新高考數(shù)學(xué)一輪復(fù)習(xí)提升練習(xí)考點(diǎn)14 導(dǎo)數(shù)的概念及應(yīng)用 (含解析)

考向14導(dǎo)數(shù)的概念及應(yīng)用1．（2021·全國(guó)高考真題）若過(guò)點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】解法一：根據(jù)導(dǎo)數(shù)幾何意義求得切線方程，再構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)圖象，結(jié)合圖形確定結(jié)果；解法二：畫(huà)出曲線SKIPIF1<0的圖象，根據(jù)直觀即可判定點(diǎn)SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時(shí)才可以作出兩條切線.【詳解】在曲線SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，由題意可知，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個(gè)交點(diǎn).故選：D.解法二：畫(huà)出函數(shù)曲線SKIPIF1<0的圖象如圖所示，根據(jù)直觀即可判定點(diǎn)SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時(shí)才可以作出兩條切線.由此可知SKIPIF1<0.故選：D.【點(diǎn)睛】解法一是嚴(yán)格的證明求解方法，其中的極限處理在中學(xué)知識(shí)范圍內(nèi)需要用到指數(shù)函數(shù)的增長(zhǎng)特性進(jìn)行估計(jì)，解法二是根據(jù)基于對(duì)指數(shù)函數(shù)的圖象的清晰的理解與認(rèn)識(shí)的基礎(chǔ)上，直觀解決問(wèn)題的有效方法.2．（2021·北京高考真題）已知函數(shù)SKIPIF1<0．（1）若SKIPIF1<0，求SKIPIF1<0在SKIPIF1<0處切線方程；（2）若函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值，求SKIPIF1<0的單調(diào)區(qū)間，以及最大值和最小值．【答案】（1）SKIPIF1<0；（2）函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0，最大值為SKIPIF1<0，最小值為SKIPIF1<0.【分析】（1）求出SKIPIF1<0、SKIPIF1<0的值，利用點(diǎn)斜式可得出所求切線的方程；（2）由SKIPIF1<0可求得實(shí)數(shù)SKIPIF1<0的值，然后利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0的單調(diào)性與極值，由此可得出結(jié)果.【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，此時(shí)，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0；（2）因?yàn)镾KIPIF1<0，則SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0增極大值減極小值增所以，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.所以，SKIPIF1<0，SKIPIF1<0.1.求函數(shù)導(dǎo)數(shù)的總原則：先化簡(jiǎn)解析式，再求導(dǎo)．注意以下幾點(diǎn)：連乘形式則先展開(kāi)化為多項(xiàng)式形式，再求導(dǎo)；三角形式，先利用三角函數(shù)公式轉(zhuǎn)化為和或差的形式，再求導(dǎo)；分式形式，先化為整式函數(shù)或較為簡(jiǎn)單的分式函數(shù)，再求導(dǎo)；復(fù)合函數(shù)，先確定復(fù)合關(guān)系，由外向內(nèi)逐層求導(dǎo)，必要時(shí)可換元2.利用導(dǎo)數(shù)研究曲線的切線問(wèn)題，一定要熟練掌握以下三點(diǎn)：(1)函數(shù)在切點(diǎn)處的導(dǎo)數(shù)值是切線的斜率，即已知切點(diǎn)坐標(biāo)可求切線斜率，已知斜率可求切點(diǎn)坐標(biāo)．(2)切點(diǎn)既在曲線上，又在切線上，切線還有可能和曲線有其它的公共點(diǎn)．(3)曲線y＝f(x)“在”點(diǎn)P(x0，y0)處的切線與“過(guò)”點(diǎn)P(x0，y0)的切線的區(qū)別：曲線y＝f(x)在點(diǎn)P(x0，y0)處的切線是指點(diǎn)P為切點(diǎn)，若切線斜率存在，切線斜率為k＝f′(x0)，是唯一的一條切線；曲線y＝f(x)過(guò)點(diǎn)P(x0，y0)的切線，是指切線經(jīng)過(guò)點(diǎn)P，點(diǎn)P可以是切點(diǎn)，也可以不是切點(diǎn)，而且這樣的直線可能有多條．