第1課時(shí)函數(shù)的單調(diào)性課后篇鞏固提升基礎(chǔ)鞏固1.下列函數(shù)在區(qū)間(0,+)上不是增函數(shù)的是()A.y=2x+1B.y=x2+1C.y=3-xD.y=x2+2x+1解析函數(shù)y=3-x在區(qū)間(0,+)上是減函數(shù).答案C2.函數(shù)f(x)=-x2+2x+3的單調(diào)減區(qū)間是()A.(-,1)B.(1,+)C.(-,2)D.(2,+)解析易知函數(shù)f(x)=-x2+2x+3是圖象開(kāi)口向下的二次函數(shù),其對(duì)稱(chēng)軸為x=1,所以其單調(diào)減區(qū)間是(1,+).答案B3.若定義在R上的函數(shù)f(x)對(duì)任意兩個(gè)不相等的實(shí)數(shù)a,b,總有f(a)-f(b)a-b0成立,則必有()A.f(x)在R上是增函數(shù)B.f(x)在R上是減函數(shù)C.函數(shù)f(x)是先增后減D.函數(shù)f(x)是先減后增解析由f(a)-f(b)a-b0知f(a)-f(b)與a-b同號(hào),即當(dāng)ab時(shí),f(a)b時(shí),f(a)f(b),所以f(x)在R上是增函數(shù).答案A4.函數(shù)f(x)=x2-2(a-1)x+1在區(qū)間(2,3)上為單調(diào)函數(shù),則實(shí)數(shù)a的取值范圍是()A.(-,34,+)B.(-,3)(4,+)C.(-,3D.4,+)解析二次函數(shù)圖象開(kāi)口向上,對(duì)稱(chēng)軸為直線x=a-1,因?yàn)楹瘮?shù)在區(qū)間(2,3)上為單調(diào)函數(shù),所以a-12或a-13,相應(yīng)解得a3或a4,故選A.答案A5.已知函數(shù)f(x)在(-,+)上是減函數(shù),若aR,則()A.f(a)f(2a)B.f(a2)f(a)C.f(a2+a)f(a)D.f(a2+1)a,f(x)在(-,+)上是減函數(shù),所以f(a2+1)f(a).而在其他選項(xiàng)中,當(dāng)a=0時(shí),自變量均是0,應(yīng)取等號(hào).故選D.答案D6.若函數(shù)f(x)=x2+3ax+5在區(qū)間(-,5)上為減函數(shù),則實(shí)數(shù)a的取值范圍是()A.-,-103B.-103,+C.-,103D.103,+解析因?yàn)楹瘮?shù)f(x)=x2+3ax+5的單調(diào)遞減區(qū)間為-,-3a2,所以(-,5)-,-3a2,所以a-103.答案A7.函數(shù)f(x)=|x-2|的單調(diào)遞增區(qū)間是.解析由圖象可知,f(x)的單調(diào)遞增區(qū)間是2,+).答案2,+)8.已知函數(shù)f(x)=2x2-mx+3,當(dāng)x-2,+)時(shí),f(x)是增函數(shù),當(dāng)x(-,-2)時(shí),f(x)是減函數(shù),則f(1)=.解析函數(shù)f(x)在(-,-2)上是減函數(shù),在-2,+)上是增函數(shù),x=-b2a=m4=-2,m=-8,即f(x)=2x2+8x+3.f(1)=13.答案139.已知函數(shù)f(x)=-2x2+mx+1在區(qū)間1,4上是單調(diào)函數(shù),則實(shí)數(shù)m的取值范圍是.解析二次函數(shù)f(x)的圖象的對(duì)稱(chēng)軸是直線x=m4.因?yàn)槎魏瘮?shù)在對(duì)稱(chēng)軸的兩側(cè)的單調(diào)性相反,即m4(1,4),所以m41或m44,即m4或m16.答案(-,416,+)10.證明函數(shù)f(x)=-x在定義域上為減函數(shù).證明函數(shù)f(x)=-x的定義域?yàn)?,+).