四川省雅安市2025屆高三數(shù)學(xué)下學(xué)期5月第三次診斷考試試題文

②當(dāng)時(shí)，單調(diào)遞減，則，當(dāng)5分綜上：..6分(2)證明：h′(x)＝eq\f(1,x)－a，因?yàn)閤1，x2是lnx－ax＋1＝0的兩個(gè)根，故lnx1－ax1＋1＝0，lnx2－ax2＋1＝0?a＝eq\f(lnx1－lnx2,x1－x2)，··················8分要證h′(x1x2)<1－a，只需證x1x2>1，即證lnx1＋lnx2>0，即證(ax1－1)＋(ax2－1)>0，即只需證明a>eq\f(2,x1＋x2)成立，即證eq\f(lnx1－lnx2,x1－x2)>eq\f(2,x1＋x2).··························9分不妨設(shè)0<x1<x2，故lneq\f(x1,x2)<eq\f(2x1－x2,x1＋x2)＝eq\f(2\b\lc\(\rc\)(\a\vs4\al\co1(\f(x1,x2)－1)),\f(x1,x2)＋1).(*)···························10分令t＝eq\f(x1,x2)∈(0,1)，φ(t)＝lnt－eq\f(2t－1,t＋1)，······································11分φ′(t)＝eq\f(1,t)－eq\f(4,t＋12)＝eq\f(t－12,tt＋12)>0，則h(t)在(0,1)上單調(diào)遞增，則φ(t)<φ(1)＝0，故（*）式成立，即要證不等式得證．······································12分22.解：（1）直線l的直角坐標(biāo)方程為：························2分曲線C的極坐標(biāo)方程為：，即，化為直角坐標(biāo)方程：.將曲線C上全部點(diǎn)的橫坐標(biāo)縮短為原來的一半，縱坐標(biāo)不變，得到曲線：.···············································5分（2）直線的極坐標(biāo)方程為，綻開可得：.可得直角坐標(biāo)方程：.可得參數(shù)方程：（為參數(shù)）.····························7分代入曲線的直角坐標(biāo)方程可得：.解得，.∴.··········································10分23.解：（1）或解出或無解,所以，原不等式的解集為[0，1]··························5分另解：（1）當(dāng)時(shí)，等價(jià)于，則∴或無解綜上，原不等式的解集為[0，1]·················································.5分（2）當(dāng)時(shí)，，因?yàn)?，所以恒成立，即恒成立，所以滿意的解集