新高考數(shù)學(xué)二輪復(fù)習(xí)重點(diǎn)突破訓(xùn)練第6講 三角函數(shù)的圖象與性質(zhì)（含解析）

第六講三角函數(shù)的圖象與性質(zhì)真題展示2022新高考一卷第六題記函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0．若SKIPIF1<0，且SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0中心對(duì)稱，則SKIPIF1<0SKIPIF1<0A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．3【解析】【解法一】(取值試驗(yàn))函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0，SKIPIF1<0中心對(duì)稱，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．SKIPIF1<0SKIPIF1<0，SKIPIF1<0，取SKIPIF1<0，可得SKIPIF1<0．SKIPIF1<0，則SKIPIF1<0．故選：SKIPIF1<0．【解法二】(解不等式)：仿法一得2<ω<3及SKIPIF1<0，k∈Z，則2<SKIPIF1<0<3，解得SKIPIF1<0，又k∈Z，∴k=4，下同法一?！驹囶}評(píng)價(jià)】本題考查SKIPIF1<0型函數(shù)的圖象與性質(zhì)，考查邏輯思維能力與運(yùn)算求解能力，是中檔題．試題亮點(diǎn)三角函數(shù)是一類重要的函數(shù)，三角函數(shù)的周期性是其基本性質(zhì)，三角函數(shù)的周期性決定了該函數(shù)的很多其他性質(zhì).刻畫(huà)三角函數(shù)周期性的是頻率ao.理解頻率a對(duì)三角函數(shù)的各種幾何性質(zhì)和代數(shù)性質(zhì)的影響，是考查和評(píng)價(jià)考生的基本要求.試題亮點(diǎn)如下：（1）試題巧妙地設(shè)計(jì)了正弦型三角函數(shù)圖像的中心對(duì)稱性，反過(guò)來(lái)要求考生經(jīng)過(guò)分析與綜合，判斷正弦型函數(shù)頻率的取值或最小正周期的取值，這是對(duì)考生全面掌握三角函數(shù)性質(zhì)及其研究方法的一次很好的檢驗(yàn).（2）在試題的求解過(guò)程中，要求考生熟練掌握基本三角函數(shù)（y=sinx）的性質(zhì)，及其與復(fù)合函數(shù)（y=sin（wx+q））的性質(zhì)之間的關(guān)系，有利于指導(dǎo)教師在高中數(shù)學(xué)教學(xué)中整體把握三角函數(shù)的教學(xué).（3）數(shù)學(xué)正向問(wèn)題的解決主要依靠形式邏輯推理思維，其解決路徑是清晰的、確定的；而數(shù)學(xué)反向問(wèn)題的解決需要建立在辯證邏輯思維的基礎(chǔ)上，其解決路經(jīng)需要分析與綜合判斷．辯證邏輯思維是考生未來(lái)進(jìn)入高等學(xué)校學(xué)習(xí)，進(jìn)一步開(kāi)展科學(xué)研究需要運(yùn)用的主要的思維方式.因此，試題有利于考查考生未來(lái)的學(xué)習(xí)潛能，有利于檢測(cè)考生的辯證邏輯思維能力，對(duì)高中數(shù)學(xué)教學(xué)具有引導(dǎo)作用.知識(shí)要點(diǎn)整理一、正弦函數(shù)、余弦函數(shù)的圖象函數(shù)y＝sinxy＝cosx圖象圖象畫(huà)法五點(diǎn)法五點(diǎn)法關(guān)鍵五點(diǎn)(0,0)，eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)，1))，(π，0)，eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)，－1))，(2π，0)(0,1)，eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)，0))，(π，－1)，eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)，0))，(2π，1)正(余)弦曲線正(余)弦函數(shù)的圖象叫做正(余)弦曲線二、正切函數(shù)的圖象與性質(zhì)解析式y(tǒng)＝tanx圖象定義域eq\b\lc\{\rc\}(\a\vs4\al\co1(x\b\lc\|\rc\(\a\vs4\al\co1(x≠\f(π,2)＋kπ，k∈Z))))值域R最小正周期π奇偶性奇函數(shù)單調(diào)性在每一個(gè)區(qū)間eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,2)＋kπ，\f(π,2)＋kπ))(k∈Z)上都單調(diào)遞增對(duì)稱性對(duì)稱中心eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(kπ,2)，0))(k∈Z)三、函數(shù)的周期性1．函數(shù)的周期性一般地，設(shè)函數(shù)f(x)的定義域?yàn)镈，如果存在一個(gè)非零常數(shù)T，使得對(duì)每一個(gè)x∈D都有x＋T∈D，且f(x＋T)＝f(x)，那么函數(shù)f(x)就叫做周期函數(shù)．非零常數(shù)T叫做這個(gè)函數(shù)的周期．2．最小正周期如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù)，那么這個(gè)最小正數(shù)叫做f(x)的最小正周期．四、正弦函數(shù)、余弦函數(shù)的周期性和奇偶性函數(shù)y＝sinxy＝cosx圖象定義域RR周期2kπ(k∈Z且k≠0)2kπ(k∈Z且k≠0)最小正周期2π2π奇偶性奇函數(shù)偶函數(shù)正弦函數(shù)、余弦函數(shù)的單調(diào)性與最值正弦函數(shù)余弦函數(shù)圖象定義域RR值域[－1,1][－1,1]單調(diào)性在每一個(gè)閉區(qū)間eq\b\lc\[\rc\](\a\vs4\al\co1(2kπ－\f(π,2)，2kπ＋\f(π,2)))(k∈Z)上都單調(diào)遞增，在每一個(gè)閉區(qū)間eq\b\lc\[\rc\](\a\vs4\al\co1(2kπ＋\f(π,2)，2kπ＋\f(3π,2)))(k∈Z)上都單調(diào)遞減在每一個(gè)閉區(qū)間[2kπ－π，2kπ](k∈Z)上都單調(diào)遞增，在每一個(gè)閉區(qū)間[2kπ，2kπ＋π](k∈Z)上都單調(diào)遞減最值x＝eq\f(π,2)＋2kπ(k∈Z)時(shí)，ymax＝1；x＝－eq\f(π,2)＋2kπ(k∈Z)時(shí)，ymin＝－1x＝2kπ(k∈Z)時(shí)，ymax＝1；x＝2kπ＋π(k∈Z)時(shí)，ymin＝－1三年真題一、單選題1．