（新教材適用）高中數(shù)學(xué)復(fù)習(xí)課第2課時(shí)一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用課后習(xí)題新人教A版選擇性

第2課時(shí)一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用課后訓(xùn)練鞏固提升1.若曲線y=x+1x-1在點(diǎn)(3,2)處的切線與直線ax+y+1=0垂直,則A.2 B.2 C.12 D.解析:y=x+1x-1=1+2x-1,y'=2因?yàn)榍€在點(diǎn)(3,2)處的切線與直線ax+y+1=0垂直,所以12×(a)=1,解得a=2,故選B答案:B2.若函數(shù)f(x)=x22x4lnx,則f(x)的單調(diào)遞增區(qū)間為()A.(1,0)B.(1,0)∪(2,+∞)C.(2,+∞)D.(0,+∞)解析:函數(shù)f(x)的定義域?yàn)?0,+∞),導(dǎo)數(shù)f'(x)=2x24x由f'(x)>0,得x>2.故f(x)的單調(diào)遞增區(qū)間是(2,+∞).答案:C3.設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f'(x).若函數(shù)y=(1x)·f'(x)的圖象如圖所示,則下列結(jié)論中一定成立的是()A.函數(shù)f(x)有極大值f(2)和極小值f(1)B.函數(shù)f(x)有極大值f(2)和極小值f(1)C.函數(shù)f(x)有極大值f(2)和極小值f(2)D.函數(shù)f(x)有極大值f(2)和極小值f(2)解析:由題圖可知,f'(2)=0,f'(2)=0,當(dāng)x<2時(shí),f'(x)>0;當(dāng)2<x<1時(shí),f'(x)<0;當(dāng)1<x<2時(shí),f'(x)<0;當(dāng)x>2時(shí),f'(x)>0.由此可以得到函數(shù)f(x)在x=2處取得極大值,在x=2處取得極小值.故選D.答案:D4.已知函數(shù)f(x)=x3+ax2+bx+a2在x=1處取得極值10,則a+b=()A.0或7 B.7C.0 D.7解析:f'(x)=3x2+2ax+b.由題意得f解得a若a=3,b=3,則f'(x)=3x26x+3=3(x1)2.當(dāng)x<1或x>1時(shí),f'(x)>0,所以x=1不是極值點(diǎn),不符合題意.經(jīng)檢驗(yàn),a=4,b=11符合題意,所以a+b=411=7.故選B.答案:B5.已知函數(shù)f(x)的定義域是R,f(0)=2,且對(duì)任意x∈R,f(x)+f'(x)>1,則不等式ex·f(x)>ex+1的解集為()A.{x|x>0}B.{x|x<0}C.{x|x<1或x>1}D.{x|x<1或0<x<1}解析:設(shè)g(x)=ex·f(x)ex,則g'(x)=ex·[f(x)+f'(x)1].∵對(duì)任意x∈R,f(x)+f'(x)>1,∴g'(x)>0在R上恒成立.∴g(x)=ex·f(x)ex在R上為增函數(shù).又f(0)=2,∴g(0)=1.故g(x)=ex·f(x)ex>1的解集為{x|x>0},即不等式ex·f(x)>ex+1的解集為{x|x>0}.答案:A6.已知函數(shù)f(x)=x33ax+b(a>0)的極大值為6,極小值為2,則f(x)的單調(diào)遞減區(qū)間為.

