2025屆高考數(shù)學(xué)二輪總復(fù)習(xí)專題1函數(shù)與導(dǎo)數(shù)專項(xiàng)突破1突破1利用導(dǎo)數(shù)求參數(shù)的值或范圍課件

突破1利用導(dǎo)數(shù)求參數(shù)的值或范圍考點(diǎn)一與單調(diào)性有關(guān)的參數(shù)范圍問題(1)當(dāng)a=-1時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若函數(shù)f(x)在(0,+∞)單調(diào)遞增,求a的取值范圍.∴m'(x)在(0,+∞)單調(diào)遞減,則m'(x)<1-ln

1-1=0,則m(x)在(0,+∞)單調(diào)遞減,∴m(x)<0-2ln

1+0=0,即g(x)在(0,+∞)單調(diào)遞減,(方法二)令g(x)=ax2+x-(x+1)ln(x+1),x>0,則g'(x)=2ax+1-ln(x+1)-1=2ax-ln(x+1).∵x+1>1,∴l(xiāng)n(x+1)>0.當(dāng)a≤0時(shí),g'(x)<0,g(x)在(0,+∞)單調(diào)遞減,∴f'(x)<0,f(x)單調(diào)遞減,不符合題意;當(dāng)a>0時(shí),令h(x)=2ax-ln(x+1),x>0,則(ⅰ)當(dāng)a≥時(shí),h'(x)>0恒成立,∴g'(x)在(0,+∞)單調(diào)遞增,∴g'(x)>0-ln

1=0,即g'(x)>0,g(x)在(0,+∞)單調(diào)遞增,∴g(x)>0+0-0=0恒成立,即f'(x)>0,f(x)在(0,+∞)單調(diào)遞增,符合題意.[對點(diǎn)訓(xùn)練1](2024河北石家莊模擬)已知函數(shù)f(x)=ax2-ln|x|+x(a∈R).(1)若a=1,求f(x)的單調(diào)區(qū)間和極值;(2)若f(x)在(-∞,0)和(0,+∞)上均為單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.考點(diǎn)二與極值有關(guān)的參數(shù)范圍問題例2(2024新高考Ⅱ,16)已知函數(shù)f(x)=ex-ax-a3.(1)當(dāng)a=1時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若f(x)有極小值,且極小值小于0,求a的取值范圍.解

(1)當(dāng)a=1時(shí),f(x)=ex-x-1,則切點(diǎn)為(1,e-2).又f'(x)=ex-1,k=f'(1)=e-1,故所求切線方程為y-(e-2)=(e-1)(x-1),整理得(e-1)x-y-1=0.(2)由題得,f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0恒成立,則f(x)在R上為增函數(shù),無極值,所以a>0.令f'(x)=0,得x=ln

a.當(dāng)f'(x)<0時(shí),x<ln

a;當(dāng)f'(x)>0時(shí),x>ln

a,故函數(shù)f(x)在區(qū)間(-∞,ln

a)上單調(diào)遞減,在區(qū)間(ln

a,+∞)上單調(diào)遞增,所以f(x)的極小值為f(ln

a)=a-aln

a-a3<0,即1-ln

a-a2<0.令g(x)=-x2-ln

x+1,x>0,g'(x)=-2x-<0,所以g(x)在區(qū)間(0,+∞)上單調(diào)遞減.又g(1)=0,所以g(a)<g(1),即a>1,故a的取值范圍為(1,+∞).[對點(diǎn)訓(xùn)練2](2024四川成都二模)已知函數(shù)f(x)=(x+a)lnx的導(dǎo)函數(shù)為f'(x).(1)當(dāng)a=1時(shí),求f'(x)的最小值;(2)若f(x)存在兩個(gè)極值點(diǎn),求a的取值范圍.當(dāng)x∈(0,1)時(shí),h'(x)<0,h(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),h'(x)>0,h(x)單調(diào)遞增,所以h(x)min=h(1)=2,即f'(x)的最小值為2.①當(dāng)a≤0時(shí),因?yàn)閤-a>0,所以g'(x)>0,即f'(x)在(0,+∞)內(nèi)單調(diào)遞增,所以至多存在一個(gè)x0∈(0,+∞),使得f'(x0)=0,故f(x)不存在兩個(gè)極值點(diǎn).②當(dāng)a>0時(shí),解g'(x)=0,得x=a,故當(dāng)x∈(0,a)時(shí),g'(x)<0,f'(x)單調(diào)遞減,當(dāng)x∈(a,+∞)時(shí),g'(x)>0,f'(x)單調(diào)遞增,所以f'(x)min=f'(a)=ln

a+2,(ⅰ)當(dāng)ln

a+2≥0,即a≥e-2時(shí),f'(x)≥f'(x)min≥0,f(x)在(0,+∞)內(nèi)單調(diào)遞增,故f(x)不存在極值點(diǎn).且f(x)在(0,x1),(x2,+∞)內(nèi)單調(diào)遞增,在(x1,x2)內(nèi)單調(diào)遞減,所以x1,x2分別是y=f(x)的極大值點(diǎn)和極小值點(diǎn).綜上,a的取值范圍為(0,e-2).考點(diǎn)三與最值有關(guān)的參數(shù)范圍問題例3(2024四川遂寧二模)已知函數(shù)f(x)=ex-ax-2.(1)若f(x)在區(qū)間(0,1)內(nèi)存在極值,求a的取值范圍;(2)若x∈(0,+∞),f(x)>x-sinx-cosx,求a的取值范圍.解

