人教A版高中數(shù)學(xué)（必修第一冊(cè)）同步講義第37講 5.4.1正弦函數(shù)、余弦函數(shù)的圖象（教師版）

第05講5.4.1正弦函數(shù)、余弦函數(shù)的圖象課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解并掌握用單位圓作正弦函數(shù)以及作余弦函數(shù)的圖象的方法.掌握數(shù)形結(jié)合的優(yōu)勢(shì)。②通過兩類函數(shù)圖象認(rèn)識(shí)函數(shù)圖象的特點(diǎn)，并能通過兩類圖象的形狀掌握兩類函數(shù)的性質(zhì)。會(huì)作正弦函數(shù)、余弦函數(shù)的圖象的同時(shí)，能認(rèn)識(shí)圖象與三角函數(shù)的密切關(guān)系，并能解決與圖象有關(guān)的三角函數(shù)問題知識(shí)點(diǎn)01：正弦函數(shù)的圖象正弦函數(shù)SKIPIF1<0,SKIPIF1<0的圖象叫做正弦曲線.知識(shí)點(diǎn)02：正弦函數(shù)圖象的畫法(1)幾何法:①在單位圓上,將點(diǎn)SKIPIF1<0繞著點(diǎn)SKIPIF1<0旋轉(zhuǎn)SKIPIF1<0弧度至點(diǎn)SKIPIF1<0,根據(jù)正弦函數(shù)的定義,點(diǎn)SKIPIF1<0的縱坐標(biāo)SKIPIF1<0.由此,以SKIPIF1<0為橫坐標(biāo),SKIPIF1<0為縱坐標(biāo)畫點(diǎn),即得到函數(shù)圖象上的點(diǎn)SKIPIF1<0.②將函數(shù)SKIPIF1<0,SKIPIF1<0的圖象不斷向左、向右平行移動(dòng)(每次移動(dòng)SKIPIF1<0個(gè)單位長(zhǎng)度).(2)“五點(diǎn)法”:在函數(shù)SKIPIF1<0,SKIPIF1<0的圖象上,以下五個(gè)點(diǎn):SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在確定圖象形狀時(shí)起關(guān)鍵作用.描出這五個(gè)點(diǎn),函數(shù)SKIPIF1<0,SKIPIF1<0的圖象形狀就基本確定了.因此,在精確度要求不高時(shí),常先找出這五個(gè)關(guān)鍵點(diǎn),再用光滑的曲線將它們連接起來,得到正弦函數(shù)的簡(jiǎn)圖.【即學(xué)即練1】（2023春·陜西西安·高二西北工業(yè)大學(xué)附屬中學(xué)?？茧A段練習(xí)）用五點(diǎn)法作出函數(shù)SKIPIF1<0的大致圖象．【答案】圖象見解析【詳解】解：因?yàn)镾KIPIF1<0，列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0描點(diǎn)、連線，函數(shù)圖象如下圖所示：知識(shí)點(diǎn)03：余弦函數(shù)的圖象余弦函數(shù)SKIPIF1<0,SKIPIF1<0的圖象叫做余弦曲線.知識(shí)點(diǎn)04：余弦函數(shù)圖象的畫法(1)要得到SKIPIF1<0,SKIPIF1<0的圖象,只需把SKIPIF1<0,SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度即可,這是因?yàn)镾KIPIF1<0.(2)用“五點(diǎn)法”:畫余弦函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象時(shí),所取的五個(gè)關(guān)鍵點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0再用光滑的曲線連接起來.【即學(xué)即練1】（2023·全國(guó)·高三專題練習(xí)）作出函數(shù)SKIPIF1<0的圖象【答案】見解析【詳解】SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0圖象后，將SKIPIF1<0軸下方的部分沿SKIPIF1<0軸翻折到SKIPIF1<0軸上方，即為函數(shù)SKIPIF1<0的圖象，如圖

題型01用“五點(diǎn)法”作三角函數(shù)的圖象【典例1】（2023·全國(guó)·高三專題練習(xí)）用“五點(diǎn)法”在給定的坐標(biāo)系中，畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像.

