【全程復(fù)習(xí)方略】2020年人教A版數(shù)學(xué)理(福建用)課時(shí)作業(yè)：第二章-第六節(jié)冪函數(shù)與二次函數(shù)

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)提升作業(yè)(九)一、選擇題1.已知冪函數(shù)y=f(x)通過(guò)點(diǎn)(2,2QUOTE),則冪函數(shù)的解析式為()(A)y= (B)y=(C)y= (D)y=2.(2021·廈門模擬)函數(shù)y=的圖象是()3.已知函數(shù)y=x2-2x+3在閉區(qū)間[0,m]上有最大值3,最小值2,則m的取值范圍是()(A)[1,+∞) (B)[0,2] (C)[1,2] (D)(-∞,2]4.(2021·湛江模擬)若f(x)=x2-x+a,f(-m)<0,則f(m+1)的值是()(A)正數(shù) (B)負(fù)數(shù) (C)非負(fù)數(shù) (D)不能確定正負(fù)5.已知P=,Q=(QUOTE)3,R=(QUOTE)3,則P,Q,R的大小關(guān)系是()(A)P<Q<R (B)Q<R<P(C)Q<P<R (D)R<Q<P6.設(shè)abc>0,二次函數(shù)f(x)=ax2+bx+c的圖象可能是()7.函數(shù)f(x)=ax2+(a-3)x+1在區(qū)間[-1,+∞)上是遞減的,則實(shí)數(shù)a的取值范圍是()(A)[-3,0) (B)(-∞,-3](C)[-2,0] (D)[-3,0]8.(2021·南平模擬)對(duì)于任意a∈[-1,1]，函數(shù)f(x)=x2+(a-4)x+4-2a的值恒大于零，那么x的取值范圍是()(A)(1,3) (B)(-∞,1)∪(3,+∞)(C)(1,2) (D)(3,+∞)9.已知函數(shù)f(x)=x2+1的定義域?yàn)閇a,b](a<b),值域?yàn)閇1,5],則在平面直角坐標(biāo)系內(nèi),點(diǎn)(a,b)的運(yùn)動(dòng)軌跡與兩坐標(biāo)軸圍成的圖形的面積為()(A)8 (B)6 (C)4 (D)210.(力氣挑戰(zhàn)題)若不等式x2+ax+1≥0對(duì)于一切x∈(0,QUOTE]恒成立,則a的最小值是()(A)0 (B)2 (C)-QUOTE (D)-3二、填空題11.(2021·莆田模擬)已知n∈{-1,0,1,2,3}，若則n=________.12.若二次函數(shù)f(x)=(x+a)(bx+2a)(a,b∈R)是偶函數(shù),且它的值域?yàn)?-∞,4],則該函數(shù)的解析式f(x)=.13.(2021·天津模擬)若關(guān)于x的不等式x2+QUOTEx-(QUOTE)n≥0對(duì)任意n∈N*在x∈(-∞,λ]上恒成立,則實(shí)常數(shù)λ的取值范圍是.14.二次函數(shù)f(x)的二次項(xiàng)系數(shù)為正,且對(duì)任意x恒有f(2+x)=f(2-x),若f(1-2x2)<f(1+2x-x2),則x的取值范圍是.三、解答題15.(力氣挑戰(zhàn)題)已知函數(shù)f(x)=3ax2+2bx+c,a+b+c=0，且f(0)·f(1)>0.(1)求證：-2<<-1.(2)若x1,x2是方程f(x)=0的兩個(gè)實(shí)根，求|x1-x2|的取值范圍.答案解析1.【解析】選C.設(shè)y=xα,則由已知得,2QUOTE=2α,即=2α,∴α=QUOTE,∴f(x)=QUOTE.2.【解析】選B.在第一象限內(nèi),類比y=的圖象知選B.3.【解析】選C.y=(x-1)2+2,由x2-2x+3=3得x=0或x=2,∴1≤m≤2,故選C.4.【解析】選B.f(x)=(x-QUOTE)2+a-QUOTE,其對(duì)稱軸為x=QUOTE,而-m,m+1關(guān)于QUOTE對(duì)稱,故f(m+1)=f(-m)<0,故選B.5.【解析】選B.由函數(shù)y=x3在R上是增函數(shù)知,(QUOTE)3<()3,由函數(shù)y=2x在R上是增函數(shù)知,>2-3=(QUOTE)3,∴P>R>Q.6.