* ·2· ·3· ·4· ·5·a+b22+b222a+a+b22+b222a+b≤ab≤≤2．對數(shù)平均不等式：若a>b>0，b<11<ab<ln--b<ab<a2b2<a，2+b2+y2-xy=1，則()A.x+y≤1B.x+y≥-2C.x2+y2≤2D.x2+y2≥1由x2+y2-xy=1可變形為，(x+y(2-1=3xy≤3(xy(2，解得-2≤x+y≤2，僅當(dāng)x=y=-1時，x+y=-2，僅當(dāng)x=y=1時，x+y=2，所以A錯B對；2+y2-xy=1可變形為(x2+y2(-1=xy≤x2y2，解得x2+y2≤2，、、例2.求證：ln(n+1(<1+++?+<ln(2n+1(.令a=n，b=n+1，則ln(n+1(-lnn<，可得ln(n+1(<1+++?+.根據(jù)b>a>0時，ab>ln--a，即lnb-lna>2+-(，令b=2n+1，a=2n-1，則ln(2n+1(-ln(2n-1(>，易證1+++?+<ln(2n+1(.·6·=b=c取等號.設(shè)ai>0(i=1,2,3,??n(：n=na1a2a3??an；n=；n=均值不等式：Hn≤Gn≤An≤Qn，等號成立的條件均為：a1=a2=a3=??=an；+的最小值.+=x2+當(dāng)且僅當(dāng)x2=2+的最小值3；(2)由均值不等式得x+y=x+y+y≥333yy(，從而xy2≤,·7·≥2+【解析】錯解：由已知得4=x≥2+【解析】錯解：由已知得4=x+y≥2y=5++y+yy=5++y+y=(x+y(?y+xy=x且x+y=4，+x≥+x≥++b的最小值為 ++b的最小值為 a=aa=aaa -c+/5+c-2-c+/5+c-2+-c=c/5+++-c=c/5++ab+5c-2c-2-+(a+b(2=a=a+a2+2ab+b2=+(a+b(2=a=a+a2+2ab+b2=5a++ab≥---52+c-c+5+ab≥c+5、5=+c-c+5+ab≥c+5、5=(c-2(+、5=c≥10+5++-c-2c-2c-2cc-2c-2c-255當(dāng)且僅當(dāng)c=2+、2 +-c/5++-c/5++5c-2·8·b則t除了滿足自身的范圍外，還要滿足a≤f(t(≤b(即解不等式)；得范圍．因為三角函數(shù)公式的變形與多項式變形的公式不同，+y2=1?(1)線性式--與縱截距相關(guān)z=ax+by(3)距離式--與距離相關(guān)z=(x-a(2+(y-b(2·9·所以x+2y+2=2xy≤,令t=x+2y>0，則t+2≤，即t2-4t-8≥0，所以x+2y的最小值為2+23.由x+2y+2=2xy，得(x-1((2y-1(=3，x+2y=(x-1(+(2y-1(+2≥2(x-1((2y-1(+2=23+2，3+13+1x=2y=所以x3+13+1x=2y=所以x+2y的最小值為2+23.令x+2y=k，所以x=k-2y，于是由x+2y+2=2xy，得k+2=2(k-2y(y，即4y2-2ky+k+2=0+23.令x+2y=t，所以y=-x+，·10·由y=-x+和(x>1(·11·形如：已知Ax2+Bxy+Cy2+Dx+Ey=F，求ax±by，mx2+ny2等的最值？依題(2x+y(2=8+6xy≤8+3(2xy(2，則(2x+y(2≤8，得-42≤2x+y≤42，+(x-y(2=8，故(2x+y(2=[3x+(y-x([2≤(32+12([(3x(2+(y-x(2[=4?法3：判別式法令2x+y=k，則y=k-2x，代入4x2-2xy+y2=8得：4x2-2x(k-2x(+(k-2x(2=8，,故f(x(在(2,+∞(減；所以x=2為f(x(的唯一極大值，也為最最大值，故f(x(max=f(2(=42；故=-+2kπ時，即當(dāng)x=2，y=22時，2x+y≤42；·12·y變量都有限制，所以相當(dāng)于二次函數(shù)在某個區(qū)間上有零點，那么如果直接用Δ≥0計算的答案就有可能出【解析】解：令x+y=k，所以x=k-y，代入等式得k-y+y+8=(k-y(y，整理得y2-ky+k+8=0，此時Δ=k2-4(k+8(≥0，即k2-4k-32≥0，解得k≤-4或k≥8，因為x>0，y>0，所以x+y=k≥8，即x+y的最小值為8代入等式得k2+(-k-c(2+c2=4，整理得c2+kc+k2-2=0此時Δ=k2-4(k2-2(≥0，即k2≤,解得k≤,所以a=k≤的最大值為≤.2+n2=k，所以m2=k-n2，代入等式得2(k-n2(+3n2=6n，整理得n2-6n+2k=0此時Δ=36-8k≥0，即k≤,所以m2+n2=k≤,即m2+n2的最大值為≤.·13··14·萬能k法并不是完全就是判別式法.令a+4b=k>0，得a=k-4b-k2b+4=0.(a+4b(2=a2+8ab+16b2=a(a+8b(+16b2=+16b2，=-+32b=，·15·故g(b(min=g=12，此時a=-2+2、3>0，·16·a>0，b>0a<0，b<0f(x(=(a≠0(；f(x(=(a≠0(；f(x(=；f(x(=(a>0(.①當(dāng)a≥5時，f(x(=a-x-+a=2a-x-最大值2a-4=5，∴a=，舍去；②當(dāng)a≤4時，f(x(=x+-a+a=x+≤5，此時命題成立；·17·③當(dāng)4<a<5時，[f(x([max=max{|4-a|+a,|5-a|+a{，·18·a>0，b<0a<0，b>0R；R；f(x(=(a≠0(；f(x(=(a≠0(；f(x(=；f(x(=(a>0(.5．重要放縮不等式：yy=-)-<lnx．y=lnxx->lnx．O1x一點的切線與直線x=0和直線y=x所圍成的三角形面積為定值.【解析】設(shè)P(x0,y0(為曲線上任一點知P(x0,y0(處的切線方程為y-y0=(1+(x-x0(，1+((x-x0(．令x=0，得y=-，從而得切線與直線x=0的交點坐標(biāo)為(0,-.·19·點P(x0,y0(處的切線與直線x=0，y=x，所圍成的三角形面積為-|x0|=6.·20·!!r{y!r{y!直線x=-，y=，直線x=-，y=，遞減區(qū)間(-∞,-遞增區(qū)間(-∞,-例1.(2021全國乙卷理4)設(shè)函數(shù)f(x(=則下列函數(shù)中為奇函數(shù)的是(A.f(x-1(-11B.f(x-1(+1C.f(x+1(-1D.f(x+1(+1例2.