3.利用導(dǎo)數(shù)的幾何意義求參數(shù)的基本方法利用切點(diǎn)的坐標(biāo)、切線的斜率、切線的方程等得到關(guān)于參數(shù)的方程(組)或者參數(shù)滿足的不等式(組)，進(jìn)而求出參數(shù)的值或取值范圍．4.求解與導(dǎo)數(shù)的幾何意義有關(guān)問(wèn)題時(shí)應(yīng)注意的兩點(diǎn)(1)注意曲線上橫坐標(biāo)的取值范圍；(2)謹(jǐn)記切點(diǎn)既在切線上又在曲線上．1．導(dǎo)數(shù)的概念(1)一般地，函數(shù)y＝f(x)在x＝x0處的瞬時(shí)變化率是eq\o(lim,\s\do4(Δx→0))eq\f(Δy,Δx)＝eq\o(lim,\s\do4(Δx→0))eq\f(fx0＋Δx－fx0,Δx)，我們稱它為函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)，記作f′(x0)或SKIPIF1<0，即f′(x0)＝eq\o(lim,\s\do4(Δx→0))eq\f(Δy,Δx)＝eq\o(lim,\s\do4(Δx→0))eq\f(fx0＋Δx－fx0,Δx).(2)如果函數(shù)y＝f(x)在開(kāi)區(qū)間(a，b)內(nèi)的每一點(diǎn)處都有導(dǎo)數(shù)，其導(dǎo)數(shù)值在(a，b)內(nèi)構(gòu)成一個(gè)新函數(shù)，這個(gè)函數(shù)稱為函數(shù)y＝f(x)在開(kāi)區(qū)間(a，b)內(nèi)的導(dǎo)函數(shù)．簡(jiǎn)稱導(dǎo)數(shù)，記作f′(x)或y′.2．導(dǎo)數(shù)的幾何意義函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)的幾何意義就是曲線y＝f(x)在點(diǎn)P(x0，f(x0))處的切線的斜率，相應(yīng)的切線方程為y－f(x0)＝f′(x0)(x－x0)．3．基本初等函數(shù)的導(dǎo)數(shù)公式基本初等函數(shù)導(dǎo)函數(shù)f(x)＝c(c為常數(shù))f′(x)＝0f(x)＝xα(α∈Q，α≠0)f′(x)＝αxα－1f(x)＝sinxf′(x)＝cosxf(x)＝cosxf′(x)＝－sinxf(x)＝ax(a>0且a≠1)f′(x)＝axlnaf(x)＝exf′(x)＝exf(x)＝logax(a>0且a≠1)f′(x)＝eq\f(1,xlna)f(x)＝lnxf′(x)＝eq\f(1,x)4.導(dǎo)數(shù)的運(yùn)算法則若f′(x)，g′(x)存在，則有[f(x)±g(x)]′＝f′(x)±g′(x)；[f(x)g(x)]′＝f′(x)g(x)＋f(x)g′(x)；eq\b\lc\[\rc\](\a\vs4\al\co1(\f(fx,gx)))′＝eq\f(f′xgx－fxg′x,[gx]2)(g(x)≠0)；[cf(x)]′＝cf′(x)．【知識(shí)拓展】復(fù)合函數(shù)的定義及其導(dǎo)數(shù)(1)一般地，對(duì)于兩個(gè)函數(shù)y＝f(u)和u＝g(x)，如果通過(guò)中間變量u，y可以表示成x的函數(shù)，那么稱這個(gè)函數(shù)為函數(shù)y＝f(u)與u＝g(x)的復(fù)合函數(shù)，記作y＝f(g(x))．(2)復(fù)合函數(shù)y＝f(g(x))的導(dǎo)數(shù)和函數(shù)y＝f(u)，u＝g(x)的導(dǎo)數(shù)間的關(guān)系為y′x＝y(tǒng)′u·u′x，即y對(duì)x的導(dǎo)數(shù)等于y對(duì)u的導(dǎo)數(shù)與u對(duì)x的導(dǎo)數(shù)的乘積．1．（2021·河南南陽(yáng)市·高二其他模擬（理））已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·千陽(yáng)縣中學(xué)高三二模（理））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0，SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0（）A．9903 B．9902 C．9901 D．99003．（2021·全國(guó)高三其他模擬（文））曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線經(jīng)過(guò)坐標(biāo)原點(diǎn)，則SKIPIF1<0___________.4．（2021·新沂市第一中學(xué)高三其他模擬）已知函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則a的值為_(kāi)__________1．（2021·河南新鄉(xiāng)市·高三三模（文））已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（）A．36 B．12 C．4 D．22．