第二十一章二次函数与反比例函数(单元测试·培优卷)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》9上多版本

                                                                           
                                                                       
第二十一章 二次函数与反比例函数·培优卷


Ⅰ
 !"#$ 10 '()* 30 *(,' 3 *-
1 #3 *-#24-25 ./01·2345·67-89:;<(𝑦=𝑥>?@AB>=#   -
A 𝑦= 2𝑥−3
B 𝑦= 𝑥2−5𝑥+ 13
C 𝑦= 𝑥2−(𝑥+ 2)(𝑥−3)
D 𝑦= 𝑥2−
1
𝑥+2
B
*CDEFG?@AB>HI(JKLM𝑦= 𝑎𝑥2 +𝑏𝑥+ 𝑐#aNbNc OPB(𝑎≠0>AB-Q
?@ABRO>S 
TU?@AB>HIVWXYZ[ 
\
A y = x >@AB(]=?@AB(^C!_]`abc
B 𝑦= 𝑥2−5𝑥+ 13=?@AB(^C!_`abc
C 𝑦= 𝑥2−(𝑥+ 2)(𝑥−3) = 𝑥+ 6(y = x >@AB(^C!_]`abc
D 𝑦= 𝑥2−
1
𝑥+2]=?@AB(^C!_]`ab 
^!
B 
2 #3 *-#24-25 ./01·def3·67-ghiAB𝑦1 = 𝑘1𝑥>jklhiAB𝑦2 =
𝑘2
𝑥>jkmn
o A(B pq(rq B >stO(−2,5)(uq A >stO#   -
A (2,−5)
B (2,5)
C (5,−2)
D (−5,2)
A
*CvwFlhiAB>jkxyNghiAB>jkzxy(JK{xy=|}>
S 
~pAB>jk*€ostq‚ƒ(uq A q B >stoq‚ƒ 
\
∵ghiAB𝑦1 = 𝑘1𝑥>jklhiAB𝑦2 =
𝑘2
𝑥>jkmno A(B pq(
∴q A q B >stoq‚ƒ(
                                                                           
                                                                       
∵q B >stO(−2,5)(
∴q A >stO(2,−5) 
^!
A 
3 #3 *-#24-25 „/08·…3†‡·67-rq𝐴(−2,𝑦1),𝐵(2,𝑦2),𝐶(3,𝑦3)ˆ‰Š‹𝑦= 2(𝑥+ 1)2 +𝑚1(
u𝑦1,𝑦2,𝑦3>Œ'=#   -
A 𝑦1 < 𝑦2 < 𝑦3
B 𝑦2 < 𝑦1 < 𝑦3
C 𝑦2 < 𝑦3 < 𝑦1
D 𝑦3 < 𝑦2 < 𝑦1
A
*CDEF?@ABjk1q>stŽ(‰Š‹‘’“8”‚ƒ•O–‹𝑥= −1—‚ƒ
•˜™š—›œ(AB›Œ(U}ž[Ÿ 
\
‰Š‹𝑦= 2(𝑥+ 1)2 +𝑚>‚ƒ•O𝑥= −1(‘’“1(
q𝐴(−2,𝑦1)>š—O|−2−(−1)| = 1(
q𝐵(2,𝑦2)>š—O|2−(−1)| = 3(
q𝐶(3,𝑦3)>š—O|3−(−1)| = 4(
o‘’“1(š—‚ƒ•›œ(y ›Œ(
∵4 > 3 > 1(
∴𝑦1 < 𝑦2 < 𝑦3 
^!
A 
4 #3 *-#2025· ¡¢£·¤¥-@AB𝑦= 𝑎𝑥+ 𝑏lhiAB𝑦= −
𝑐
𝑥>jkMj¦§(u?@AB
𝑦= 𝑎𝑥2 +𝑏𝑥+ 𝑐>jk[¨=#    -
A 
B 
C 
D 
                                                                           
                                                                       
D
*CDEFG?@ABjk(@ABjkzlhiABjk©a(TU@ABzlhiA
Bjkª«>k¬[Ÿ­𝑎> 0(𝑏< 0(𝑐< 0(u−
𝑏
2𝑎> 0(u[Ÿ­?@AB𝑦= 𝑎𝑥2 +𝑏𝑥+ 𝑐>jk‘’
“8( y •no®¯•(‚ƒ• z ˆ y •°v(U}[Ÿ 
\
∵@AB𝑦= 𝑎𝑥+ 𝑏>jkª«N¤N±k¬(
∴𝑎> 0(𝑏< 0(
∵lhiAB𝑦= −
𝑐
𝑥>jkª«N¤k¬(
∴−𝑐> 0(
∴𝑐< 0(
∴−
𝑏
2𝑎> 0(
∴?@AB𝑦= 𝑎𝑥2 +𝑏𝑥+ 𝑐>jk‘’“8( y •no®¯•(‚ƒ• z ˆ y •°v(
∴只有D !_<>ABjk`ab(
^!
D 
5 #3 *-#2025·dee ·¥²³´-~‰Š‹𝑦= 𝑎(𝑥+ 3)2 +2#a OPB(𝑎≠0-(µ‰Š‹“8
™¶ 4 W·¸†¹ºŸ­>‰Š‹ x •pWnq»>š—O 4(u a >O#   -
A −2
B 2
C −
1
2
D 
1
2
D
*CF‰Š‹>™¶(‰Š‹ x •>nq µ‰Š‹“8™¶ 4 W·¸ºŸ­4‰Š‹(
ž¼{;½¾H x •>nq(¿Ànq»šO 4 ÁÂÃĞ a > 
\
‰Š‹O𝑦= 𝑎(𝑥+ 3)2 +2(“8™¶ 4 W·¸ºŸ­4‰Š‹𝑦= 𝑎(𝑥+ 3)2−2 
Å𝑦= 0(u𝑎(𝑥+ 3)2−2 = 0(Ÿ𝑥= −3 ±
2
𝑎(
∴4‰Š‹ x •>pWnqstO −3 +
2
𝑎,0 ( −3−
2
𝑎,0 (
∵‰Š‹ x •pWnq»>š—O 4(
∴−3 +
2
𝑎−−3−
2
𝑎
= 4(
∴𝑎=
1
2 
^!
D 
                                                                           