函數(shù)f(x(=的值域為.·21··22·導(dǎo)法作為備選方案.例1.函數(shù)y=(x>1(的最小值為.令t=x-1，則t>0，x=t+1，將變形為y(x2-x+1(=x2-2x+2，整理得(y-1(x2+(2-y(x+y-2=0，其判別式Δ=(2-y(2-4(y-1((y-2(≥0，解得：≤y≤2(y≠1(，設(shè)f(x(=(x>1(，則f/(x(=，令f/(x(=0得x=2，所以f/(x(>0?x>2，f/(x(<0?1<x<2，從而f(x(在上↘,在(2,+∞(上↗,故f(x(min=f(2(=.·23·a>0a<0RR對稱軸x=h對稱軸x=hf(x(=a|x-b|+2=，D.y=1-|x-1|(0≤x≤2(【解析】可類似于y=a(x-1(2+的圖象，由函數(shù)圖像可得選B.·24·f(x(=|x-a|+|x-b|f(x(=m|x-a|+n|x-b|(m,n>0(圖像m>n>0時0<m<n時特點值f(x(=|x-a|-|x-b|f(x(=m|x-a|-n|x-b|(m,n>0(圖像特“Z”字型在x≤a，x≥a上取得點備注：若是f(x(=-|x-a|+|x-b|，f(x(=-m|x-a|+n|x-b|(m,n>0(，只是如上雙絕對值差的相反情補充：三絕對值函數(shù)：f(x(=|x-a|+|x-b|+|x-c|(其中a<b<c(，A.5或8B.-1或5C.-1或-4D.-4或8-b|+|x-4|-|2x-5|≥0，則()A.a≤1，b≥3B.a≤1，b≤3C.a≥1，b≥3D.a≥1，b≤3-b|≥|2x-5|-|x-4|恒成立.·25·設(shè)f(x(=a|x-b|，g(x(=|2x-5|-·26·1．定義：如y=max{f(x(,g(x({，y=min{f(x(,g(x({，稱為最大值函數(shù)、最小值函數(shù)；2．圖像：y=max{f(x(,g(x({是保留兩者最高的圖象，最后得到最大值函數(shù)的圖象；y=min{f(x(,g(x({例1.用max{a,b}表示a，b兩個數(shù)中的最大值，設(shè)函數(shù)f(x(=max|x|,(x>0(，則f(x(的最小值是.當(dāng)x>0時，若x≥,則x2≥1，解得x≥1或x≤-1(舍去)，2<1所以f(x(=max|x|,=max作出函數(shù)圖象，如圖：當(dāng)0<x<1時，函數(shù)f(x(=單調(diào)遞減，所以f(x(>f(1(=1；例2.已知函數(shù)y=max{a,b,c}，設(shè)f(x(=max{x2,|x-1|,3x{，則f(x(的最小值為.【解析】在同一直角坐標(biāo)系中作出函數(shù)y=x2，y=|x-1|，y=3x，根據(jù)題意可得函數(shù)f(x(=max{x2,|x-1|,3x{為圖中黑線表示部分，·27·根據(jù)圖像可得，點A為函數(shù)y=x2與y=|x-1|，(x<1(的交點，所以x2=1-x解得x=，故點A的橫坐標(biāo)為，點B為y=3x與y=|x-1|，(x<1(的交點，故3x=1-x，得x=，故點B橫坐標(biāo)為，所以函數(shù)f(x(=·28·例1.(2024年九省聯(lián)考14題)以maxM表示數(shù)集M中最大的數(shù)．設(shè)0<a<b<c<1，已知b≥2a或a+b≤1，則max{b-a,c-b,1-c}的最小值為.max{b-a,c-b,1-c}=max{a,b-a,c-b,1-c}≥=，(2)當(dāng)a+b≤1時，max{b-a,c-b,1-c}≥=≥=，綜上可知max{b-a,c-b,1-c}的最小值為.(1)若b≥2a，則b=1-n-p≥2(1-m-n-p(，故2m+n+p≥1，(2)若a+b≤1，則1-n-p+1-m-n-p≤1，即m+2n+2p≥1，僅當(dāng)m+2n+2p=1且max{m,n,p時等號成立，如取m=n=p=時可滿足等號成立，綜上可知max{b-a,c-b,1-c}的最小值為-a,c-b,1-c}=max{|AB|,|BC|,|CD|{.因為|AB|+|BC|+|CD|=|AD|，所以max{|AB|,|BC|,|CD|{取到最小值的必要條件為|AD|最小，由|OA|+|AD|=1可知|AD|最小時，|OA|最大．在|OA|最大時，|AD|為定值，當(dāng)|AB|=|BC|=|CD|時，max{|AB|，·29·|BC|,|CD|{最小.此時，當(dāng)|OA|=|AB|=|BC|=|CD|=時，max{|AB|,|BC|,|CD|{最小值為；當(dāng)|AB|=|BC|=|CD|時，|OA|=|BD|=2|AB|，此時|OD|=5|AB|=1，即此時|BC|=|CD|=時，max{|AB|,|BC|,|CD|{最小值為；綜合(1)(2)可知，max{|AB|,|BC|,|CD|{最小值為，即max{b-a,c-b,1-c}最小值為.析可得max{|AB|,|BC|,|CD|{取得最小值的充要條件：|AD|(|AB|+|BC|+|CD|(最小且|AB|=|BC|=|CD|.·30·①x=[x[+{x}；②x-1<[x[≤x<[x[+1；x[+[y[≤[x+y[.雙曲正弦函數(shù):sinh(x(=雙曲正切函數(shù):tanh(x(=x(-sinh2(x(=1，tanh(x(=①sinh(x±y(=sinh(x(cos·31·②值域：{-1,0,1}；它的周期是任意負(fù)有理數(shù)和正有理數(shù).②值域：{-1,0,1}；③關(guān)于x=對稱.·32··33·模型1：若f(x±y(=f(x(±f(y(，則f(x(=f(1(x，(即f(x(為奇函數(shù))；模型2：若f(x±y(=f(x(?f(y(+m，則f(x(=[f(1(±m(xù)[x?m；模型1：若f(x+y(=f(x(f(y(，則f(x(=[f(1([x，f(x(>0；模型2：若f(x-y(=，則f(x(=[f(1([x，f模型3：若f(x+y(=mf(x(f(y(，則f(x(= 模型4：若f(x-y(=m，則f(x(=m模型1：若f(xy(=f(x(+f(y(，則f(x(=f(a(logax；模型=f(x(-f(y(，則f(x(=f(a(logax；模型3：若f(xy(=f(x(+f(y(+m，則f(x(=[f(a(+m[logax-m；=f(x(-f(y(+m，則f(x(=[f(a(-m[logax+m；模型1：若f(xy(=f(x(f(y(，則f(x(=[f(a([logx，代入f(a(則可化簡為冪函數(shù)；f(x(=cosωx；模型2：若f(x+y(+f(x(-f(y(=2f(x(f(y(，則f(x(=cosωx；模型3：若f(x+y(+f(x(-f(y(=kf(x(f(y(，則f(x(=cosωx；f(x(=tanωx.