數(shù)學(xué)-江蘇揚(yáng)州2023-2024學(xué)年高三上學(xué)期11月期中檢測(cè)帶答案

12023－2024學(xué)年度第一學(xué)期期中檢測(cè)試題(全卷滿分150分，考試時(shí)間120分鐘)一、單項(xiàng)選擇題(本大題共8小題，每小題5分，共40分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)符合要求)1．已知集合M＝{x|0≤log3x≤1}，B＝{x||x－3|≤1}，則A∩B＝()．A．充分不必要條件B．必要不充分條件C．充要條件D．既不充分也不必要條件3．古希臘數(shù)學(xué)家泰特托斯(Theaetetus，公元前417－公元前369年)詳細(xì)地討論了無理數(shù)的理論，他通過右圖來構(gòu)造無理數(shù)…，則sin∠BAD＝()．A．(－4，2)∪(3，＋∞)B．(－3，2)∪(4，＋∞)C．(－∞3)∪(2，4)D．(－∞4)∪(2，3)AB1C．1D．36．已知a＝log34，b＝()，c＝3則a，b，c的大小關(guān)系為(A．a(chǎn)＜b＜cB．b＜c＜aC．c＜a＜bD．c＜b＜a7．已知函數(shù)f(x)＝sin2x－4x，g(x)＝ex＋e－x，則下圖所對(duì)應(yīng)的函數(shù)可能是()．A．f(x)＋g(x)B．f(x)－g(x)C．f(x)g(x)D．28．有兩個(gè)點(diǎn)在y軸上移動(dòng)，t時(shí)刻的位置分別由函數(shù)y1＝(t-)sint和y2cost確定，在(0，2π)時(shí)段內(nèi)兩點(diǎn)重合的時(shí)刻t有().二、多項(xiàng)選擇題(本大題共4小題，每小題5分，共20分．在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求．全部選對(duì)的得5分，有選錯(cuò)的得0分，部分選對(duì)的得2分)9．下列函數(shù)中，最小值是4的有().A．y＝2x＋B．y＝lnx＋C．y10．下列選項(xiàng)中，能說明“?x∈(-∞,2)，都有x2＜4”為假命題的x取值有().A4B2C．0D．311．在△ABC中，角A、B、C所對(duì)的邊分別為a、b、C，則能推出A=的有().A．a(chǎn)sinC－ccosA＝0B．bsinA－acosC＝(c－b)cosAC．tan(A＋B)(1－tanAtanB)＝D．bsinA－acosC＝(c＋b)cosA12．已知函數(shù)f(x)及其導(dǎo)函數(shù)f′(x)的定義域均為R，且滿足f(x)f(6－x)，f′(x)＝2－f′(4-x)，f′(3)1，記g(x)＝2f(3－x)－1，則下列說法中正確的有().A．函數(shù)f′(x)的圖象關(guān)于(5，1)對(duì)稱B．函數(shù)y＝f′(2x＋4)－1為奇函數(shù)C．函數(shù)g′(x)的圖象關(guān)于x＝1對(duì)稱D．?dāng)?shù)列{g′(n)}的前2023項(xiàng)之和為－4050三、填空題(本大題共4小題，每小題5分，共20分)13．已知函數(shù)f(x)＝xα的圖象過點(diǎn)(2，8)，則f′(1)=▲.14．將函數(shù)y＝sinx圖象上每一個(gè)點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼谋?縱坐標(biāo)不變)，然后再向右平移個(gè)單位長度，得到函數(shù)y＝f(x)的圖象，則函數(shù)f(x)的解析式為▲.15．已知正數(shù)a，b滿足a＋3b＝4，則＋的最小值為▲.16．若函數(shù)f(x)＝x3＋bx2＋cx＋d(b，c，d∈R)恰有兩個(gè)零點(diǎn)x1，x2，且|x2－x1|＝1，則函數(shù)f(x)所有可能的極大值為▲.四、解答題(本大題共6小題，共70分，解答應(yīng)寫出必要的文字說明、證明過程或演算步驟)317．(本小題滿分10分)已知函數(shù)f(x)＝4sin(ωx+φ)(ω>0，|φ|<)的圖象過點(diǎn)(0，2)、(x1，0)、(x2，0)，且|x1-x2|的最小值為.(1)求函數(shù)f(x)的解析式，并求出該函數(shù)的單調(diào)遞增區(qū)間；(2)若f(α)α∈(,)，求sin2α的值.18．(本小題滿分12分)如圖，直三棱柱ABC－A1B1C1中，BA＝BC＝BB1＝1，BA⊥BC.(1)記平面AB1C1∩平面A1BC＝l，證明：l∥平面A1B1C1；(2)點(diǎn)Q是直線BB1上的點(diǎn)，若直線QC1與CA1所成角的余弦值為，求線段BQ長.19．(本小題滿分12分)4定義域?yàn)镽的函數(shù)f(x)＝是奇函數(shù).(1)求實(shí)數(shù)a的值；(2)若存在θ∈[-，0]，使得f(sinθcosθ)＋f(k－cos2θ)＞0成立，求實(shí)數(shù)k的取值范圍.20．(本小題滿分12分)某品牌方便面每袋中都隨機(jī)裝入一張卡片(卡片有A、B、C二種)，規(guī)定：如果集齊A、B、C卡片各一張，便可獲得一份獎(jiǎng)品.(1)已知該品牌方便面有兩種口味，為了了解這兩種口味方便面中C卡片所占比例情況，小明收集了以下調(diào)查數(shù)據(jù)：合計(jì)C卡片2030非C卡片7545合計(jì)9555根據(jù)以上數(shù)據(jù)，判斷是否有99%的把握認(rèn)為“該品牌方便面中C卡片所占比例與方便面口(2)根據(jù)《中華人民共和國反不正當(dāng)競(jìng)爭法》，經(jīng)營者舉辦有獎(jiǎng)銷售，應(yīng)當(dāng)向購買者明示獎(jiǎng)品種類、中獎(jiǎng)概率、獎(jiǎng)品金額或者獎(jiǎng)品種類、兌獎(jiǎng)時(shí)間和方式．經(jīng)小明查詢，該方便面中A卡片、B卡片和C卡片的比例分別為若小明一次購買3袋該方便面.