                                                                       
6 #3 *-#2025·ÆÇÈÉÊ·<Ë-Mj(lhiAB𝑦=
𝑘
𝑥ª«𝐴N𝐶pq(«q𝐴Ì𝐴𝐵⊥𝑦•oq
𝐵(«q𝐶Ì𝐶𝐷⊥𝑥•oq𝐷(ÍÎ𝑂𝐴N𝑂𝐶N𝐴𝐶 r𝑆△𝐴𝐶𝑂= 4(𝐶𝐷:𝑂𝐵= 1:3(u𝑘>=#    -
A −12
B −9
C −6
D −3
D
*CFGlhiAB>ÏÐbI(ÑL>XHxy(ÒÓJK𝑘ÏÐbI=S Ԇ𝐷𝐶,𝐵𝐴
noq E(Õ𝐶𝐷= 𝑎(𝑎> 0)(u𝑂𝐵= 3𝑎(ž¼𝑂𝐷= −
𝑘
𝑎(𝐴𝐵= −
𝑘
3𝑎(Öן­𝑆△𝐷𝑂𝐶= 𝑆△𝐴𝑂𝐵= −
𝑘
2(Ø
Ù±ÚL𝑂𝐵𝐸𝐷=ÑL(۞¼𝐴𝐸= −
2𝑘
3𝑎,𝐶𝐸= 2𝑎(Ÿ­𝑆△𝐴𝐸𝐶= −
2𝑘
3 (TU𝑆矩形𝑂𝐵𝐸𝐷−𝑆△𝐷𝑂𝐶−𝑆△𝐴𝑂𝐵−𝑆△𝐴𝐸𝐶
= 𝑆△𝐴𝐶𝑂(ÁÂÃĞZ[ 
\
Ԇ𝐷𝐶,𝐵𝐴noq E(
Õ𝐶𝐷= 𝑎(𝑎> 0)(
∵𝐶𝐷:𝑂𝐵= 1:3(
∴𝑂𝐵= 3𝑎(
∵𝐴𝐵⊥𝑦•(𝐶𝐷⊥𝑥•(
∴q𝐴>ÜstO3𝑎(q𝐶>ÜstO𝑎(
∴𝑎=
𝑘
𝑥𝐶,3𝑎=
𝑘
𝑥𝐴(
∴𝑥𝐶=
𝑘
𝑎,𝑥𝐴=
𝑘
3𝑎(
                                                                           
                                                                       
∴𝑂𝐷= −
𝑘
𝑎(𝐴𝐵= −
𝑘
3𝑎(
∵lhiAB𝑦=
𝑘
𝑥ª«𝐴N𝐶pq(
∴𝑆△𝐷𝑂𝐶= 𝑆△𝐴𝑂𝐵= −
𝑘
2(
∵∠𝐸𝐷𝑂= ∠𝐷𝑂𝐵= ∠𝐸𝐵𝑂= 90°(
∴±ÚL𝑂𝐵𝐸𝐷=ÑL(
∴𝐵𝐸= 𝑂𝐷= −
𝑘
𝑎,𝐷𝐸= 𝑂𝐵= 3𝑎(
∴𝐴𝐸= 𝐵𝐸−𝐴𝐵= −
2𝑘
3𝑎,𝐶𝐸= 𝐷𝐸−𝐶𝐷= 2𝑎(
∴𝑆△𝐴𝐸𝐶=
1
2𝐴𝐸·𝐶𝐸= −
2𝑘
3 (
∴𝑆矩形𝑂𝐵𝐸𝐷= 𝑂𝐷·𝑂𝐵= −
𝑘
𝑎× 3𝑎= −3𝑘(
∵𝑆△𝐴𝐶𝑂= 4(
∴𝑆矩形𝑂𝐵𝐸𝐷−𝑆△𝐷𝑂𝐶−𝑆△𝐴𝑂𝐵−𝑆△𝐴𝐸𝐶= 𝑆△𝐴𝐶𝑂(Z−3𝑘−−
𝑘
2 −−
𝑘
2 −−
2𝑘
3
= 4(
∴𝑘= −3(
^!
D 
7 #3 *-#24-25 ./01·ÝÁÞß·àáÓâ-r𝑎(𝑏(𝑎< 𝑏)=o𝑥>ÃÄ(𝑥−𝑚)(𝑥−𝑛) +2022 = 0
(𝑚< 𝑛)>pWãBT(uãB𝑎(𝑏(𝑚(𝑛>Œ'=#   -
A 𝑎< 𝑏< 𝑚< 𝑛B 𝑚< 𝑛< 𝑎< 𝑏
C 𝑎< 𝑚< 𝑛< 𝑏
D 𝑚< 𝑎< 𝑏< 𝑛
D
*CFGä?@ÃÄ>TBå»>(?@ABä?@ÃÄ(ʼnŠ‹;𝑦1 =
(𝑥−𝑚)(𝑥−𝑛)(Ÿ­‰Š‹𝑥•nq>æstO𝑚(𝑛(ÛçajkŸ‰Š‹𝑦1 = (𝑥−𝑚)(𝑥−𝑛)𝑦2 = −2022
nq(ZnqæstO𝑎(𝑏(è×¾H¼𝑎(𝑏(𝑚(𝑛>Œ'(ÒÓJK‰Š‹>xy=>S 
\
ʼnŠ‹;𝑦1 = (𝑥−𝑚)(𝑥−𝑛)(
é𝑦1 = 0((𝑥−𝑚)(𝑥−𝑛) = 0(
Ÿ
𝑥1 = 𝑚(𝑥2 = 𝑛(
∴‰Š‹𝑥•nq>æstO𝑚(𝑛(
∴‰Š‹𝑦1 = (𝑥−𝑚)(𝑥−𝑛)𝑦2 = −2022nq(æstO𝑎(𝑏(
∵𝑚< 𝑛(𝑎< 𝑏(
                                                                           
                                                                       
∴Mj(
∴𝑚< 𝑎< 𝑏< 𝑛(
^!
D 
8 #3 *-#24-25 „/08·êë†ì·67-~q(𝑥1,𝑦1)((𝑥2,𝑦2)ˆlhiAB𝑦=
𝑘
𝑥#𝑘≠0(k OPB-
>jk1(r𝑥1 < 𝑥2(”𝑥1 + 𝑥2 < 0(u#    -
A 𝑦1 < 𝑦2
B |𝑦1| < |𝑦2|
C 𝑦1 > 𝑦2
D |𝑦1| > |𝑦2|
B
*CFGlhiAB>jkzxy(*𝑘> 0z𝑘< 0píîï(TUlhiAB>jkzxy
Z[ž(JKlhiAB>jkzxy=>S 
\
é𝑘> 0(lhiABjk*ðˆN¤k¬(ˆ,k¬ñ(𝑦ò𝑥>óŒ×ô'(”𝑥> 0õ
𝑦> 0(𝑥< 0õ𝑦< 0(
∵𝑥1 < 𝑥2(”𝑥1 + 𝑥2 < 0(
∴é𝑥1 < 𝑥2 < 0õ(𝑦2 < 𝑦1 < 0(u|𝑦1| < |𝑦2|c
é𝑥1 < 0 < 𝑥2õ(|𝑥1| > |𝑥2|(u0 < 𝑥2 < −𝑥1(
∴0 < −𝑦1 < 𝑦2(u|𝑦1| < |𝑦2|(
∴|𝑦1| < |𝑦2|c
é𝑘< 0(lhiABjk*ðˆ?N±k¬(ˆ,k¬ñ(𝑦ò𝑥>óŒ×óŒ(”𝑥> 0õ𝑦< 0(𝑥< 0
õ𝑦> 0(
∵𝑥1 < 𝑥2(”𝑥1 + 𝑥2 > 0(
∴é𝑥1 < 𝑥2 < 0õ(0 < 𝑦1 < 𝑦2(u|𝑦1| < |𝑦2|(
é𝑥1 < 0 < 𝑥2õ(|𝑥1| > |𝑥2|(u0 < 𝑥2 < −𝑥1(
∴𝑦2 < −𝑦1 < 0(u|𝑦1| < |𝑦2|c 
∴|𝑦1| < |𝑦2|c
                                                                           
