资阳市2014年高中阶段教育学校招生统一考试数学试题及参考答案（(3)

320．（1）因为函数y?kx?b图象过点P(?,0)和A(?2,1)，

2?3??k?b?0,所以?2 ··············································· 1分

???2k?b?1,?k??2,解得?所以一次函数的解析式为y??2x?3． ··· 3分

?b??3,又反比例函数的图象过点A(?2,1)，所以

m?1，所以m??2， ?22故反比例函数的解析式为y??． ····································································· 4分

x1??y??2x?3,?x1??2,?x2?,1?（2）联立?解得?或?·························· 6分 2所以点B(,?4)． ·2y?1,2y??,?1??x??y2??4.由图可知，当?2?x?0或x?

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1时，一次函数的值小于反比例函数的值． ··················· 8分 221．（1）因为AB是⊙O的直径，所以?ADB?90?. 所以?ABD??BAD?90?. ····································· 1分 又因为AC是⊙O的切线，则AB⊥AC，即?BAC?90?. 所以?CAD??BAD?90?， ··································· 2分 所以?ABD??CAD. ············································ 3分

因为?ABD??BDO??CDE，所以?CAD??CDE， ············································ 4分 又∠C＝∠C，所以△CDE∽△CAD．································································ 5分 （2）在Rt△OAC中，∠OAC＝90?，所以OA2?AC2?OC2，

即12?(22)2?OC2，所以OC?3，则CD?2， ··················································· 6分 222CDCA?，即，所以CE?2， ······················ 8分 ?CE2CECD所以AE?AC?CE?22?2?2． ································································· 9分

又由△CDE∽△CAD，得

22．设空调的采购数量为x台，则冰箱的采购数量为(20?x)台．

?11?x?(20?x),（1）根据题意可得? ······························································ 2分 9???20x?1500?1200,解得11?x?15， ···························································································· 3分 因为x为整数，所以x可取的值为11，12，13，14，15．

所以该商家共有5种进货方案． ········································································· 4分 （2）设总利润为W（元），

则W?(1760?y1)x1?(1700?y2)x2 ······································································· 5分 ?1760x?(?20x?1500)x?1700(20?x)?[?10(20?x)?1300](20?x) ?1760x?(?20x?1500)x?1700(20?x)?(10x?1100)(20?x)

?30x2?540x?12000?30(x?9)2?9570． ··························································· 7分

当x?9时，W随x的增大而增大，

因为11?x?15，所以当x?15时，W最大值?30(15?9)2?9570?10650（元），

所以采购空调15件时，获得总利润最大，最大利润值为10650元． ·························· 9分

?AB?CB,?23．（1）易知??ABP??CBE?90?,所以△ABP≌△CBE． ······································ 3分

?BP?BE,?（2）延长AP交CE于点H，

①因为△ABP≌△CBE，所以∠PAB＝∠ECB， 则∠PAB＋∠AEH＝∠ECB＋∠AEH＝90?，

所以AP⊥CE， ················································ 4分

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BC?2，即P是BC的中点， BP易得四边形BECD是平行四边形，······················· 5分 因为

则BD∥CE，所以AP⊥BD． ·············································································· 6分

BC②因为········································ 7分 ?n，即BC?n?BP，所以CP?(n?1)?BP， ·

BPPDPC因为CD∥BE，易得△CPD∽△BPE，所以····································· 8分 ??n?1， ·

PEPBSPC?n?1， 设△PBE的面积S?PBE?S，则△PCE的面积S?PCE满足?PCE?S?PBEPB即S2?(n?1)?S， ··························································································· 9分 又S?PAB?S?BCE?n?S，所以S?PAE?(n?1)?S，

SPD?n?1，所以S1?(n?1)?S?PAE，即S1?(n?1)(n?1)?S， ·又因为?PAD?················ 10分

S?PAEPES(n?1)(n?1)S?n?1． ·所以1?········································································ 11分

S2(n?1)S 24．（1）由题知抛物线与x轴另一个交点为(?1,0)， ?9a?3b?c?0,?a??1,??由?a?b?c?0,解得?b?2,所以抛物线的解析式为y??x2?2x?3． ···················· 2分 ?c?3,?c?3,??（2）①当MA＝MB时，得M(0,0)； ②当AB＝AM时，得M(0,?3)；

③当BA＝BM时，得M(0,3?32)或M(0,3?32)．

所以点M的坐标为(0,0)、(0,?3)、(0,3?32)、(0,3?32)． ································ 6分 （3）平移后的三角形记为△PEF，设直线AB的解析式为y?kx?b， ?3k?b?0,?k??1,所以?解得?所以直线AB的解析式为y??x?3．

b?3,b?3,??△AOB沿x轴向右平移m个单位长度（0＜m＜3）得到△PEF，

易得直线EF的解析式为y??x?3?m， ····························································· 7分 设直线AC的解析式为y?k?x?b?， ?3k??b??0,?k???2,则?解得?所以直线AC的解析式为

???k?b?4,b?6,??y??2x?6，

*………………………………………………………………8分

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3连结BE，直线BE交AC于G，则G(,3)，

2在△AOB沿x轴向右平移的过 …… 此处隐藏：747字，全部文档内容请下载后查看。喜欢就下载吧 ……