已知SKIPIF1<0，關(guān)于該函數(shù)有下列四個(gè)說(shuō)法：①SKIPIF1<0的最小正周期為SKIPIF1<0；②SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍為SKIPIF1<0；④SKIPIF1<0的圖象可由SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度得到．以上四個(gè)說(shuō)法中，正確的個(gè)數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的最小正周期為SKIPIF1<0，①不正確；令SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，②正確；因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，③不正確；由于SKIPIF1<0，所以SKIPIF1<0的圖象可由SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到，④不正確．故選：A．2．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】A【詳解】令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，排除BD；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除C.故選：A.3．已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增【答案】C【詳解】因?yàn)镾KIPIF1<0.對(duì)于A選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，A錯(cuò)；對(duì)于B選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上不單調(diào)，B錯(cuò)；對(duì)于C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，C對(duì)；對(duì)于D選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上不單調(diào)，D錯(cuò).故選：C.4．在SKIPIF1<0中，SKIPIF1<0．P為SKIPIF1<0所在平面內(nèi)的動(dòng)點(diǎn)，且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】解：依題意如圖建立平面直角坐標(biāo)系，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0在以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓上運(yùn)動(dòng)，設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；故選：D

5．設(shè)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0恰有三個(gè)極值點(diǎn)、兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：依題意可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，要使函數(shù)在區(qū)間SKIPIF1<0恰有三個(gè)極值點(diǎn)、兩個(gè)零點(diǎn)，又SKIPIF1<0，SKIPIF1<0的圖象如下所示：則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0．故選：C．6．如圖是下列四個(gè)函數(shù)中的某個(gè)函數(shù)在區(qū)間SKIPIF1<0的大致圖像，則該函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，故排除B;設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故排除C;設(shè)SKIPIF1<0，則SKIPIF1<0，故排除D.故選：A.7．將函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到曲線C，若C關(guān)于y軸對(duì)稱，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由題意知：曲線SKIPIF1<0為SKIPIF1<0，又SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱，則SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0.故選：C.8．記函數(shù)SKIPIF1<0的最小正周期為T(mén)．若SKIPIF1<0，且SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】A【詳解】由函數(shù)的最小正周期T滿足SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，又因?yàn)楹瘮?shù)圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，所以SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故選：A9．函數(shù)SKIPIF1<0是A．奇函數(shù)，且最大值為2 B．偶函數(shù)，且最大值為2C．奇函數(shù)，且最大值為SKIPIF1<0 D．偶函數(shù)，且最大值為SKIPIF1<0【答案】D【分析】由函數(shù)奇偶性的定義結(jié)合三角函數(shù)的性質(zhì)可判斷奇偶性；利用二倍角公式結(jié)合二次函數(shù)的性質(zhì)可判斷最大值.【詳解】由題意，SKIPIF1<0，所以該函數(shù)為偶函數(shù)，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值SKIPIF1<0.故選：D.10．函數(shù)SKIPIF1<0的最小正周期和最大值分別是（

