高考數(shù)學(xué)：不等式恒成立、能成立、恰成立問(wèn)題匯編

fxminABfxAf(x)f(x)DD上，AfxfxBDD上,Amax例當(dāng)xaax22xafx,x,fx0xa;例fxfcos2msinf2m2022例R恒m.f(x)axlnxbxc(x0)x13c44abab、、例在f(x)x0f(x)2c2cx例x10對(duì)aa1x(a4)x42a02例xa3f(x)xx(ax132＞f(x)x2xa1a(0)32a中例xfxgfxg）fxxD）求在gf(x)gfx）或)maxmin））xmx40x2m.例1f(x)axbxx332a0a,bf(x)a0,3例,f(x)(0,1].ab且例xR|xa(x2logx例x<aafxmaxAfxABxDD上；fxfxBminxDD.x4x3aa例Rx例x2a3a．1fxlnxax22xa0a2例（1x|1x103ab則例ax2x2xa2fx,x,fxxa.例當(dāng)例kxkxkk不等式恒成立、能成立、恰成立問(wèn)題專(zhuān)項(xiàng)練習(xí)(m1)x(m1)x3(m1)02xmk622xx2xR29f(x)xx6xa32f(x)m2x，mxpx1p2x22xx22xax3aaf(x)x(a4)x42a2x1xlogx022、m在m。x(4x)x在akx2k20kx2x1ax[05]xaaMN，．x3x2aax3x2aaaxyc0cx3x2ax(y1)122、x(y1)1xyc022，c，2f(x)11f(x)xax2xb(xR)，其中a,bRa432在b1f(x)xa)x4ax24a321x0af(x)0a3f(x)abax2b，=t不等式恒成立、能成立、恰成立問(wèn)題參考答案例ax2xa0x例、解：等價(jià)于x2x1,x0于.x1a1x2則在,x1a3a3a3mint·fcos2msinf2m202o例3、解：由得到：1圖1fxfcos2msinf2m22fcos2msinf2m22fx22m2m2為R2對(duì)0設(shè)sint2t22m10ttgtt2mt2m1tm2.·o1g02m10圖2tm0，11mm0m0∴22即tm0m1,t4m4m2m102m10·2m2即,o1m0圖3∴12m12又∴0m1g112m2m120tm1∴m11m2.f(x)x1f3c例412）略（32在.f(x)2c2(x0)3c2ccc302即2，323c(,[,)(2cc0.或c.1.c21ax(,1)(3,)2例例ax3x(a1)xxa1a(0)對(duì)a(x2)x2x02222例7對(duì)a(0)g(a)g(a)(x2)ax2xaR22（x20xRg(a)R為所a則2，a(0)g(a)0g0，，x2x02x0x2范圍是{x|2x0}。x4x4422mf(x)xxf(x)x402xxx例:當(dāng)?shù)昧钤谏蟲(chóng)42f(1)5()5xf(x)m5x時(shí)max∴.ab2（2）f(x)(0,1]f'(x)ax2bx10(0,1]在2例91）ax11b,x(0,1]22x()bx(0,1]22xmax，。1a(x)2ax1a1g'(x)ag(x)22x22x22x2設(shè)，，11xxg'(x)0aa令當(dāng)當(dāng)?shù)没颍?1ax1g(x)22xx)01g'(x)0a1aa,時(shí)，1ax122xx(g(x)a時(shí)g'(x)0，,1g()ag(x)maxaab。1ax11g(x)a1g'(x)0(0,1](0,1]0a22x當(dāng)a1a1g(1)bg(x)max22。a1yby|x|a1,ba；當(dāng)0a1。2yaxyaxxR|x例x1a11a1。O例a1例fxx2axaxx2a3例.設(shè)fx3,fx3min在,4aa2fxa6a2或4即2x1.112,h(x)lnxax2xh(x)2b22xx例hxhxh(x)0,,12xahx0hx0xx而在,即2,,12uxauxxx2.121112uxux1x,a1,xx由2,0,a0a例6x2xa2fx0xx例.x2xaa2fxx23當(dāng)a0x1,,xxx2xaax2x2fxfxf10當(dāng)a0時(shí),是,,f1,令,即x1a2a3.例xhmin令或。由hminx在xhmaxh。max2?f(x)g(x),x[3g(x)g(3，由，得或，易得min，又，x[.故f(x)f120k.令專(zhuān)項(xiàng)練習(xí)：13(,)11、解：[f(x)3x9x6Rf(x)m3x9x(6m)0xR'2x'23、解析：,對(duì),,即在上恒成立,34348112(6)0,得mm。設(shè)則4x30f(0x或x3x12x10或x1xf(2)1或2即1[16ya0x(,0)(4,)yyx(4x)x和在303xay3x3時(shí)，333aax(4x)當(dāng)x3322k2kk(x2x11kx2k202x229、解：不等式有解有解有解max，所以k(。2x1(x1)，f(x)x2x13(1x2)≤≤，af(x)af(x)3min2x1(x2).又，M{ax2x，x[0]agx)ag(x)g(5)9g(x)所以．令恒成立．所以maxNaaa[c③②c[12,12]a5a5②21，2f(x)4x3ax4xx(4x3ax4)a322f(x)0f(x)09a04x40x0x0f(x)2而21f(1)f(1)在與1a2f(x)1f(x)1，max在f(1)1b(2a)b2aa2a21)1b(2a)f(b2a即在，4b4．bx0f(x)在x2ax0或14f(2a)(2a)a)(2a)4a2a24aa4a24a3232f(0)24a33；aa14a(aa6)