22022222222a22a222022222222a22a212壓軸大題破練習

函數與導已知函f()＝x－＋.函數fx)在x＝0處切線方程為x＋y＋＝0求，值；當＞0時若曲線＝f(x)存在三條斜率為k的線，求實數的取值范.解(1)()(xax2)ef(0)2e

0f′x)(x2axx2[(22a)]ef′(0)2a2∴abf(x)[x(22ax2]ehx)′())k′)(x2ax2x2x2[(42a)]eh(x[x

(4ax4a]ex2x2a2.1axxf(x(∞(22∞(212.x→∞f(x→x→∞′()→∞∴′(x)f′2)(2a2)f′x)f′2)e

2)∴e(22)ke

2

(22).設函數f()＝－alnx其中∈R.討論fx)的單調性；確定的所有可能取值使fx)>－x

在區間(＋)內恒成立(e＝…為自然對數的底數)2ax1解(1)′(x2ax＝(>0).xa≤f′(f()(∞).f′(x)x

2

x∈

a

f(x()

12312222222221231222222222x∈

∞a

f(x)>0f().(x)－()e.xes′)e1.x>1s′xs)(∞).ss(xx>1)>0.a≤x>1f(x)axlnf)>()(∞)a1<2(<f0>02afx()(1∞).a時h(xfxg(≥1x1xx1x>1h(x)2ax＋>x＋＝xxxx()(∞.hx>1h(x()gx)>0f(x)>gx∈∞已知函f()＝x－x.求曲線fx)點(，f處的切線方程；求函數fx)單調遞減區間；設函數g(x＝f(x)x＋＞0，若x∈e]g)的最小值是，實數a的e為自然對數的底數).解(1)∵f)xln∴′(x)2∴f(1)x∵(1)1∴yf)1(1))11∵f(xxln(0∞)′()x＜xx∴f(x)x

ln(

).∵()axln

22222222222222∴g(x′x0.x①≥0≤時e′)≤0(xgx)(e](x)ga3a(mine1②＜e時ex′(x)(x

↘

1

(e)↗

ee1(x))1ln3aminx∈(0g(x3.已知函f()＝＋ax－ax若曲線＝()在點(1，f(1))處的切線與直線＝x＋垂直，求函數y＝f(x的單調區間；若對∈(0，＋∞都有fx)>2(－成立，試求實數的值范圍；記()＝(x)＋x－b(∈)，當＝1時函數(x)在區間

上兩個零點，求實數b的取值范圍.a解(1)2()(0∞)f′)，xa∴′(1)1a∴(x)＋lnx2f(xx′(>2f(x0<x<2∴(x)(2∞)(02).af(x)＝(a>0)x2′(>f(x)<00<x<a∴(x)(∞)

22112211(0x時f)fx)f).min∵x∈(0∞)()>2(a∴()>2(1)＋ln2>2(a2∴a>aln0<aae∴(0e1g(x)＋lnxx2bxxxx′)g(x)>0>1xg(x0<∴gx)(1∞(01)x()g∵(x)∴

1<b＋e1.e∴b1＋ee.已知函數f)＝－x，x∈.試求函數fx)遞減區間；試求函數fx)區間－，上最值.解(1)′()－′(π＋2<k∈Z

3323622222233236222222ππ∴f(x)(＋kk∈上πππ(1)f)(π)(π)(，)ππ3πππ∴fx)x處()x處f()∵π)π，f(∴(x，(π).已知函f()＝alnx－1.求曲線＝()在點(1，(1))處的切線方程；若f)＞(a＋1)ln＋ax(1，＋∞)恒成立，求a的值范圍.解(1)′()＋x(0)x∴′(1)af(1)∴y(a2)(1)(2)yaf(x)(a1)lnxxlnln∵x∴ax.xlngx)xxln1g(xxhx)xlnxh(x2＞0x∴hx)(∞)h∴x∈(1∞)()∴′(x∴gx)(∞)∴gx)g1∴≤1a(x∴a(∞1].已知函f()＝x－＋a(x－1)有兩個零.求取值范圍；設x，是f)兩個零點，證明：x＋12解

f′(x(1)ea(x1)(xa①0()(

(x②x∈∞f′

22222x2]22222x2]上點的個數222x∈∞)f′(f)(∞1)(1∞)(1)(2)ab<0<ln(b)>(bab

2

>0(x)③′()0x1xaea≥，ln(a≤1∈∞)′()>0f((∞).x≤1(f)e<ln(2a∈(1)))<0∈(ln(2a)f)(2))(ln(2a)∞.x≤1(f)(0∞).證明x<x(1)∈(∞x∈∞2x∈∞1)fx)(12∞1)x(x)>(2))<0.12fxe2a(x1)2(x(xx(x1)02222fxe2(2)ex22(x)e(x2)e′)(x1)(ee)x′(x)<0g0x>1(x)<0g)f)<0<2.2212已知函f()＝－ax(a∈).若f)在點(，處的切線與直線x－2＋1＝0垂，求實數的值；求函數fx)單調區間；討論函數fx)區間，解(1)()(0∞∵(x)lnxax

1∴′(x)－x

222222224224222424222222224224222424xy14∴×∴241(1)f′(x－ax.xa≤f′(∴()(0∞.f′(xx<

′(x

∴(x)(0

)(

∞≤0f(x)∞)fx)(0

)(

∞).(2)a<0時f)[

]∵a>0∴fx[]afx)[1e]∵a∴(x)[1e]①

≤1a≥1()[

]∵a<0∴fx[1

]②1<

<efx)[e

][

]∵a<0(

1)lna2f)ae.1ln<0a(x)[1]2e1ln＝0a時fx)[]21114ln>0<<fx)[1]2e2.

24