（2021·千陽(yáng)縣中學(xué)高三其他模擬（理））已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足：（1）SKIPIF1<0，（2）SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A． B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·全國(guó)高三月考（文））拉格朗日中值定理又稱拉氏定理，是微積分學(xué)中的基本定理之一，它反映了函數(shù)在閉區(qū)間上的整體平均變化率與區(qū)間某點(diǎn)的局部變化率的關(guān)系，其具體內(nèi)容如下：若SKIPIF1<0在SKIPIF1<0上滿足以下條件：①在SKIPIF1<0上圖象連續(xù)，②在SKIPIF1<0內(nèi)導(dǎo)數(shù)存在，則在SKIPIF1<0內(nèi)至少存在一點(diǎn)SKIPIF1<0，使得SKIPIF1<0（SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)）．則函數(shù)SKIPIF1<0在SKIPIF1<0上這樣的SKIPIF1<0點(diǎn)的個(gè)數(shù)為（）A．1 B．2 C．3 D．44．（2021·云南紅河哈尼族彝族自治州·高三三模（文））丹麥數(shù)學(xué)家琴生是19世紀(jì)對(duì)數(shù)學(xué)分析做出卓越貢獻(xiàn)的巨人，特別是在函數(shù)的凹凸性與不等式方面留下了很多寶貴的成果.定義：函數(shù)SKIPIF1<0在SKIPIF1<0上的導(dǎo)函數(shù)為SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上的導(dǎo)函數(shù)為SKIPIF1<0，若在SKIPIF1<0上SKIPIF1<0恒成立，則稱函數(shù)SKIPIF1<0在SKIPIF1<0上的“嚴(yán)格凸函數(shù)”，稱區(qū)間SKIPIF1<0為函數(shù)SKIPIF1<0的“嚴(yán)格凸區(qū)間”.則下列正確命題的序號(hào)為_(kāi)_____.①函數(shù)SKIPIF1<0在SKIPIF1<0上為“嚴(yán)格凸函數(shù)”；②函數(shù)SKIPIF1<0的“嚴(yán)格凸區(qū)間”為SKIPIF1<0；③函數(shù)SKIPIF1<0在SKIPIF1<0為“嚴(yán)格凸函數(shù)”，則SKIPIF1<0的取值范圍為SKIPIF1<0.5．（2021·江蘇高二專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0______．6．（2021·合肥市第六中學(xué)高三其他模擬（理））已知SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是___________.7．（2021·河北饒陽(yáng)中學(xué)高三其他模擬）曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則該切線的方程為_(kāi)_________.8．（2021·吉林松原市·高三月考）已知SKIPIF1<0，則SKIPIF1<0最小值為_(kāi)__________.9．（2021·廣東佛山市·高三其他模擬）已知函數(shù)SKIPIF1<0，則SKIPIF1<0所有的切線中斜率最小的切線方程為_(kāi)________.10．（2021·全國(guó)高三其他模擬）函數(shù)SKIPIF1<0在SKIPIF1<0處的切線與坐標(biāo)軸圍成的圖形面積為_(kāi)__________.11．（2021·全國(guó)高三其他模擬（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0___.12．（2021·四川省綿陽(yáng)南山中學(xué)高三其他模擬（文））設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，曲線SKIPIF1<0在SKIPIF1<0處切線的傾斜角的正切值為SKIPIF1<0．（1）求SKIPIF1<0的值；（2）證明：SKIPIF1<0．1．（2013·全國(guó)高考真題（文））已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則a的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2020·全國(guó)高考真題（理））若直線l與曲線y=SKIPIF1<0和x2+y2=SKIPIF1<0都相切，則l的方程為（）A．y=2x+1 B．y=2x+SKIPIF1<0 C．y=SKIPIF1<0x+1 D．y=SKIPIF1<0x+SKIPIF1<03．（2019·全國(guó)高考真題（理））已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2016·四川高考真題（文））設(shè)直線l1，l2分別是函數(shù)f(x)=SKIPIF1<0圖象上點(diǎn)P1，P-2處的切線，l1與l2垂直相交于點(diǎn)P，且l1，l2分別與y軸相交于點(diǎn)A，B，則△PAB的面積的取值范圍是A．