【答案】答案見解析【詳解】列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0120SKIPIF1<001描點(diǎn)，連線，畫出SKIPIF1<0在SKIPIF1<0上的大致圖像如圖：【典例2】（2023·全國(guó)·高一課堂例題）（1）作出函數(shù)SKIPIF1<0的簡(jiǎn)圖；（2）作出函數(shù)SKIPIF1<0的簡(jiǎn)圖．【答案】（1）答案見解析；（2）答案見解析【詳解】（1）列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0010SKIPIF1<00SKIPIF1<0020SKIPIF1<00描點(diǎn)并用光滑的曲線連接起來，可得函數(shù)SKIPIF1<0的圖象，如圖所示：

（2）列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<001SKIPIF1<001SKIPIF1<001210描點(diǎn)并用光滑的曲線連接起來，可得函數(shù)SKIPIF1<0的圖象，如圖所示：

【變式1】（2023春·北京·高一北京市第三十五中學(xué)?？茧A段練習(xí)）用五點(diǎn)法畫出函數(shù)SKIPIF1<0一個(gè)周期的圖象.【答案】答案見解析【詳解】令SKIPIF1<0，則SKIPIF1<0.列表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0函數(shù)SKIPIF1<0在一個(gè)周期內(nèi)的圖象如下圖所示：【變式2】（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．在用“五點(diǎn)法”作函數(shù)SKIPIF1<0的圖象時(shí)，列表如下：SKIPIF1<0xSKIPIF1<0完成上述表格，并在坐標(biāo)系中畫出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象；

【答案】填表見解析；作圖見解析【詳解】由題意列出以下表格：SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0x0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0020SKIPIF1<0SKIPIF1<0函數(shù)圖象如圖所示：

【變式3】（2023·全國(guó)·高一隨堂練習(xí)）用五點(diǎn)法分別畫下列函數(shù)在SKIPIF1<0上的圖象:(1)SKIPIF1<0;(2)SKIPIF1<0.【答案】（1）見解析（2）見解析【詳解】解:xSKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0010-10SKIPIF1<032123

題型02利用圖象解三角不等式【典例1】（2023·全國(guó)·高一假期作業(yè)）不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：SKIPIF1<0SKIPIF1<0SKIPIF1<0函數(shù)圖象如下所示：

SKIPIF1<0SKIPIF1<0，SKIPIF1<0不等式的解集為：SKIPIF1<0．故選：SKIPIF1<0．【典例2】（2023秋·江西撫州·高二黎川縣第二中學(xué)?？奸_學(xué)考試）不等式SKIPIF1<0的解集為．【答案】SKIPIF1<0【詳解】畫出SKIPIF1<0時(shí)，SKIPIF1<0的圖象．

令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0又SKIPIF1<0的周期為SKIPIF1<0，所以SKIPIF1<0的解集為SKIPIF1<0．用SKIPIF1<0代替SKIPIF1<0解出SKIPIF1<0．可得SKIPIF1<0則SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·全國(guó)·高一假期作業(yè)）不等式SKIPIF1<0SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】如圖所示，不等式SKIPIF1<0，SKIPIF1<0的解集為SKIPIF1<0故選：A【變式2】（2023春·高一課時(shí)練習(xí)）在(0，2π)內(nèi)使sinx>|cosx|的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)閟inx>|cosx|且x∈(0，2π)，所以sinx>0，所以x∈(0，π)，在同一平面直角坐標(biāo)系中畫出y=sinx，x∈(0，π)與y=|cosx|，x∈(0，π)的圖象，觀察圖象易得x∈SKIPIF1<0．故選：A.【變式3】（2023春·上海嘉定·高一?？计谥校┎坏仁絊KIPIF1<0的解集為.【答案】SKIPIF1<0【分析】畫出SKIPIF1<0的圖象，由圖象即可求解.【詳解】

畫出SKIPIF1<0的圖象，如圖所示，由圖可知，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0題型03利用圖象求方程的解或函數(shù)零點(diǎn)的個(gè)數(shù)問題【典例1】（2023·全國(guó)·高一假期作業(yè)）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為.【答案】3【詳解】由SKIPIF1<0，則函數(shù)SKIPIF1<0零點(diǎn)個(gè)數(shù)為SKIPIF1<0圖象交點(diǎn)個(gè)數(shù)，在同一坐標(biāo)系中畫出兩函數(shù)圖象如下，則交點(diǎn)有3個(gè)，即SKIPIF1<0有3個(gè)零點(diǎn).故答案為：3【典例2】（2023秋·河南新鄉(xiāng)·高一校聯(lián)考期末）已知函數(shù)SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】由SKIPIF1<0且SKIPIF1<0，得SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上無零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0，無解；若SKIPIF1<0在SKIPIF1<0上恰有1個(gè)零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上恰有1個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0；若SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則SKIPIF1<0在SKIPIF1<0上無零點(diǎn)，則SKIPIF1<0，無解，所以SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0【典例3】（2023春·新疆塔城·高一塔城地區(qū)第一高級(jí)中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為．【答案】7【詳解】依題意求函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，可以轉(zhuǎn)化為求函數(shù)SKIPIF1<0與SKIPIF1<0的交點(diǎn)個(gè)數(shù)，SKIPIF1<0，如圖，對(duì)于函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以在SKIPIF1<0軸非負(fù)半軸上兩個(gè)函數(shù)圖像有4個(gè)交點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以在SKIPIF1<0軸負(fù)半軸上兩個(gè)函數(shù)圖像有3個(gè)交點(diǎn)，