例1.(2022年新課標(biāo)2卷8)已知函數(shù)f(x(的定義域為R，且f(x+y(+f(x-y(=f(x(f(y(，f(1(=1，則f(k(=()·34·A.-3B.-2C.0D.1cos(x+y(+cos(x-y(=2cosxcosy，可設(shè)f(x(=acosωx，f(x(=2cos，所以f(x(=2cos符合條件，因此f(x(的周期，f(0(=2，f(1(=1，且f(2(=-1，f(3(=-2，f(4(=-1，f(5(=1，f(6(=2，所以f(1(+f(2(+f(3(+f(4(+f(5(+f(6(=0，f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選A.·35·(1)??-2，-1，0，1，2??等特殊值代入求解；(2)通過f(x1(-f(x2(的變換判定單調(diào)性；(3)令式子中出現(xiàn)f(x(及f(-x(判定抽象函數(shù)的奇偶性；(4)換x為x+T確定周期性.①若給出的是“和型”抽象函數(shù)f(x+y(=??,判斷符號時要變形為：f(x2(-f(x1(=f((x2-x1(+x1(-f(x1(或f(x2(-f(x1(=f(x2(-f((x1-x2(+x2(；@若給出的是“積型”抽象函數(shù)f(xy(=??,判斷符號時要變形為：f(1(等，①如f(x+y(，可令y=-x；@如f(xy(，可令y=-1等等；例1.已知函數(shù)f(x(的定義域為R，f(xy(=y2f(x(+x2f(y(，則()A.f(0(=0B.f(1(=0C.f(x(是奇函數(shù)D.f(x(是偶函數(shù)f(xy(=y2f(x(+x2f(y(中，令x=y=0得，f(0(=0，A正確；f(xy(=y2f(x(+x2f(y(中，令x=y=1得，f(1(=f(1(+f(1(，得f(1(=0，B正確；f(xy(=y2f(x(+x2f(y(中，令x=y=-1得，f(1(=f(-1(+f(-1(，解得f(-1(=0，f(xy(=y2f(x(+x2f(y(中，令y=-1得，f(-x(=f(x(+x2f(-1(=f(x(+0=f(x(，例2.(2022年新課標(biāo)2卷8)已知函數(shù)f(x(的定義域為R，且f(x+y(+f(x-y(=f(x(f(y(，f(1(=1，則Σk1f(k(=()A.-3B.-2C.0D.1·36·因f(x+y(+f(x-y(=f(x(f(y(，令x=1，y=0可得2f(1(=f(1(f(0(，f(0(=2，令x=0可得，f(y(+f(-y(=2f(y(，即f(y(=f(-y(，所以函數(shù)f(x(為偶函數(shù)，令y=1得，f(x+1(+f(x-1(=f(x(f(1(=f(x(，即有f(x+2(+f(x(=f(x+1(，從而可知f(x+2(=-f(x-1(，f(x-1(=-f(x-4(，故f(x+2(=f(x-4(，即f(x(=f(x+6(，所以函數(shù)f(x(的一個周期為6.因為f(2(=f(1(-f(0(=1-2=-1，f(3(=f(2(-f(1(=-1-1=-2，f(4(=f(-2(=f(2(=-1，f(5(=f(-1(=f(1(=1，f(6(=f(0(=2，所以一個周期內(nèi)的f(1(+f(2(+…+f(6(=0．由于22除以6余4，所以f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選：A.·37·①f(x+a(+f(b-x(=2c?y=f(x(圖像關(guān)于直線,c(對稱；@f(x+a(=f(b-x(?y=f(x(圖像關(guān)于直線x=對稱.如f(x(=對稱中心(-,=ax3+bx2+cx+d對稱中心為(-,f(-；f(x(=對稱中心(loga|t|,=loga對稱中心(-+loga等.形.f(x(=ln-+ax+b(x-1(3的定義域為(0,2)，設(shè)P(m,n(為y=f(x(圖象上任意一點，故n=ln-+am+b(m-1(3，而f(2-m(=ln+a(2-m(+b(2-m-1(3=-ln-+am+b(m-1(3+2a=-n+2a=-f(m(+2a，即f(2-m(=-f(m(+2a，例3.已知f(x(是定義在R上的增函數(shù)，且f(x(+f(-x(=-2，函數(shù)g(x(=f(x(++x，則g(x(圖像關(guān)于對稱.【解析】因為f(x(+f(-x(=-2，所以f(x(關(guān)于(0，-1)對稱，由g(x(=0得f(x(=-x-1，記h(x(=-x-1，因為h(x(+h(-x(=(-x-1(+(+x-1(-x-1(+(+x-1(=-2，所以h(x(關(guān)于對稱，所以g(x(=f(x(+的零點關(guān)于(0，-1)對稱.·38·(2)抽象函數(shù)下的結(jié)論：主要型如f(x+T(=f(x(等變形結(jié)構(gòu)的函數(shù)；①y=f(x(有兩條對稱軸x=a和x=b(b>a(，則f(x(的周期是T=2(b-a(；②y=f(x(有兩個對稱中心(a,0)和(b,0((b>a(，則f(x(的周期是T=2(b-a(；③y=f(x(有一條對稱軸x=a和一個對稱中心(b,0((b>a(，則f(x(的周期是T=4(b-a(；推論1：偶函數(shù)y=f(x(滿足f(a+x(=f(a-x(?f(x(周期T=2a；推論2：奇函數(shù)y=f(x(滿足f(a+x(=-f(a-x(?f(x(周期T=2a.例1.(2024·河北滄州·一模)已知定義在R上的函數(shù)f(x(滿足：f(x(+f(2-x(=2，f(x(-f(4-x)=0，且f(i(=()又∵f(x(-f(4-x(=0，∴f(2-x(=f(4-(2-x((=f(2+x(，②即函數(shù)f(x(的圖象關(guān)于直線x=2對稱，f(3(=f(1(=1且由①和②,得f(x(+f(2+x(=2?f(2+x(+f(4+x(=2，所以f(x(=f(4+x(，則函數(shù)f(x(的一個周期為4，則f(4(=f(0(=2，例2.