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代 0.1.2
rliegA021ifhtt
4433ii
iz
⼀
偶
ˇXD
c e pex i
turhplieegarcxxferfrhfootrwte rrfflliirreeggpp22xaaffffhhiicctttt
ˇ
rlieg35P2crfhtt
rleiQPgfht mtin
fr6955d111ac证
A的 细
rrslliiseeggq2baayxxrfiffnhhrttttt
偶 rrrllliiieeeggg0xfffffhhhtttttt sqb22art rsfliregq322baxaffhrcttt
max
vraflisregpwr31aipfnhicthti
Rigfhrpt22ykaaircrow
vaspr11ipni hi
⽽ vsafcrqr022baaoaprscthi
即 np005beaamq ⽭盾
tthiemrAeefosre
此 题题 ⼲ n0beaq 改为 n0b22eaq 的话
单选选Dfrb222ea1ac
ˇ
fr62e1ac
fr452e1ac
fr627e1ac fr827e1ac
ABD
cfrd32xaoct时取最⼩值
偶 0a ⽭盾
b⽭盾
奇rlieg00ffhtt
关于frp3xaic对称
关于5s对称
x
rliegnnaatiifhtt
恒成t
取 n1i
ther0nnaae1tfore
ˇ 等差数列
there必fore 要
前推后若0d 推不出不充分
对称
图像关于原点
rliegThneeeeaxxxxxfhtt 为奇函数 therefore
函数分离常数后易证单增
berrclliieeggTT2hhnnaaxxa11ffhhuttttse therefo在re原点
函数涨幅最⼤
得
亦可
求反函数
严谨
ffree tmmmmbb l h i err e eeg 05T aaaa 2222hneeeeeennaxxxxxxx11111111111 r aa e ggtttt f h frrrro dd ii cc iiii t nn rxxxx t e
2222eexx111
theDrefore
ll32ex1nn
02x111oovveerrrriiOggAahhtthhaarrppoooonn oovveerrrrOiiBbgghhtthhaarrppoooonn
soovveeirrrrgiiggcrhhttmhhaarrppooaoonn
downarrow
rrrrlllliiiieeeeggggOOOOCCCAABB22zffffhhhhtttttttt
ˇ oooooovvvvvveeeeeerrrrrrrrrrrriiiiiigggggghhhhhhtttttthhhhhhaaaaaarrrrrrppppppoooooooooooonnnnnn
borveelrirceiggAB2hatfhhuartpsotoen
⼀
therefore
C在射线AD上
成射线AE上
cBd1ot
oavrerlliriepggOC2htfhhhatrpatoon
2c
the9refore
CL为两个成⽴的点
A
c1
B2 cdDot
fr22yx1ac
ffdrr42211aaivcc
f41rac
rcliedgC65ax1fohttt
2
a11 fr322aac fr8332aac
cffrrd8833522aaaocct
theldr35onaaaetfsore an
是⾸项为1 公⽐为
f
98
rac
的
rrcflliireedgg99681affohhcttttt
军⽐悉
到
thlearrrrccffflllliiiirnrrreeeeddgggg699996888Sae11111aaagffffoohhhhfcccttttlottttttereAIT
边为2的正
补
成 三棱柱
ffrrMCCAVVBBS322111aacc 横截
c3dot
tttssiiiffmmmrrqq443322221aarreeeccttsss
A1
a
B1
naxt 佐域为R
IN
六
当Righma1tarrow
rliegxxx1ffhttlemmb 00ex eaangttqrrdiiinxx
1
如古显然不存在减区间
f
x
30
aconf 如不 a为
极⼤ 值点
i
rrrllliiieeeggg2xxfffffhhhtttttt
a
a
rliegx1ffhtt x11f
Rrigliehg0xtfafhrrtotw
当aeons 钞有6个 交点舍
由 y1 y1
rlieg05yxffhtt
舍
x0ai 时 由
有64交点
当 g0neasq rlieg32ea1ffhtt
有三交点
bercliegl0ayyyyxza11nfhuttse
32aa11 只需 le032c11q
le32a1q
sflreq033aaqrct
ˇ
therefore1
ffssrrCA65aaiinncc
sssCAA6552iiinnn
thteifffcccmrrrrCAAA935552222227oooose11aaaefssscccosre
b1 z 2c
2 cC2oas
zeb
thferr3b1aefocre
frA4nsac
tifmr6355552222bb7aecs tcciffffmsrrrrddA6455522bc111aaaainooecccctts
ther0sefore
ABC
35221
ldX55bbots rlieg45bb2111fhtt
40555bb21
Z
rrlliieegg055bb11ffhhtttt
ACIAAI
AAII⾯ D
ABCyAAI