設(shè)x1,x2是0,+)上的任意兩個(gè)實(shí)數(shù),且0x10,f(x2)-f(x1)=(-x2)-(-x1)=x1-x2=(x1-x2)(x1+x2)x1+x2=x1-x2x1+x2.x1-x20,f(x2)-f(x1)0,即f(x2)f(x1).函數(shù)f(x)=-x在定義域0,+)上為減函數(shù).能力提升1.函數(shù)f(x)=|x|與g(x)=x(2-x)的單調(diào)遞增區(qū)間分別為()A.(-,0,1,+)B.(-,0,(-,1C.0,+),1,+)D.0,+),(-,1解析由函數(shù)圖象(圖略)可知選D.答案D2.若函數(shù)y=ax與y=-bx在區(qū)間(0,+)上都是減函數(shù),則函數(shù)y=ax2+bx在區(qū)間(0,+)上是()A.增函數(shù)B.減函數(shù)C.先增后減D.先減后增解析由于函數(shù)y=ax與y=-bx在區(qū)間(0,+)上都是減函數(shù),所以a0,即a0,b0.因?yàn)閽佄锞€y=ax2+bx的對(duì)稱(chēng)軸為x=-b2af(0),解得a0,0a1.答案D5.給出下列三個(gè)結(jié)論:若函數(shù)y=f(x)的定義域?yàn)?0,+),且滿足f(1)f(2)f(3),則函數(shù)y=f(x)在(0,+)上是增函數(shù);若函數(shù)y=f(x)在(-,+)上是減函數(shù),則f(a2+1)a2,又y=f(x)在(-,+)上是減函數(shù),f(a2+1)f(a2),正確;取x1=-1,x2=1,f(-1)=-1,f(1)=1,f(-1)x2-2,則f(x1)f(x2),則實(shí)數(shù)a的取值范圍是.(用區(qū)間來(lái)表示)解析由“若x1x2-2,則f(x1)f(x2)”可知函數(shù)f(x)在(-2,+)上單調(diào)遞增.而f(x)=ax+1x+2=a+1-2ax+2,故有1-2a12,即a的取值范圍為12,+.答案12,+7.若函數(shù)f(x)=x2+2ax+3,x1,ax+1,x1是減函數(shù),則實(shí)數(shù)a的取值范圍為.解析由題意可得-a1,a0,12+2a1+3a1+1,解得-3a-1,則實(shí)數(shù)a的取值范圍是-3,-1.答案-3,-18.討論函數(shù)f(x)=ax+1x+2a12在區(qū)間(-2,+)上的單調(diào)性.解f(x)=ax+1x+2=a+1-2ax+2,設(shè)任意的x1,x2(-2,+),且x1x2,則f(x1)-f(x2)=1-2ax1+2-1-2ax2+2=(1-2a)x2-x1(x2+2)(x1+2).-2x10,(x2+2)(x1+2)0.當(dāng)a0,f(x1)-f(x2)0,即f(x1)f(x2),故f(x)在區(qū)間(-2,+)上為減函數(shù).當(dāng)a12時(shí),1-2a0,f(x1)-f(x2)0,f(x1)f(x2),故f(x)在區(qū)間(-2,+)上為增函數(shù).綜上,當(dāng)a12時(shí),f(x)在區(qū)間(-2,+)上為增函數(shù).9.某市一家報(bào)刊攤點(diǎn),從該市報(bào)社買(mǎi)進(jìn)該市的晚報(bào)價(jià)格是每份0.40元,賣(mài)出價(jià)格是每份0.60元,賣(mài)不掉的報(bào)紙以每份0.05元的價(jià)格退回報(bào)社.在一個(gè)月(按30天計(jì)算)里,有18天每天可賣(mài)出400份,其余12天每天只能賣(mài)出180份.則攤主每天從報(bào)社買(mǎi)進(jìn)多少份晚報(bào),才能使每月獲得的利潤(rùn)最大(設(shè)攤主每天從報(bào)社買(mǎi)進(jìn)晚報(bào)的份數(shù)是相同的)?解設(shè)攤主每天從報(bào)社買(mǎi)進(jìn)x(180x400,xN)份晚報(bào),每月獲利為y元,則有y=(0.60-0.40)(18x+12180)-(0.40-0.05)12(x-180)=-0.6