第七節(jié)專題導(dǎo)函數(shù)與原函數(shù)的奇偶周期性講義-高二下學(xué)期數(shù)學(xué)人教A版選擇性（答案詳解）

0第七節(jié)專題:導(dǎo)函數(shù)與原函數(shù)的奇偶周期性【1】ABD解析:A選項(xiàng),gx關(guān)于點(diǎn)(1,0)對(duì)稱,故g令x=?1得gB選項(xiàng),fx是偶函數(shù),故f?x=f又函數(shù)fx與導(dǎo)函數(shù)f′x的定義域均為故f′x為奇函數(shù),C選項(xiàng),g2?x=?gxg故g′x關(guān)于直線x=1對(duì)稱,無法得到g′xD選項(xiàng),由C選項(xiàng)知,g′2?a=g′a【2】ABD解析:對(duì)A:令x=1,則令x=0,則f0+f2=2對(duì)B:因?yàn)閒x兩邊求導(dǎo),得gx?g2?因數(shù)gx?所以g?所以gx=g?4+x,故對(duì)C:因?yàn)間x故fx+4+c1=fx+c2?fx+4=fx+c,c未必為故選:ABD【3】ABD解析:因?yàn)閒′x為偶函數(shù),所以f′?x而f?1+f1=2,故又fx+1為偶函數(shù),所以f?x所以f2?x+f?x=2,故fx+f2+x=對(duì)A,由f?x+1=令x=0得?f′1=對(duì)C,由上知fx+f?x所以f1012=對(duì)D,因?yàn)閒x故f2?x+f2+x=2,故fx的圖象關(guān)于(2,1)對(duì)稱,因?yàn)?是【4】A解析:因?yàn)閒x+3為奇函數(shù),所以f?所以fx+2為偶函數(shù),故又對(duì)f?x+3=?即f′?x+3=f′x+由f?x+3=?由fx+2=f所以fx+6=?fx+4所以fx是周期為4的函數(shù),則fx+2?4=f2?x?由f?x+3=?即f4又由fx+6=?f所以f4?x=f4+x,即fx+4【5】C解析:因?yàn)閒x=f即f′因?yàn)閒x為定義在R上的奇函數(shù),則f所以兩邊求導(dǎo),得f′x=f′?x,所以所以f′2?x=所以f′x?2+所以f′x是周期為所以f′由①式,令x=1,得f′1=2【6】ABD解析:對(duì)于A,由fx+2為奇函數(shù),得f2?x+f2+x由g′x=f′x?2又fx?g4?x=2,于是fx即fx則fx+2=?fx,fx對(duì)于B,由gx=fx?2對(duì)于C,因?yàn)楹瘮?shù)fx是周期函數(shù),周期為顯然函數(shù)gx是周期為4數(shù),g1+g3=C錯(cuò)誤;對(duì)于D,f1+f3=0,f【7】ACD解析:由題意可得f1?g′所以g1+x=g1?x+所以g1所以gx的圖象關(guān)于x=1因?yàn)間x+2為奇函數(shù),所以所以gx+g4?x=0結(jié)合g′1+x所以g′4?x=?g′2?所以g′t=g′t?4因?yàn)閒x=6?g′x即fx+4=fx,兩邊求導(dǎo)可得f′xfx是周期為4的周期函數(shù),所以f因?yàn)間′1+x=?g′1?x,令又g′t=?g′t?2,所以g則g′3=g′?1=0所以f1+f3=12,所以【8】BC解析:對(duì)于選項(xiàng)A,因?yàn)閒′1?2x即f′1+x=f′1?x若fx+1為偶函數(shù),則f?x所以f′x關(guān)于點(diǎn)(1,0)對(duì)稱,這與f′x關(guān)于A錯(cuò)誤;對(duì)于選項(xiàng)B,因?yàn)閒x?x為偶函數(shù),所以所以當(dāng)x≠0時(shí),fxx?1=?f對(duì)于選項(xiàng)C,因?yàn)閒x?x為偶函數(shù),即所以?f′?x+由f′1+x=所以f′x+2+f對(duì)于選項(xiàng)D,由f′?x+f′x=2,得f′0【9】BD解析:對(duì)于A,∵f即f′x關(guān)于(1,0)對(duì)稱,故對(duì)于B,f′x+2為偶函數(shù),故f′x+2=由f′x關(guān)于x=2對(duì)稱,知f′3對(duì)于C,因?yàn)閒′1+x+因?yàn)閒′x+2=則f′x+4=?f′x+所以f′x的周期為4,則f′2025=f對(duì)于D,由f′x+2=即fx+2+f?x+2故f2+x+f2?x=2f2【10】ACD解析:對(duì)于A,由fx+2為偶函數(shù),得f?x+2=fx而fx為實(shí)數(shù)集上的奇函數(shù),因此f4對(duì)于B,由fx為奇函數(shù),得f兩邊求導(dǎo)得:?f′?x=?f′x,即又f?x+2=函數(shù)f′x圖象關(guān)于(2,0)對(duì)稱,得不到B選項(xiàng)說法,B對(duì)于C,由f′x關(guān)于(2,0)對(duì)稱,得f對(duì)于D,由?f′?x+2=f因此f′x是以f而f′2=f′綜上f故i=12025f′【11】ABD解析:由f2x+2=f又函數(shù)fxx∈R因此可得fx+2=?fx,由f2x+2=f∴f2022=f由函數(shù)fxx∈R故f′又gx是fx的導(dǎo)函數(shù),則gx=g?x由fx+2=f?x,則因此y=fx在x=1左右附近的單調(diào)性發(fā)生改變,即x=故g1=f′1=由fx+4=fx,則因此函數(shù)gx的周期為4,因此D正確.故選:【12】ABD解析:由f′x+gx=2,得f′?x+g?x=2,則即f′?x=?f′x,則對(duì)于B,在g?x+gx=4中,令在?f′4?x+gx?1f′3對(duì)于C,在?f′4?x+gx?1?與f′x+gx=2求和,可得g1+對(duì)于D,由g?x+gx=4以及從而g3+x=gx,則數(shù),g3i=1故選:ABD【13】BCD解析:對(duì)A,因?yàn)閒2所以?1所以函數(shù)y=12x對(duì)B,因?yàn)閥=f3x+1f所以f′1+x=?f′1?xg所以g1=0,故對(duì)C,因?yàn)閒2+x?f即g2+x+g2?x=所以g2+x=gxg4所以函數(shù)gx的圖象關(guān)于點(diǎn)(3,1)對(duì)稱,故C對(duì)D,因?yàn)間2+x+g2?x=所以g2=12,又g2g4=g2+1=3確.故選:BCD.【14】3解析:因?yàn)閒x+1為奇函數(shù),所以則fx的圖象關(guān)于(1,0)中心對(duì)稱,則f因?yàn)閒′x+2即f′?x+2設(shè)?f?x+2+fx+則f?x+2=fx+2又因fx可得fx則fx+4=fx,故函數(shù)因?yàn)閒x+2=?f所以f0+f1+故答案為:3【15】0解析:∵fx+1∴fx=f2?x,∵fx+2為奇函數(shù),∴fx+2又∵f∴f′x∴f′1=?f由fx=fx+4∴f2026+【16】2解析:因?yàn)閒2x+1為偶函數(shù),則即fx又因?yàn)閒′x+2由fx+1=f即f′所以f′x=?f′所以f′x是以由f′x=?f′?x+2ff′2