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1、新編人教版精品教學(xué)資料學(xué)業(yè)分層測(cè)評(píng)(十四) 指數(shù)函數(shù)及其性質(zhì)的應(yīng)用 (建議用時(shí):45分鐘)學(xué)業(yè)達(dá)標(biāo)一、選擇題1函數(shù)f(x)的圖象()A關(guān)于原點(diǎn)對(duì)稱(chēng)B關(guān)于直線yx對(duì)稱(chēng)C關(guān)于x軸對(duì)稱(chēng)D關(guān)于y軸對(duì)稱(chēng)【解析】f(x)f(x),f(x)是偶函數(shù),圖象關(guān)于y軸對(duì)稱(chēng),故選D.【答案】D2指數(shù)函數(shù)f(x)ax(a0,且a1)在R上是減函數(shù),則函數(shù)g(x)(a2)x2在R上的單調(diào)性為()【導(dǎo)學(xué)號(hào):97030090】A單調(diào)遞增B單調(diào)遞減C在(,0)上遞減,在(0,)上遞增D在(,0)上遞增,在(0,)上遞減【解析】因?yàn)橹笖?shù)函數(shù)f(x)ax在R上是減函數(shù),則0a1,所以2a2abBbacCabcDacb【解析】a4
2、0.921.8,b80.4821.44,c1.521.5,因?yàn)楹瘮?shù)y2x在R上是增函數(shù),且1.81.51.44,所以21.821.521.44,即acb.【答案】D5設(shè)函數(shù)yf(x)在(,)內(nèi)有定義,對(duì)于給定的正數(shù)K,定義函數(shù)fK(x)取函數(shù)f(x)2|x|,當(dāng)K時(shí),函數(shù)fK(x)的單調(diào)遞增區(qū)間為()A(,0)B(0,)C(,1)D(1,)【解析】由f(x)2|x|及K,得fK(x)函數(shù)fK(x)的單調(diào)遞增區(qū)間是(,1)【答案】C二、填空題6已知y21ax在R上是減函數(shù),則a的取值范圍是_【解析】y21ax22ax在R上是減函數(shù),a,則實(shí)數(shù)x的取值范圍為_(kāi). 【導(dǎo)學(xué)號(hào):97030091】【解析
3、】函數(shù)f(x)a是奇函數(shù),可得f(x)f(x),即aa,即2a1,解得a,f(x),4x1,解得x0.【答案】x08已知函數(shù)f(x),g(x)(其中e2.718),有下列命題:f(x)是奇函數(shù),g(x)是偶函數(shù);對(duì)任意xR,都有f(2x)f(x)g(x);f(x)有零點(diǎn),g(x)無(wú)零點(diǎn)其中正確的命題是_(填上所有正確命題的序號(hào))【解析】f(x)(exex)(exex)f(x),故f(x)為奇函數(shù),g(x)(exex)g(x),故g(x)為偶函數(shù),故命題正確;f(2x)(e2xe2x)(exex)(exex),f(x)g(x)(exex)(exex)(exex)(exex),故命題不正確;函數(shù)y
4、ex,yex在實(shí)數(shù)集上均為增函數(shù),f(x)在R上單調(diào)遞增,設(shè)x1x20,則g(x1)g(x2)(ex1ex1)(ex2ex2),x1x20,g(x1)g(x2)0,即g(x1)g(x2)g(x)在(,0)上單調(diào)遞減,當(dāng)x0時(shí),g(x)有最小值1,且函數(shù)是偶函數(shù),g(x)無(wú)零點(diǎn),由f(x)0,即(exex)0,得x0,f(x)有零點(diǎn)0,故命題正確【答案】三、解答題9比較下列各組數(shù)的大小:(1)1.9與1.93;(2)0.72與0.70.3;(3)0.60.4與0.40.6.【解】(1)由于指數(shù)函數(shù)y1.9x在R上單調(diào)遞增,而3,所以1.91.93.(2)因?yàn)楹瘮?shù)y0.7x在R上單調(diào)遞減,而20.
5、267 90.70.3.(3)因?yàn)閥0.6x在R上單調(diào)遞減,所以0.60.40.60.6;又在y軸右側(cè),函數(shù)y0.6x的圖象在y0.4x的圖象的上方,所以0.60.60.40.6,所以0.60.40.40.6.10已知函數(shù)f(x)3x,f(a2)81,g(x).(1)求g(x)的解析式并判斷g(x)的奇偶性;(2)用定義證明:函數(shù)g(x)在R上是單調(diào)遞減函數(shù);(3)求函數(shù)g(x)的值域【解】(1)由f(a2)3a281,得a24,故a2,則g(x),又g(x)f(x),故g(x)是奇函數(shù)(2)證明:設(shè)x1x2R,f(x1)f(x2).x1x2,2 x10,2x20,f(x1)f(x2)0,即f
6、(x1)f(x2),則函數(shù)g(x)在R上是單調(diào)遞減函數(shù) (3)g(x)1,2x0,2x11,01,02,111.11.2,3131.131.2,acb.【答案】acb3函數(shù)f(x)在R上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為_(kāi)【解析】函數(shù)f(x)在R上單調(diào)遞增,求得4a8.【答案】4,8)4(2016承德高一檢測(cè))已知函數(shù)f(x)1,x(b3,2b)是奇函數(shù)(1)求a,b的值;(2)證明:f(x)是區(qū)間(b3,2b)上的減函數(shù);(3)若f(m1)f(2m1)0,求實(shí)數(shù)m的取值范圍【解】(1)函數(shù)f(x)1,x(b3,2b)是奇函數(shù),f(0)10,且b32b0,即a2,b1.(2)證明:由(1)得f(x)1,x(2,2),設(shè)任意x1,x2(2,2)且x1x2,f(x1)f(x2),x1x2,5 x10,又5 x110,5x210,0,f(x1)f(x2)f(x)是區(qū)間(2,2)上的減函數(shù) (3)f(m1)f(2m1)0,f(m1)f(2m1)f(x)是奇函數(shù),f(m1)f(2m1),f(x)是區(qū)間(2,2)上的減函數(shù),即有1m0, 則實(shí)數(shù)m的取值范圍是(1,0).