(2024·陜西榆林·一模改編)定義在R上的周期函數(shù)f(x(，g(x(，滿足f(3-x(=f(1+x(，g(2-x(+【解析】由f(3-x(=f(1+x(可得f(2-x(=f(2+x(，所以f(x(關(guān)于直線x=2對稱，所以f(2x(關(guān)于直線x=1對稱，即g(x+關(guān)于直線x=1對稱，·39·又由g(2-x(+g(x(=2可得g(x(關(guān)于點(1,1)對稱，(x(的周期為T2=2；由已知f(2x(=g(x+周期為2，所以f(x(的周期T1=4.則T+T2=6.·40·周期函數(shù)的圖象可以看成將函數(shù)f(x(一個周期的圖象向左或向右不斷平移來得到．若在每次平移的同時，下面以兩個具體的實例來給出常見的仿周期函數(shù)f(x(滿足的條件形式及其對應(yīng)的圖象．定義在[1,+∞)上的函數(shù)f(x(滿足當(dāng)x∈[1,2)時，f(x(=-x2+3x-2；A.(-B.C.D.如圖所示：當(dāng)2<x≤3時，f(x(=4f(x-2(=4(x-2((x-3(，令4(x-2((x-3(=-整理得9x2-45x+56=0，∴x1=，x2=·41··42·設(shè)y=g(t(，t=f(x(，且函數(shù)g(x(的值域為g(t(定義域的子集，那么y通過t的聯(lián)系而得到自變量x的函f(x(=2x，g(x(=x2-2x，計算g(f(2((．解答時先求f(2(=4，所值．例如已知f(x(=2x，g(x(=x2-2x，計算g(f(x((=0，求x？解答時由g(t(=t2-2t=0，解得t=0或t=2；再由f(x(=0和f(x(=2分別解得x∈?和x=1.(1)第一層是解關(guān)于f(x(的方程，觀察有幾個f(x((2)第二層是結(jié)合著第一層f(x(的值求出每一個f(x(被幾個x對應(yīng)；將x的個數(shù)匯總后即為g(f(x((=0的根的個數(shù).類型2：已知g(f(x((零點個數(shù)求參數(shù)的范圍(1)先估計關(guān)于f(x(的方程g(f(x((=0中f(x(解的個數(shù)；(2)再根據(jù)個數(shù)與f(x(的圖像特點，分配每個函數(shù)值fi(x(被幾個fi(x(所對應(yīng)，從而確定fi(x(的取值范圍，進(jìn)而決定參數(shù)的范圍.例1.若f(x(=x-則方程f2(x(-f(x(-6=0的實根個數(shù)為.由方程f2(x(-f(x(-6=0，得f(x(=3或f(x(=-2，所以方程f2(x(-f(x(-6=0的實根個數(shù)為3.·43·例2.若f(x(=|log2|>3，若m[f(x([2-3f(x(+4m=0有8個不相等的實根，則實數(shù)m的取【解析】若關(guān)于x的方程m[f(x([2-3f(x(+4m=0有8個不相等的實根，若m<0，則t1t2=4>0，t1+t2=不符合，所以m>0，·44·導(dǎo)函數(shù)f/(x(=3ax2+2bx+c，把Δ=4b2-12a2．圖象：三次函數(shù)f(x(=ax3+bx2+cx+d(a≠0(有以下6種可能的圖象：a>0a<0f/(x(f(x(有唯一的零點.①f(x(有一個零點)?f(x1(f(x2(>0，如下圖所示：②f(x(有兩個零點?f(x1(f(x2(=0，如下圖所示：③f(x(有三個零點?f(x1(f(x2(<0，如下圖所示：4．對稱中心：f(x(=ax3+bx2+cx+d(a≠0(f(-·45·(2)三次函數(shù)對稱中心橫坐標(biāo)為x0，兩極值點為x1，x2，則f’(x0(=-(x1-x2(2；選填相關(guān)題.·46·*過點P(m,n(可以作出三次函數(shù)f(x(=ax3+bx2+cx+d(a≠0(圖象的幾條切線，本質(zhì)上是研究方程n-f(x0(=f/(x0((m-x0(根的個數(shù)．結(jié)論如下圖所示：設(shè)f(x(=ax3+bx2+cx+d(a≠0(的三個零點分別為x1，x2，x3，則：0=k1+k2+k3；三次函數(shù)中心對稱點(x0,f(x0((且處切線的斜率為k0，f(x(在這三點處的切線的斜率分別為k1，k2，k3，定理1：在三次函數(shù)f(x(=ax3+bx2+cx+d(a≠0(的圖像上任取點P，過P作一條切線，切=2xr；推論：在三次函數(shù)f(x(=ax3+bx2+cx+d(a≠0(的圖像上任取點P，過點P作一條切線，切點為T，過點2xT=xM+xN=xR+xS定理2：三次函數(shù)f(x(=ax3+bx2+cx+d(a>0(的中心對稱點為點P(x0,f(x0((，極大值為m，且f(x(=2(x1<x2(，f(x(的極小值點為x3，區(qū)間[x1,x2[被x0與x3平分，即x2-x3=x3-x0=x0-x1(四段坐標(biāo)差相等).·47··48·函數(shù)f(x(的四性有單調(diào)性，奇偶性，對稱性，周期性，分析可見一輪復(fù)習(xí)材料．近年高考熱衷考察f(x(與已知函數(shù)f(x(連續(xù)且可導(dǎo)：1．若f(x(為奇函數(shù)，則f(x(為偶函數(shù)；若f(x(為偶函數(shù)，則f(x(為奇函數(shù).若f(x(的圖像關(guān)于x=a對稱，則f(x(的圖像關(guān)于(a,0)對稱.3．若f(x(的周期為T，則f(x(的的周期為T.備注：以上的性質(zhì)都是從原函數(shù)f(x(到推理導(dǎo)函數(shù)f(x(，逆向不一定成立.例1.已知函數(shù)f(x(的定義域為R，且滿足f(x(+f(3-x(=4，f(x(的導(dǎo)函數(shù)為g(x(，函數(shù)y=g(1+3x(-1為奇函數(shù)，則g(2024(=A.-3B.3C.-1D.1【解析】根據(jù)題意f(x(滿足f(x(+f(3-x(=4，兩邊同時求導(dǎo)可得：f(x(-f(3-x(=0，即g(x(=g(3-x(，即g(x(關(guān)于x=對稱①,又因為函數(shù)y=g(1+3x(-1為奇函數(shù)，所以g(x(的圖象關(guān)于點(1,1)對稱，即函數(shù)g(x(是周期為4×-1(=2的周期函數(shù)，例2.