kejian123.com/gzsx/gkshuxue/200711/18837.html

这是耍人呐...
你让大家弄200题什么意思呢?

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