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26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案

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26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案
26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案
26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案
26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案
26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案
26数学皖南八校一联卷DA_2025年10月_251023原版:安徽省2026届“皖南八校”高三第一次大联考(全科)_答案

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2026)“*+,-”./01234&·!" %&’(、56789:; 1.A *+,-./0123:*+,-p:x∈M,p(x),4./0瓙p:x 0∈M,瓙p(x 0 ).5678,- p:x>0,9:x(x+2)>2,;<*+,-./012,4./0=:x 0>0,>?x 0 (x 0 +2)≤2.@AA. 2.D B3A=[ -槡2,槡2 ],B={x|x2-4>0}=(-∞,-2)∪(2,+∞),瓓 R A=( -∞,-槡2 )∪( 槡2,+∞ ), 瓓B=[-2,2],@A瓓B.@AD. R R 1 x 1 1 3.B Cy=lnx+1?y′= x ,CDEy= e +aFGEy=lnx+1HI,JIK3(x 0 ,y0 ),L? x = e ,? 0 e x=e,@ +a=lne+1,?a=1.@AB. 0 e 13 4.C CMN0OL?c2=a2+b2-2abcosC=49+64-2×7×8× =9,@c=3,CMN0OL?cosA= 14 b2+c2-a2 64+9-49 1 π = = ,B30<A<π,@A= .@AC. 2bc 2×8×3 2 3 (1 1) (1)α 1 5.C C8K , PQRSf(x)=xα.TUV,56 = ,α=3,56f(x)=x3,2h(x)=x3+ 2 8 2 8 lgx-18.WB3RSy=lgx,y=x3P(0,+∞)VXYZ[\,2RSh(x)P0]^(0,+∞)VYZ[\, h(2.5)=lg2.5+(2.5)3-18=lg2.5-2.375,B3lg2.5<lg10,_lg2.5<1,56h(2.5)<0,h(3)=lg3 +33-18=lg3+9,B3lg3>lg1,_lg3>0,56h(3)>0,B3h(x)P(0,+∞)VYZ[\,f(2.5)· h(3)<0,56P(2.5,3):‘K.@AC. 烄f(-1)+1=-[f(1)+1], 6.A C8RSy=f(x)+x2 =aRS,RSy=f(x)+2x3bRS,56烅 _ 烆f(-1)+2-1=f(1)+2, 烄f(-1)+1=-1-f(1), 1 烅 f(-1)+ 1 =f(1)+2, c?f(-1)=- 4 .@AA. 烆 2 AB AF DE cos2x 7.D ∠DEA=2x,P△ADEd,?DE=cos2x,ef = =cosx,?AB=cos2x,2 = =1- AF AE AB cos2x tan2x.@AD. 烄2ax(x-b),x≥b, 8.A gf(x)=2axx-b =烅 h>i 烆-2ax(x-b),x<b, jka≤f(x)≤b.cl3{xa≤x≤2b},ma<0n,o 烄f(a)=b, 烄-2a2(a-b)=b, 1 T1,@烅 烅 c?b= ;ma 烆f(2b)=a 烆2a2b(2b-b)=a, 2 >0 n,oT 2,pq rsa≤f(x)≤b.cl 3 {xa≤x≤2b},@tu.@AA. 9.BD a3<b3a<b,@“a<b”=“a3<b3”.vhwx,@Ayz;C槡b-a>1?b>a+1>a,{|}a<b, ~(cid:127)i(cid:128)(cid:129),56“槡b-a>1”=“a3<b3”.v(cid:130)i(cid:131)hwx,@B(cid:132)(cid:133);(cid:134)a=1,b=-2,a2<b2,a3<b3i(cid:128) (cid:129),@v(cid:130)(cid:135)i(cid:128)(cid:129),(cid:134)a=-2,b=-1,a3<b3,a2<b2i(cid:128)(cid:129),@(cid:131)h(cid:135)i(cid:128)(cid:129),56“a2<b2”=“a3<b3”. 