福建省福州市八縣（市）協(xié)作校2024-2025學(xué)年高二下學(xué)期期中聯(lián)考數(shù)學(xué)試卷（含答案）

1．若CEQ\*jc3\*hps13\o\al(\s\up5(2x),20)=CEQ\*jc3\*hps13\o\al(\s\up5(3x),20)-5，則實(shí)數(shù)x的值為()假設(shè)從中依次不放回地隨機(jī)抽取兩個(gè)節(jié)目參加福州市元宵晚會(huì)，求類節(jié)目的條件下，第二次抽到語(yǔ)言類節(jié)目的概率為()3.函數(shù)f(x)在x=x0處可導(dǎo)，若A．1B．C．3D．-1推薦官必須由不同性別的主播擔(dān)任，且小李（男）和小紅（女）至少不同的安排方案有()6．定義f(x)=f'(x)的實(shí)數(shù)根x叫函數(shù)f(x)的“躺平點(diǎn)”．若函數(shù)g(x)=ex-2x，φ(x)=2024x+2024，h(x)=lnx的“躺平點(diǎn)”分別為a,b,c則a,b,c大小為()7．已知函數(shù)f(x)=4x3-ax2+3在[0,2]上的最大值為3，則實(shí)數(shù)a的范圍是()9．已知二項(xiàng)式展開式的二項(xiàng)式系數(shù)的和為6C．展開式常數(shù)項(xiàng)為-160D．展開式中各項(xiàng)系數(shù)和為-110．隨機(jī)變量X分布列如右表，下列說法正確的是()A．D．D(x)≤E(x)11．函數(shù)f(x)=x3+3ax+2存在3個(gè)零點(diǎn)，則實(shí)數(shù)a的取值可以是()A．-2B．-1C．-3D．-5在的展開式中，求．（2）求函數(shù)f(x)的極值．答一個(gè)知識(shí)問題，回答正確的概率為p(0<p<1)，回答錯(cuò)誤的概率為1-p，每輪比賽的結(jié)果是獨(dú)立的，即每輪比賽甲回答正確的概率不受其他光潔曲線"。曲線對(duì)應(yīng)的函數(shù)g(x)為“光潔函數(shù)”，即函數(shù)g(x)在(0,+∞)上單調(diào)遞增，此時(shí)稱f(x)=xg(x)為“超光潔函數(shù)”．若是“超光潔函數(shù)”，求a的取值范圍；（2）已知光潔曲線y=ex-1的一條基準(zhǔn)線l與曲線y=lnx+1相切，求l的方程；（3）若f(x)=2ex-x2lnx-2，a>0，b>0，c>0.證明：f(a)+f(b)+f(c)<f(a+b+c).（可用結(jié)論：當(dāng)x→0+時(shí)，x2lnx→0）高二數(shù)學(xué)參考答案及評(píng)分細(xì)則一、單項(xiàng)選擇題：本題共8小題，每小題5分，滿分40分．1.A2.D3.B4.B5.C6.C7.D8.C二、多項(xiàng)選擇題：本題共3小題，每小題6分，滿分18分．9.AC10.BCD11.ACD二、填空題：每小題3分，滿分15分．三、解答題：本大題共5小題，共77分．解答應(yīng)寫出文字說明，證明過程或演算步驟．15.(1)由題知展開式的通項(xiàng)為 k=0,1,2,…,6....................................................................................................2分令,解得k=2.................................................................................3分所以展開式中含x-2項(xiàng)為T3=CEQ\*jc3\*hps14\o\al(\s\up5(2),6).22.x-2=60x-2.........................................5分所以展開式中含x-2項(xiàng)的系數(shù)為60..................................................................6分(2)令3-k∈Z,k=0,1,2,…,6.∴k=0,2,4,6......................................................................................................8分,...........................................................................................9分EQ\*jc3\*hps14\o\al(\s\up5(2),6)2.x-2=60x-2,............................................................10分EQ\*jc3\*hps14\o\al(\s\up5(4),6)4.x-7=240x-7,.......................................................11分EQ\*jc3\*hps14\o\al(\s\up4(6),6)6.x-12=64x-12.........................................................12分綜上所述，展開式中所有的有理項(xiàng)分別為x3,60x-2,240x-7,64x-12.13分所以a=0.42........................................................................................................3分所以該100名居民的每日睡眠時(shí)長(zhǎng)的第50百分位數(shù)：所以50%分位數(shù)一定位于[6,8)內(nèi)，.................................................................5分由，..............................................................7分所以a=0.42........................................................................................................3分所以該100名居民的每日睡眠時(shí)長(zhǎng)的第50百分位數(shù)：設(shè)第50百分位數(shù)為x，則0.09+(x-6)×0.42=0.5，................................5分解得x≈6.98........................................................................................................7分所以X可取0，1，2，3.......................................................................................9分所以X的分布列為X0123P數(shù)學(xué)期望分17.解：(1)由f(x)=a(x-7)2+12lnx(a∈R)，得f分令x=1，則f(1)=36a,f(1)=12-12a................................................3分所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-36a=(12-12a)(x-1)..................................................................................................................................5分∵點(diǎn)(0,12)在切線上，可得12-36a=12a-12.6分解得a=............................................................................................................7分.................................................................................................................................8分則f分令f(x)=0解得x1=3,x2=4..........................................................................10