2022高考数学萃取精华试题（9）新人教A版

2022高考数学萃取精髓30套（9）冀州一模20、已知数列{an}知足a11，a21，且[3(1)n]an22an2[(1)n1]，2（n=1,2,3,）（1）求a3,a4,a5,a6的值及数列{an}的通项公式；（2）令bna2n1a2n，记数列{bn}的前n项的和为Tn，求证：Tna33,a41,a55,a612mmN*a2m1a2m12{a2m1}a2m12mann2m48mN*2a2m2a2m0{a2m}a2m1(1)m11an(1)2n222m2n(n2m1mN*)1annbna2n1a2n(2n1)(1)2（n2mmN*)2n2Tn113151(2n1)1222232n111112Tn122323(2n3)2n(2n1)2n11111111114(12n1)12Tn22(22232n)(2n1)2n1211(2n1)2n12Tn32nn3Tn3a3,a4,a5,a6y3xy3xAB23ABCQ(1,0)llC233xx1x2,M、NRMMQRNNQP(x,y)A(x1,y1)B(x2,y2)AB2y1y2.y23xy3xy13x13x2x1x223y,A、Byy2y1y223x.AB2333333(x1x2)2(y1y2)21212y24x212Cx2y21llyk(x1)M(x3,y3)39yk(x1),9k2)x218k2x9k2N(x4,y4)R(0,y5)M、Nx2y21.(1909x3x418k29k29RMMQ(x3,y3)(0,y5)(1,0)(x3,y3)19k2x3x419k2x3(1x3),x3(1x3)lx31x3x4x3x4y3y51x31x41x31x4y3.(x3x4)2x3x49a0,且a1x)f(x)f(x)1(x3x4)x3x44f(x)loga(1an*求limaf(n);aeh(x)(1ef(x))(x2m1)h(x)h(x)1xN,na0naa0a1时，f(x)的定义域是（0，）；当a1时，f(x)的定义域是（，0）f(x)=-axlnalogaeax1axax10a1时，x(0,).因为ax10,ax0,故f(x)<0,所以f(x)是减函数a1时，x(,0),因为ax10,ax0,故f(x)0,所以f(x)是减函数f(n)loga(1an),所以af(n)1an1an0,因为n是正整数，故0<a<1所以limaf(n)lim1an16分anaanaann（Ⅲ）（hx)ex(x2m1)(x0),所以h(x)ex(x22xm1)令h(x)0,即x22xm10,由题意应有0，即m0①当m=0时，h(x)0有实根x1，在x1点左右两侧均有h(x)0故无极值②当0m1时，h(x)0有两个实根x11m,x21m当变化时，h(x)、h(x)的变化情况如下表所示：(,x1)x1(x1,x2)x2(x2,0)h(x)0-0h(x)↗极大值↘极小值↗h(x)的极大值为2e1m(1m)，h(x)的极小值为2e1m(1m)③当m1时，h(x)0在定义域内有一个实根，x1m同上可得h(x)的极大值为2e1m(1m)10分综上所述，m（0，）时，函数h(x)有极值；当0m1时h(x)的极大值为2e1m(1m)，h(x)的极小值为2e1m(1m)当m1时，h(x)的极大值为2e1m(1m)分12温州三模2115CF01A2tI

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AQPQ215Ix24y3Ax24yt16IIAPAQAPy1k(x2)即ykx(2k1)ykx(2k1)x24yx24kx4(21)09kxAxp4(2k1)xp2(2k1)4k2ypkxp(2k1)4k24k1xq42,yQ4k24k1k12yqyP8k115kPQxp8kxQ2215f(x)ax2lnx2IIIf(x)1a22215Ia1f(x)1x2lnx(x0)21f'(x)x3xf'(x)0011∞5fmin(x)f(1)1072II1ax21f'(x)axx9xif'(x)01f'(x)11x0,aaa,0,f(1)11lnaa22f(x)111lna1122222iif'(x)00,f(1)a02f(x)114215f(x)1a12lnx92x2(x)12lnx(x0)'(x)4lnx11x2x3'(x)0max(x)(1)113a(x)15湛江二模2014C1yax2(a0)C2x2(y1)25l1:y2xmm0C1C2FC11ma2AC1AC1lyBFMFAFBM20141C2x2(y1)25C2(0,1)r51l1:y2xmd|1m|322(1)2|1m|5m6m4422(1)2l1A0(x0,y0)y/2ax52ax02x01y016aa121aa6a6m6a17621C1y1x2F(0,3)862A(x1,6x11Al12)y1x1(xx1)1x1210361x0lyB2)(0,6x111FA(x1,1x123)FB(0,1x123)126262FMFAFB(x1,3)13FMy31422114f(x)xalnxax1aRyf(x)2a1yf(x)3a(0,m]yf(x)m21141lnx0x1f(1)1yf(x)(1,1)22a1f(x)xlnxf/(x)11lnxx2lnx14xx2x2)x2lnx1g(x)0g(x)0x1(g/(x)2x1x0g/(x)0yg(x)(0,)xg(x)g(x)0x1x(0,1)f/(x)0f(x)(0,1)g(x)x2x(1,)f/(x)0f(x)(1,)x27x1f(x)f(1)1ln1181x23f/(x)1aalnxalnxah(x)x2alnxax2x2a(0,m]h(x)≥0x(0,)2x2a2(xa)(x