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26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)

  • 2026-03-15 23:30:42 2026-02-08 11:54:45

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26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)
26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)
26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)
26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)
26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)
26高三皖八二联-数学答案_2025年12月_251221安徽省2026届皖南八校高三第二次大联考(全科)

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2026)“*+,-”./01234&·!" %&’(、56789:; 1.C z 1= a-i = (a-i)(1-2i) = a-2-(2a+1)i ,*+ z 1+,-.,/0a-2=012a+1≠0,/0a=2. z 1+2i (1+2i)(1-2i) 5 z 2 2 23C. 2.D A={x|log2 x<1}={x|0<x<2},B={x||x|<2}={x|-2<x<2},4瓓 R B={xx≤-25x≥2}, A∩(瓓B)=.23D. R 3.C *+x2-2x+2=(x-1)2+1≥1,2f(x)6789+R,f(x)+1=ln(x2-2x+2)+1,:;<=.,2 A>?;f(x)-1=ln(x2-2x+2)-1,:;<=.,2B>?;f(x+1)=ln((x+1)2-2(x+1)+2)= ln(x2+1),+<=.,2C@A;f(x-1)=ln((x-1)2-2(x-1)+2)=ln(x2-4x+5),:;<=.,2 D>?.23C. 1 a 1 1 1 1 1 4.A *+a 1 = 2 ,a 4 =4,/0 a 4=q3=8,/0q=2,a n =2n-2,/0 a + a +…+ a =2+1+ 2 +…+ 26-2 = 1 1 2 6 [ (1)6] 2× 1- 2 63 = .23A. 1 16 1- 2 5.D *+f(x)+B=.,Cx<0D,f(x)=x+sinx+1,4Cx>0D,f(x)=-f(-x)=x+sinx-1, π π π π π f′(x)=1+cosx.EFf′( )=1,Gf( )= ,4HIy=f(x)Jx= K6LIMN;y- =1× 2 2 2 2 2 ( π) x- .Ox-y=0.23D. 2 (π) π π 1 槡3 6.A PQR,Sf =sin +acos = a+ =0,Sa=-槡3,/0f(x)=sin2x-槡3cos2x= 6 3 3 2 2 ( π) ( π) [( 5π) π] 2sin2x- ,y=2cos2x=2sin2x+ =2sin2x+ - ,TUV=.y=f(x)6WXYZ[\ 3 2 12 3 5π ]^_‘a.23A. 12 7.C 0A+bcde,AB/JfI+xg,AD/JfI+yg,hi[jfkbc → → lmW/n,4A(0,0),E(2,1).o DF =x,4F(x,2),0≤x≤2,2AF=(x,2), → → → → → AE=(2,1)./0AE·AF=(2,1)·(x,2)=2x+2,Cx=2D,AE·AFpSqr 1 1- (π ) 2 1 s,tDtan∠EAF=tan -∠BAE = = .23C. 4 1 3 1+ 2 8.B ug(x)=-f(4-x)=-(4-x-a)(4-x-b)(4-x-c)=(x+a-4)(x+b-4)(x+c-4),*+ f(x)=(x-a)(x-b)(x-c)6ve+a,b,c,wxy=f(x-1)6ve+a+1,b+1,c+1,y=g(x)6ve +4-a,4-b,4-c,G*+0<a<b<c,41<a+1<b+1<c+1,4>4-a>4-b>4-c,yf(x-1)·f(4 -x)≤0,Of(x-1)·[-g(x)]≤0,4f(x-1)·g(x)≥0,wxy=f(x-1)6vezy=g(x)6ve 烄a+1=4-c 3 {|,4烅b+1=4-b,wS2b=a+c=3,b= .23B. 2 烆c+1=4-a 【“!"”#$%&·!"