(0,1) B．(0,2) C．(0,+∞) D．(1,+∞)5．（2021·全國(guó)高考真題）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0的兩條切線互相垂直，且分別交y軸于M，N兩點(diǎn)，則SKIPIF1<0取值范圍是_______．6．（2021·全國(guó)高考真題（理））曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為_(kāi)_________．7．（2019·江蘇高考真題）在平面直角坐標(biāo)系SKIPIF1<0中，點(diǎn)A在曲線y=lnx上，且該曲線在點(diǎn)A處的切線經(jīng)過(guò)點(diǎn)（-e，-1)(e為自然對(duì)數(shù)的底數(shù)），則點(diǎn)A的坐標(biāo)是____.8．（2019·江蘇高考真題）在平面直角坐標(biāo)系SKIPIF1<0中，P是曲線SKIPIF1<0上的一個(gè)動(dòng)點(diǎn)，則點(diǎn)P到直線x+y=0的距離的最小值是_____.9．（2017·天津高考真題（文））已知SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0的圖象在點(diǎn)（1，SKIPIF1<0）處的切線為l，則l在y軸上的截距為_(kāi)_______.10．（2021·全國(guó)高考真題（理））已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，且SKIPIF1<0與圓SKIPIF1<0上點(diǎn)的距離的最小值為SKIPIF1<0．（1）求SKIPIF1<0；（2）若點(diǎn)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0是SKIPIF1<0的兩條切線，SKIPIF1<0是切點(diǎn)，求SKIPIF1<0面積的最大值．1．【答案】B【分析】依題意求出函數(shù)的導(dǎo)函數(shù)，再解方程即可；【詳解】解：由題意可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0．故選：B2．【答案】C【分析】求出前幾項(xiàng)的導(dǎo)數(shù)，計(jì)算數(shù)列SKIPIF1<0，找到規(guī)律，代入數(shù)值計(jì)算.【詳解】解：因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0,數(shù)列SKIPIF1<0為1,3,7,13，SKIPIF1<0，每一項(xiàng)為上一項(xiàng)的常數(shù)與上一項(xiàng)的一次項(xiàng)的系數(shù)之和，即SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0.故選：C.【點(diǎn)睛】思路點(diǎn)睛：本題考查數(shù)列的應(yīng)用：計(jì)算前幾項(xiàng)的導(dǎo)數(shù)，發(fā)現(xiàn)每一項(xiàng)的常數(shù)都為上一項(xiàng)的常數(shù)與上一項(xiàng)中一次項(xiàng)的系數(shù)的和，寫(xiě)出遞推關(guān)系式，然后求得通項(xiàng)公式，代入計(jì)算.3．【答案】SKIPIF1<0【分析】利用導(dǎo)數(shù)的幾何意義即可求解.【詳解】由SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，化簡(jiǎn)整理可得SKIPIF1<0.故答案為：SKIPIF1<04．【答案】SKIPIF1<0【分析】根據(jù)點(diǎn)Ｐ在函數(shù)的圖象上，求得ｂ的值，得到SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義和直線垂直的條件求得SKIPIF1<0．【詳解】由已知可得SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0.則函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率SKIPIF1<0，因?yàn)榍芯€與直線SKIPIF1<0垂直，所以SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.1．【答案】C【分析】根據(jù)函數(shù)SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)的定義將SKIPIF1<0變形為SKIPIF1<0即可求解.【詳解】解：根據(jù)題意，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0，故選：C.2．【答案】C【分析】根據(jù)題意構(gòu)造函數(shù)SKIPIF1<0與SKIPIF1<0，利用二者的單調(diào)性即可得到結(jié)果.