新高考數(shù)學(xué)二輪復(fù)習(xí)函數(shù)培優(yōu)專題04 函數(shù)的解析式(含解析)_第1頁
新高考數(shù)學(xué)二輪復(fù)習(xí)函數(shù)培優(yōu)專題04 函數(shù)的解析式(含解析)_第2頁
新高考數(shù)學(xué)二輪復(fù)習(xí)函數(shù)培優(yōu)專題04 函數(shù)的解析式(含解析)_第3頁
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新高考數(shù)學(xué)二輪復(fù)習(xí)函數(shù)培優(yōu)專題04 函數(shù)的解析式(含解析)_第5頁
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專題04函數(shù)的解析式專項(xiàng)突破一待定系數(shù)法1.設(shè)SKIPIF1<0為一次函數(shù),且SKIPIF1<0.若SKIPIF1<0,則SKIPIF1<0的解析式為(

)A.SKIPIF1<0或SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.冪函數(shù)SKIPIF1<0的圖象經(jīng)過函數(shù)SKIPIF1<0SKIPIF1<0且SKIPIF1<0所過的定點(diǎn),則SKIPIF1<0的值等于(

)A.8 B.4 C.2 D.13.已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的增函數(shù),且SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.34.已知二次函數(shù)f(x)的圖象經(jīng)過點(diǎn)(-3,2),頂點(diǎn)是(-2,3),則函數(shù)f(x)的解析式為___________.5.已知函數(shù)SKIPIF1<0恒過定點(diǎn)P,點(diǎn)P恰好在冪函數(shù)SKIPIF1<0的圖象上,則SKIPIF1<0___________.6.(1)已知SKIPIF1<0是一次函數(shù),且SKIPIF1<0,求SKIPIF1<0;(2)已知SKIPIF1<0是二次函數(shù),且滿足SKIPIF1<0,求SKIPIF1<0.7.已知SKIPIF1<0是二次函數(shù),且滿足SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的解析式;(2)當(dāng)SKIPIF1<0時(shí),表示出函數(shù)SKIPIF1<0的最小值SKIPIF1<0,并求出SKIPIF1<0的最小值.8.已知函數(shù)SKIPIF1<0為二次函數(shù),不等式SKIPIF1<0的解集是SKIPIF1<0,且SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0.(1)求SKIPIF1<0的解析式;(2)設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0,求SKIPIF1<0的表達(dá)式.9.已知一次函數(shù)SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0.(1)求實(shí)數(shù)a?b的值;(2)令SKIPIF1<0,求函數(shù)SKIPIF1<0的解析式.專項(xiàng)突破二換元法1.設(shè)函數(shù)SKIPIF1<0,則SKIPIF1<0的表達(dá)式為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.已知SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.若SKIPIF1<0,則SKIPIF1<0的解析式為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<04.若SKIPIF1<0,則SKIPIF1<0等于(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<05.已知SKIPIF1<0是定義域?yàn)镾KIPIF1<0上的單調(diào)增函數(shù),且對(duì)任意SKIPIF1<0,都有SKIPIF1<0,則SKIPIF1<0的值為(

)A.12 B.14 C.SKIPIF1<0 D.186.已知函數(shù)SKIPIF1<0,那么SKIPIF1<0的表達(dá)式是___________.7.已知SKIPIF1<0,則SKIPIF1<0______.8.若SKIPIF1<0,則SKIPIF1<0______.9.若函數(shù)SKIPIF1<0,則SKIPIF1<0______.10.已知定義在SKIPIF1<0上的單調(diào)函數(shù)SKIPIF1<0,若對(duì)任意SKIPIF1<0都有SKIPIF1<0,則SKIPIF1<0_____專項(xiàng)突破三配湊法1.已知函數(shù)SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.已知函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.已知SKIPIF1<0求SKIPIF1<0=(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<04.若函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的解析式是__________5.已知SKIPIF1<0,則SKIPIF1<0______.6.已知函數(shù)SKIPIF1<0,則函數(shù)SKIPIF1<0的解析式為SKIPIF1<0______.7.若SKIPIF1<0,則SKIPIF1<0______.8.已知函數(shù)y=f(x)滿足SKIPIF1<0,求函數(shù)y=f(x)的解析式.專項(xiàng)突破四構(gòu)造方程組法1.已知SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.若函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.已知函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0___________.4.若SKIPIF1<0,則SKIPIF1<0______.5.設(shè)函數(shù)SKIPIF1<0是SKIPIF1<0→SKIPIF1<0的函數(shù),滿足對(duì)一切SKIPIF1<0,都有SKIPIF1<0,則SKIPIF1<0的解析式為SKIPIF1<0______.6.已知函數(shù)SKIPIF1<0對(duì)SKIPIF1<0的一切實(shí)數(shù)都有SKIPIF1<0,則SKIPIF1<0______.專項(xiàng)突破五利用奇偶性1.已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù),且當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.設(shè)SKIPIF1<0為奇函數(shù),且當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<04.若SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù),且SKIPIF1<0是偶函數(shù),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0時(shí)SKIPIF1<0的解析式為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<05.若定義在R上的偶函數(shù)SKIPIF1<0和奇函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的解析式為SKIPIF1<0___________.6.已知函數(shù)SKIPIF1<0是SKIPIF1<0上的奇函數(shù),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0

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