                                                                       
©1(|𝑦1| < |𝑦2|(
^!
B 
9 #3 *-#2025·ÝÁÞß·¥²³´-~–‹𝑦= 𝑥+ 3‰Š‹𝑦= 𝑥2 +(𝑚−2)𝑥+ 2𝑚−1no ANB p
q#q A ˆq B >öv-( x •noq C(r‰Š‹>‚ƒ•= y •(u𝑆Δ𝐴𝐵𝑂:𝑆Δ𝐴𝐶𝑂÷o#-
A 1﹕2
B 1﹕3
C 1﹕4
D 3﹕4
B
*CFG?@AB@AB©a 
‰Š‹>‚ƒ•O y •([žŸ𝑚= 2(ø–‹‰Š‹ÃÄ(Ÿnq𝐴(0,3)N𝐵(1,4)(–‹ x •
nq𝐶(−3,0) ¿À¤ùLúû;*€üý △𝐴𝐵𝑂z △𝐴𝐶𝑂>úû(۞hZ[ 
\
‰Š‹‚ƒ•O y •(Zþqæst−
𝑚−2
2
= 0(Ÿ𝑚= 2 
ÿĀŸ‰Š‹ÃÄ𝑦= 𝑥2 +(𝑚−2)𝑥+ 2𝑚−1Ÿ𝑦= 𝑥2 +3 
øÂÃÄ𝑦= 𝑥+ 3z𝑦= 𝑥2 +3(Ÿ𝑥2−𝑥= 0(
Ÿ𝑥= 0ā𝑥= 1 
∴𝐴(0,3)z𝐵(1,4) 
Å𝑦= 0(ÿĀ𝑦= 𝑥+ 3Ÿ𝑥= −3(
Z𝐶(−3,0) 
∵𝐴(0,3)N𝐵(1,4)N𝑂(0,0) 
∴𝑆△𝐴𝐵𝑂=
1
2 × 3 × 1 =
3
2c
∵𝐴(0,3)N𝐶(−3,0)N𝑂(0,0) 
∴𝑆△𝐴𝐶𝑂=
1
2 × 3 × 3 =
1
2 × 9 =
9
2c
𝑆△𝐴𝐵𝑂:𝑆△𝐴𝐶𝑂=
3
2:
9
2 = 1:3 
^! B 
10 #3 *-#2024 ./01·Ăă·ĄÓâ-ˆlhiAB𝑦=
10
𝑥(𝑥> 0)>jk1(ą9q𝐴1(𝐴2(
𝐴3(…(𝐴𝑛(𝐴𝑛+1(r𝐴1>æstO 2(”Ćº,q>æstćĈWq>æst>ĉĊO 2(ċ*€«
q𝐴1(𝐴2(𝐴3(…(𝐴𝑛(𝐴𝑛+1Ì𝑥•𝑦•>ȋá(čRrĎWÑLMj¦§(µj<ďĐđ*>úûè
ö­°Ē@ēO𝑆1(𝑆2(𝑆3(…(𝑆𝑛(u𝑆1 + 𝑆2 + 𝑆3 +… + 𝑆𝑛= #ÀĔ𝑛>ÿB;ĕ§-#   -
                                                                           
                                                                       
A 
10𝑛
𝑛+1
B 
10
𝑛
C 
10
𝑛+1
D 
𝑛
𝑛+1
A
*CDEFGlhiAB©aĖÀ(𝐴1>æstO 2(”Ćº,q>æstćĈWq>æ
st>ĉĊO 2(ÛTUq𝐴1N𝐴2N𝐴3N…N𝐴𝑛N𝐴𝑛+1ˆlhiAB1(ž¼:qst(ېúû;ž
¼𝑆𝑛>ĕė;(ÒÓJKlhiAB>xy½¨ž¼𝐴𝑛>st>ĕė;(ې}ž¼𝑆𝑛>ĕė;=|}
>S 
\
 ∵q𝐴1N𝐴2N𝐴3N…N𝐴𝑛N𝐴𝑛+1ˆlhiAB𝑦=
10
𝑥(𝑥> 0)>jk1(”,q>æstć
ĈWq>æst>ĉĊO 2(
Ę ∵q𝐴1>æstO 2(
∴𝐴1(2,5)(𝐴2(4,
5
2)(𝐴3stO(6,
5
3) 
jk(𝐴𝑛(2𝑛,
10
2𝑛)(𝐴𝑛+1(2𝑛+ 2,
10
2𝑛+2)(
∴𝑆1 = 2 × (5−
5
2) = 5(
𝑆2 = 2 × (
5
2−
5
3) = 2 ×
5
6 =
5
3(
…(
𝑆𝑛= 2 × (
10
2𝑛−
10
2𝑛+2) =
10
𝑛(𝑛+1)(𝑛= 1(2(3(…)(
∵ 
1
𝑛(𝑛+1) =
1
𝑛−
1
𝑛+1(
∴𝑆1 + 𝑆2 + 𝑆3 + … + 𝑆𝑛
= 10( 1
2 + 1
6 + … +
1
𝑛(𝑛+ 1) )
= 10(1−1
2 + 1
2 −1
3 + … + 1
𝑛−
1
𝑛+ 1 )
=
10𝑛
𝑛+1 
                                                                           
                                                                       
^!
A 
? ęĚ#$ 6 '()* 18 *(,' 3 *-
11 #3 *-#2025·ĝeĞ𷤥-µ‰Š‹𝑦= 𝑥2−6𝑥+ 1“ö™¶ 2 W·¸†¹(ۓ1™¶ 3 W·¸
†¹(Ÿ­>‰Š‹>þqst=        
 
(1,−5)
*CFG?@AB>™¶(ĆĠ?@ABġ;Êþq;(>Sˆog¾JKAB™¶
>Ģģ Ĥĥ𝑦= 𝑥2−6𝑥+ 1ĦRþq;(ÛĥABĤ“ö™¶ 2 W·¸†¹(“1™¶ 3 W·¸†¹(Ÿ
­™¶º>þq;(Z[Ÿ­™¶º>‰Š‹>þqst 
\
µ‰Š‹𝑦= 𝑥2−6𝑥+ 1ÊOþq;ą𝑦= (𝑥−3)2−8(
ۓö™¶ 2 W·¸†¹(“1™¶ 3 W·¸†¹(
Ÿ𝑦= (𝑥−3 + 2)2−8 + 3 = (𝑥−1)2−5(
^™¶º>‰Š‹>þqst=(1,−5)(
^O
(1,−5) 
12 #3 *-#24-25 ./01·Ăă·67-~?@AB𝑦= −𝑥2 +𝑏𝑥+ 𝑐>đ*jkMj¦§ r𝑦> 0(
u𝑥>ħĨĩ=      
 
−3 < 𝑥< 1/1 > 𝑥> −3
*CFG‰Š‹𝑥•>nq(‚ƒ•nqst>(¿ÀBLça>Īī(g¾žŸ‰Š
‹𝑥•>ĬWnq>st=>S 
TU‰Š‹>‚ƒ•O𝑥= −1(WnqO(1,0)([ĭ¼ĬnqO(−3,0)(çajkž¼𝑦> 0õ(𝑥>
Ĩĩ 
\
TU‰Š‹>jk[
‰Š‹>‚ƒ•O𝑥= −1(WnqO(1,0)(
TU‚ƒx(uĬnqO(−3,0)(
¦Ć𝑦> 0(𝑥>ħĨĩ=−3 < 𝑥< 1(
                                                                           