【詳解】SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0.故選：C【點(diǎn)睛】方法點(diǎn)睛：本題主要考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，需要構(gòu)造函數(shù)，一般：（1）條件含有SKIPIF1<0，就構(gòu)造SKIPIF1<0,（2）若SKIPIF1<0，就構(gòu)造SKIPIF1<0，（3）SKIPIF1<0，就構(gòu)造SKIPIF1<0，（4）SKIPIF1<0就構(gòu)造SKIPIF1<0，等便于給出導(dǎo)數(shù)時(shí)聯(lián)想構(gòu)造函數(shù).3．【答案】A【分析】用已知定義得到存在點(diǎn)SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，轉(zhuǎn)化為研究函數(shù)數(shù)SKIPIF1<0和SKIPIF1<0圖象的交點(diǎn)個(gè)數(shù)，作出函數(shù)圖象即可得到答案．【詳解】函數(shù)SKIPIF1<0，則SKIPIF1<0，由題意可知，存在點(diǎn)SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，如圖所示，由圖象可知，函數(shù)SKIPIF1<0和SKIPIF1<0的圖象只有一個(gè)交點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0只有一個(gè)解，即函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上SKIPIF1<0點(diǎn)的個(gè)數(shù)為1個(gè)．故選：A4．【答案】①②【分析】根據(jù)題干中給出的定義逐項(xiàng)檢驗(yàn)后可得正確的選項(xiàng).【詳解】SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上恒成立，所以函數(shù)SKIPIF1<0在SKIPIF1<0上為“嚴(yán)格凸函數(shù)”，所以①正確；SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的“嚴(yán)格凸區(qū)間”為SKIPIF1<0，所以②正確；SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0為SKIPIF1<0上的“嚴(yán)格凸函數(shù)”，故SKIPIF1<0在SKIPIF1<0上恒成立，所以SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，故SKIPIF1<0，所以③不正確.所以正確命題為：①②.故答案為：①②.5．【答案】2【分析】先對(duì)SKIPIF1<0求導(dǎo)，將SKIPIF1<0代入SKIPIF1<0即可求解.【詳解】由SKIPIF1<0可得，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：2.【點(diǎn)睛】本題主要考查導(dǎo)數(shù)的運(yùn)算，屬于基礎(chǔ)題.6．【答案】SKIPIF1<0【分析】由條件求得當(dāng)SKIPIF1<0時(shí)的函數(shù)解析式，求導(dǎo)，通過(guò)導(dǎo)數(shù)幾何意義求得在點(diǎn)SKIPIF1<0處的切線方程.【詳解】由題知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0則SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0則在點(diǎn)SKIPIF1<0的切線方程為：SKIPIF1<0，即SKIPIF1<0故答案為：SKIPIF1<07．【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義，先求切線斜率SKIPIF1<0，而直線SKIPIF1<0的斜率SKIPIF1<0，根據(jù)兩條直線垂直則SKIPIF1<0，代入即可得解.【詳解】由題意得SKIPIF1<0，則SKIPIF1<0，所以切線的斜率SKIPIF1<0.直線SKIPIF1<0的斜率SKIPIF1<0.因?yàn)閮芍本€相互垂直，所以SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0.所以SKIPIF1<0，則SKIPIF1<0，故該切線的方程為SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<08．【答案】4【分析】將SKIPIF1<0看作兩點(diǎn)SKIPIF1<0，SKIPIF1<0之間距離的平方，然后根據(jù)幾何意義進(jìn)行求解即可.