數(shù)學(xué)-廣東省廣州市2025屆高三一模試題和解析

【答案】B是【答案】DEQ\*jc3\*hps23\o\al(\s\up2(---→),AF)EQ\*jc3\*hps23\o\al(\s\up2(--),A)EQ\*jc3\*hps23\o\al(\s\up2(--),A)【答案】B【答案】D22)2【答案】B 3232EQ\*jc3\*hps23\o\al(\s\up2(–),B)為32nx+3f2x)fx>fx2C.f(2+x)=f(2-x)2【解析】fx定義域2):fx<fx2)B錯(cuò).21<4-x2):fx1<f(4-x2)fx1+fx2)<f(4-x2)+對(duì).22:223EQ\*jc3\*hps23\o\al(\s\up2(–––),Q)EQ\*jc3\*hps23\o\al(\s\up2(–––→),pt)EQ\*jc3\*hps23\o\al(\s\up2(–––→),pt)EQ\*jc3\*hps23\o\al(\s\up2(––––→),tpt)EQ\*jc3\*hps23\o\al(\s\up2(––––→),tQt)EQ\*jc3\*hps23\o\al(\s\up2(–––),Q)EQ\*jc3\*hps23\o\al(\s\up2(–––→),pt)下面解答各個(gè)選項(xiàng). 2代入四個(gè)點(diǎn)可知是最大值是最小值)____種.9BD1212EQ\*jc3\*hps16\o\al(\s\up4(2),1)EQ\*jc3\*hps16\o\al(\s\up4(2),2)EQ\*jc3\*hps16\o\al(\s\up4(2),3)D運(yùn)動(dòng)路徑的長度為____.EQ\*jc3\*hps16\o\al(\s\up4(2),1)EQ\*jc3\*hps16\o\al(\s\up4(2),2)EQ\*jc3\*hps16\o\al(\s\up4(2),3)22222222)2:軌跡長為應(yīng)填:2【解析】:2-22 【解析】:EQ\*jc3\*hps23\o\al(\s\up2(–),D)EQ\*jc3\*hps23\o\al(\s\up0(–),n)EQ\*jc3\*hps23\o\al(\s\up0(–→),1)EQ\*jc3\*hps23\o\al(\s\up2147483647(–),n)EQ\*jc3\*hps23\o\al(\s\up2147483647(–→),2)*52525353；【解析】【解析】1)2)12222.y1y2而EQ\*jc3\*hps16\o\al(\s\up4(4),2)EQ\*jc3\*hps16\o\al(\s\up4(2),2)22EQ\*jc3\*hps16\o\al(\s\up4(3),2)EQ\*jc3\*hps16\o\al(\s\up4(2),2)EQ\*jc3\*hps16\o\al(\s\up4(3),2)EQ\*jc3\*hps16\o\al(\s\up4(2),2)32x2:2fy2=0.1212*1【解析】21且{fai→至少要有兩個(gè)不等的正數(shù)根.:1+12EQ\*jc3\*hps16\o\al(\s\up4(2),2)EQ\*jc3\*hps19\o\al(\s\up4(n),2)21EQ\*jc3\*hps16\o\al(\s\up5(2),1)3EQ\*jc3\*hps16\o\al(\s\up5(3),1)EQ\*jc3\*hps19\o\al(\s\