                                                                       
^O
−3 < 𝑥< 1 
13 #3 *-#24-25 ./08·2Įį˜·àáÓâ-Mj=İàı>IJ‹§bj(,0“İà”>ijzĴĊ=
1(“İà”>ĵijqO𝐴(1,4)(rlhiAB𝑦=
𝑘
𝑥(𝑥> 0)>jkĶIJ‹ą$q(u k >ķBą     W 
4
*CDEFGlhiAB>ĖÀ(j[Ÿ(élhiABjk«q B ‘ĸİàąnq(–
­lhiABjk«q C OĹ(ÖמZ[ 
\
∵,0“İà”>ijzĴĊ= 1(“İà”>ĵijqO𝐴(1,4)(
∴𝐵(1,3)N𝐶(2,3)N𝐷(2,2)N𝐸(3,2)N𝐹(3,1)N𝐺(4,1)(
Mj(élhiABjk«q B ‘ĸİàąnq(–­lhiABjk«q C OĹ(
∴3 ≤𝑘≤6(
∴k ħ 3(4(5(6(
∴k >ķBą 4 W(
^O
4 
14 #3 *-Mj(‰Š‹y = −2x2 +8x−6x•noqA(B(ĥ‰Š‹ˆx•Ġ{1Ã>đ*ēÌC1(µC1
“°™¶ŸC2(C2x•noqB(D(r–‹y = x + mC1(C2$ą3W]ĺ>nq(um>ħĨĩ
=        
 
                                                                           
                                                                       
−3 < m < −
15
8
*ĻĤž¼q A zq B >st(ļºž¼ C2 ;(*€ž¼–‹ yĽxľm ‰Š‹ C2 mĿõ m
>ĆĠ–‹ yĽxľm «q B õ m >(çajLZ[Ÿ­.
\Å yĽŀ2x2ľ8xŀ6Ľ0(Z x2ŀ4xľ3Ľ0(Ÿ xĽ1 ā xĽ3(uq A#1,0-(B(3,0)o C1 “°
™¶pW†¹·¸Ÿ C2,u C2 ;O yĽŀ2#xŀ4-2ľ2#3≤x≤5-(é yĽxľm1  C2 mĿõ(Å yĽxľ
m1ĽyĽŀ2#xŀ4-2ľ2(Z 2x2ŀ15xľ30ľm1Ľ0(△Ľŀ8m1ŀ15Ľ0(Ÿ m1Ľŀ
15
8 (é yĽxľm2 «q B
õ(Z 0Ľ3ľm2(m2Ľŀ3(éŀ3ŁmŁŀ
15
8 õ–‹ yĽxľm  C1NC2 $ą 3 W]ĺnq(^=ŀ3Ł
mŁŀ
15
8 .
qłCDEF‰Š‹ x •nqĆĠ?@ABjkÏÐnŃ>ń(C>S=g¾Ņ
¼jL(¿ÀBLçaÖņ(}ąH>Ň¹.
15 #3 *-#24-25 „/08·ÈňʼnŊ·67-Mj(ˆ™ú–ùst<(AB𝑦= 𝑚𝑥(𝑚< 0)lhi
AB𝑦=
𝑘
𝑥(𝑘≠0)no𝐴N𝐵pq(q𝐶ˆ𝑥•1(”𝐴𝐶= 𝐴𝑂(r𝑆△𝐴𝐵𝐶= 4(u𝑘=      
 
−2
*CFlhiAB@ABnqst(lhiABB𝑘>ÏÐbI(÷ŋ¤ùL>xy(
>S=JKlhiABB𝑘>ÏÐbI «q𝐴Ì𝐴𝐷⊥𝐶𝑂oq𝐷(TUghiABzlhiABno
𝐴N𝐵pq(Ÿ¼𝐴,𝐵pq>stoq‚ƒ(u[Ÿ­𝑆△𝐴𝑂𝐶= 𝑆△𝐵𝑂𝐶=
1
2𝑆△𝐴𝐵𝐶= 2(÷ŋ¤ùL>x
y[Ÿ𝑆△𝐴𝑂𝐷= 𝑆△𝐴𝐶𝐷=
1
2𝑆△𝐴𝑂𝐶= 1(ÛTUlhiABhiB>ÏÐbI[Ÿ 
                                                                           
                                                                       
\
Mj¦§(«q𝐴Ì𝐴𝐷⊥𝐶𝑂oq𝐷(
∵ghiABzlhiABno𝐴N𝐵pq(
∴𝐴,𝐵pq>stoq‚ƒ(Z𝑦𝐴= −𝑦𝐵(
∵𝑆△𝐴𝑂𝐶=
1
2𝑂𝐶⋅𝑦𝐴(𝑆△𝐵𝑂𝐶=
1
2𝑂𝐶⋅|𝑦𝐵|(𝑆△𝐴𝐵𝐶= 4(
∴𝑆△𝐴𝑂𝐶= 𝑆△𝐵𝑂𝐶=
1
2𝑆△𝐴𝐵𝐶= 2(
∵𝐴𝐶= 𝐴𝑂(
∴𝐶𝐷= 𝐶𝑂(
∴𝑆△𝐴𝑂𝐷= 𝑆△𝐴𝐶𝐷= 1
2 𝑆△𝐴𝑂𝐶= 1
∴|𝑘| = 2𝑆△𝐴𝑂𝐷= 2
∵lhiAB>jk*ðˆ?N±k¬(
∴𝑘= −2
^O
−2 
16 #3 *-#24-25 ./01·ŌōÈŎ·6<-Mj(‰Š‹𝑦= 𝑎𝑥2 +𝑏𝑥+ 𝑐>‚ƒ•O–‹𝑥= −1(”
«q
1
2 ,0 (ą89çŏ①𝑎𝑏𝑐> 0c②𝑎−2𝑏+ 4𝑐> 0c③25𝑎−10𝑏+ 4𝑐= 0c④3𝑏+ 2𝑐> 0c{<¦ą
g¾>çŏ=        
 
①②③
*CF?@AB>jkB>÷ń(>S=ŐőjkŒ(œŔŕÀ¦ń
                                                                           
                                                                       
|Ŗ(ŗo<PŘ 
TU?@ABjřĠ{xy‚Ś	Ē@XYZ[ 
\jř[𝑎< 0(𝑏< 0(𝑐> 0(
∴𝑎𝑏𝑐> 0(^①g¾ 
é x=
1
2õ(y=0(
Z
1
4𝑎+
1
2𝑏+ 𝑐= 0
∴𝑎+ 2𝑏+ 4𝑐= 0
∴𝑎+ 4𝑐= −2𝑏
∴𝑎−2𝑏+ 4𝑐= −4𝑏> 0(^②g¾ 
‚ƒ•O𝑥= −1( x •WnqO(
1
2,0)[ x •ĬWnqO(−
5
2,0)
Z
25
4 𝑎−
5
2𝑏+ 𝑐= 0
ÊśŸ25𝑎−10𝑏+ 4𝑐= 0(^③g¾ 
∵‚ƒ•O𝑥= −1
∴−
𝑏
2𝑎= −1
∴𝑏= 2𝑎(𝑎=
1
2𝑏
µ𝑎=
1
2𝑏ÿĀ𝑎+ 2𝑏+ 4𝑐= 0ą
1
2 𝑏+ 2𝑏+ 4𝑐= 0
Z𝑏= −
8
5𝑐
∴3𝑏+ 2𝑐= 3(−
8
5𝑐) + 2𝑐= −
14
5 𝑐< 0(^④Ŝŝ 
©1¦Ş①②③g¾ 
^O①②③ 
Ⅱ
¤ #$ 8 '()* 72 *-
17 #6 *-~AB yĽx2ľ(mŀ3)xľ1ŀ2m#m OPB- 
#1-žØ
]ŏ m OН(ĶAB>jř x •šąpW$q 
#2-]ŏ m OН(ĶAB>jřĊŢª«WHq(žHq>st 
                                                                           