分則x、f(x)、f(x)的變化情況如下：x34f(x)+0-0+f(x)單調(diào)遞增8+12ln3單調(diào)遞減+12ln4單調(diào)遞增.................................................................................................................................12分所以f(x)在(0,3)和(4,+∞)上單調(diào)遞增，f(x)在(3,4)上單調(diào)遞減；.................................................................................................................................13分所以f(x)的極大值為f(3)=8+12ln3，f(x)的極小值為.................................................................................................................................15分法21）同解法一；由知且f的定義域?yàn)?0,+∞)，.................................................................................................................................8分則f分令f(x)=0解得x1=3,x2=4.10分當(dāng)0<x<3或x>4時(shí)，f(x)>0,故f(x)的單調(diào)遞增區(qū)間是(0,3)和(4,+∞)；.................................................................................................................................12分當(dāng)3<x<4時(shí)，f(x)<0,故f(x)的單調(diào)遞減區(qū)間是(3,4)；.................13分所以f(x)的極大值為f(3)=8+12ln3，f(x)的極小值為f(4)=+12ln4..................................................................................................................................15分18.(1)記“甲最終以2:1獲勝”為事件A，記“甲最終以2:0獲勝”為事件B，“甲最終獲勝”為事件C........................................................................................................................1分于是C=AUB，A與B為互斥事件，............................................................2分由于分 ...................................................................................................6分則即甲最終獲勝的概率為，..................7分（2）由（1）可知，P(C)=P(A)+P(B)=3p2-2p3.........................8分若選用方案一，記甲最終獲得積分為X分，則X可取5,2.......................9分P(X=5)=P(C)=3p2-2p3,P(X=2)=1-3p2+2p3....................11分則X的分布列為：X52p3p2-21-3p2+2則E(X)=15p2-10p3+2-6p2+4p3=-6p3+9p2+2....................12分，若選用方案二，記甲最終獲得積分為Y分，則Y可取3，0.......................13分P(Y=3)=P(C)=3p2-2p3，P(Y=0)=1-3p2+2p3....................14分 則Y的分布列為：303p2-21-3p2+2則E(Y)=9p2-6p3......................................................................................15分所以E(X)-E(Y)=2...................................................................................16分答：方案一始終更優(yōu).........................................................................................17分 由題意可知g,(x)≥0恒成立，.......................................................................2分故x2-x-a≥0，即分故解得a≤1........................................................................4分解法由題意可知g’(x)≥0恒成立，.......................................................................2分a≤(4x24x)min=1....................................................................................4分則k=ea.............................................................................................................設(shè)直線l與曲線y=lnx+1相切于點(diǎn)Q(b,lnb+1)則.............................................................................................................因此有當(dāng)a=1時(shí)，l的方程為y=ex一1.............................................................................當(dāng)a=0時(shí)，l的方程為y=x....................................................................................設(shè)直線l與曲線y=lnx+1相切于點(diǎn)Q(b,lnb+1).........................................................................................................................................7分整理得(a-1)(ea-1)=0，解得a=1或a=0......................................................8分當(dāng)a=1時(shí)，l的方程為y=ex-1.............................................................................9分當(dāng)a=0時(shí)，l的方程為y=x....................................................................................10分令h(x)=2(x-1)ex-x2lnx-x2+2，通過求導(dǎo)易證：lnx≤x-1,ex≥x+1......................................................................13分所以所以h(x)在(0,+∞)上單調(diào)遞增，又x→0+時(shí)，x2lnx→0所以x→0+時(shí)，h(x)→0，所以h(x)>h(0)=0所以g'(x)>0，即g(x)=在(0,+∞)上單調(diào)遞增..........