上海市宝山区行知中学2020-2021学年高一上学期期中考试数学试题

645A1,2,3,4，Bxx4AB132()()0384xy0xy2log2alog27记a34x13x2xbx6axb00x)x112x2xx3x3x2x3kxRkxm1112xm3ax1axa0xlgx,0x10,a,b,cf(a)f(b)f(c)abc(x)f1x6,x10.2IAI,②Amin()A表A示Amin()为的AIIAb若a2）2a0bab0ababAxxP）00AAxAxP”P”P”00xAP”x001x2(x)f）x32abab0,b0240且ab（）a6655xax2(a1)x10(aR).18.（a,b,c,dRac2(bd)20xaxb和20xcxd.1问72问7分）2“”()x()3pp.(x10xaa，102p件产品的销售价格定为).pxy;?.(x)x2axb(a,bR)1问4分2问62问6f.1f(x)0(,(,)xba、20,2f(x)g(x)12,b0,g(x)kx（2ak数.(x)0(1,2)f(1).f1问42问62问8f(x)(x)fIyIxyf(x)I()fxxex，gxx()242”x1()2()xb（m、b，hxxm2求、bm(x)|x|x2|k|x（kk0Ifyf(x)Ik.645A1,2,3,4，Bxx4AB【答案】{4}1323()()08420132381415.384xy0xy2【答案】{(1,1)}log2alog27记a3433133332222242a23x13x2xRbx6axb00x)x,6,112x2xx3x311x0x30且，|x2|x2得x3x3,33,2.x2x3kxRk,5xm111mx32|xm1得xm11，m1312因为不等式|xm1x，1312m1m114m，2314,m.23ax1axxa0a01当a0xmax,aa11x0a1，a(a0)当aaaxa,0,.algx,0x10,a,b,cf(a)f(b)f(c)abc(x)f1x6,x10.2f(a)f)fc)kk，01令f0a1bc，Aakbk，ab01，c10,12abc.IAI,②Amin()A表示Amin()为的AIIAmin()，当当当当当min()1|A1A，min()2|A2A，min()3|A3min()4|A4A，A{4},{4,5}，min()5|A5A{5}，14421.Ib若a2C）2a0bab0ababAxxD）P00AAxAxP”P”P”00xAP”x001x2(x)f）x31x2()fx()()fxfx，x30f(x)0f(x)0C，当x，x2ab0,b0240且ab（）aab65564044a2bba2aba2a，aaaa，abaa4abaa4222abaa111453333，4baba2a44aa121aaaab8xax2(a1)x10(aR).ax2(1)10(0，axx1a0当,1，a0(，当a10a1(,1),，当a1当a1,).a18.（a,b,c,dRac2(bd)20xaxb和20xcxd.4040，a2b且c2d12abc4dac4(bd)0，222212bd)acacac)02(22212.1问72问72“”()x()3pp.(x10xaa，102p件产品的销售价格定为).px;y?.242)3pxyppx1p4y16y17x(0xa)；x144x1172(x1)13，）1x1x4x1x1x1a114a1y17a]1在x当1x1a.a(x)x2axb(a,bR)1问4分2问62问6f.1(x)0xf(,(,)a、b20,2f(x)g(x)12,b0,g(x)kx（2ak数.(x)0(1,2)f(1).f1(x)0(x)0f(,2),，2121f和，2(()a25,1ab，12(()b2202a(x)g(x)1，bf(x)x2x，f在1221xx在0101当x当x110,222xkx在xx112kx2，即xxx1122在yx在yxxx1在1,2k0；2(x)0f(x)(xx)(xx)x(xx)xxxf，2121212,xxxx且，1212与()fxx2axbaxxbxx，(1212(1)11()abxxxx，(fxx121212,xx.且xxf1212.1问42问62问8f(x)(x)fIyIxyf(x)I()fx”xe，gxx2x()42x1()2()，hxxmxb（m、b2求m、b(x)|x|x2|k|x（kk0Ifyf(x)Ik.f(x)efx，x()xg(x)1(x)2x1)2)在g在xx(x))g是；12(x)x2mxbhh(x)b1xm且xx21m1220,b1m1，b；23kk)x,x11k,1x2(x)()，ffx3kk)x,2x33kk)x,3设yf(x)I(,)I(I(1,2),II(3,)，1234IIIyf(x)若III*1*111f(x)3k(x)3kk)x(2k)故fxxk00IIIIIyf(x)I若*2*222f(x)1k(x)1kk故fxxk010；1k03IIIyf(x)I若II*3*333f(x)3k(x)3kk)x(2k)fxx20，解得2；30IIIIIyf(x)I若*4*444f(x)3k(x)3kk)x(2k)fxx20＜1；301(2,1)(,0)(1,2).k312(x)x2mxbhh(x)b1xm且xx21m1220,b1m1，b；23kk)x,x11k,1x2(x)()，ffx3kk)x,2x33kk)x,3设yf(x)I(,)I(I(1,2),II(3,)，1234IIIyf(x)若III*1*111f(x)3k(x)3kk)x(2k)故fxxk00IIIIIyf(x)I若*2*222f(x)