#$%&’( ’ 1(()6() W】 书书书(1)m (1)m 9.CD }QR,olog1a=log1b=m,4a= ,b= ,Cm>0D,0<b<a<1,Cm=0D,a=b=1, 3 4 3 4 Cm<0D,1<a<b.23CD. 10.ABD ~A,mW,(cid:127)(cid:128)AC,BD(cid:129)(cid:130)eO,(cid:127)(cid:128)PO,*+J@(cid:131)(cid:132)(cid:133)P-ABCD(cid:134),(cid:135)jABCD+@M(cid:136), /0AC⊥BD,G*+PB=PD,O+BD(cid:134)e,/0PO⊥BD,G*+AC∩PO=O,AC,PO[jPAC, 1 /0BD⊥[jPAC,A@A;~B,(cid:127)(cid:128)QE,QF,EF,*+λ= ,/0QE∥AB,QF∥AD,GQE,QF[ 2 jQEF,AB,AD[jABCD,QE∩QF=Q,AB∩AD=A,/0[jQEF∥[j ABCD,B@A;~(cid:130)C,*+PA=PC,O+AC(cid:134)e,/0PO⊥AC,*+(cid:131)(cid:137)(cid:136) ABCD+@M(cid:136),/0AC⊥BD,G*+BD∩PO=O,BD,PO[jPBD,/0AC ⊥[jPBD,4AC⊥[jPEF,/0Cλ=0,OeQzP(cid:138)(cid:139)D,AC⊥[jQEF, 1 → 1→ → → → → C>?;~D,oEFzPO6(cid:129)e+H,Cλ= D,PQ= PA,QC=PC-PQ=PC 3 3 1→ → → → 1 → → 1→ 1→ 1→ → → - PA,QH=PH-PQ= (PA+PC)- PA= PC- PA,4QC=4QH,/0Q,H,C)I,/0 3 4 3 4 12 Q,E,F,C(cid:131)e)j,D@A.23ABD. 烄2sinC-sinA=槡2sinB, 烄4sin2C-4sinCsinA+sin2A=2sin2B, 11.BCD }烅 S烅 (cid:140)(cid:141){(cid:142)S4+4cos(C+ 烆2cosC+cosA=槡6cosB, 烆4cos2C+4cosCcosA+cos2A=6cos2B, 1 A)+1=2+4cos2B,S4cos2B+4cosB-3=0,S(2cosB-1)(2cosB+3)=0,ScosB= ,5cosB= 2 烄 槡6 烄 槡6 2sinC-sinA= , 2sinC= +sinA, 2 2 3 π π - ((cid:143)(cid:144)),*+0<B< ,/0B= ,2 A>?;4烅 S烅 S 2 2 3 槡6 槡6 2cosC+cosA= , 2cosC= -cosA, 烆 2 烆 2 4sin2C+4cos2C= (槡6 +sinA )2 + (槡6 -cosA )2 ,S4= 6 +槡6sinA+sin2A+ 6 -槡6cosA+cos2A,S 2 2 4 4 π π π π 5π sinA=cosA,OtanA=1,*+0<A< ,/0A= ,SC=π- - = ,4sinC-cosC= 2 4 3 4 12 ( π) (5π π) π 槡2 槡2 槡2sinC- =槡2sin - =槡2sin = =sinA= ,2B(cid:145)@A;}(cid:146)k△ABC6j(cid:147)+3+ 4 12 4 6 2 2 1 槡3,S acsinB=3+槡3,Sac=4( 槡3+1 ),o△ABC6(cid:148)(cid:128)(cid:149)(cid:150)(cid:151)+R,42RsinA·2RsinC= 2 槡2 5π 7π (π π) 槡6+槡2 槡2 槡6+槡2 4( 槡3+1 ),FsinA= ,sinC=sin =sin =sin + = ,44R2· × = 2 12 12 3 4 4 2 4 槡2 槡3 4( 槡3+1 ),SR=2,Sa=2RsinA=2×2× =2槡2,b=2RsinB=2×2× =2槡3,c=2RsinC=2×2× 2 2 槡6+槡2 =槡6+槡2,Sb2-a2=( 2槡3 )2-( 2槡2 )2=4,2C(cid:145)@A;△ABC6(cid:152)‘+a+b+c=2槡2+2槡3+ 4 槡6+槡2=3槡2+2槡3+槡6,2D@A.23BCD. 【“!"”#$%&·!"#$%&’( ’ 2(()6() W】a·b 12.槡5 *+a=(3,-1),b=(2,1),aJbMY(cid:153)6(cid:154)(cid:155)Y(cid:156)6(cid:157)+ acos〈a,b〉= =槡5. b 槡3 13. mW/n,(cid:158)(cid:159)(cid:160)](cid:135)j;(cid:137)‘+16¡(cid:137)$k(cid:136),¢(cid:132)‘+46@$(cid:132)£ABC-DEF, 2 ⁄(cid:134),BE=4-1=3=AA,AD=4-3=1=BB,CF=4-2=2=CC,*tV =V ,O 1 1 1 1 1 1 ABC-A1B1C1 A1B1C1-DEF 1 1 V = V ,¥ƒ$(cid:132)£§(cid:147)¤(cid:141),V = ×1×1×sin60°×4=槡3,2'“«§6§(cid:147) ABC-A1B1C1 2 ABC-DEF ABC-DEF 2 1 槡3 ;V = ×槡3= . ABC-A1B1C1 2 2 1 2π [ 2π ] (2π 2π) 14.- 2 a n =a 1 +(n-1)d=a 1 + 3 (n-1),4cosa n =cosa 1 + 3 (n-1)=cos 3 n+a 1 - 3 ,⁄(cid:152)‹+ 2π =3,Fn∈N,Ocosa q›3]:|ps,}Qwxfi(cid:139)A={x|x=cosa,n∈N}fl1(cid:176)fl(cid:140)]– 2π n n 3 †,A={x,x},4Jcosa,cosa ,cosa (cid:134),cosa=cosa ≠cosa 5cosa≠cosa = 1 2 n n+1 n+2 n n+1 n+2 n n+1 cosa ,5cosa=cosa ≠cosa ,Gcosa=cosa ,Ocosa =cosa ≠cosa ,(cid:160)7‡fl{· n+2 n n+2 n+1 n n+3 n+3 n+2 n+1 ( 2π) ( 2π) ( 2π) 6(cid:140)(cid:145){¡,o(cid:181)(cid:140)(cid:145)¶•+cosθ,cosθ+ ,(cid:130);flcosθ=cosθ+ ,Oflθ+θ+ =2kπ,k∈Z, 3 3 3 π ( 4π) ( π) [( π) 4π] ‚Sθ=kπ- ,k∈Z,:{¡6(cid:140)(cid:145)+cosθ,cosθ+ ,2xx=coskπ- cos kπ- + = 3 3 1 2 3 3 3 ( π) π 1 -coskπ- coskπ=-cos2kπcos =- . 