(2022新課標(biāo)卷12)已知函數(shù)f(x(及其導(dǎo)函數(shù)f(x(的定義域均為R，記g(x(=f(x(，若f(2+x(均為偶函數(shù)，則()A.f(0(=0B.g(-=0C.f(-1(=f(4(D.g(-1(=g(2(對于g(x(，因g(2+x(為偶函數(shù)，得g(2+x(=g(2-x(，即g(x(關(guān)于x=2對稱；對于f(x(，因f-2x(為偶函數(shù)，得f-2x(=f+2x(，即f-x(=f+x(，所以f(3-x(=f(x(，所以f(x(關(guān)于x=對稱；因g(x(=f(x(，所以g(x(關(guān)于,0(對稱；由上分析，從f(x(關(guān)于x=對稱，得f(-1(=f(4(，故C正確；從g(x(關(guān)于,0(對稱，得g=0，故g(-=g(-+2(=g=0，故B正確；從g(x(周期T=2，得g(-1(=g(-1+2(=g(1(=-g(2(，D錯誤；·49·若f(x(滿足題設(shè)，則f(x(+C(C為常數(shù))也滿足題設(shè)，故無法確定f(x(的值，故A錯誤．故選：BC．法·50·原函數(shù)f(x(到推理導(dǎo)函數(shù)f(x(，函數(shù)的奇偶對稱性質(zhì)發(fā)生變化，但是從導(dǎo)函數(shù)f(x(到推理原函數(shù)f(x(，性質(zhì)2：若f(x(為奇函數(shù)，若f(x0(=f(-x0(時(x0任意)，則f(x(為偶函數(shù).推論：若f(x(為偶函數(shù)，則原函數(shù)f(x(關(guān)于(0,f(0((對稱.0∈D使得f(x0(=f(2m-x0(，此時f(x(圖像關(guān)于x=m對稱.∈R，使得f(x0+T(=f(-x0(，則函數(shù)f(x(是周期期為T.例1.設(shè)定義在R上的函數(shù)f(x(與g(x(的導(dǎo)數(shù)分別為f(x(與g(x(，已知f(x(=g(3-x(-1，f(x+1(=A.f(x(+f(2-x(=0B.f(2(=0C.g(1-x(=g(1+x(D.g(x(+g(2-x(=0【解析】因為f(x(=g(3-x(-1，求導(dǎo)可得f(x(=-g(3-x(①,因為f(x+1(=g(x(，所以f(x(=g(x-1(②,所以①②得-g(3-x(=g(x-1(，由f(x(=g(x-1(，所以f(x(關(guān)于點(2,0)和直線x=1對稱，f(x(有周期4；可以取f(x(=cos+1，可得A選項錯誤.·51·選項B是利用f,(x(與g,(x(關(guān)系，并由g,(x(的對稱性得到；選項A是利用f,(x(的雙對稱得到f,(x(有周期，并利用性質(zhì)5的要點去舉反例.·52·ex≥x+1ex≥exex≥e2(x-1(.ex≥en+1(x-n(.lnx≤x-1.lnx≤x+n.sinx≤xx≤tanxcosx≥1-x2例1.函數(shù)f(x(=，若g(x(=f(x(-x+a只一個零點，則a的取值范圍是【解析】作出函數(shù)f(x(=如圖所示：若g(x(=f(x(-x+a只一個零點，則f(x(=與y=x-a只有一個交點，·53·y=x-a與y=ex-1只有一個交點，則-a≥0，即a≤0；·54·合.若直線l:y=kx+b與曲線y=f(x(和y=g(x(均相切，則求切線常見的方法有：例1.直線l:y=kx+b與函數(shù)曲線f(x(=2lnx，g(x(=-x2+4x-3均相切，求直線l的方程.法2：設(shè)直線l與y=f(x(的切點為(x1,y1(，與y=g(x(的切點為(x2,y2(，令F(x(=2lnx+x2-4x+3，F(xiàn),(x(=+2x-4，當(dāng)x>0，F(xiàn),(x(=+2x-4≥2-4=0，僅當(dāng)=2x即x=1時等號成立，函數(shù)F(x(在(0,+∞(上單調(diào)遞增，且F(1(=2ln1+12-4×1+3=0，代入y=x+2lnx1-2得y=2x-2，所以直線l的方程為y=2x-2.法3：設(shè)直線l與曲線y=f(x(的切點為(x1,2lnx1(，·55·因為fl(x(=所以曲線y=f(x(在點(x1,2lnx1(的切線方程為x+2lnx1-2，(-4(x+2lnx1+1=0，因為直線l與y=g(x(均相切，得Δ=0，得(-4(2-4(2lnx1+1(=0，·56·(2)表示切線y=ax+b=a|x-(-|在x軸上截距-的相反數(shù).設(shè)y=ax+b和f(x(的圖象相切于點P(x0,lnx0((x0>0(，因為f,(x(=，所以f(x(的圖象在點P處的切線方程為y-lnx0=(x-x0(，即x+lnx0-1，從而a=，b=lnx0-1，所以a+b=+lnx0-1， 所以φ,(x(>0?x>1，φ,(x(<0?0<x<1，易得f(x(在x=1處的切線方程為y=x-1，對于這條切線，a+b=1+(-1(=0，變式：若直線y=ax+b和f(x(=lnx的圖象相切，則的最小值為．解：y=ax+b?y=·57··58·，則斜率k=f,(x0(；(2)利用切點和斜率寫出切線方程為：y-y0=f,(x0((x-x0(；(3)又因為切線方程過點A(a,b(，點入切線得b-y0=f,(x0((a-x0(．(*(.【解析】∵y=(x+a(ex，∴y,=(x+1+a(ex，0=(x0+a(ex，切線斜率k=(x0+1+a(ex，切線方程為：y-(x0+a(ex=(x0+1+a(ex(x-x0(，∵切線過原點，∴-(x0+a(ex=(x0+1+a(ex(-x0(，2+4a>0，解得a<-4或a>0，xA.eb<aB.ea<bC.0<a<ebD.0<b<ea對函數(shù)y=ex求導(dǎo)得y,=ex，(1-t(et=(a+1-t(et.令f(t(=(a+1-t(et，則f,(t(=(a-t(et.所以f(t(max=f(a(=ea，f(t(→-∞,·59·畫出函數(shù)曲線y=ex的圖象如圖所示，·60·如圖，y=x+1是y=ex在(0,1)處的切線，有ex≥x+1恒成立；y處的切線，有l(wèi)nx≤x-1恒成立．在不等式“改造”或證明的過程中，有時借助于ex，lnx有關(guān)x≥x+1引出的放縮：(2)由lnx≤x-1(也可以記為lnx≤x，切點為(1,0()引出的放縮：圖②x--lnx+恒成立.