LBC BCI⾯AAD
AsIBC
i AIDLBC
x
建系
选
Xˇ rliegA0021fhtt rrlliieeggCB0002221ffhhtttt rlieg002cfhtt
rliegD111fhtt
overrriligeghDA1111tfhhatrtpoon
法ovlearrrliingeghnyxztgfhhatrlpteoon
ˇ llrr oove ee vii r GC00000B r 22222 gg cyx ei z ghrth ff lhh airnp tt o tteon
oocvveerrrrdiiggCB0nhhtthhoaarrppootoonn
oocvveerrrrdiiggC40n1hhtthhoaarrppootoonn
otvhererrilgiregh0ne11tfhhftaortprooen
oovveerrrrriitarggcsssfhclihhrnedgqqqA633P22nosse1attigfohhhrrrtsctaatttlatterrppoooonn⼠
了
ftrav1aidoutcs
3
bk与数学期望逻辑相同结果寻河平均表
可取
a X 0.1.2.3 MN
rfliregCC000044333P2x11afhctt
rcfliredgCC00043322px111afohcttt
frMCC0422222x111ac
rrflliireegg660003332acxx1affhhctttt
分布列
mmmmmmbbbtriffffeeelieeemaaaaaarrrreg60083Pnnn222111111aaaagggttttttfhedddrrrrrrcccctiiiiiiiiinnntsxxxxxx
ttttiiiifffffmmmmrrrrrX90003333522E11111aaaaaeeeecccccssss4a 2c
ffrr6222yx111aacc ffrr2eca1aacc
M
2 由对称性 城 TS 过 轴上定点 S
设TS myxt i
F1
A B
448322yx
r
rrlliieeggST22yyxx11ffhhtttt
Dttiimmeleetass
rrrrrllllliiiiieeeeegggggmmmmmmm32424832426324803622122163248124824824816481222160yyyytttttttfffffhhhhhtttttttttt
mmbeeaammm64433n2222yyyy11tgttdrriiinxx
AS rrcfflliirreedggM0444yyyyxxx1111aaffohhccttttt
BT rrcfflliirreedggN044442222yyyxxxaaffohhccttttt
oobvveeerrrriicggMhhN35ooatthhuaarrppsooeoonn
rrrrflllliiiireeeeggggmmmm0044433355522222222222yyyyyyyyyyyyx11111111attttffffhhhhctttttttt
therefore
rarrrrrrcflllllliiiiiirneeeeeedggggggmmmm6000448888335522222222yyyyyy11111attttttgffffffohhhhhhcttttttlttttttte
rliegm443522ttfhtt
80y11t
m432
rrrrlllliiiieeeeggggm440332y111tttffffhhhhtttttttt
rrlliieeggm043322y11111ttffhhtttt
不恒为0 必成主
ther0e1tfore
1t 城 TS 必过 01a1
1 rliegexxxffhtt rrlliieeggexxx11fffhhtttt
令 rlieg0xffhtt x1 H11f
与 轴 的 切线 1
平⾏ y
rrfflliirreegg222geeaaxxxxxxxxx1111aafffhhcctttt
n.in
零点
900 为 ⼀ 个
rflireg2ecxxx11alfhctt
增
rflireg2heaxx1afhctt lreihgxfh单tt
rflireg02ha1afhctt
rflireg02hea111afhctt
rlieg00hxfhtt fr002eax1ac 00x
使 即
erxliegi0aaix1nsfhttts
减
rrrccinllliiieeeddgggi0hhxxxxfnfffoohhhttttyttttt 单
增
rlieg0hxfhtt
tiMm1e单s
rinliegi0xxfnfhttyt
therrfflliirrreegg00022hhxxxe111aaffhhfccttott re
只有0 个 零 点 即 rlieg00fhtt ⼀个 交点
therefore
rflireg2ea11attttfhctt
iii Rirrglliiheegggtfttaffhhrttrttow
rliegea11tttfhtt
rrlliieegggfttffhhtttt
ea1t
不成⽴
若 0t 则
thenr0eetfqore
fr2ea11att c
rliegea11tttfhtt
therrflliirreegg2e1111atttffhhfcttottre
fr2ea11ac
2221ttt
thecrdl332aentofotre点移
xx11ii 时 0yy1ii 右移1
mXX0p1ii yy11ii 或1 上移1或下移 1
相当于由1D 1按上述⾛法起深
不重不漏
是
A不
2
如左 从A到B⾛ 完 所有 点
步
共需15
⽽
BA