                                                                       
#1-ţc#2-(2(ŀ1)
*#1-Å yĽ0 Ÿ­o x >ä?@ÃÄ x2ľ(mŀ3)xľ1ŀ2mĽ0(ļºÀT>X€;Z[ 
#2-*—¼ m(Å m >BO 0(Ĥž¼ x(۞¼ y(Z[¾H m >Ť>Hq.
\#1-ØÙ
Å yĽ0(u x2ľ(mŀ3)xľ1ŀ2mĽ0 
ťO aĽ1(bĽmŀ3(cĽ1ŀ2m(
¦Ć b2ŀ4acĽ(mŀ3)2ŀ4(1ŀ2m)Ľm2ľ2mľ5Ľ(mľ1)2ľ4Ŧ0 
¦ĆÃÄąpW]m÷>ãBT 
¦Ć]ŏ m OН(ĶAB>jř x •šąpW$q 
#2-
yĽx2ľ(mŀ3)xľ1ŀ2mĽ(xŀ2)mľx2ŀ3xľ1 
ťOĶAB>jřĊŢª«WHq(
¦Ć xŀ2Ľ0(Ÿ xĽ2 
é xĽ2 õ(yĽŀ1 
¦ĆĶABjřĸŧ«Hq#2(ŀ1- 
qłCDEFGä?@ÃÄÃÄ?@AB>N?@ABjř x •>nqŖ÷ńq(
JK?@AB y=ax2+bx+c#a(b(c =PB(a≠0-jř x •>nqä?@ÃÄ ax2+bx+c=0 Tå»>
=C>S 
18 #6 *-#24-25 ./01·Ăă··ä´-‰Š‹𝑦= 𝑚𝑥2−4𝑚𝑥+ 3𝑥•>nqO𝐴(1,0)(𝐵(𝑦•
noq𝐶 
(1)ž‰Š‹>;c
(2)𝑃O‰Š‹k¬1>q(r∠𝑃𝐴𝐶= ∠𝑂𝐴𝐶(žq𝑃>stc
(1)𝑦= 𝑥2−4𝑥+ 3
(2)𝑃
15
4 ,
33
16
*CF>=?@AB©aŕÀ(ŨĠ­–ù¤ùLNũHBŪžABĕė;(À–ù¤
ùL>ÃŪž¼q H >st=>S 
                                                                           
                                                                       
#1-ũHBŪZ[žc
#2-ˆ𝐴𝑃1ħq𝑀(ūŸ𝑂𝐴= 𝐴𝑀= 1(ÍÎ𝐶𝑀(«q𝑀Ì𝐻𝑁⊥𝑥•oq𝑁(«q𝐶Ì𝐶𝐻⊥𝐻𝑁oq𝐻(
u±ÚL𝑂𝐶𝐻𝑁=ÑL(Ÿ𝐻𝐶= 𝐴𝑁(𝐻𝑁= 𝑂𝐶= 3(∠𝐻= ∠𝐴𝑁𝑀= ∠𝐶𝑂𝐴= 90°(Ö×ØÙ
△𝑂𝐴𝐶≌△𝑀𝐴𝐶(SAS)，得𝐶𝑀= 𝑂𝐶= 3(∠𝐴𝑀𝐶= ∠𝐴𝑂𝐶= 90°(ĘØÙ △𝐴𝑀𝑁∽△𝑀𝐶𝑃(Ÿ
3−𝑀𝑁
𝐴𝑁
=
1+𝐴𝑁
𝑀𝑁
=
3
1(èמŸ𝑀
9
5 ,
3
5 (¿ÀũHBŪžŸ–‹𝐴𝑀O𝑦=
3
4𝑥−
3
4(øÂ𝑦=
3
4𝑥−
3
4𝑦= 𝑥2−4𝑥+ 3žZ
[Ÿ 
\#1-
ĥ𝐴(1,0)ÿĀ𝑦= 𝑚𝑥2−4𝑚𝑥+ 3Ÿ
0 = 𝑚−4𝑚+ 3(
Ÿ𝑚= 1(
∴𝑦= 𝑥2−4𝑥+ 3c
#2-
ˆ𝐴𝑃1ħq𝑀(ūŸ𝑂𝐴= 𝐴𝑀= 1(ÍÎ𝐶𝑀(«q𝑀Ì𝐻𝑁⊥𝑥•oq𝑁(«q𝐶Ì𝐶𝐻⊥𝐻𝑁o
q𝐻(
é𝑥= 0õ(𝑦= 02−4 × 0 + 3 = 3(
∴𝐶(0,3)(
∴𝑂𝐶= 3
∵𝑂𝐴⊥𝑂𝐶(𝐻𝑁⊥𝑥•(𝐶𝐻⊥𝐻𝑁(
∴±ÚL𝑂𝐶𝐻𝑁=ÑL(∠𝐻𝐶𝑀+ ∠𝐻𝑀𝐶= 90°(
∴𝐻𝐶= 𝑂𝑁(𝐻𝑁= 𝑂𝐶= 3(∠𝐻= ∠𝐴𝑁𝑀= ∠𝐶𝑂𝐴= 90°(
∵𝐴𝐶= 𝐴𝐶(∠𝑂𝐴𝐶= ∠𝑀𝐴𝐶(𝑂𝐴= 𝐴𝑀= 1(
∴△𝑂𝐴𝐶≌△𝑀𝐴𝐶(SAS)，
∴𝐶𝑀= 𝑂𝐶= 3(∠𝐴𝑀𝐶= ∠𝐴𝑂𝐶= 90°(
∴∠𝐴𝑀𝑁+ ∠𝐶𝑀𝐻= 90°(
∴∠𝐴𝑀𝑁= ∠𝑀𝐶𝐻(
∴△𝐴𝑀𝑁∽△𝑀𝐶𝐻(
                                                                           
                                                                       
∴
𝐻𝑀
𝐴𝑁=
𝐻𝐶
𝑀𝑁=
𝐶𝑀
𝐴𝑀=
3
1(
∴
3−𝑀𝑁
𝐴𝑁
=
1+𝐴𝑁
𝑀𝑁=
3
1(
∴𝐴𝑁=
4
5(𝑀𝑁=
3
5(
∴𝑂𝑁= 𝑂𝐴+ 𝐴𝑁=
9
5(
∴𝑀
9
5 ,
3
5 (
Ֆ‹𝐴𝑀O𝑦= 𝑎𝑥+ 𝑏(
ĥ𝑀
9
5 ,
3
5 (𝐴(1,0)ÿĀ𝑦= 𝑎𝑥+ 𝑏Ÿ(
∴
0 = 𝑎+ 𝑏
3
5 = 9
5 𝑎+ 𝑏
 