令f(x(=2ex--2x+3(x>0(，則f,(x(=2ex--2(x>0(，所以f(x(≥f((=0，所以2ex-≥2x-3(當(dāng)且僅當(dāng)時等號成立)①.易知g(x(在(0,1]上是減函數(shù)，在[1,+∞)上是增函數(shù)，所以g(x(≥g(1(=0，所以2x-3≥lnx-當(dāng)且僅當(dāng)x=1時等號成立)②.因為①和②中的等號不能同時成立，所以由①和②,得2ex->lnx-，所以2ex--lnx+.·61·隱零點法不行可嘗試用凹凸反轉(zhuǎn)．如要證明f(x(>0，可把f(x(拆分成兩個函數(shù)g(x(，h(x(，放在不等式兩個單峰函數(shù)，然后利用導(dǎo)數(shù)分別求兩個函數(shù)的最值并進(jìn)行比較．f(x(=xexf(x(=xlnx例1.(2014·全國I卷改編)設(shè)函數(shù)f(x(=exlnx+(x>0(．證明：f(x(>;(2(f(x(=exlnx+(x>0(，從而f(x(>1等價于xlnx>xe-x-.構(gòu)造g(x(=xlnx，則(0,+∞(上的最大值為+2ex≤0≤,因為f(x(=elnx-ax，所以f(x(max=f(1(=-e.·62·構(gòu)造g(x(=-2e(x>0(，所以g(x(min=g(1(=-e.綜上，當(dāng)x>0時，f(x(≤g(x(，即f(x(≤-2e，即xf(x(-ex+2ex≤0.法2：由題意知，即證exlnx-ex2-ex+2ex≤0，從而等價于lnx-x+2≤.構(gòu)造g(x(=lnx-x+·63·x，則f(x(=aex-lnx-1≥-lnx-1；(3)當(dāng)x>0時，若a<0時，顯然ax<0，則g(x(=x-+sinx<sinx-；?≥0.因為a≥,所以f,(x(=aex-在區(qū)間(0,+∞(內(nèi)單調(diào)遞增.f,(x0(=aex-=0x即aex0=，f,(x(<0,+∞(，f,(x(>0，故在區(qū)間(0,x0(內(nèi)f(x(單調(diào)遞減，在區(qū)間(x0,+∞(內(nèi)當(dāng)a≥時，f(x(≥-lnx-1．下證-lnx-1≥0即可.·64·我們討論帶來了極大的困難.必要條件．這樣相當(dāng)于對參數(shù)增加了一個條件，對問題解決有很好2)例1.(2022屆武漢九月調(diào)考)已知函數(shù)f(x(=2(x-2(lnx+ax2-1.故f(x(=2(x-2(lnx+ax2-1≥2(x-2(lnx+x2-1，下只須證2(x-2(lnx+x2-1≥0，省略.例2.(2020全國1卷)已知函數(shù)f(x(=ex+ax2-x.當(dāng)x≥0時，f(x(≥x3+1，求a的取值范圍.下只須證x2-x≥x3+1，省略.h(x(與g(x(相切于點P(2,這正是取x=的原因所在.·65·往是使結(jié)論成立的臨界條件，這種觀察區(qū)間端點值來解決問題的做利用“端點效應(yīng)”解決問題的一般步驟可分為以(1)如果函數(shù)f(x(在區(qū)間[a,b[上，f(x(≥0恒成立，則f(a(≥0或f(b(≥0.(2)如果函數(shù)f(x(在區(qū)間[a,b[上，f(x(≥0恒成立，且.a(=0?φn(a(≥(≤(0，且函數(shù)即其各階導(dǎo)函數(shù)在給定區(qū)間內(nèi)單調(diào)f(x(+sinx<0，求a的取值范圍.令(2(g(x(=f(x(+sinx=ax-+sinx(0<x<(，則g,(x(=a-+cosx，①當(dāng)a=0時，因為g(x(=sinx-=sinx(1-所以g(x(=f(x(+sinx=sinx-滿足題意；②當(dāng)a<0時，由于0<x<，顯然ax<0，所以g(x(=f(x(+sinx=ax-+sinx<sinx-滿足題意；·66·例2.(2020全國1卷)已知函數(shù)f(x(=ex+ax2-x.當(dāng)x≥0時，f(x(≥x3+1，求a的取值范圍.誤解：設(shè)g(x(=ex-x3+ax2-x-1≥0，且g(0(=0，·67·進(jìn)而把握整個問題，促成問題轉(zhuǎn)化，直至徹底解決．它是唯物辯證化.到解決問題的目的.【解析】將原不等式看成關(guān)于變量a得不等式，即a2-(x2+2x(a+x3-1≤0，其判別式為Δ=(x2+2x(2-4(x3-1(=x4+4x2+4，因此不等式的解為x-1≤a≤x2+x+1，-1,1[，所以(x-1(max=0，(x2+x+1(min=故答案為.令h(a(=exa-lnx-1，a≥,則h,(a(=ex>0，x>0，故h(a(在a≥上遞增，那么h(a(≥h((=ex-1-lnx-1，x-1≥x?ex-1-1≥x-1，等號成立當(dāng)且僅當(dāng)x=1，綜上，h(a(≥ex-1-lnx-1≥0，等號成立當(dāng)且僅當(dāng)x=1．證畢.·68·在證明或處理含對數(shù)函數(shù)不等式時，如f(x(為可導(dǎo)函數(shù)，則有(f(x(lnx(,=f,(x(lnx+，若f(x(為非(1)設(shè)f(x(>0，f(x(lnx+g(x(>0?lnx+>0，則(lnx+,=+,，不含超越函數(shù)，求解過程簡單．或者f(x(lnx+g(x(>0?f(x((lnx+>0，即將前面部分提出，就留下lnx這個單f(x(lnx+g(x(=0?lnx+=0，則(lnx+,=+求解過程簡單．或者f(x(lnx+g(x(=0?f(x((lnx+=0，即將前面部分提出，就留下lnx這個單例1.(2016·全國II卷)已知函數(shù)f(x(=(x+1(lnx-a(x-1(.f(x(=(x+1(lnx-a(x-1(>0等價于lnx->0.①當(dāng)a≤2，x∈(1,+∞(時，x2+2(1-a(x+1≥x2-2x+1>0，故g,(x(>0，g(x(在(1,+∞(上單調(diào)遞增，因此g(x(>0；②當(dāng)a>2時，令g,(x(=0得x1=a-1-(a-1(2-1，x2=a-1+(a-1(2-1.2>1和x1x2=1得0<x1<1，例2.(2018．全國III卷)已知函數(shù)f(x(=(2+x+ax2(?ln(1+x(-2x.【解析】當(dāng)a=0時，f(x(=(2+x(ln(1+x(-2x(x>-1(，由于2+x>0.-1,+∞(，g,(0(>0．所以g(x(在(-1,+∞(上單調(diào)遞增.