1 1 1 1 (cid:136)iv(cid:130)(cid:137)i(cid:131)hwx,@Cyz;0< < 0<a<b,56“0< < ”=“a3<b3”.v(cid:130)i(cid:131)hwx,@ b a b a D(cid:132)(cid:133).@ABD. 【“!"”#$%&·!"#$%&’( ’ 1(()6() W】 书书书π 10.ACD 78A(cid:138)A,B3f(x)=sin(2x+φ )=bRS,_RSTU(cid:139)8y(cid:140)7+,562×0+φ= +kπ, 2 π ( π ) k∈Z,_ φ= +kπ(k∈Z),56A(cid:138)A(cid:132)(cid:133);78A(cid:138)B,f(x)=sin2x+ +kπ ,mk=2m,m∈Zn, 2 2 ( π) f(x)=cos2x,mk=2m+1,m∈Zn,f(x)=-cos2x,56f(x)P(cid:141)(cid:142) 0, VL{YZ[\(cid:143)(cid:144)Y 8 ( π ) Z[(cid:145),@A(cid:138) B yz;78A(cid:138) C,CA(cid:138) A fg(x)=sin 2x+ +kπ +sin(2x+π+2kπ)= 2 ( π ) ( π) sin2x+ +kπ -sin2x,mk=2m,m∈Zn,g(x)=cos2x-sin2x=-槡2sin2x- ,mk=2m+1, 2 4 m∈Zn,g(x)=-cos2x-sin2x=-槡2sin(2x+ π ),g(x)∈[ -槡2,槡2 ],56A(cid:138)C(cid:132)(cid:133);78D,CA 4 ( π) (cid:138)Cf,mk3aSn,g(x)=-cos2x-sin2x=-槡2sin2x+ ,mk3bSn,g(x)=cos2x- 4 ( π) ( π π) π ( π) π ( π ) sin2x=-槡2sin2x- ,mx∈ - , n,2x+ ∈ 0, ,2x- ∈ - ,0 ,56p(cid:146)k3aS 4 8 8 4 2 4 2 (cid:147)=bS,g(x)(cid:148)YZ[(cid:145),56A(cid:138)D(cid:132)(cid:133).@AACD. 4 11.BCD 78A,B3a,b,c3(cid:132)(cid:149)S,a2+b2+ab=4,(cid:150)(cid:151)(cid:152)(cid:153)ijkL?4≥2ab+ab,c?ab≤ ,m(cid:154)(cid:155) 3 2槡3 2槡3 4 ma=b= nj(cid:156)(cid:128)(cid:129),ma=b= n,(cid:157)(cid:158)b2+3c2=1? +3c2=1,(cid:159)n(cid:160)¡jki(cid:128)(cid:129),56a≠ 3 3 3 ( )2 4 b+槡3c b,56ab< ,@Ayz;78B,b2+3c2=1,C(cid:152)(cid:153)ijkL?b2+3c2≥ ,c?b+槡3c≤槡2, 3 2 槡2 槡6 槡2 槡6 1 槡6 m(cid:154)(cid:155)mb=槡3c,_b= ,c= nj(cid:156)(cid:128)(cid:129),mb= ,c= n,(cid:157)(cid:158) a2+c2+ac=1,L?a2+ a- 2 6 2 6 3 2 5 槡2 7 =0,¢(cid:157)(cid:158)a2+b2+ab=4,?a2+ a- =0,apc,56b≠槡3c,56b+槡3c<槡2,@B(cid:132)(cid:133);78 2 2 2 C,C 1 a2+c2+ac=1,L?a2+3c2+3ac=3,£⁄(cid:128)MN0O?a2+( 槡3c )2-2a·槡3c·cos 5π =3,Ca2 3 6 +b2+ab=4,(cid:137)£⁄(cid:128)MN0O?a2+b2-2abcos 2π =4,Cb2+3c2=1,£⁄(cid:128)¥ƒ0O?b2+( 槡3c )2= 3 1,ga=MB,b=MC,槡3c=MA,oT2:a2+( 槡3c )2-2a·槡3c·cos 5π =3=AB2,a2 6 +b2-2abcos 2π =4=BC2,b2+( 槡3c )2=1=AC2,LfAB2+AC2=BC2,2∠CAB= 3 3c2+3-a2 90°,sin∠CAM=cos∠MAB,2b= ,6bc=3c2+3-a2=4-a2-b2,ab=6bca=6c,@C(cid:132)(cid:133); 6c 1 槡3 1 2S = ×槡3×1= ,WCS =S +S +S ,§S +S +S = ·槡3c·a· △ABC 2 2 △ABC △ABM △MBC △AMC △ABM △MBC △AMC 2 5π 1 2π 1 槡3 sin + ·b·a·sin + ·槡3c·b= ,56:ab+2bc+ac=2,@D(cid:132)(cid:133).