【詳解】SKIPIF1<0看作兩點(diǎn)SKIPIF1<0，SKIPIF1<0之間距離的平方，點(diǎn)A在直線SKIPIF1<0上，點(diǎn)B在曲線SKIPIF1<0上，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，取點(diǎn)SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0最小值為4.故答案為：4.9．【答案】SKIPIF1<0【分析】求得函數(shù)導(dǎo)數(shù)，由基本不等關(guān)系求得導(dǎo)數(shù)的最小值，即函數(shù)SKIPIF1<0所有切線中斜率最小值，進(jìn)而求得切線方程.【詳解】由SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，則函數(shù)SKIPIF1<0所有切線中斜率最小為3，且過(guò)點(diǎn)SKIPIF1<0，則切線方程為SKIPIF1<0故答案為：SKIPIF1<010．【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義可求得切線方程，進(jìn)而確定與坐標(biāo)軸的交點(diǎn)坐標(biāo)，從而求得面積.【詳解】切點(diǎn)SKIPIF1<0，SKIPIF1<0，切線：SKIPIF1<0，即SKIPIF1<0，與SKIPIF1<0軸交點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交點(diǎn)SKIPIF1<0，故SKIPIF1<0，故答案為：SKIPIF1<0.11．【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義可知SKIPIF1<0，又SKIPIF1<0在切線上，可解得SKIPIF1<0的值，進(jìn)而可求SKIPIF1<0的值.【詳解】由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又切線方程為：SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0.12．【答案】（1）SKIPIF1<0；（2）證明見(jiàn)解析．【分析】（1）求出函數(shù)的導(dǎo)函數(shù)，再代入計(jì)算可得；（2）依題意即證SKIPIF1<0，即SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)說(shuō)明其單調(diào)性與最值，即可得到SKIPIF1<0，從而得證；【詳解】解：（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0．（2）由（1）可得SKIPIF1<0即證SKIPIF1<0．令SKIPIF1<0，SKIPIF1<0，于是SKIPIF1<0在SKIPIF1<0上是減函數(shù)，在SKIPIF1<0上是增函數(shù)，所以SKIPIF1<0（SKIPIF1<0取等號(hào)）．又令SKIPIF1<0，則SKIPIF1<0，于是SKIPIF1<0在SKIPIF1<0上是增函數(shù)，在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0（SKIPIF1<0時(shí)取等號(hào)）．所以SKIPIF1<0，即SKIPIF1<0．1．【答案】D【分析】作出函數(shù)SKIPIF1<0的圖像，和函數(shù)SKIPIF1<0的圖像，結(jié)合圖像可知直線SKIPIF1<0介于SKIPIF1<0與SKIPIF1<0軸之間，利用導(dǎo)數(shù)求出直線SKIPIF1<0的斜率，數(shù)形結(jié)合即可求解.【詳解】由題意可作出函數(shù)SKIPIF1<0的圖像，和函數(shù)SKIPIF1<0的圖像.由圖像可知：函數(shù)SKIPIF1<0的圖像是過(guò)原點(diǎn)的直線，當(dāng)直線介于SKIPIF1<0與SKIPIF1<0軸之間符合題意，直線SKIPIF1<0為曲線的切線，且此時(shí)函數(shù)SKIPIF1<0在第二象限的部分的解析式為SKIPIF1<0，求其導(dǎo)數(shù)可得SKIPIF1<0，因?yàn)镾KIPIF1<0，故SKIPIF1<0，故直線SKIPIF1<0的斜率為SKIPIF1<0，故只需直線SKIPIF1<0的斜率SKIPIF1<0SKIPIF1<0.故選：D【點(diǎn)睛】本題考查了不等式恒成立求出參數(shù)取值范圍，考查了數(shù)形結(jié)合的思想，屬于中檔題.2．【答案】D【分析】根據(jù)導(dǎo)數(shù)的幾何意義設(shè)出直線SKIPIF1<0的方程，再由直線與圓相切的性質(zhì)，即可得出答案.【詳解】設(shè)直線SKIPIF1<0在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，則直線SKIPIF1<0的斜率SKIPIF1<0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，由于直線SKIPIF1<0與圓SKIPIF1<0相切，則SKIPIF1<0，兩邊平方并整理得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0（舍），則直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.