【解析】選D.對(duì)于選項(xiàng)A,C,都有QUOTE∴abc<0,故排解A,C.對(duì)于選項(xiàng)B,D,都有QUOTE>0,即ab<0,則當(dāng)c<0時(shí),abc>0,故選D.7.【解析】選D.當(dāng)a=0時(shí),f(x)=-3x+1明顯成立,當(dāng)a≠0時(shí),需QUOTE解得-3≤a<0,綜上可得-3≤a≤0.【誤區(qū)警示】本題易忽視a=0這一狀況而誤選A,失誤的緣由是將關(guān)于x的函數(shù)誤認(rèn)為是二次函數(shù).8.【解析】選B.f(x)=x2+(a-4)x+4-2a=(x-2)a+x2-4x+4,令g(a)=(x-2)a+x2-4x+4,由題意知解得x>3或x<1,故選B.9.【思路點(diǎn)撥】對(duì)于函數(shù)f(x)=x2+1而言,當(dāng)x=±2時(shí),y=5,從而結(jié)合題意得出a,b的取值范圍,點(diǎn)(a,b)的運(yùn)動(dòng)軌跡是兩條線段,與兩坐標(biāo)軸圍成的圖形是一個(gè)邊長(zhǎng)為2的正方形,從而得出結(jié)果.【解析】選C.如圖,對(duì)于函數(shù)f(x)=x2+1,當(dāng)x=±2時(shí),y=5.故依據(jù)題意得a,b的取值范圍為:-2≤a≤0且b=2或a=-2且0≤b≤2.∴點(diǎn)(a,b)的運(yùn)動(dòng)軌跡與兩坐標(biāo)軸圍成的圖形是一個(gè)邊長(zhǎng)為2的正方形,面積為4.10.【解析】選C.方法一:設(shè)g(a)=ax+x2+1,∵x∈(0,QUOTE],∴g(a)為單調(diào)遞增函數(shù).當(dāng)x=QUOTE時(shí)滿足:QUOTEa+QUOTE+1≥0即可,解得a≥-QUOTE.方法二:由x2+ax+1≥0得a≥-(x+QUOTE)在x∈(0,QUOTE]上恒成立,令g(x)=-(x+QUOTE),則知g(x)在(0,QUOTE]為增函數(shù),∴g(x)max=g(QUOTE)=-,∴a≥-QUOTE.11.【解析】依據(jù)冪函數(shù)的性質(zhì)，冪函數(shù)y=x-1和y=x2在(-∞,0)上是減函數(shù),故滿足的n值只有-1和2.答案：-1或212.【思路點(diǎn)撥】化簡(jiǎn)f(x),函數(shù)f(x)為偶函數(shù),則一次項(xiàng)系數(shù)為0可求b.值域?yàn)?-∞,4],則最大值為4,可求2a2,即可求出解析式.【解析】∵f(x)=(x+a)(bx+2a)=bx2+(2a+ab)x+2a2是偶函數(shù),則其圖象關(guān)于y軸對(duì)稱.∴2a+ab=0,∴b=-2或a=0(舍去).∴f(x)=-2x2+2a2,又f(x)的值域?yàn)?-∞,4],∴2a2=4,f(x)=-2x2+4.答案:-2x2+413.【解析】∵n∈N*時(shí),(QUOTE)n≤QUOTE,∴x2+QUOTEx≥QUOTE在x∈(-∞,λ]上恒成立,又x2+QUOTEx=(x+QUOTE)2-,∴QUOTE解得λ≤-1.答案:(-∞,-1]14.【思路點(diǎn)撥】由題意知二次函數(shù)的圖象開口向上,且關(guān)于直線x=2對(duì)稱,則距離對(duì)稱軸越遠(yuǎn),函數(shù)值越大,依此可轉(zhuǎn)化為不等式問(wèn)題.【解析】由f(2+x)=f(2-x)知x=2為對(duì)稱軸,由于二次項(xiàng)系數(shù)為正的二次函數(shù)中距對(duì)稱軸越遠(yuǎn),函數(shù)值越大,∴|1-2x2-2|<|1+2x-x2-2|,即|2x2+1|<|x2-2x+1|,∴2x2+1<x2-2x+1,∴-2<x<0.答案:(-2,0)15.【解析】(1)當(dāng)a=0時(shí)，f(0)=c,f(1)=2b+c,又b+c=0,則f(0)·f(1)=c(2b+c)=-c2<0與已知沖突.因