故當(dāng)-1<x<0時，f(x(<0；當(dāng)x>0時，f(x(>0.·69·運算.例1.(2018·全國II卷)已知函數(shù)f(x(=ex-ax2.≥1；≥1等價于(x2+1(e-x-1≤0．設(shè)g(x(=(x2+1(e-x-1，下省略.例2.(2020·全國I)已知函數(shù)f(x(=ex+ax2-x.【解析】(2(f(x(≥x3+1等價于x3-ax2+x+1(e-x≤1．設(shè)函數(shù)g(x(=x3-ax2+x+1(e-x(x≥0(，(ii)若0<2a+1<2，即-<a<，則當(dāng)0<x<2a+1或x>2時，g’(x(<0；當(dāng)2a+1<x<2時，(x(>0.由于g(0(=1，所以g(x(≤1僅當(dāng)g(2(=(7-4a(e-2≤1，即a≥.所以當(dāng)≤a<時，(iii)若2a+1≥2，即a≥,則g(x(≤x3+x+1(e-x.3-ax2+x+1(e-x≤1．故當(dāng)時，g(x(≤1..·70·xlnx(,=exlnx+并不能簡便，處理時經(jīng)常要分離，也就是所稱的“指對在一例1.(2014·全國I卷改編)設(shè)函數(shù)f(x(=exlnx+(x>0(．證明：f(x(>1.解：f(x(=exlnx+等價于xlnx>xe-x-．下省略備注：另可見凹凸反轉(zhuǎn)技巧.例2.已知f(x(=ex-alnx-a，其中常數(shù)a>0.2x2-ex1lnx-x≥0.fx(=ex+>0，∴f,(x(在(0,+∞(單調(diào)遞增.f,(x(<f,(1(=0，x∈(1,+∞(，f,(x(>f,(1(=0.x-elnx-e≥0?ex-elnx≥e，所證不等式e2x-2-ex-1lnx-x≥0?ex-elnx≥.設(shè)g(x(==xe2-x，g,(x(=e2-x-xe2-x=(1-x(e2-x，令g,(x(>0解得x<1，令g,(x(<0解得x>1.·71·:g(x(max=g(1(=e．:ex-elnx≥e≥g(x(，x-elnx≥,:e2x-2-ex-1lnx-x≥0.·72·x=ex+lnx，=ex-lnx，=elnx-x(4)x+lnx=ln(xex(，x-lnx=ln，lnx-x=lnx2ex=ex+2lnx，=ex-2lnx，=e2lnx-x(6)x+2lnx=ln(x2ex(，x-2lnx=ln2lnx-x=ln??再結(jié)合常用切線不等式：ex≥x+1，ex≥ex，lnx≤x+1，lnx≤等，可以得到更多x=ex+lnx≥x+lnx+1，xex=ex+lnx≥e(x+lnx(，x+lnx=ln(xex(≤xex-1，x+lnx=ln(xex(≤=xex-1.=ex-lnx≥x-lnx+1，=ex-lnx≥e(x-lnx(，x-lnx=ln≤-1，x-lnx=ln≤(9)=elnx-x≥lnx-x+1，=elnx-x≥e(lnx-x(，lnx-x=ln≤-1，lnx-x=ln≤例1.已知f(x(=lnx+x-xex+1，則函數(shù)f(x(的最大值為.f(x(=lnx+x-xex+1=x+lnx-ex+lnx+1≤x+lnx-(x+lnx+2(=-2.(當(dāng)且僅當(dāng)x+lnx+1=0取等號).例2.不等式xex-ax-lnx-1≥0恒成立，則實數(shù)a的最大值是.x-ax-lnx-1≥0恒成立?a≤((min當(dāng)且僅當(dāng)x+lnx=0等號成立.·73·=xex，3lnx≥r|a(lnx< alnea<lnba-lna<lnb-ln(lnb(是→f(x(==xlnx=lnx x|a±a>elnb±lnbaa>b±lnb________構(gòu)造函數(shù)a±a>elnb±lnbax+ax>ln(x+1(+x+1?eax+ax>eln(x+1(+ln(x+1(?ax>ln(x+1(.ax>lnx-同乘x(無中生有(→axeax>xlnx，后面的轉(zhuǎn)化同2(1(；x>aln(ax-a(-a?ex>lna(x-1(-1?ex-lna-lna>ln(x-1(-1問加x(無中生有)x-lna+x-lna>ln(x-1(+x-1=eln(x-1(+ln(x-1(?x-lna>ln(x-1(；x>logax?exlna>?(xlna(exlna>xlnx，后面的轉(zhuǎn)化同2(1(.lnxlnax>logax?exlnxlna(xlna(exlna>(lnx(elnx→f(x(=xexxlna+ln(xlna(>lnx+ln(xlna(exlna>(lnx(elnx→f(x(=xexxlna+ln(xlna(>lnx+ln(lnx(→f(x(=x+(2)f(x(=aex1-lnx+lna=elnax1-lnx+lna≥1等價elnax1+lna+x-1≥lnx+x=elnx+lnx，令g(x(=ex+x，上述不等式等價于g(lna+x-1(≥g(lnx(，顯然g(x(為單調(diào)遞增函數(shù)，∴又等價于lna+x-1≥lnx，即lna≥lnx-x+1，令h(x(=lnx-x+1，則h,(x(=，·74·x2的具有輪換對稱性的不等式恒成立，求參數(shù)變x12或pf(x1(-kx1<f(x2(-kx2?y=f(x(-kx為增函數(shù)；?f(x1(+>f(x2(+?y=f(x(+為減函數(shù)；例1.已知函數(shù)f(x(=ex+mlnx(m∈R(，若對任意正數(shù)x1，x2，當(dāng)x1>x2時，都有f(x1(-f(x2(>x1-x2【解析】由f(x1(-f(x2(>x1-x2得，f(x1(-x1>f(x2(-x2，令g(x(=f(x(-x，∴g(x1(>g(x2(，∴g(x(在(0,+∞(單調(diào)遞增，又∵g(x(=f(x(-x=ex+mx-x，∴g,(x(=ex+-1≥0在(0,+∞(上恒成立，即m≥(1-ex(x，令h(x(=(1-ex(x，則h,(x(=-ex(x+1(+1<0，f(x1(<x2f(x2(，令g(x(=xf(x(，則g(x(在(0,+,(x(=ex-2ax≥0在(0,+∞(上恒成立，即a≤,·75·雙變量問題是導(dǎo)數(shù)綜合題的一個難點，其困難之處是如何消元1+x2=1時，則可利用x2=1-x1消去x2；1x2=12=消去x2；例1.(2018?全國I卷)已知函數(shù)f(x(=-x+alnx.<a-2.