）A．SKIPIF1<0和SKIPIF1<0 B．SKIPIF1<0和2 C．SKIPIF1<0和SKIPIF1<0 D．SKIPIF1<0和2【答案】C【詳解】由題，SKIPIF1<0，所以SKIPIF1<0的最小正周期為SKIPIF1<0，最大值為SKIPIF1<0.故選：C．11．下列函數(shù)中最小值為4的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】對(duì)于A，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以其最小值為SKIPIF1<0，A不符合題意；對(duì)于B，因?yàn)镾KIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，等號(hào)取不到，所以其最小值不為SKIPIF1<0，B不符合題意；對(duì)于C，因?yàn)楹瘮?shù)定義域?yàn)镾KIPIF1<0，而SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以其最小值為SKIPIF1<0，C符合題意；對(duì)于D，SKIPIF1<0，函數(shù)定義域?yàn)镾KIPIF1<0，而SKIPIF1<0且SKIPIF1<0，如當(dāng)SKIPIF1<0，SKIPIF1<0，D不符合題意．故選：C．12．已知命題SKIPIF1<0﹔命題SKIPIF1<0﹐SKIPIF1<0，則下列命題中為真命題的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由于SKIPIF1<0，所以命題SKIPIF1<0為真命題；由于SKIPIF1<0在SKIPIF1<0上為增函數(shù)，SKIPIF1<0，所以SKIPIF1<0，所以命題SKIPIF1<0為真命題；所以SKIPIF1<0為真命題，SKIPIF1<0、SKIPIF1<0、SKIPIF1<0為假命題.故選：A．13．下列區(qū)間中，函數(shù)SKIPIF1<0單調(diào)遞增的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，取SKIPIF1<0，可得函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，A選項(xiàng)滿足條件，B不滿足條件；取SKIPIF1<0，可得函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，CD選項(xiàng)均不滿足條件.故選：A.二、多選題14．已知函數(shù)SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱，則（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減B．SKIPIF1<0在區(qū)間SKIPIF1<0有兩個(gè)極值點(diǎn)C．直線SKIPIF1<0是曲線SKIPIF1<0的對(duì)稱軸D．直線SKIPIF1<0是曲線SKIPIF1<0的切線【答案】AD【詳解】由題意得：SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0．對(duì)A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由正弦函數(shù)SKIPIF1<0圖象知SKIPIF1<0在SKIPIF1<0上是單調(diào)遞減；對(duì)B，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由正弦函數(shù)SKIPIF1<0圖象知SKIPIF1<0只有1個(gè)極值點(diǎn)，由SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0為函數(shù)的唯一極值點(diǎn)；對(duì)C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0不是對(duì)稱軸；對(duì)D，由SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0,從而得：SKIPIF1<0或SKIPIF1<0,所以函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為SKIPIF1<0，切線方程為：SKIPIF1<0即SKIPIF1<0．故選：AD．三、填空題15．記函數(shù)SKIPIF1<0的最小正周期為T(mén)，若SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的零點(diǎn)，則SKIPIF1<0的最小值為_(kāi)___________．【答案】SKIPIF1<0【詳解】解：因?yàn)镾KIPIF1<0，（SKIPIF1<0，SKIPIF1<0）所以最小正周期SKIPIF1<0，因?yàn)镾KIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0為SKIPIF1<0的零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0；故答案為：SKIPIF1<016．已知函數(shù)SKIPIF1<0的部分圖像如圖所示，則SKIPIF1<0_______________.【答案】SKIPIF1<0【詳解】由題意可得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0可得：SKIPIF1<0，據(jù)此有：SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】已知f(x)＝Acos(ωx＋φ)(A＞0，ω＞0)的部分圖象求其解析式時(shí)，A比較容易看圖得出，困難的是求待定系數(shù)ω和φ，常用如下兩種方法：(1)由ω＝SKIPIF1<0即可求出ω；確定φ時(shí)，若能求出離原點(diǎn)最近的右側(cè)圖象上升(或下降)的“零點(diǎn)”橫坐標(biāo)x0，則令ωx0＋φ＝0(或ωx0＋φ＝π)，即可求出φ.