20xx屆新高考二輪-專題二-函數(shù)與導(dǎo)數(shù)-單元測試(更新版)
【正文】 ex(e1),所以當(dāng)x1時,u39。(x)g39。ln1x0,令φ(x)=xex,在(0,+∞)上,φ39。(x)0,f(x)在定義域上單調(diào)遞增,不可能有兩個零點(diǎn)。(x)0,h(x)在(0,+∞)上單調(diào)遞增,h(1)=10,h(e12a)=e12a10,?x1∈(e12a,1),使得h(x1)=0.∴當(dāng)x∈(0,x1)時,f39。(a)=lna0.故h(a)在(1,+∞)上單調(diào)遞減,所以h(a)1b,所以1b≤0,即b≥1.綜上,實(shí)數(shù)b的取值范圍為(1,+∞).20.(1)解當(dāng)a=1,x1時,f(x)=x(lnx)2,x1.f39。(x)0,所以n(x)在(0,ln2)上單調(diào)遞減。4x(2x2+1)2=4(2x1)(x+1)(2x2+1)2,由h39。(x)0,所以f(x)在(0,+∞)上單調(diào)遞增。=1ax1(x+1)2=(x+1)2axax(x+1)2,當(dāng)a0時,F39。(n)=4n+7+1n20,可得g(n)在正奇數(shù)集上遞增,可得g(n)的最小值為g(1)=8,可得λ≤8.①當(dāng)n為偶數(shù)時,2n37n2λn1≥0,即λ≤2n27n1n恒成立,等價于λ≤2n27n1nmin,設(shè)h(n)=2n27n1n,h39。3π4=e3π4+cos3π40,f39。∵f(x)=f(6x),∴f(3)=f(63)=f(3),∴f(3)=0,故A對.∵f(x+1)=f(x+1),∴x=1為對稱軸,. 解析f(x)=x,x≤0,lnx,x0,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),A錯。(2)f(x)≤xex恒成立,求a的取值范圍.22.(2020浙江,22)已知1a≤2,函數(shù)f(x)=exxa,其中e= 28…是自然對數(shù)的底數(shù).(1)證明:函數(shù)y=f(x)在(0,+∞)上有唯一零點(diǎn)。若不等式f(n)≥0恒成立,則λ的最大值為 .16.(2020山西太原三模,文16)對任意正整數(shù)n,函數(shù)f(n)=2n37n2cos nπλn1,若f(2)≥0,則λ的取值范圍是 。(3)若f(x)在x=x0處有極大值,證明1f(x0)e2.21.(2020山西太原二模,理21)已知函數(shù)f(x)=ln x+ax+1.(1)若函數(shù)f(x)有兩個零點(diǎn),求a的取值范圍。f(a)=f(2020)=f(2528+4)=f(4)=f(3+1)=f(2)=[f(6(2))]=f(8),又a∈[5,9]且f(x)在[5,9]上單調(diào),易得a=。(x)為(π,+∞)上的增函數(shù),又f39。=4x+44x+1,∵x014,∴4x0+10,則tanα=4x0+44x0+1=4x0+1+44x0+11≥2(4x0+1)44x0+11=41=3.當(dāng)且僅當(dāng)4x0+1=44x0+1,即x0=14時,等號成立.即x0=14時,tanα最小,α取最小值.∞,132 132 解析由函數(shù)f(n)=2n37n2cosnπλn1,若f(2)≥0,則1628cos2π2λ1≥0,即15282λ≥0,解得λ≤132.不等式f(n)≥0恒成立,即2n37n2cosnπλn1≥0恒成立,當(dāng)n為奇數(shù)時,2n3+7n2λn1≥0,即λ≤2n2+7n1n恒成立,等價于λ≤2n2+7n1nmin,設(shè)g(n)=2n2+7n1n,g39。(x)1g(x)39。(x)=a+1x+2ax=2ax2+a+1x.當(dāng)a≥0時,f39。(x)=4(2x2+1)(4x+1)(x)=ex2,所以當(dāng)x∈(0,ln2)時,n39。(x)0得0xlna,所以F(x)在(0,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.所以F(x)min=F(lna)=aalnab.又F(x)x2ax+1b,故當(dāng)x→+∞時,F(x)→+∞,所以F(x)的值域?yàn)閇aalnab,+∞).故F(x)在(0,+∞)上有零點(diǎn)的充要條件為aalnab≤0.令h(a)=aalnab(a1),則h39。(x)=12ax=x2ax.①當(dāng)a=0時,f(x)=x無極大值.②當(dāng)a0時,h39。(x)=1x+a.①當(dāng)a≥0時,f39。(x)=2xex+x2ex+1x0,∴h(x)在(0,+∞)上單調(diào)遞增,而h(1)=e0,h1e=e1ee210,∴?x0∈1e,1使得h(x0)=0,即x02ex0+ln x0=0,∴x0ex0=1x0lnx0=1x0ln1x0=eln1x0(x)在[0,+∞)上單調(diào)遞增,故當(dāng)x0時,g39。(x)0,故函數(shù)u(x)在區(qū)間[1,+∞)上單調(diào)遞增,因此u(x)≥u(1)=0.由ex0=x0+a可得x0f(ex0)=x0f(x0+a)=(ea1)x02+a(ea2)x0≥(e1)ax02,由x0≥a1,得x0f(ex0)≥(e1)(a1)a.14
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