故選：D.【點(diǎn)睛】本題主要考查了導(dǎo)數(shù)的幾何意義的應(yīng)用以及直線與圓的位置的應(yīng)用，屬于中檔題.3．【答案】D【分析】通過(guò)求導(dǎo)數(shù)，確定得到切線斜率的表達(dá)式，求得SKIPIF1<0，將點(diǎn)的坐標(biāo)代入直線方程，求得SKIPIF1<0．【詳解】詳解：SKIPIF1<0SKIPIF1<0，SKIPIF1<0將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，故選D．【點(diǎn)睛】本題關(guān)鍵得到含有a，b的等式，利用導(dǎo)數(shù)幾何意義和點(diǎn)在曲線上得到方程關(guān)系．4．【答案】A【詳解】試題分析：設(shè)SKIPIF1<0（不妨設(shè)SKIPIF1<0），則由導(dǎo)數(shù)的幾何意義易得切線SKIPIF1<0的斜率分別為SKIPIF1<0由已知得SKIPIF1<0切線SKIPIF1<0的方程分別為SKIPIF1<0，切線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0．分別令SKIPIF1<0得SKIPIF1<0又SKIPIF1<0與SKIPIF1<0的交點(diǎn)為SKIPIF1<0，故選A．考點(diǎn)：1.導(dǎo)數(shù)的幾何意義；2.兩直線垂直關(guān)系；3.直線方程的應(yīng)用；4.三角形面積取值范圍.5．【答案】SKIPIF1<0【分析】結(jié)合導(dǎo)數(shù)的幾何意義可得SKIPIF1<0，結(jié)合直線方程及兩點(diǎn)間距離公式可得SKIPIF1<0，SKIPIF1<0，化簡(jiǎn)即可得解.【詳解】由題意，SKIPIF1<0，則SKIPIF1<0，所以點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，同理SKIPIF1<0,所以SKIPIF1<0.故答案為：SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：解決本題的關(guān)鍵是利用導(dǎo)數(shù)的幾何意義轉(zhuǎn)化條件SKIPIF1<0，消去一個(gè)變量后，運(yùn)算即可得解.6．【答案】SKIPIF1<0【分析】先驗(yàn)證點(diǎn)在曲線上，再求導(dǎo)，代入切線方程公式即可．【詳解】由題，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故點(diǎn)在曲線上．求導(dǎo)得：SKIPIF1<0，所以SKIPIF1<0．故切線方程為SKIPIF1<0．故答案為：SKIPIF1<0．7．【答案】SKIPIF1<0.【分析】設(shè)出切點(diǎn)坐標(biāo)，得到切線方程，然后求解方程得到橫坐標(biāo)的值可得切點(diǎn)坐標(biāo).【詳解】設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0.又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，點(diǎn)A在曲線SKIPIF1<0上的切線為SKIPIF1<0，即SKIPIF1<0，代入點(diǎn)SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，考查函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，注意到SKIPIF1<0，故SKIPIF1<0存在唯一的實(shí)數(shù)根SKIPIF1<0，此時(shí)SKIPIF1<0，故點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0.【點(diǎn)睛】導(dǎo)數(shù)運(yùn)算及切線的理解應(yīng)注意的問(wèn)題：一是利用公式求導(dǎo)時(shí)要特別注意除法公式中分子的符號(hào)，防止與乘法公式混淆．二是直線與曲線公共點(diǎn)的個(gè)數(shù)不是切線的本質(zhì)，直線與曲線只有一個(gè)公共點(diǎn)，直線不一定是曲線的切線，同樣，直線是曲線的切線，則直線與曲線可能有兩個(gè)或兩個(gè)以上的公共點(diǎn)．8．【答案】4.【分析】將原問(wèn)題轉(zhuǎn)化為切點(diǎn)與直線之間的距離，然后利用導(dǎo)函數(shù)確定切點(diǎn)坐標(biāo)可得最小距離【詳解】當(dāng)直線SKIPIF1<0平移到與曲線SKIPIF1<0相切位置時(shí)，切點(diǎn)Q即為點(diǎn)P到直線SKIPIF1<0的距離最小.由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，即切點(diǎn)SKIPIF1<0，則切點(diǎn)Q到直線SKIPIF1<0