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20xx屆山東省師大附中高三上學(xué)期第二次模擬考試數(shù)學(xué)理試題解析版-資料下載頁(yè)

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【正文】 輔助角公式等,一般要將解析式化為fx=Asin(ωx+φ)的形式后再求解最值、周期、對(duì)稱軸、余弦定理對(duì)邊角轉(zhuǎn)化.19.m≤12+1e【解析】【分析】圖像上的任意點(diǎn)P在函數(shù)y=g(x)上存在點(diǎn)Q,使得P與Q關(guān)于坐標(biāo)原點(diǎn)對(duì)稱,等價(jià)于函數(shù)y=f(x)關(guān)于原點(diǎn)對(duì)稱的函數(shù)圖像與y=g(x)恒有交點(diǎn),即可以通過(guò)參數(shù)分離求m的范圍.【詳解】先求關(guān)于原點(diǎn)對(duì)稱的函數(shù),問(wèn)題等價(jià)于 ,與有交點(diǎn) ,即方程有解,即有解,設(shè),,當(dāng)時(shí) ,方程有解.解法二:函數(shù)是奇函數(shù),其圖象關(guān)于原點(diǎn)對(duì)稱,問(wèn)題等價(jià)于函數(shù)的圖象與函數(shù)的圖象有交點(diǎn),即有解,設(shè)函數(shù),遞增;遞減,,當(dāng) 時(shí),函數(shù)的圖象與函數(shù)的圖象有交點(diǎn).【點(diǎn)睛】本題考查函數(shù)中參數(shù)的取值范圍,注意運(yùn)用參數(shù)分離法和轉(zhuǎn)化的數(shù)學(xué)思想,解題中將g(x)存在點(diǎn)Q使其與P對(duì)稱問(wèn)題轉(zhuǎn)化為關(guān)于原點(diǎn)對(duì)稱的函數(shù)與g(x)恒有交點(diǎn)是本題的難點(diǎn)和關(guān)鍵突破點(diǎn).20.(1)fx的單調(diào)遞增區(qū)間為π,0,π6,5π6,fx的單調(diào)遞減區(qū)間為0,π6,5π6,π;(2)π2.【解析】【分析】(1)求函數(shù)fx的導(dǎo)數(shù),利用導(dǎo)函數(shù)判斷原函數(shù)fx的單調(diào)區(qū)間.(2)結(jié)合(1)知函數(shù)單調(diào)性,即可確定出在區(qū)間π2,π2上的最值.【詳解】(1),0+0_0+0_fx的單調(diào)遞增區(qū)間為,fx的單調(diào)遞減區(qū)間為.(2)由第一問(wèn)的單調(diào)性可知.【點(diǎn)睛】本題考查了導(dǎo)數(shù)的應(yīng)用,在解題中首先要準(zhǔn)確求解導(dǎo)函數(shù),也是關(guān)鍵的步驟,其次是列表確定函數(shù)的單調(diào)性,利用函數(shù)單調(diào)性確定函數(shù)的最值.21.(1)函數(shù)fx在∞,+∞上遞增;(2)a≥1.【解析】【分析】(1)利用導(dǎo)函數(shù)確定函數(shù)單調(diào)區(qū)間;(2)fx≥0恒成立,確定a的范圍可以先分離參數(shù),然后求解新構(gòu)造函數(shù)的最大值.【詳解】(1) ,函數(shù)在上遞增 .(2)對(duì)于任意的實(shí)數(shù),所以,下面證明充分性:即當(dāng)當(dāng) ,設(shè)且,所以,綜上:.解法二:,可化為,設(shè),102+0+0極大極大,所以.解法三當(dāng),與題設(shè)矛盾,當(dāng),設(shè),單調(diào)遞減;單調(diào)遞增;單調(diào)遞減,當(dāng),綜上:.【點(diǎn)睛】本題考查導(dǎo)數(shù)的應(yīng)用和求參數(shù)范圍;導(dǎo)數(shù)應(yīng)用是每年高考必考題型,在解題中,首先要準(zhǔn)確求解導(dǎo)函數(shù),這是解題的關(guān)鍵,因此必須熟練掌握基本函數(shù)導(dǎo)數(shù)公式和和差積商的導(dǎo)數(shù)以及復(fù)合函數(shù)導(dǎo)數(shù),其次參數(shù)范圍問(wèn)題也是高考熱點(diǎn)之一,常用的方法是分離參數(shù)法和構(gòu)造函數(shù)法.22.(1)當(dāng)a≥14時(shí),fx無(wú)極值;當(dāng)0a14時(shí),fx有兩個(gè)極值點(diǎn); 當(dāng)a≤0時(shí),fx有一個(gè)極值點(diǎn);(2)證明見(jiàn)解析.【解析】【分析】(1)分類討論判斷導(dǎo)函數(shù)對(duì)應(yīng)的方程根的個(gè)數(shù)來(lái)確定極值點(diǎn)個(gè)數(shù);(2)由(1)可知當(dāng)0a14時(shí),fx有兩個(gè)極值點(diǎn),利用韋達(dá)定理可以構(gòu)造出fx1+fx2關(guān)于a的函數(shù),利用導(dǎo)數(shù)求最大值.【詳解】(1) ,設(shè),①若即,上單調(diào)遞減 ,無(wú)極值 .②,,在上,單調(diào)遞減;在上單調(diào)遞增,函數(shù)有兩個(gè)極值點(diǎn).③當(dāng),在上,單調(diào)遞增;上單調(diào)遞減,函數(shù)有一個(gè)極值點(diǎn),綜上,當(dāng),函數(shù)無(wú)極值;當(dāng),函數(shù)有兩個(gè)極值點(diǎn);當(dāng)時(shí),函數(shù)有一個(gè)極值點(diǎn) .(2)由(1)知,當(dāng)時(shí),有兩個(gè)極值點(diǎn),x1,x2且,設(shè)遞增,, ,.【點(diǎn)睛】本題考查利用導(dǎo)數(shù)求解極值點(diǎn)個(gè)數(shù)、證明不等式;求解極值點(diǎn)個(gè)數(shù)其方法是利用導(dǎo)函數(shù)零點(diǎn)的個(gè)數(shù)結(jié)合原函數(shù)的單調(diào)性來(lái)確定,要注意導(dǎo)函數(shù)的零點(diǎn)并不一定是函數(shù)的極值點(diǎn),要成為極值點(diǎn)其左右兩邊的單調(diào)性必須相異;不等式的證明其實(shí)質(zhì)還是利用函數(shù)的單調(diào)性確定最值,當(dāng)需要構(gòu)造合理的函數(shù),這是解題的難點(diǎn)和關(guān)鍵點(diǎn).
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