@ABCD. 6 2 3 2 2 1 4 (4 1) 16x y 16x y 12.16 ¤fx,y(cid:148)3(cid:132)S, + =1,24x+y=(4x+y) + = +4+4+ ≥8+2槡 × = x y y x y x y x 16,_x=2,y=8nj(cid:156)(cid:128)(cid:129),'“«316. sin15°+cos15° tan15°+1 tan15°+tan45° 13.-槡3 efsin15°-cos15°≠0,∴tanα= = =- = sin15°-cos15° tan15°-1 1-tan45°tan15° -tan(15°+45°)=-tan60°=-槡3. 【“!"”#$%&·!"#$%&’( ’ 2(()6() W】3槡5 14. J2x-y=t(t>0),e2x-y-1-ln(2x-y)≤1et-1-lnt-1≤0,Jf(t)= 5 1 et-1-lnt-1,f′(t)=et-1- ,m0<t<1n,f′(t)<0,f(t)YZ[(cid:145),mt>1n, t f′(t)>0,f(t)YZ[\,56f(t) min =f(1)=0,56et-1-lnt-1≤0t=2x-y =1,_KPPDE2x-y-1=0V‹›.JF2x-y-1=0fifl.DEFy=2ex HI8KQ(x 0 ,2ex0),gy′ x=x0 =2ex0=2,?x 0 =0,@IK3Q(0,2),CTf(cid:176)– 3槡5 DE2x-y-1=0.†‡d= ,_3 PQ .'“«. 5 15.c:(1)f(x)=22x2-mx+1=22(x-m 4 )2-m 8 2 +1, 2RSf(x)=22x2-mx+1(m>0).'“«3f (m) =2-m 8 2 +1= 1 ,··············· 4(cid:130) 4 2 ?–m=±4,Wm>0,2m=4.····························· 6(cid:130) (2)f(x)=12x2-4x+1=0,22x2-4x+1=0.(cid:181);3logc a,logb a, 烄logc a+logb a=2, 烅 1 ·································· 8(cid:130) 烆 logc a·logb a= 2 , 烄 1 + 1 =2, 烅 log a c log a b log a c+log a b=4, ························· 11(cid:130) 烆logc·logb=2 a a _logbc=4.···································· 13(cid:130) a 槡6 槡2 π 16.c:(1)f(x)= sinωx+ cosωx=槡2sin(ωx+ ),···················· 1(cid:130) 2 2 6 2π B3f(x).'“(cid:132)¶•3 =π,2ω=2, ························ 3(cid:130) ω ( π) π kπ π 56f(x)=槡2sin 2x+ ,g2x+ =kπ,k∈Z,c?x= - ,k∈Z, 6 6 2 12 (kπ π ) 56f(x).7+d‚3K - ,0 ,k∈Z.······················· 7(cid:130) 2 12 5π (( 5π) π) ( π) (2)f(x)TUV.5:K„”fi» ¡Y…‰(cid:190)¿,?–y=槡2sin 2x- + =槡2sin 2x- , 24 24 6 4 ( π) ¢(cid:192)5?TUV5:K.`´ˆ˜3¯˘.2˙,?–g(x)=槡2sin x- .········· 9(cid:130) 4 ( π) y=槡2sin x- +sin2x=sinx-cosx+2sinxcosx=sinx-cosx+1-(sinx-cosx)2, ·· 11(cid:130) 4 ( π) gt=sinx-cosx,x∈[0,π],2t=槡2sin x- ∈[ -1,槡2 ], ·············· 13(cid:130) 4 ( 1)2 5 y=t+1-t2=-t- + ,····························· 14(cid:130) 2 4 56mt=-1n,y=-t2+t+1¨?'“«,'“«3-1,················· 15(cid:130) 1 1 17.c:(1)ma=0n,f(x)=- -x-1,2f′(x)= -1,·················· 1(cid:130) ex ex mx<0n,f′(x)>0,f(x)YZ[\;mx>0n,f′(x)<0,f(x)YZ[(cid:145),·········· 3(cid:130) 56f(x)Px=0n¨?'(cid:201)«,_f(x) =f(0)=-2.················· 4(cid:130) max 【“!"”#$%&·!"#$%&’( ’ 3(()6() W】1 ae2x-(a+1)ex+1 (aex-1)(ex-1) (2)C-?f′(x)=aex+ -(a+1)= = . ex ex ex gf′(x)=0,L?x=0(cid:143)x=-lna.··························· 5(cid:130) ma∈(0,1)n,Cf′(x)>0?x<0(cid:143)x>-lna,Cf′(x)<0?0<x<-lna, @f(x)P(-∞,0),(-lna,+∞)VYZ[\,P(0,-lna)VYZ[(cid:145);··········· 6(cid:130) ma=1n,f′(x)≥0,f(x)PRVYZ[\; ······················· 7(cid:130) ma∈(1,+∞)n,Cf′(x)>0?x<-lna(cid:143)x>0,Cf′(x)<0,?-lna<x<0, @f(x)P(-∞,-lna),(0,+∞)VYZ[\,P(-lna,0)VYZ[(cid:145).··········· 8(cid:130) ˚V,ma∈(0,1)n,f(x)P(-∞,0),(-lna,+∞)VYZ[\,P(0,-lna)VYZ[(cid:145); Pa=1n,f(x)PRVYZ[\; ma∈(1,+∞)n,f(x)P(-∞,-lna),(0,+∞)VYZ[\,P(-lna,0)VYZ[(cid:145). ···· 9(cid:130) (3)C-LfRSy=f(x)-t.0]^3R,(cid:154)(cid:139)8K(0,0)d‚7+, 2f(0)-t=0,_t=a-2①.······························ 10(cid:130) 56y=f(x)-t3aRS, 56f(-x)+f(x)-2t=0,······························ 11(cid:130) a 1 56 -ex+(a+1)x-1-t+aex- -(a+1)x-1-t=0, ex ex a-1 ¸O? +(a-1)ex-2-2t=0②.·························· 12(cid:130) ex a-1 C①(cid:157)(cid:158)②? +(a-1)ex-2(a-1)=0, ex (1 ) _(a-1) +ex-2 =0.······························· 13(cid:130) ex 1 B3 +ex-2≥2-2=0,m(cid:154)(cid:155)mx=0n,j(cid:156)(cid:128)(cid:129), ex 1 _ +ex-2i(cid:204)30, ex 56a-1=0,_a=1, 56t=-1.····································· 15(cid:130) 18.c:(1)C2acosB=b+2c,C(cid:132)N0O?2sinAcosB=sinB+2sinC, ∵A+B+C=π,∴sinC=sin(A+B)=sinAcosB+cosAsinB, ∴2sinAcosB=sinB+2sinAcosB+2cosAsinB,∴sinB+2sinBcosA=0, ········· 2(cid:130) 1 ∵B∈(0,π),sinB>0,∴cosA=- ,·························· 3(cid:130) 2 2π ∵A∈(0,π),∴A= .································· 4(cid:130) 3 1 1 槡3 (2)B3S = bcsinA= a·AD,_a= bc, △ABC 2 2 2 槡6 Wa= c,56b=槡2,································· 6(cid:130) 2 2 2槡3 CMN0O?a2=b2+c2-2bccosA=b2+c2+bc=2+ a2+ a, 3 3 ˝˛L?a2-2槡3a-6=0, 【“!"”#$%&·!"#$%&’( ’ 4(()6() W】c?a=3+槡3,···································· 8(cid:130) 1 3+槡3 56S = a·AD= .····························· 10(cid:130) △ABC 2 2 2π (3)B3AD3∠BAC.ˇfi(cid:130)E,(cid:154)AD=1,A= 3 ,S △ABC =S △ABD +S △ADC , 1 2π 1 π 1 π 56 bcsin = ·c·1·sin + ·b·1·sin , 2 3 2 3 2 3 56bc=b+c≥2槡bc, 56bc≥4,m(cid:154)(cid:155)mb=c=2n,“=”(cid:128)(cid:129),························ 12(cid:130) 3bsinB+3csinC 562asinA+3bsinB+3csinC=2asinA+ ·sinA sinA 3(b2+c2) 3·2bc 24 =2asinA+ ·sinA≥2asinA+ ·sinA≥2asinA+ ·sinA a a a 24 3 ≥2槡2asinA· ·sinA=2槡48sin2A=2槡48× =12, a 4 24 m(cid:154)(cid:155)m2asinA= ·sinA,_a=2槡3n¨j(cid:156),···················· 15(cid:130) a 2π π Wmb=c=2,A= n,a=槡b2+c2-2bccosA=2槡3,B=C= , 3 6 2asinA+3bsinB+3csinC=12, @2asinA+3bsinB+3csinC.'“«312.······················· 17(cid:130) 19.c:(1)—(cid:209):C-L?f′(x)=cosx, f′(2x)-f′(4x)=2f(x)f(3x)cos2x-cos4x=2sinxsin3x, cos2x-cos4x=cos(3x-x)-cos(3x+x) ························ 2(cid:130) =cos3xcosx+sin3xsinx-cos3xcosx+sin3xsinx =2sinxsin3x. @f′(2x)-f′(4x)=2f(x)f(3x).··························· 4(cid:130) x (2)g(x)=sinx- ,@g(0)=0, ex x-1 x-1 2-x g′(x)=cosx+ ,Jh(x)=cosx+ ,2h′(x)=-sinx+ , ex ex ex 2-x mx∈(-π,0)n, >0,-sinx>0,@h′(x)>0,g′(x)YZ[\, ex 56g′(x)<g′(0)=1-1=0,@RSg(x)YZ[(cid:145),g(x)>g(0)=0, @RSg(x)P(-π,0)Vp‘K; ···························· 6(cid:130) x 1 mx∈(0,π)n,g(x)=sinx- = (exsinx-x), ex ex JF(x)=exsinx-x,2F′(x)=ex(sinx+cosx)-1, Jk(x)=ex(sinx+cosx)-1,2k′(x)=2excosx, ( π) mx∈ 0, n,k′(x)=2excosx>0,k(x)YZ[\, 2 (π ) mx∈ ,π n,k′(x)=2excosx<0,k(x)YZ[(cid:145), 2 (cid:154)k(0)=0,k (π) =e2 π -1>0,k(π)=-eπ-1<0, 2 【“!"”#$%&·!"#$%&’( ’ 5(()6() W】(π ) @(cid:210)Px 0∈ 2 ,π ,>k(x 0 )=0, mx∈(0,x 0 )n,k(x)>0,F(x)YZ[\, mx∈(x 0 ,π)n,k(x)<0,F(x)YZ[(cid:145), B3F(0)=0, @F(x 0 )>0,WB3F(π)=-π<0, @RSg(x)Px∈(x 0 ,π)V:1¡‘K.························· 9(cid:130) ˚V5(cid:211),g(x)P(cid:141)(cid:142)(-π,π)(cid:212).‘K¡S32.····················· 10(cid:130) n sinx sin3x sin[(2n-1)x] (3)—(cid:209):gH(x)=f(x)+ i ∑ =1 h n (x)=sinx+ 3 + 5 +…+ 2n+1 , 8sin2x 2sinxsin3x 2sinxsin[(2n-1)x] 56(2sinx)H(x)= + +…+ 3 5 2n+1 4(1-cos2x) cos2x-cos4x cos[(2n-2)x]-cos2nx = + +…+ ·············· 12(cid:130) 3 5 2n+1 4 (4 1) (1 1) ( 1 1 ) cos2nx = - - cos2x- - cos4x-…- - cos[(2n-2)x]- 3 3 5 5 7 2n-1 2n+1 2n+1 4 (4 1) (1 1) ( 1 1 ) 1 ≥ - - - - -…- - - =0, ·············· 15(cid:130) 3 3 5 5 7 2n-1 2n+1 2n+1 m(cid:154)(cid:155)mcos2kx=1(k∈N)n,Vk¨?j(cid:156),(cid:213)iL{(cid:128)(cid:129), n 56mx∈(0,π)n,f(x)+∑h n (x)>0. ························ 17(cid:130) i=1 【“!"”#$%&·!"#$%&’( ’ 6(()6() W】