由于f(x(的兩個極值點x1，x2滿足x2-ax+1=0，所以<a-2等價于-x2+2lnx2<0.所以-x2+2lnx2<0，即<a-2.例2.已知函數(shù)f(x(=ax2-x-ln.(2)若函數(shù)f(x(在定義域內(nèi)有兩個極值點x1，x2，求證：f(x1(+f(x2(<2ln2-3.f(x(=2ax-1+，2-x+1=0在(0,+∞(上有兩個不等實根x1，x2，∴Δ=1-8a>0，x1+x2=>0，x1x2=>0，∴0<a<.∴f(x1(+f(x2(=ax+ax-(x1+x2(+lnx1+lnx2=a(x+x(-(x1+x2(+ln(x1x2(=a[(x1+x2(2-2x1x2[-(x1+x2(+ln(x1x2(=ln--1，·76·t(在(4,+∞(上單調(diào)遞減.∴g(t(<ln4-3=2ln2-3，即f(x1(+f(x2(<2ln2-3.·77·若兩個變量不存在確定的關(guān)系，有時可以將兩個變量之間x2；-x2；+x2；例1.已知函數(shù)f(x(=x2+xlnx．(1)省略；(2)求證：當(dāng)n>m>0時，lnn-lnm>-.【解析】(2)要證lnn-lnm>-，即證ln>-，只需證ln-+>0.令構(gòu)造函數(shù)g(x(=lnx-+x(x≥1(，則g,(x(=x(=+1>0，故g(x(在(1,+∞(上單調(diào)遞增.即證得ln-+>0成立，得證.例2.已知函數(shù)f(x(=lnx-ax2+x，a∈R.2滿足f(x1(+f(x2(+x1x2=0，證明：x1+x2≥.【解析】(2)當(dāng)a=-2時，f(x(=lnx+x2+x，x>0．由f(x1(+f(x2(+x1x2=0，得lnx1+x+x1+lnx2+x+x2+x1x2=0，從而(x1+x2(2+(x1+x2(=x1x2-ln(x1x2(，φ(t(=t-lnt，得φ,(t(=1-，由φ,(t(>0得t>1，φ,(t(<0得t<1；所以φ(t(≥φ(1(=1，所以(x1+x2(2+(x1+x2(≥1，因為x1>0，x2>0，所以x1+x2≥成立.·78·A.b<a<cB.b<c<aC.c<b<aD.a<c<b22.8>log22=1，b=log0.82.8<log0.81=0，0<c=2-0.8<20=1，A.c<b<aB.b<a<cC.a<c<bD.a<b<cA.a<b<cB.b<a<cC.b<c<aD.c<a<b==?<2=2=2<1，∴a<b；b=55<84c=84<85綜上所述，a<b<cA.a<b<cB.c<b<aC.c<a<bD.a<c<b·79·②a-c=0.1e0.1+ln(1-0.1(，令g(x(=xex+ln(1-x(，x∈,則g,(x(=xex+ex-，令k(x(=(1+x((1-x(ex-1，所以k,(x(=(1-x2-2x(ex>0，故c<a<b.·80·等比較大?。袝r也要在相等關(guān)系中去放縮尋找不等關(guān)系，再構(gòu)A.a>b>cB.c>b>aC.c>a>bD.a>c>b【解析】根據(jù)題意，構(gòu)造函數(shù)y=xe2-x，所以y=(1-x(e2-x，當(dāng)0<x<1時y=(1-x(e2-x>0，所以y=xe2-x在(0,1)上遞增，A.c>b>aB.c>a>bC.b>c>aD.a>b>c構(gòu)造函數(shù)f(x(=(x>0(，則f(x(=，當(dāng)0<x<e2時，f(x(>0，則f(x(在(0,e2(上單調(diào)遞增，所以f(3(<f(4(<f(5(，即，所以lna<lnc<lnb，又函數(shù)y=lnx在(0,+∞(上單調(diào)遞增，所以b>c>a，故選C.·81·(1)0<x<，tanx>x>sinx，sinx≥x-x2，cosx≥1-x2；2-x≥xlnx≥x-1≥lnx≥1-≥,x+1≤ex≤-(x<1(；A.c>a>bB.a>c>bC.b>a>cIa>b>cA.c>b>aB.b>a>cC.a>b>cD.a>c>b設(shè)f(x(=cosx+x2-1，x∈(0,+∞(，f,(x(=-sinx+x>0，所以f(x(在(0,+∞(單調(diào)遞增，則 當(dāng)2>1-22=，故b>a此時sin=cosφ=，cos=sinφ=，故cos=<=sin<4sin，故b<c所以b>a．所以c>b>a，故選A·82·eπ=-≈1+x+x2+x3?,≈1-x+x2-x3， ex≈1+x+x2+x3，ln(1+x(≈x-x2+x3，sinx≈x-，cosx≈1-+，tanx≈x+@帕德逼近-2≤x≤2(，ln(x(≈,(0<x<2(，、1+x≈1+x-x2，(-≤x≤A.c>b>aB.b>a>cC.a>b>cD.a>c>bb=cos≈1-+，計算得c>b>a，故選A.A.a<b<cB.c<b<aC.c<a<bD.a<c<ba=0.1e0.1=0.1c=-ln0.9=-≈0.1.53，·83··84·x≥1+x(取等?x=0)x≥ex(取等?x=1)(3)x≥0?ex≥1+x+x2(取7(?x=0(x<0?ex<1+x+x2x≥、e(x+取?x=x>x2+1x≥0?ex≥ex+(x-1(x<0?ex<ex+(x-1(2(8(x≥-?ex≥(x+1(2(取等?x=1(x+e-x≥x2+2(即?x=0){x<0?ex>-+(10)0≤x<2?ex≤{x<0?ex>-+x=ex-lnx≥x-lnx+1x=ex+lnx≥x+lnx+x=ex-lnx≥x-lnx+1(1)x>0?1-≤lnx≤x-1≤x2-x(取等?x=1)(2)x>0?x-x2<ln(x+1(<x(取等?x=0((3)x>0?lnx≤(取等?x=e)(4)x>1?1-<2+-11(<x32(1<lnx<x-1x<(x-(<x-10<x<1?1-<(x-(<x-1x<lnx<x32(1<2+-11(|x>1?lnx>r|x>1?lnx>x(x+1|0<x<1?lnx<4((x+1(6)x>1?lnx>62(+-(0<x<1?lnx<62(+-(·85·x>0?lnx≤取等?x=1)x>0?lnx≤取等?x=1)x∈R?1-x2≤cosx≤1-即等?x=0)?cosx≤1-取等?x=0)(10(x>-1?x+sinx≥≥2ln(1+x((取等?x=0)·86··87·y=ex+x過定y=lnx+xy=ex-x函數(shù)y=lnx-x函y=xex函數(shù)極y=x