(
Ÿ
𝑎= 3
4
𝑏= −3
4
 
(
–‹𝐴𝑀O𝑦=
3
4𝑥−
3
4(
øÂ𝑦=
3
4𝑥−
3
4𝑦= 𝑥2−4𝑥+ 3Ÿ
𝑦= 𝑥2−4𝑥+ 3
𝑦= 3
4 𝑥−3
4
 
(
Ÿ
𝑥= 1
𝑦= 0
 
ā
𝑥= 15
4
𝑦= 33
16
 
(
∴𝑃
15
4 ,
33
16 
qłCDEFGũHBŪž?@AB(mŬ¤ùL>XHĠxy(ÑL>XHĠxy(Ă÷¤
ùL>XHĠxy(ÒÓJKmŬ¤ùL>XHĠxy(ÑL>XHĠxy(Ă÷¤ùL>XHĠxy=
>S 
19 #8 *-#24-25 „/08·ŭÈŮů·67-ÕAB𝑦1 =
𝑘
𝑥(𝑦2 =
𝑘+1
𝑥(𝑘> 0)(é1 ≤𝑥≤3õ(AB𝑦1>
ĵ'= a(AB𝑦2>ĵŒ=6𝑎 
                                                                           
                                                                       
(1)ž k > 
(2)rq𝑃(𝑚,𝑛)ˆAB𝑦1 =
𝑘
𝑥>jk1(”q P ­ y •>š—Œo 3(ž n >ħĨĩ 
(3)@AB𝑦=
3
11𝑥+
8
11AB𝑦1 =
𝑘
𝑥>jkˆk¬>nqOq A(” x •noq B(q C ˆAB𝑦1 =
𝑘
𝑥¸ok¬>jk1(r𝑆△𝐴𝑂𝐶= 𝑆△𝐴𝑂𝐵(–ÎŰ¼q C >æst 
(1)𝑘= 1
(2)−
1
3 < 𝑛< 0z0 < 𝑛<
1
3
(3)3 ā
1
3
*#1-TUˆ,k¬ñ(𝑦1ò x >óŒ×ô'(𝑦2ò x >óŒ×ô'žZ[c
#2-TUb[Ÿ𝑚< −3ā𝑚> 3(ÿĀlhiAB;[Ÿ n >ħĨĩc
#3-*píîïűŏ①éq C ˆ A q>°v(②éq C ˆ A q>öv(TUúû9¼mĖ>ÃĞ
¼ m Z[ 
\#1-
∵𝑘> 0,𝑘+ 1 > 0(1 ≤𝑥≤3(
∴ˆ,k¬ñ(𝑦1ò x >óŒ×ô'(𝑦2ò x >óŒ×ô'(
∴é𝑥= 3õ(𝑦1ĵ'O
1
3 = 𝑎①(
é𝑥= 1õ(𝑦2ĵŒO𝑘+ 1 = 6𝑎②(
①(②Ÿ
𝑘= 1 
#2-∵𝑃(𝑚,𝑛)­ y •>š—Œo 3(
∴𝑚< −3ā𝑚> 3(
∵𝑦1 =
1
𝑥(
∴−
1
3 < 𝑛< 0ā0 < 𝑛<
1
3c
#3-
𝑦= 3
11 𝑥+ 8
11
𝑦= 1
𝑥
 
(Ÿ
𝑥1 = 1
𝑦1 = 1
 
(
𝑥2 = −11
3
𝑦2 = −3
11
 
(
∴𝐴(1,1) 
0 =
3
11𝑥+
8
11(Ÿ𝑥= −
8
3(
∴𝐵−
8
3 ,0 (
                                                                           
                                                                       
∴𝑆△𝐴𝑂𝐵=
4
3
∴𝑆△𝐴𝑂𝐶= 𝑆△𝐴𝑂𝐵=
4
3
①éq C ˆ A q>°v
Õ𝐶𝑚,
1
𝑚(« A(C *€o x •Ìȋnoq MNN(
∵𝑆△𝐴𝑂𝑀= 𝑆△𝐶𝑂𝑀(
∴𝑆△𝐴𝑂𝐶= 𝑆直角梯形𝐴𝐶𝑁𝑀=
4
3(
∴
1
2 1 +
1
𝑚(𝑚−1) =
4
3(
∴𝑚1 = 3(𝑚2 = −
1
3#Ųų-(
②éq C ˆ A q>öv(
Õ𝐶𝑚,
1
𝑚(« A(C *€o x •Ìȋnoq MNN(
∵𝑆△𝐴𝑂𝐶= 𝑆直角梯形𝐴𝐶𝑁𝑀=
4
3(
∴
1
2 1 +
1
𝑚(1−𝑚) =
4
3(
∴𝑚1 = −3#Ų-(𝑚2 =
1
3(
¦Ćq C >æstO 3 ā
1
3 
qłCFGlhiAB>jkxy(@ABst•>nq(lhiAB@AB>nq
Ŗ(lhiAB k ŸÏÐbI(ÒÓJKĶńq=S 
20 #8 *-#24-25 ./01·ĝōĞŴ·67-~lhiAB𝑦=
𝑘
𝑥(𝑘≠0)(q(2,𝑚)((1,3𝑚−2)ĊˆĶ
                                                                           
                                                                       
lhiABjk1 
(1)žlhiAB>;c
(2)rq𝐴(𝑥1,𝑦1)(𝐵(𝑥2,𝑦2)ĊˆĶlhiABjk1c
①é𝑦2 = 𝑦1 +8(”q𝐴zq𝐵oqR<ŵ‚ƒ(žq𝐵>stc
②é𝑥1 = 2(𝑦1 + 𝑦2 < 0õ(ž𝑥2>ħĨĩ 
(1)𝑦=
4
𝑥
(2)①𝐵(1,4)c②−2 < 𝑥2 < 0
*#1-TUlhiABjkxy(¿ÀũHBŪ9ÃĞZ[Ÿ­c
#2-①¿ÀlhiABjkxy(çabž¼𝐵(𝑥2,4)(¿ÀũHBŪ9ÃĞZ[Ÿ­c
②¿ÀlhiABjkxy(¿ÀũHBŪž¼𝐴(𝑥1,𝑦1)(9]÷;žZ[Ÿ­ 
\#1-
∵lhiAB𝑦=
𝑘
𝑥(𝑘≠0)(q(2,𝑚)((1,3𝑚−2)ĊˆĶlhiABjk1(
∴𝑘= 2𝑚= 3𝑚−2(Ÿ𝑚= 2(
∴𝑘= 2𝑚= 4c
∴lhiAB>;O
𝑦=
4
𝑥c
#2-
q𝐴(𝑥1,𝑦1)(𝐵(𝑥2,𝑦2)ĊˆĶlhiABjk1(q𝐴zq𝐵oq<ŵ‚ƒ(
∴𝑦1 + 𝑦2 = 0(
∵𝑦2 = 𝑦1 +8(u𝑦1 + (𝑦1 + 8) = 0(Ÿ𝑦1 = −4(
∴𝑦2 = 4(
µ𝐵(𝑥2,4)ÿĀ𝑦=
4
𝑥ŸŸ𝑥2 = 1(
∴𝐵(1,4)c
②∵𝑥1 = 2(u𝑦1 =
4
2 = 2(
∵𝑦1 + 𝑦2 < 0(
∴𝑦2 < −2(q𝐵ˆ¤k¬(𝑥2 < 0
∴
4
𝑥2 < −2(
∴−2 < 𝑥2 < 0 
qłCFlhiABjkxy(ŨĠũHBŪ¾H kNq>‚ƒxyN]÷;÷ń(ÒÓ
JKlhiABjkxy=|Ŗ>S 
                                                                           