(1)由f(x)=ex-ax-2,得f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,則f(x)單調(diào)遞增,f(x)不存在極值.當(dāng)a>0時(shí),令f'(x)=0,則x=ln

a,若x<ln

a,則f'(x)<0,f(x)單調(diào)遞減;若x>ln

a,則f'(x)>0,f(x)單調(diào)遞增,所以x=ln

a是f(x)的極小值點(diǎn).因?yàn)閒(x)在區(qū)間(0,1)存在極值,則0<ln

a<1,即1<a<e,所以f(x)在區(qū)間(0,1)內(nèi)存在極值時(shí),a的取值范圍是(1,e).(2)由f(x)>x-sin

x-cos

x當(dāng)x∈(0,+∞)時(shí)恒成立,即ex+cos

x+sin

x-(a+1)x-2>0當(dāng)x∈(0,+∞)時(shí)恒成立.設(shè)g(x)=ex+cos

x+sin

x-(a+1)x-2,則g(x)>0當(dāng)x∈(0,+∞)時(shí)恒成立.則g'(x)=ex-sin

x+cos

x-(a+1),令m(x)=g'(x)=ex-sin

x+cos

x-(a+1),則m'(x)=ex-cos

x-sin

x,令n(x)=m'(x)=ex-cos

x-sin

x,則n'(x)=ex+sin

x-cos

x,當(dāng)x∈(0,1)時(shí),ex+sin

x>1,則n'(x)=ex+sin

x-cos

x>0,當(dāng)x∈[1,+∞)時(shí),ex≥e,則n'(x)>0,所以當(dāng)x∈(0,+∞)時(shí),n'(x)>0,則n(x)即m'(x)單調(diào)遞增,所以m'(x)>m'(0)=0,則m(x)即g'(x)單調(diào)遞增,所以g'(x)>g'(0)=1-a,①當(dāng)a≤1時(shí),g'(0)=1-a≥0,故當(dāng)x∈(0,+∞)時(shí),g'(x)>0,則g(x)單調(diào)遞增,所以g(x)>g(0)=0,所以f(x)>x-sin

x-cos

x在x∈(0,+∞)時(shí)恒成立.②當(dāng)a>1時(shí),g'(0)=1-a<0,故在區(qū)間(0,ln(a+3))內(nèi)函數(shù)g'(x)存在零點(diǎn)x0,即g'(x0)=0,由于函數(shù)g'(x)在(0,+∞)內(nèi)單調(diào)遞增,則當(dāng)x∈(0,x0)時(shí),g'(x)<g'(x0)=0,故函數(shù)g(x)在區(qū)間(0,x0)內(nèi)單調(diào)遞減,所以,當(dāng)x∈(0,x0)時(shí),函數(shù)g(x)<g(0)=0,不合題意.綜上所述,a的取值范圍為(-∞,1].[對點(diǎn)訓(xùn)練3](2024江蘇常州模擬)已知函數(shù)f(x)=ax2-lnx+(2a-1)x,其中a∈R.(1)討論f(x)的單調(diào)區(qū)間;(2)設(shè)a>0,若不等式f(x)+≥0對x∈(0,+∞)恒成立,求a的取值范圍.考點(diǎn)四與函數(shù)零點(diǎn)(方程的根)有關(guān)的參數(shù)范圍問題例4(2024遼寧錦州模擬)已知函數(shù)f(x)=x3-alnx.(1)當(dāng)a=1時(shí),求函數(shù)f(x)的極值;(2)函數(shù)f(x)在區(qū)間(1,e]上存在兩個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍.[對點(diǎn)訓(xùn)練4](2024山東濰坊一模)已知函數(shù)f(x)=2mlnx-x+(m>0).(1)討論f(x)的單調(diào)性;設(shè)k(x)=-x2+2mx-1,則Δ=4(m2-1),①當(dāng)0<m≤1時(shí),Δ≤0,f'(x)≤0恒成立,且f'(x)至多在一點(diǎn)處為0,函數(shù)f(x)在(0,+∞)內(nèi)單調(diào)遞減;則當(dāng)0<x<x1或x>x2時(shí),k(x)<0,即f'(x)<0;當(dāng)x1<x<x2時(shí),k(x)>0,即f'(x)>0,即函數(shù)f(x)在(0,x1),(x2,+∞)內(nèi)單