綜上，函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為7.故答案為：7.【典例4】（2023春·貴州遵義·高一統(tǒng)考期中）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有10個(gè)零點(diǎn)，則ω的取值范圍是.【答案】SKIPIF1<0【詳解】由SKIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有10個(gè)零點(diǎn)，所以由余弦函數(shù)的性質(zhì)可知：SKIPIF1<0解得SKIPIF1<0，故答案為：SKIPIF1<0.【典例5】（2023春·高一單元測(cè)試）方程SKIPIF1<0的解的個(gè)數(shù)是．【答案】7【詳解】由正弦函數(shù)值域可得SKIPIF1<0，又因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以，分別畫出SKIPIF1<0和SKIPIF1<0在SKIPIF1<0上的圖象如下圖所示：

根據(jù)圖像并根據(jù)其對(duì)稱性可知，在SKIPIF1<0上兩函數(shù)圖象共有7個(gè)交點(diǎn)；由函數(shù)與方程可知，方程SKIPIF1<0有7個(gè)解.故答案為：7【變式1】（2023·四川綿陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)數(shù)為．【答案】2【詳解】令SKIPIF1<0，可得SKIPIF1<0，原題意等價(jià)于求SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的交點(diǎn)個(gè)數(shù)，∵SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，有余弦函數(shù)可知SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上有2個(gè)交點(diǎn)所以SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上有2個(gè)交點(diǎn).故答案為：2.【變式2】（2023春·湖北武漢·高一華中科技大學(xué)附屬中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0），若方程SKIPIF1<0在SKIPIF1<0上恰有5個(gè)實(shí)數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰好有5個(gè)x，使得SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上恰有5條對(duì)稱軸．令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恰有5條對(duì)稱軸，如圖：所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.【變式3】（2023春·江西南昌·高一?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若存在實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0SKIPIF1<0互不相等SKIPIF1<0，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的圖象如下圖所示：存在實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0SKIPIF1<0互不相等SKIPIF1<0，不妨設(shè)SKIPIF1<0，則由圖可知SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0解得SKIPIF1<0或SKIPIF1<0，故而SKIPIF1<0，SKIPIF1<0，且由圖可得SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，綜上所述SKIPIF1<0；故答案為：SKIPIF1<0.【變式4】（2023春·四川廣安·高一?？茧A段練習(xí)）已知關(guān)于x的方程SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，則m的取值范圍是．【答案】SKIPIF1<0【詳解】由題設(shè)SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)數(shù)根，又SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0的圖象如下，只需SKIPIF1<0與SKIPIF1<0在給定區(qū)間內(nèi)有兩個(gè)交點(diǎn)即可，如圖，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0【變式5】（2023秋·河北衡水·高二衡水市第二中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，令SKIPIF1<0在區(qū)間SKIPIF1<0上恰有2個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.【答案】SKIPIF1<0/SKIPIF1<0SKIPIF1<0/0.75【詳解】由題意得SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，即SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，畫出SKIPIF1<0在SKIPIF1<0時(shí)的函數(shù)圖象，

SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0A夯實(shí)基礎(chǔ)B能力提升A夯實(shí)基礎(chǔ)1．（2023秋·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的圖象中與y軸最近的最高點(diǎn)的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】用五點(diǎn)法畫出函數(shù)SKIPIF1<0的部分圖象如圖所示，由圖易知與y軸最近的最高點(diǎn)的坐標(biāo)為SKIPIF1<0.

故選：B2．（2023·全國(guó)·高三專題練習(xí)）用“五點(diǎn)法”作SKIPIF1<0的圖象，首先描出的五個(gè)點(diǎn)的橫坐標(biāo)是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由“五點(diǎn)法”作圖知：令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，即為五個(gè)關(guān)鍵點(diǎn)的橫坐標(biāo).故選：B.3．（2023·全國(guó)·高三專題練習(xí)）三角函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖像為（）A． B．C． D．【答案】C【詳解】解：∵SKIPIF1<0為奇函數(shù)，∴SKIPIF1<0的圖像關(guān)于原點(diǎn)對(duì)稱，故排除A、D選項(xiàng)，三角函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，故排除B選項(xiàng).故選：C.4．（2023·全國(guó)·高三專題練習(xí)）從函數(shù)SKIPIF1<0的圖象來看，當(dāng)SKIPIF1<0時(shí)，對(duì)于SKIPIF1<0的x有（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】C【詳解】先畫出SKIPIF1<0，SKIPIF1<0的圖象，即A與D之間的部分，再畫出SKIPIF1<0的圖象，如下圖：由圖象可知它們有2個(gè)交點(diǎn)B、C，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的x的值有2個(gè).故選：C5．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)的個(gè)數(shù)為（