(2)代入點(diǎn)的坐標(biāo)，利用一些已知點(diǎn)(最高點(diǎn)、最低點(diǎn)或“零點(diǎn)”)坐標(biāo)代入解析式，再結(jié)合圖形解出ω和φ，若對(duì)A，ω的符號(hào)或?qū)Ζ盏姆秶幸?，則可用誘導(dǎo)公式變換使其符合要求.17．已知函數(shù)SKIPIF1<0的部分圖像如圖所示，則滿足條件SKIPIF1<0的最小正整數(shù)x為_(kāi)_______．【答案】2【詳解】由圖可知SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；由五點(diǎn)法可得SKIPIF1<0，即SKIPIF1<0；所以SKIPIF1<0.因?yàn)镾KIPIF1<0，SKIPIF1<0；所以由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0；因?yàn)镾KIPIF1<0，所以，方法一：結(jié)合圖形可知，最小正整數(shù)應(yīng)該滿足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，可得SKIPIF1<0的最小正整數(shù)為2.方法二：結(jié)合圖形可知，最小正整數(shù)應(yīng)該滿足SKIPIF1<0，又SKIPIF1<0，符合題意，可得SKIPIF1<0的最小正整數(shù)為2.故答案為：2.四、解答題18．設(shè)函數(shù)SKIPIF1<0.（1）求函數(shù)SKIPIF1<0的最小正周期；（2）求函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.【詳解】（1）由輔助角公式得SKIPIF1<0，則SKIPIF1<0，所以該函數(shù)的最小正周期SKIPIF1<0;（2）由題意，SKIPIF1<0SKIPIF1<0SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，函數(shù)取最大值SKIPIF1<0.19．小明同學(xué)用“五點(diǎn)法”作某個(gè)正弦型函數(shù)SKIPIF1<0在一個(gè)周期內(nèi)的圖象時(shí)，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0030-30根據(jù)表中數(shù)據(jù)，求：（1）實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的值；（2）該函數(shù)在區(qū)間SKIPIF1<0上的最大值和最小值.【答案】（1）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；（2）最大值是3，最小值是SKIPIF1<0.【詳解】（1）由表可知SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，因?yàn)楹瘮?shù)圖象過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0.（2）由（1）可知SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因此，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，SKIPIF1<0.所以該函數(shù)在區(qū)間SKIPIF1<0上的最大值是3，最小值是SKIPIF1<0.20．在銳角△ABC中，角A，B，C的對(duì)邊分別為a，b，c，且SKIPIF1<0．（I）求角B的大?。唬↖I）求cosA+cosB+cosC的取值范圍．【答案】（I）SKIPIF1<0；（II）SKIPIF1<0【詳解】（I）[方法一]：余弦定理由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0.結(jié)合余弦定SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0為銳角三角形，∴SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0，又B為SKIPIF1<0的一個(gè)內(nèi)角，故SKIPIF1<0.[方法二]【最優(yōu)解】：正弦定理邊化角由SKIPIF1<0，結(jié)合正弦定理可得：SKIPIF1<0SKIPIF1<0為銳角三角形，故SKIPIF1<0.（II）[方法一]：余弦定理基本不等式因?yàn)镾KIPIF1<0，并利用余弦定理整理得SKIPIF1<0，即SKIPIF1<0.結(jié)合SKIPIF1<0，得SKIPIF1<0.由臨界狀態(tài)（不妨取SKIPIF1<0）可知SKIPIF1<0.而SKIPIF1<0為銳角三角形，所以SKIPIF1<0.由余弦定理得SKIPIF1<0，SKIPIF1<0，代入化簡(jiǎn)得SKIPIF1<0故SKIPIF1<0的取值范圍是SKIPIF1<0.[方法二]【最優(yōu)解】：恒等變換三角函數(shù)性質(zhì)結(jié)合(1)的結(jié)論有：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.由SKIPIF1<0可得：SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.即SKIPIF1<0的取值范圍是SKIPIF1<0.三年模擬1．函數(shù)SKIPIF1<0的圖象如圖所示，將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】結(jié)合圖像，易得SKIPIF1<0，則SKIPIF1<0，所以由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又因?yàn)镾KIPIF1<0落在SKIPIF1<0上，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0滿足要求，所以SKIPIF1<0，因?yàn)閷⒑瘮?shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0的圖象，所以SKIPIF1<0.