①求小明中獎(jiǎng)的概率；②若小明未中獎(jiǎng)，求小明未獲得C卡的概率.(a＋b)(c＋d)(a＋c)((a＋b)(c＋d)(a＋c)(b＋d)，5P(χ2≥x0)0.0500.0100.001x03.8416.63510.82821．(本小題滿分12分)在△ABC中，AC＝AB，且BC邊上的中線AD長為1.(1)若BC＝2AB，求△ABC的面積；(2)若∠ABC＝2∠DAC，求BC的長.22．(本小題滿分12分)已知函數(shù)f(x)＝x2－px－qlnx(p，q∈R)(1)若p＝a，q＝a2，f(x)的最小值為0，求非零實(shí)數(shù)a的值；(2)若p＝a2，q＝a，f(x)≥0恒成立，求實(shí)數(shù)a的取值范圍.4_4_5高三數(shù)學(xué)參考答案2023.11123456789CACBABDCABACD3f(x)=sin(2x_)2417.【答案】(1)因?yàn)閒(x1)=0，f(x2)=0，且x_xπ2π2,所以Tπ則T=π,所以負(fù)==2，············································································2分π2,所以Q=π3所以f(x)=4sin(2x+).令_π+2kπ<2x+π<π232+2kπ,解得5ππ+kπ,keZ,所以函數(shù)f(x)的單調(diào)遞增區(qū)間為_+kπ,+kπ,keZ.···································5分因?yàn)镃e(π,π)，所以2C+πe(π,4π)，_1_(5)所以sin2C=sin[(2C+π)_π]=sin(2C+π)cosπ_cos(2C+π)sinπ則平面AB1C1和平面A1BC交線為EF，所以E為AB1中點(diǎn)，F(xiàn)為AC1中點(diǎn)，所以EFB1C1，··········································4分所以EF平面A1B1C1，即l平面A1B1C1.························································6分(2)直三棱柱ABC_A1B1C1中，BA」BC，所.····································································7分A1AzB1FBEFBExyy2. ·····································3解得t=，所以線段BQ長為.································2xx2x+1，則f(x)是奇函數(shù). xx+1)=0對(duì)vxeR恒成立，所以a=1.················································(2)由(1)知f(x)則f(x)在R上為減函數(shù)，··························6分又f(x)是奇函數(shù)，由f(sinθcosθ)+f(k一cos2θ)>0得：f(sinθcosθ)>一f(k一cos2θ)=f(一k+cos2θ)，2θ,即3sinθcosθcos2θ<k在θe,0上有解，·············································9分 因?yàn)棣萫,0，則2θe,，X22又P(X2≥0.010)心6.635，(2)①記“小明一次購買3袋該方便面，中獎(jiǎng)”為事件A，P(A)=2x2x1xA=24.··············································································8分②記“小明一次購買3袋該方便面，未獲得C卡”為事件B. 64 64P(BA)=(5)=P(A)1101 64;②小明為中獎(jiǎng)，未獲得C卡的概率為64··························2.BA.BC2xcx2c2又經(jīng)ABCe(0,π)，所以經(jīng)ABC=，··············································又D為BC的中點(diǎn)，所以BD=BC=c，則BD所以S△ABC=.BA.BC.sin經(jīng)ABC=x1x2x(2)方法一：由題可知AB=c，AC=c，AD 2sin經(jīng)ADBsin經(jīng)ABDsinCsin2θa在△ABD和△ADC中，由余弦定理得2c2c)2所以a2+4=8c2，②·························22_2223c將a22_4代入得cosθ=，④22故BC的長為2.················································································AθθBDC(ABAM|BCCMABAMABAMBCCMAMMθBDCACBC所以△ACD∽△BCM，所以=，ACBCCDCM2所以a=_3c(舍)或a=2c.··············································································9分在△ABD和△ADC中，由余弦定理得2_(c)2故BC的長為2.················································································方法三：延長CB到E，使EB=BA，連接EA.a,2ACECACECCDAC2所以a=3c(舍)或a=2c.············································································在△ABD和△ADC中，由余弦定理得2c22x2x12(c)22xx1故BC的長為2.·········································