解析:令f'(x)=3x23a=0,解得x=±a.易得f(x)在x=a處取得極大值,在x=a處取得極小值.由題意知f(a)=6,f(a)=2,解得a=1,b=4.令f'(x)=3x23<0,得1<x<1.故函數(shù)f(x)的單調(diào)遞減區(qū)間為(1,1).答案:(1,1)7.已知函數(shù)f(x)=x3+ax24x+3(x∈R).(1)當(dāng)a=2時(shí),求f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若函數(shù)f(x)在區(qū)間(1,2)內(nèi)單調(diào)遞減,求實(shí)數(shù)a的取值范圍.解:(1)當(dāng)a=2時(shí),∵f(x)=x3+2x24x+3,∴f'(x)=3x2+4x4.∴f'(1)=3,即切線的斜率為3.∵f(1)=2,∴切點(diǎn)為(1,2).故所求切線方程為y2=3(x1),即3xy1=0.(2)∵f(x)=x3+ax24x+3,∴f'(x)=3x2+2ax4.∵函數(shù)f(x)在區(qū)間(1,2)內(nèi)單調(diào)遞減,∴f'(x)≤0對(duì)x∈(1,2)恒成立,即3x2+2ax4≤0對(duì)x∈(1,2)恒成立,從而a≤2x-32x,設(shè)h(x)=2x-32x,則h'(x∵當(dāng)x>0時(shí),h'(x)<0,∴函數(shù)h(x)在區(qū)間(0,+∞)上單調(diào)遞減.∴當(dāng)x∈[1,2]時(shí),h(x)min=h(2)=2.∴a≤2.故實(shí)數(shù)a的取值范圍為-∞8.已知函數(shù)f(x)=(x+a)ex(a∈R).(1)若函數(shù)f(x)在區(qū)間[3,+∞)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)若f(x)≥e2對(duì)x∈[0,2]恒成立,求實(shí)數(shù)a的取值范圍.解:(1)f'(x)=(x+a+1)ex,x∈R.因?yàn)楹瘮?shù)f(x)在區(qū)間[3,+∞)上單調(diào)遞增,所以f'(x)≥0,即x+a+1≥0對(duì)x∈[3,+∞)恒成立.因?yàn)閥=x+a+1是增函數(shù),所以3+a+1≥0,解得a≥2.故實(shí)數(shù)a的取值范圍為[2,+∞).(2)因?yàn)閒(x)=(x+a)ex≥e2對(duì)x∈[0,2]恒成立,即a≥e2xx對(duì)x∈[0,2]恒成立.設(shè)g(x)=e2xx,x∈[0,2],則g'(x)=e2x1.因?yàn)間'(x)<0,所以函數(shù)g(x)在區(qū)間[0,2]上單調(diào)遞減.所以a≥g(x)max=g(0)=e2.因此,實(shí)數(shù)a的取值范圍是[e2,+∞).9.已知函數(shù)f(x)=x2mlnx,h(x)=x2x+a,(1)當(dāng)a=0時(shí),f(x)≥h(x)在區(qū)間(1,+∞)上恒成立,求實(shí)數(shù)m的取值范圍;(2)當(dāng)m=2時(shí),若函數(shù)k(x)=f(x)h(x)在區(qū)間[1,3]上恰有兩個(gè)不同零點(diǎn),求實(shí)數(shù)a的取值范圍.解:(1)由f(x)≥h(x)在區(qū)間(1,+∞)上恒成立,得m≤xlnx在區(qū)間(1,+∞設(shè)g(x)=xlnx,則g'(x)=令g'(x)=0,解得x=e.當(dāng)x∈(1,e)時(shí),g'(x)<0,函數(shù)g(x)在區(qū)間(1,e)上單調(diào)遞減;當(dāng)x∈(e,+∞)時(shí),g'(x)>0,函數(shù)g(x)在區(qū)間(e,+∞)上單調(diào)遞增.所以當(dāng)x=e時(shí),函數(shù)g(x)取得極小值也是最小值,且最小值為g(e)=e.故m≤e.(2)由題意得k(x)=x2lnxa,函數(shù)k(x)在區(qū)間[1,3]上恰有兩個(gè)不同零點(diǎn)等價(jià)于函數(shù)φ(x)=x2lnx的圖象與直線y=a在區(qū)間[1,3]上有兩個(gè)不同的交點(diǎn).φ'(x)=12x令φ'(x)=0,解得x=2.當(dāng)x∈[1,2)時(shí),φ'(x)<0,函數(shù)φ(x)在區(qū)間[1,2)內(nèi)單調(diào)遞減;當(dāng)x∈(2,3]時(shí),φ'(x)>0,函數(shù)φ(x)在區(qū)間(2,3