）A．0 B．1 C．2 D．3【答案】C【詳解】作出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象，并作出直線SKIPIF1<0，如圖：觀察圖形知：函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象與直線SKIPIF1<0有兩個(gè)公共點(diǎn)，所以函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)的個(gè)數(shù)為2.故選：C6．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的簡(jiǎn)圖是（

）A．B．C． D．【答案】B【詳解】由SKIPIF1<0知，其圖象和SKIPIF1<0的圖象相同，故選B．7．（2023秋·安徽合肥·高一校聯(lián)考期末）函數(shù)SKIPIF1<0，SKIPIF1<0的圖象在區(qū)間SKIPIF1<0的交點(diǎn)個(gè)數(shù)為（

）A．3 B．4 C．5 D．6【答案】A【詳解】分別作出SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象，如圖所示，

由圖象可知：SKIPIF1<0，SKIPIF1<0的圖象在區(qū)間SKIPIF1<0的交點(diǎn)個(gè)數(shù)為3.故選：A.8．（2023春·遼寧·高一鳳城市第一中學(xué)校聯(lián)考階段練習(xí)）華羅庚說：“數(shù)缺形時(shí)少直觀，形少數(shù)時(shí)難入微，數(shù)形結(jié)合百般好，隔離分家萬事休.”所以研究函數(shù)時(shí)往往要作圖，那么函數(shù)SKIPIF1<0的部分圖像可能是（

）A．

B．

C．

D．

【答案】B【詳解】因?yàn)镾KIPIF1<0，所以ACD錯(cuò)誤.故選：B二、多選題9．（2023秋·高一課時(shí)練習(xí)）（多選）函數(shù)SKIPIF1<0與SKIPIF1<0有一個(gè)交點(diǎn)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．0C．1 D．SKIPIF1<0【答案】BD【詳解】畫出SKIPIF1<0的圖象.如圖：直線SKIPIF1<0和SKIPIF1<0與SKIPIF1<0的圖象只有一個(gè)交點(diǎn)，

故SKIPIF1<0或SKIPIF1<0.故選：BD.10．（2023·全國(guó)·高一假期作業(yè)）函數(shù)SKIPIF1<0，SKIPIF1<0的圖像與直線SKIPIF1<0（t為常數(shù)，SKIPIF1<0）的交點(diǎn)可能有（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】ABC【詳解】作出SKIPIF1<0，SKIPIF1<0的圖像觀察可知，

當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為0；當(dāng)SKIPIF1<0或SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為l；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0的圖像與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)為2.故選：ABC.三、填空題11．（2023秋·湖南邵陽(yáng)·高三湖南省邵東市第一中學(xué)?？茧A段練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0的最小值為.【答案】SKIPIF1<0【詳解】令SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，依題意，SKIPIF1<0在SKIPIF1<0上有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.12．（2023春·上海嘉定·高一?？计谥校┎坏仁絊KIPIF1<0的解集為.【答案】SKIPIF1<0【詳解】

畫出SKIPIF1<0的圖象，如圖所示，由圖可知，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0四、解答題13．（2023·全國(guó)·高三專題練習(xí)）用“五點(diǎn)法”在給定的坐標(biāo)系中，畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖像.

【答案】答案見解析【詳解】列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0120SKIPIF1<001描點(diǎn)，連線，畫出SKIPIF1<0在SKIPIF1<0上的大致圖像如圖：14．（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0，用五點(diǎn)作圖法畫出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象；（先列表，再畫圖）【答案】答案見解析【詳解】SKIPIF1<0，按五個(gè)關(guān)鍵點(diǎn)列表：SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0010SKIPIF1<00SKIPIF1<003010描點(diǎn)并將它們用光滑的曲線連接起來如下圖所示：15．（2023·高一課時(shí)練習(xí)）作函數(shù)SKIPIF1<0的圖象.【答案】圖象見解析.【詳解】SKIPIF1<0SKIPIF1<0故SKIPIF1<0的圖象實(shí)際就是SKIPIF1<0的圖象在x軸下方的部分翻折到x軸上方后得到的圖象，如圖16．（2023秋·山東泰安·高一泰山中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0（1）作出該函數(shù)的圖象；（2）若SKIPIF1<0，求SKIPIF1<0的值；（3）若SKIPIF1<0，討論方程SKIPIF1<0的解的個(gè)數(shù)．【答案】（1）圖見解析；（2）SKIPIF1<0或SKIPIF1<0或SKIPIF1<0；（3）當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為0；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，解的個(gè)數(shù)為1；當(dāng)SKIPIF1<0時(shí)，解的個(gè)數(shù)為3．【詳解】（1）SKIPIF1<0的函數(shù)圖象如下：（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF