六、解答题（共2道小题，21题5分，22题4分，共9分）
21．解：设小强乘公交车的平均速度是每小时x千米，则小强乘自家车的平均速度是每小时(x?36)千米． 1分 依题意，得
205x?(x?36)． ······································································ 2分 6060解得x?12． ································································································ 3分
?20?12?4． ···························································································· 4分 60答：从小强家到学校的路程是4千米． ······························································· 5分 22．解：（1）AP?BP的值为32． ································································ 2分 （2）AP?BP的值为5． ··············································································· 3分 （3）?2m?3??1?2···································· 4分 ?8?2m??4的最小值为34． ·
2七、解答题（本题满分7分）
23．解：（1）∵一元二次方程kx?2x?2?k?0有实数根，
2??k?0,??2 ·············································································· 1分
2?4?k?2?k?0.??????k?0,?? 2??4?k?1??0.2∴当k?0时，一元二次方程kx?2x?2?k?0有实数根． ·································· 2分
（2）①由求根公式，得x??1?(k?1)．
k?x1?k?22?1?，x2??1．???????3分 kk2要使x1，x2均为整数，必为整数，
ky24y =k321-4-3-2-1O-1-2-3-4y = k - 11234K?2时，x1，x2均为整数． 所以，当k取?1、??????????????5分
2，x2??1代入方程 k2x1?x2?k?1?0中，得?k?1．
k22设y1?，y2?k?1，并在同一平面直角坐标系中分别画出y1?与y2?k?1的图象（如图所示）． 6分
kk②将x1?1?由图象可得，关于k的方程x1?x2?k?1?0的解为k1??1，k2?2．??????7分 八、解答题（本题满分7分）
24．解：（1）∵点C?2,3?在直线y?kx?1上，
?2k?1?3．
解得k?1．
··································································· 1分 ?直线AC的解析式为y?x?1． ·∵点A在x轴上，
?A(?1，0)．
?抛物线y??x2?bx?c过点A、C， ??1?b?c?0， ????4?2b?c?3．?b?2，解得?
c?3．?···························································· 2分 ?抛物线的解析式为y??x2?2x?3． ·（2）由y??x?2x?3???x?1??4，
22ly4P21EC，B(3，0)． 可得抛物线的对称轴为x?1?E?1,2?．???????3分
-3-2AOD-1-2-32B45x
根据题意，知点A旋转到点B处，直线l过点B、E． 设直线l的解析式为y?mx?n．
将B、E的坐标代入y?mx?n中，联立可得m??1,n?3．
∴直线l的解析式为y??x?3． ······································································ 4分
?P?0,3?．
过点E作ED?x轴于点D.
?S1?APE?S?APB?S?2?AB?PO?12?AB?ED?1EAB?2?4??3?2??2． ················ 5分 （3）存在，点F的坐标分别为?3?2,0?、?3?2,0?、??1?6,0?、??1?6,0?．九、解答题（本题满分8分）
25．解：（1）PC与PD的数量关系是相等 ． ··················································· 1分 证明：过点P作PH?OA，PN?OB，垂足分别为点H、N． ∵?AOB?90?，易得?HPN?90?．
??1??CPN?90?，
A而?2??CPN?90?，
PM??1??2．
H1∵OM是?AOB的平分线，
RC2?PH?PN， G又??PHC??PND?90?，
3?△PCH≌△PDN．
ONDB?PC?PD． ····················································································S·········· 2分 （2）?PC?PD，?CPD?90?， ??3?45?， ??POD?45?， ??3??POD．
又??GPD??DPO， ?△POD∽△PDG． ··················································································· 3分
?GDOD?PGPD． 7分
∵PG?3PD， 2?GDPG3． ····················································································· 4分 ??ODPD2（3）如图1所示，若PR与射线OA相交，则OP?1； ········································ 6分 如图2所示，若PR与直线OA的交点C与点A在点O的两侧，则OP?2?1．
·················································································································· 8分
AMAPCGREO图1DMPOEDBSSBC
R图2