                                                                       
21 #10 *-#23-24 ./01·deÔ ·6<-ąŶ‰Š‹ŘŷŸ(ˆgP˜¸õ˜úĴ𝐴𝐵= 20m(é
˜¸1Ź3mõ(˜úĴ𝐶𝐷= 10m źMj¦§Á™ú–ùst 
(1)ž}‰Š‹>ABĕė;c
(2)ąŻżĆ6km/h>Ž¹“}Ÿž–ſƀ(鿚—}Ÿ36kmõ(Ÿ8˜¸gƁˆ𝐴𝐵Ƃ(庘¸,'
õ1ƃ0.3m(OƄØ Ă(阸ė­šŷŸĵijq2mõ(µƅĹżƆƇņ MƈĶż>Ž¹]Ɖ(ƊƋ
ć¨ƌ ĂƇ«}Ÿƍ
(1)𝑦= −
1
25𝑥2 +
4
5𝑥
(2)MƈĶż>Ž¹]Ɖ(ƊƋ樠ĂƇ«}Ÿ
*CDEFG?@AB>ãƎĖÀ(g¾Əb=>S 
#1-TUb[Ÿ𝐵20(0 (𝐶5(3 (ļº¿ÀũHBŪžZ[c
#2-Ĥž¼ż­ėŸ8˜ú>ij¹(۞¼‰Š‹þqst(Öן­ż­ėŸ8õ˜úš—ĵijq>ij
¹(}Z[Ÿ­ 
\#1-
bŸ(𝐵20(0 (𝐶5(3 (
ՉŠ‹;O𝑦= 𝑎𝑥(𝑥−20)(
∴5𝑎(5−20) = 3(
∴𝑎= −
1
25(
∴‰Š‹;O𝑦= −
1
25𝑥(𝑥−20) = −
1
25𝑥2 +
4
5𝑥c
#2-
żņſ­Ÿ8>õ»O
36 ÷ 6 = 6'õ(
˜¸1Ź>ij¹O
0.3 × 6 = 1.8m 
∵‰Š‹;O𝑦= −
1
25𝑥2 +
4
5𝑥= −
1
25(𝑥−10)2 +4(
∴‰Š‹þqstO 10(4 (
∴éż­ėŸ8õ(}õ˜úš—ŷŸĵijq>š—O4−1.8 = 2.2m > 2m(
∴MƈĶż>Ž¹]Ɖ(ƊƋ樠ĂƇ«}Ÿ 
                                                                           
                                                                       
22 #10 *-#2025·Ɛ3Ƒƒ·¥-Mj(lhiAB𝑦=
𝑘
𝑥>jkª«q𝐴(2,𝑚)(«q A Ì𝐴𝐵Ȗ y •
oq B( △𝐴𝑂𝐵>úûO 5  
(1)ž k z m >c
(2)~q𝐶(−5,𝑛)ˆlhiABjk1(–‹𝐴𝐶n x •oq M(ž △𝐴𝑂𝑀>úûc
(3)«q C Ì𝐶𝐷⊥𝑥•oq D(Íç𝐵𝐷(ØÙ
±ÚL𝐴𝑀𝐷𝐵=™ņ±ÚL 
(1)𝑘= 10(𝑚= 5
(2)7.5
(3)ţ
*}FGlhi©a(ŨĠ>ńą
stjLxy(™ņ±ÚL>XHxy(ũH
BŪ¾HAB;(ĆĠ¤ùL>úûžŪ(œŔŕÀũHBŪ=C>S 
#1- △𝐴𝑂𝐵>úûž¼ m >( m >¾H¼ A >st(µ A >stÿĀlhi;<(Z[ž
¼ k >c
#2-Ĥž¼𝐶(−5,−2)(ÛTUũHBŪž¼–‹𝐴𝐶>;O𝑦= 𝑥+ 3(Ö×¾H𝑀(−3,0)(Z[žc
#3-ĭ¼𝐷𝑀∥𝐴𝐵(𝐷𝑀= 𝐴𝐵(Z[ØÙ 
\#1-
∵𝑆△𝐴𝑂𝐵=
1
2 × 2 ⋅𝑚= 5(
∴𝑚= 5(
∴𝑘= 2 × 5 = 10 
#2-
#1-ŸlhiAB;O𝑦=
10
𝑥,
q𝐶(−5,𝑛)ÿĀ𝑦=
10
𝑥Ÿ𝑛=
10
−5
Ÿ𝑛= −2(
∴𝐶(−5,−2),
Ֆ‹𝐴𝐶>;O𝑦= 𝑎𝑥+ 𝑏(
µ𝐴(2,5)(𝐶(−5,−2)ÿĀ𝑦= 𝑎𝑥+ 𝑏Ÿ
2𝑎+ 𝑏= 5
−5𝑎+ 𝑏= −2
 
(
                                                                           
                                                                       
Ÿ
𝑎= 1
𝑏= 3
 
(
∴𝑦= 𝑥+ 3(
Å𝑦= 0Ÿ𝑥= −3(
∴𝑀(−3,0)(
∴𝑂𝑀= 3,
∴𝑆△𝐴𝑂𝑀=
1
2 × 3 × 5 = 7.5 
#3-ØÙ
∵𝐴𝐵⊥𝑦•(
∴𝐷𝑀∥𝐴𝐵(
∵𝐶(−5,−2),
∴𝑂𝑀= 5
∴𝐷𝑀= 5−3 = 2(𝐴𝐵= 2(
∴𝐷𝑀= 𝐴𝐵(
∴±ÚL𝐴𝐵𝐷𝑀=™ņ±ÚL 
23 #12 *-Mj(‰Š‹ª«q𝐴(−3,0)N𝐵(1,0)N𝐶(0,3) 
#1-ž‰Š‹>;c
#2-q𝑃(𝑚,𝑛)=‰Š‹1>Ɠq(é−3 < 𝑚< 0õ(¾H m >(ūŸ △𝑃𝐴𝐶>úûĵŒc
#3-‰Š‹1=ƌƔˆ]ĺoq B >q D()ƕ𝐷𝐴2−𝐷𝐶2 = 6(rƔˆ(Ɩž¼q D >stcr]Ɣˆ(
ƖƗÙƏ 
#1-𝑦= −𝑥2−2𝑥+ 3c#2-𝑚= −
3
2c#3-𝐷(−2,3)
*#1-Ub[ՉŠ‹>;O𝑦= 𝑎(𝑥+ 3)(𝑥−1)(µqÿĀ𝐶(0,3)¼ a(Z[ž¼‰Š‹
>;c
#2-Ĥž¼–‹ AC >;(ļºTUé−3 < 𝑚< 0õ(q𝑃(𝑚,𝑛)ˆ–‹𝐴𝐶1Ã(«q P Ì x •>ȋ
                                                                           
