2025屆高考數(shù)學(xué)一輪復(fù)習(xí)專項練習(xí)課時規(guī)范練6函數(shù)的單調(diào)性與最值

課時規(guī)范練6函數(shù)的單調(diào)性與最值基礎(chǔ)鞏固組1.已知函數(shù)f(x)=k(x+2),x≤0,2A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件2.已知函數(shù)f(x)=ax,x>1,(4A.(1,+∞) B.[4,8)C.(4,8) D.(1,8)3.已知函數(shù)f(x)=x2-2A.(-∞,1] B.[3,+∞)C.(-∞,-1] D.[1,+∞)4.若2x+5y≤2-y+5-x,則有()A.x+y≥0 B.x+y≤0C.x-y≤0 D.x-y≥05.函數(shù)f(x)在(-∞,+∞)上單調(diào)遞減,且為奇函數(shù),若f(1)=-1,則滿足-1≤f(x-2)≤1的x的取值范圍是 ()A.[-2,2] B.[-1,1]C.[0,4] D.[1,3]6若2x-2y<3-x-3-y,則()A.ln(y-x+1)>0 B.ln(y-x+1)<0C.ln|x-y|>0 D.ln|x-y|<07.函數(shù)f(x)=12

-x2+2A.-2 B.2 C.-1 D.18.(多選)下列四個說法,其中不正確的是()A.函數(shù)f(x)在(0,+∞)上單調(diào)遞增,在(-∞,0]上單調(diào)遞增,則f(x)在R上是增函數(shù)B.若函數(shù)f(x)=ax2+bx+2與x軸沒有交點,則b2-8a<0且a>0C.當(dāng)a>b>c時,則有ab>ac成立D.y=|1+x|和y=(1+9.(多選)已知函數(shù)f(x)=x-2x,g(x)=acosπx2+5-2a(a>0).A.若?x∈[1,2],使得f(x)<a成立,則a>-1B.若?x∈R,使得g(x)>0恒成立,則0<a<5C.若?x1∈[1,2],?x2∈R,使得f(x1)>g(x2)恒成立,則a>6D.若?x1∈[1,2],?x2∈[0,1],使得f(x1)=g(x2)成立,則3≤a≤410.設(shè)函數(shù)f(x)=1,x>0,0,x=0,-1,x<011.函數(shù)f(x)=2xx+112.已知函數(shù)f(x)=x+2x-3,x≥1,lg(x2+1),綜合提升組13.(多選)函數(shù)f(x)在[a,b]上有定義,若對隨意x1,x2∈[a,b],有fx1+x22≤12[f(x1)+f(x2)],則稱f(x)在[a,b]上具有性質(zhì)P.設(shè)f(x)在[1,3]上具有性質(zhì)A.f(x)在[1,3]上的圖像是連綿起伏的B.f(x2)在[1,3]上具有性質(zhì)PC.若f(x)在x=2處取得最大值1,則f(x)=1,x∈[1,3]D.對隨意x1,x2,x3,x4∈[1,3],有fx1+x2+x3+x44≤12[f(x1)+f(x2)14.已知f(x)=1-lnx,0<x≤1,-1+lnx創(chuàng)新應(yīng)用組15.假如函數(shù)y=f(x)在區(qū)間I上單調(diào)遞增,且函數(shù)y=f(x)x在區(qū)間I上單調(diào)遞減,那么稱函數(shù)y=f(x)是區(qū)間I上的“緩增函數(shù)”,區(qū)間I叫做“緩增區(qū)間”.若函數(shù)f(x)=12x2-x+32是區(qū)間A.[1,+∞) B.[0,3]C.[0,1] D.[1,3]16.已知P(m,n)是函數(shù)y=-x2-2x圖像上的動點,則|4m+3n-A.25 B.21 C.20 D.4參考答案課時規(guī)范練6函數(shù)的單調(diào)性與最值1.D若f(x)單調(diào)遞增,則k>0且k(0+2)≤20+k,解得0<k≤1,因為“k<1”與“0<k≤1”沒有包含的關(guān)系,所以充分性和必要性都不成立.2.B由f(x)在R上單調(diào)遞增,則有a>1,43.B設(shè)t=x2-2x-3,由t≥0,即x2-2x-3≥0,解得x≤-1或x≥3.所以函數(shù)的定義域為(-∞,-1]∪[3,+∞).因為函數(shù)t=x2-2x-3的圖像的對稱軸為x=1,所以函數(shù)t在(-∞,-1]上單調(diào)遞減,在[3,+∞)上單調(diào)遞增.所以f(x)的單調(diào)遞增區(qū)間為[3,+∞).4.B設(shè)函數(shù)f(x)=2x-5-x,易知f(x)為增函數(shù),又f(-y)=2-y-5y,由已知得f(x)≤f(-y),∴x≤-y,∴x+y≤0.5.D由題意f(-1)=-f(1)=1,-1≤f(x-2)≤1等價于f(1)≤f(x-2)≤f(-1).又f(x)在(-∞,+∞)上單調(diào)遞減,所以-1≤x-2≤1,即1≤x≤3.所以x的取值范圍是[1,3].6.A∵2x-2y<3-x-3-y,∴2x-3-x<2y-3-y.∵f(t)=2t-3-t在R上為增函數(shù),且f(x)<f(y),∴x<y,∴y-x>0,∴y-x+1>1,∴l(xiāng)n(y-x+1)>ln1=0.故選A.7.B∵-x2+2mx-m2-1=-(x-m)2-1≤-1,∴12

