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第1页（共5页）2020-2021学年度第二学期期末学业水平诊断高二数学参考答案一、单选题DBAACBBD二、多选题9.AC10.BCD11.ACD12.BD三、填空题13.(0,2]14.[2,3]15.116.3四、解答题17.解：（1）当0x<时，0x−>，()()sin()

sinfxxxxx−=−−−=−+，··········2分又()fx为偶函数，所以()()sinfxfxxx=−=−+.····································4分（2）当0x≥时，()(sin)1cos0fxxxx′′=−=

−≥，所以()fx在[0,)+∞单调递增.····························································6分又()fx为偶函数，所以(2)(1)(2)(1)fmfm

fmfm>−⇔>−.所以21mm>−,··············································································8分两边平方，整理得(31)(1)0mm−+>，解得1m<−或13m

>.·······································································10分18.解：（1）2()4fxx′=−.······················································

·················2分令()0fx′=，解得2x=−或2x=.·······················································3分x(,2)−∞−2−

(2,2)−2(2,)+∞()fx′+0−0+()fx单调递增极大值单调递减极小值单调递增·························································5分因此，当2x=−时，()fx有极大值，且极大值为19(2)

3f−=.··················6分当2x=时，()fx有极小值，且极小值为13(2)3f=−.·····························7分第2页（共5页）xyf(2)f(-2)O（2）方程()fxa=的实数解的个数，

即为函数()yfx=的图象与直线ya=的交点的个数.················································9分当x→−∞时，()fx→−∞，当x→+∞时，()fx→+∞，结合（1）知()fx的大致图象如右图所示.所以，当193a

>或133a<−时，解为1个；··············································10分当193a=或133a=−时，解为2个；······················································11分当1

31933a−<<时，解为3个.······························································12分19.解：（1）要使()fx的定义域为R，只需4210xxk+⋅+>在R上恒成立.·······2分令2

0xt=>，只需210ytkt=++>在0t>上恒成立.当02k−≤，即0k≥时，()yt在(0,)+∞单增，恒有()(0)10yty>=>，因此，对任意0k≥均成立.···························

·········································3分当02k−>，即0k<时，()yt在(0,)2k−单减，(,)2k−+∞单增，只需()02kf−>，即221042kk−+>，解得22k−<<，所以20k−<<.························

·······5分综上，k的取值范围为(2,)−+∞.····························································6分（2）若不等式()()fxgx<有解

，即ln(421)ln2ln2xxxkx+⋅+<=，可得04212xxxk<+⋅+<有解.·····························································7分因为当x→

+∞时，421xxk+⋅+→+∞，所以，对任意实数k，总存在00x>，使得004210xxk+⋅+>，即4210xxk+⋅+>有解.···········································8分由421

2xxxk+⋅+<可得，11(2)2xxk−<−+.··········································9分令20xt=>，1ytt=−−，221(1)(1)1ttytt

−+′=−+=，··························10分第3页（共5页）显然当(0,1)t∈时，函数单调递增，当(1,)t∈+∞时，函数单调递减，所以当1t=时，y取最大值2−，········

··················································11分所以12k−<−，即1k<−.·································································12分20.解：（1）由题意知

，长方体容器的长、宽、高分别为2ax−，528ax−，x，容器的体积5(2)(2)8Vaxaxx=−−.·······················································

2分令20ax−>，5208ax−>，0x>，可得5016xa<<.····························4分故函数3225135()(2)(2)4848Vxaxaxxxaxax=−−=−+，5016xa<<.······5分（2）令22135()1228

Vxxaxa′=−+.····························································6分令()0Vx′=，得118xa=，2512ax=（舍去）.····································

····8分x1(0,)8a18a15(,)816aa()Vx′+0−()Vx单调递增极大值单调递减·································9分因此，18xa=是函数()Vx的极大值点，相应的极大值39

()8256aaV=,也是()Vx在区间5(0,)16a上的最大值.·······················································11分答：截去的小正方形边长为18a时，容器的容积最大，最大容积3

9256a.···············12分21.解：（1）()fx的定义域为(0,)+∞，1()ln1lnfxxxxx′=+⋅−=.···············1分第4页（共5页）当01x<<时，()0fx′<，()fx单调递减；当1x>时，()0fx′>，()fx单调递增.···

···································3分因此，当1x=时，min()(1)1fxfm==−.···············································4分

由题意，min()0fx>，·········································································5分即10m−>，解得1m>.··

····································································6分（2）由（1）及()fx的单调性知，1201xx<<<.·········································7分构造函数()()

(2)gxfxfx=−−，01.x<<·············································8分则2()lnln(2)ln[1(1)]gxxxx′=+−=−−，··········································

···9分当01x<<时，21(1)1x−−<，2ln[1(1)]0x−−<，即()0gx′<，所以()gx在区间(0,1)上单调递减.························································10

分因为11x<，所以1()(1)0gxg>=，即11()(2)fxfx>−.由题意21()()fxfx=，所以21()(2)fxfx>−.····································

··11分因为()fx在(1,)+∞，且单调递增,21x>，121x−>,所以212xx>−，即122xx+>.··························································

··12分22.解：（1）1()exxfx−′=，······································································1分令()0fx′>，得1x<；令()0fx′<，得1x>.·

········································3分所以()fx的单调增区间为(,1)−∞，单调减区间为(1,)+∞.···························4分（2）由题意知e()lnxagxxxx=−+.·

·························································5分于是22e(1)1(1)e()1()exxxaxxxgxaxxx−−′=−+=−，·····················

··········6分由（1）知，在[1,)+∞上，()fx单调递减，且1()(0,]efx∈，当0a≤时，()0gx′≤，函数()gx在[1,)+∞上单调递减，取0ex=，显然e1>，但e1(e)ee11e<1ga−=−+≤−−，因此，0a≤不合题意.·······

·················8分第5页（共5页）当10ea<<时，结合（1）中()fx的单调性知，存在0(1,)x∈+∞，使得00exax=，此时()gx在0(0,)x上单调递减，在0(,)x+∞上单调递增，所以0min0000e()()lnxagxgxxxx==−+001ln

(e)1ln1xxaa=−+=+≥−，解得21ea≥，即211eea≤<；····························································10分当1ea≥时，(

)0gx′≥，函数()gx在[1,)+∞上单调递增，min()(1)e11gxga==−≥−，解得0a≥，即1ea≥；综上所述，a的取值范围21[,)e+∞.·····················

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