                                                                       
‹á𝐴𝐶mnoq Q([µ𝑥= 𝑚*€ÿĀ𝑦= −𝑥2−2𝑥+ 3z𝑦= 𝑥+ 3Ÿ𝑃(𝑚,−𝑚2−2𝑚+ 3)(
𝑄(𝑚,𝑚+ 3)(èן¼ PQ >ÿB;(è×[ž¼ m >c
#3-b[Ÿ𝐴𝐵= 4,𝑂𝐵= 1,𝐶𝑂= 3(TU𝐵𝐶2 = 10(∠𝐶𝐴𝑂= 45°([ž¼𝐵𝐴2−𝐵𝐶2 = 6(ÍÎ𝐵𝐶(
« B Ì𝐴𝐶>ȋn‰Š‹oq D(n𝐴𝐶oq H([Ÿ𝐷𝐴2−𝐷𝐶2 = 𝐻𝐴2−𝐻𝐶2 = 𝐵𝐴2−𝐵𝐶2 = 6(TU
∠𝐶𝐴𝑂= ∠𝐷𝐵𝐴([Ÿ𝐵𝐷𝐴𝐶o𝐴𝐵>Ȗ™*‹‚ƒ(Zo‰Š‹>‚ƒ•𝑥= −1‚ƒ(Zq D q
C o‰Š‹>‚ƒ•𝑥= −1‚ƒ(è×[ž¼q D >st 
\
#1-Ub[ՉŠ‹>;O𝑦= 𝑎(𝑥+ 3)(𝑥−1)(
µq𝐶(0,3)ÿĀ([Ÿ𝑎= −1
∴‰Š‹>;O𝑦= −𝑥2−2𝑥+ 3c
#2-Ֆ‹ AC >;O
𝑦= 𝑘𝑥+ 𝑏(
µ𝐴(−3,0)N𝐶(0,3)ÿĀŸ{
0 = −3𝑘+ 𝑏
3 = 𝑏
(
Ÿ{
𝑘= 1
𝑏= 3(
∴–‹𝐴𝐶>;
𝑦= 𝑥+ 3(
é−3 < 𝑚< 0õ(q𝑃(𝑚,𝑛)ˆ–‹𝐴𝐶1Ã(
«q P Ì x •>ȋ‹á𝐴𝐶mnoq Q(
µ𝑥= 𝑚*€ÿĀ𝑦= −𝑥2−2𝑥+ 3z𝑦= 𝑥+ 3Ÿ𝑃(𝑚,−𝑚2−2𝑚+ 3)(𝑄(𝑚,𝑚+ 3)(
∴𝑃𝑄= −𝑚2−2𝑚+ 3−(𝑚+ 3)
= −𝑚2−3𝑚
= −(𝑚+ 3
2 )
2
+ 9
4
∵−3 < 𝑚< 0(
∴é”Ƙé𝑚= −
3
2õ(𝑃𝑄ħŸĵŒ(
                                                                           
                                                                       
}õ𝑆△𝑃𝐴𝐶=
1
2𝑃𝑄× 𝐴𝑂=
3
2𝑃𝑄ĵŒ(
∴𝑚= −
3
2c
#3-𝐴(−3,0)N𝐵(1,0)N𝐶(0,3)Ÿ𝐴𝐵= 4,𝑂𝐵= 1,𝐶𝑂= 3(
∵𝐵𝐶2 = 10(∠𝐶𝐴𝑂= 45°(
∴𝐵𝐴2−𝐵𝐶2 = 6(
ÍÎ𝐵𝐶(« B Ì𝐴𝐶>ȋn‰Š‹oq D(n𝐴𝐶oq H(
u∠𝐴𝐻𝐵= 90°,∠𝐷𝐵𝐴= ∠𝐶𝐴𝑂= 45°(
𝐷𝐴2−𝐷𝐶2 = 𝐻𝐴2−𝐻𝐶2 = 𝐵𝐴2−𝐵𝐶2 = 6(
∵∠𝐶𝐴𝑂= ∠𝐷𝐵𝐴(
∴𝐵𝐷𝐴𝐶o𝐴𝐵>Ȗ™*‹‚ƒ(Zo‰Š‹>‚ƒ•𝑥= −1‚ƒ(
∴q D q C o‰Š‹>‚ƒ•𝑥= −1‚ƒ(
Ę∵𝐶(0,3)(
∴q D >stO#-2(3- 
qłC=?@AB>©a(F?@AB>xy(ž@AB;(çab(g¾ƙƚƛƜ
‹(œŔŕÀńq=S 
24 #12 *-#24-25 „/08·ĝeĞğ·67-©aãƝ
Ŗîƞ
Mj(Ɵ=Ơ>bơƢtB y òõ» x#·¸
*ƣ->ƉÊĢģ>jk({<𝐴𝐵(𝐵𝐶=
‹á(𝐶𝐷OƤƥ‹ˆk¬ñ>đ* 
                                                                           
                                                                       
Ŗ|
(1)ž‹á𝐴𝐵zƤƥ‹𝐶𝐷¦ĕ§>ABĕė;(½*€Ű¼ƦƉƧ x >ħĨĩ 
(2)ƨƩƪ(ġîï8(ˆƫ 45 *ƣ>Ƭ<(Ơ>bơòõ»>ƉÊ×ƉÊ(Ơ>bơƢt
B›Œ(bơƭ›Ʈ< Ƈ«üý‚h1Ƭº> 3 *ƣz 30 *ƣ(ƠbơƯWưƚƮ< 
(3)~ƱƲEƳWwEńqcOGūƠƴƬƵƈưƁ(EžƠ>bơƢtB]Ÿƶo 40(ƱƲ
ƷƸˆƠ>bơė­¦ƹıƺ8Ƴƻ(Ɩ–ÎŰ¼ƱƲƳƟWńqĵƁ ƼˆƽƋõ»á #ƾ
ƿOˆõ»áñ¨Ƴƻ-
(1)𝑦= 2𝑥+ 30(0 ≤𝑥≤10)c𝑦=
1000
𝑥(𝑥≥20)
(2)Ơ1Ƭº> 3 *ƣh1Ƭº> 30 *ƣbơưƚƮ<
(3)5 ≤𝑥≤25
*CDEFG@ABlhiAB>ãƎĖÀ(g¾Əb=>S 
#1-Õ¼‚Ė>;(¿ÀũHBŪžZ[c
#2-*€ž¼𝑥= 3z𝑥= 30õ>ABZ[Ÿ­c
#3-*€ž¼pWAB>AB÷o 40 õ x >çajkZ[Ÿ­ 
\#1-
Ջá𝐴𝐵>ABĕė;O𝑦= 𝑘𝑥+ 𝑏(
µ𝐴(0,30)(𝐵(10,50)ÿĀ(Ÿ
30 = 𝑏
50 = 10𝑘+ 𝑏
 
(
Ÿ
𝑘= 2
𝑏= 30
 
(
∴‹á𝐴𝐵>ABĕė;O𝑦= 2𝑥+ 30(0 ≤𝑥≤10) 
Õƥ‹𝐶𝐷>ABĕė;O𝑦=
𝑚
𝑥(µ𝐶(20,50)ÿĀ(Ÿ𝑚= 50 × 20 = 1000(
∴ƥ‹𝐶𝐷>ABĕė;O𝑦=
1000
𝑥(𝑥≥20) 
#2-ĥ𝑥= 3ÿĀ𝑦= 2𝑥+ 30(Ÿ𝑦= 2 × 3 + 30 = 36(
ĥ𝑥= 30ÿĀ𝑦=
1000
𝑥(Ÿ𝑦=
1000
30 =
100
3 
                                                                           
                                                                       
∵36 >
100
3 (
∴Ơ1Ƭº> 3 *ƣh1Ƭº> 30 *ƣbơưƚƮ< 
#3-
é2𝑥+ 30 = 40(Ÿ𝑥= 5(
é
1000
𝑥
= 40(Ÿ𝑥= 25(
çajk(EžƠ>bơƢtB]Ÿƶo 40(u x >ħĨĩ=5 ≤𝑥≤25(
∴ Ƽˆ 5 *ƣǀ 25 *ƣ