故選：A.2．下列四個(gè)函數(shù)中，在區(qū)間SKIPIF1<0上為增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】對(duì)A，因?yàn)镾KIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A錯(cuò)誤；對(duì)B，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故B錯(cuò)誤；對(duì)C，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；對(duì)D，由C知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D錯(cuò)誤.故答案為：C3．函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B． C． D．【答案】D【詳解】易知函數(shù)SKIPIF1<0為偶函數(shù)，所以其圖象關(guān)于y軸對(duì)稱，排除A，B項(xiàng)；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除C選項(xiàng).故選：D．4．已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象由SKIPIF1<0圖象向右平移SKIPIF1<0個(gè)單位得到，則下列關(guān)于函數(shù)SKIPIF1<0的圖象說(shuō)法正確的是（

）A．關(guān)于y軸對(duì)稱 B．關(guān)于原點(diǎn)對(duì)稱C．關(guān)于直線SKIPIF1<0對(duì)稱 D．關(guān)于點(diǎn)SKIPIF1<0對(duì)稱【答案】D【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，所以函數(shù)是非奇非偶函數(shù)，故A，B項(xiàng)錯(cuò)誤；因?yàn)镾KIPIF1<0，既不是SKIPIF1<0的最大值也不是最小值，所以SKIPIF1<0不是SKIPIF1<0的對(duì)稱軸，故C項(xiàng)錯(cuò)誤；因?yàn)镾KIPIF1<0，所以SKIPIF1<0是SKIPIF1<0的一個(gè)對(duì)稱中心，故D項(xiàng)正確.故選：D.5．對(duì)于函數(shù)SKIPIF1<0，給出下列四個(gè)命題：（1）該函數(shù)的值域是SKIPIF1<0；（2）當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，該函數(shù)取最大值SKIPIF1<0；（3）該函數(shù)的最小正周期為SKIPIF1<0；（4）當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；其中所有正確命題個(gè)數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)镾KIPIF1<0，所以，SKIPIF1<0，對(duì)于（3），SKIPIF1<0SKIPIF1<0，所以，函數(shù)SKIPIF1<0為周期函數(shù)，作出函數(shù)SKIPIF1<0的圖象（圖中實(shí)線）如下圖所示：結(jié)合圖形可知，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，（3）對(duì)；對(duì)于（1），由圖可知，函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，（1）錯(cuò)；對(duì)于（2），由圖可知，當(dāng)且僅當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最大值SKIPIF1<0，（2）錯(cuò)；對(duì)于（4），由圖可知，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，（4）對(duì).故選：B.6．對(duì)于函數(shù)SKIPIF1<0，給出下列五個(gè)命題：（1）該函數(shù)的值域是SKIPIF1<0；（2）當(dāng)且僅當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，該函數(shù)取最大值1；（3）該函數(shù)的最小正周期為2π；（4）當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；（5）當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增；其中所有正確命題的個(gè)數(shù)有（

）A．1 B．2 C．3 D．4【答案】C【詳解】函數(shù)SKIPIF1<0的圖象如下圖所示：對(duì)于（1），由圖象可知，該函數(shù)的值域是SKIPIF1<0，所以（1）錯(cuò)誤；對(duì)于（2），當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；此外再無(wú)其他等于1的值，所以當(dāng)且僅當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，該函數(shù)取最大值1.即（2）正確.對(duì)于（3），觀察圖像可知，該函數(shù)的最小正周期為2π，故（3）正確；對(duì)于（4），由圖可知，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以（4）正確；對(duì)于（5），根據(jù)圖像可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0也是單調(diào)遞增的，故（5）錯(cuò)誤；因此，正確的命題有（2）（3）（4）共3個(gè).故選：C.7．已知SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的一個(gè)對(duì)稱中心為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SK