∴f(x)的值域為[2,+∞),∵y=12x是減函數(shù),y=-(x-m)2-1的單調(diào)遞減區(qū)間為[m,+∞),∴f(x)的單調(diào)遞增區(qū)間為[m,+∞).由條件知m=2.8.ABCf(x)=x,x≤0,lnx,x>0,滿足在(0,+∞)上單調(diào)遞增,在(-∞,0]上單調(diào)遞增,但f(x)在R上不是增函數(shù),故A錯誤;當(dāng)a=b=0時,f(x)=2,它的圖像與x軸無交點,不滿足b2-8a<0且a>0,故B錯誤;當(dāng)a>b>c,但a=0時,ab=ac,不等式ab>ac不成立,故C錯誤;y=9.ACD對于A,由f(x)在[1,2]上單調(diào)遞增,則f(x)min=f(1)=-1,所以a>-1,故A正確;對于B,只需g(x)min>0,由g(x)min=-a+5-2a=5-3a>0,得0<a<53,故B錯誤;對于C,只需在給定的范圍內(nèi)f(x)min>g(x)max,即-1>5-a,解得a>6,故C正確;對于D,只需g(x)min≤f(x)min,g(x)max≥f(x)max,f(x)max=f(2)=2-22=1,所以x1∈[1,2],f(x1)∈[-1,1],當(dāng)x∈[0,1]時,πx2∈0,π2,所以g(x)在[0,1]上單調(diào)遞減,g(x)min=g(1)=5-2a,g(x)max=g(0)=5-a,所以g(x)∈[5-2a,5-a],由題意,可得5-210.[0,1)∵g(x)=x2,x>1,0,11.1,43∵f(x)=2xx+1=2(x+1)-2x+1=2-2x+1,∴f(x)在區(qū)間[1,2]上單調(diào)遞增,即f(x)max=f(2)=12.022-3因為f(-3)=lg[(-3)2+1]=lg10=1,所以f[f(-3)]=f(1)=1+2-3=0.當(dāng)x≥1時,x+2x-3≥2x·2x-3=22-3,當(dāng)且僅當(dāng)x=2x,即x=2時,等號成立,此時f(x)min=2當(dāng)x<1時,lg(x2+1)≥lg(02+1)=0,此時f(x)min=0.所以f(x)的最小值為22-3.13.ABD對于A,函數(shù)f(x)=x2,1≤x<3,11,x=3在[1,3]上具有性質(zhì)P,但f(x)在[1,3]上的圖像不連續(xù),故A錯誤;對于B,f(x)=-x在[1,3]上具有性質(zhì)P,但f(x2)=-x2在[1,3]上不滿足性質(zhì)P,故B錯誤;對于C,因為f(x)在x=2處取得最大值1,所以f(x)≤1,由性質(zhì)P可得1=f(2)≤12[f(x)+f(4-x)],即f(x)+f(4-x)≥2,因為f(x)≤1,f(4-x)≤1,所以f(x)=1,x∈[1,3],故C正確;對于D,fx1+x2+x3+x44=fx1+x22+x3+x422≤114.2e因為f(x)=1-ln由f(a)=f(b),得1-lna=-1+lnb,0<a≤1,b>1,所以lnab=2,即ab=e2.設(shè)y=1a+1b=be2+1b,令y'=1e215.D因為函數(shù)f(x)=12x2-x+32的對稱軸為x=1,所以函數(shù)y=f(x)在區(qū)間[1,+∞)上單調(diào)遞增.又因為當(dāng)x≥1時,f(x)x=12x-1+32x