3 3 2 ( π) 15.‚:(1)f(x)=sin2x+槡3cos2x=2sin2x+ ,„„„„„„„„„„„„„„„„„„„„„„ 3¶ 3 π π u2x+ = +kπ,k∈Z,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 4¶ 3 2 π kπ ‚Sx= + ,k∈Z, 12 2 π kπ 2=.f(x)6~”g+fIx= + ,k∈Z. „„„„„„„„„„„„„„„„„„„„„„ 6¶ 12 2 (x) ( π) 6 ( π) 3 (2)*+f 0 =2sinx+ = ,Osinx+ = ,„„„„„„„„„„„„„„„„„„ 8¶ 2 0 3 5 0 3 5 π ( 2π π) 1x∈(-π,0),4x+ ∈ - , , 0 0 3 3 3 π ( π) ( π) ( π) 4 wSx+ ∈ 0, ,4cosx+ =槡1-sin2 x+ = ,„„„„„„„„„„„„„„ 10¶ 0 3 3 0 3 0 3 5 ( π) [( π) π] ( π) 4fx+ =2sin2x+ + =2sin2x+ 0 6 0 6 3 0 3 ( π) ( π) 3 4 48 =4sinx+ cosx+ =4× × = , 0 3 0 3 5 5 25 ( π) 48 /0fx+ = .„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 13¶ 0 6 25 【“!"”#$%&·!"#$%&’( ’ 3(()6() W】16.‚:(1)¥ƒQR2b=bcosA+槡3asinB, 4}@»7…S,2sinB=cosAsinB+槡3sinAsinB,„„„„„„„„„„„„„„„„„„„„ 2¶ *+sinB≠0,/02=cosA+槡3sinA, „„„„„„„„„„„„„„„„„„„„„„„„„ 3¶ ( π) /0sinA+ =1,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 5¶ 6 π }A∈(0,π),/0A= . „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 7¶ 3 → → → AB → AC → → (2)uAE= → ,AF= → ,4 AE = AF =1. AB AC → → → GAD=AE+AF,4(cid:131)(cid:137)(cid:136)AEDF+‰(cid:136),AD+∠BAC6k[¶I.„„„„„„„„„„„„„ 8¶ → → → → → → → π AD 2=(AE+AF)2=AE2+AF2+2AE·AF=1+1+2×1×1×cos =3, 3 AD=槡3, „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 10¶ 1 π 1 π S = bcsin = (b+c)·AD·sin ,Obc=b+c,„„„„„„„„„„„„„„„„„„ 12¶ △ABC 2 3 2 6 π }(cid:190)»7…wS:a2=b2+c2-2bccos =4, 3 O(b+c)2-3bc=(bc)2-3bc=4,‚Sbc=4,„„„„„„„„„„„„„„„„„„„„„„„ 14¶ 1 π /0S = bcsin =槡3.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 15¶ △ABC 2 3 17.‚:(1)*+∠BAF=90°,/0AB⊥AF, *+[jABCD⊥[jABEF,[jABCD∩[jABEF=AB,AF[jABEF, /0AF⊥[jABCD.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 2¶ *+BC[jABCD,/0BC⊥AF. „„„„„„„„„„„„„„„„„„„„„„„„„„ 3¶ (2)}(1)xAF⊥[jABCD,AD[jABCD,/0AF⊥AD, *+∠BAD=∠BAF=90°,/0AB⊥AF,AB⊥AD, 0eA+bcde,mWhi¿(cid:192)fkbcl, *+AB∥CD,AB∥EF,CD=EF=1,AB=AD=AF=2, /0A(0,0,0),D(2,0,0),F(0,0,2),C(2,1,0),E(0,1,2),B(0,2,0), → → CE=(-2,0,2),CB=(-2,1,0),„„„„„„„„„„„„„„„„„„„ 5¶ o[jBCE6(cid:160)]`Y(cid:156)+m=(x,y,z), → 烄CB·m=0, {-2x+y=0, 4烅 O ux=1,Sm=(1,2,1),„„„„„„„„„„„„„„„„„„ 6¶ 烆C → E·m=0, -2x+2z=0, |…´x[jACF6(cid:160)]`Y(cid:156)+n=(-1,2,0),„„„„„„„„„„„„„„„„„„„„„ 7¶ m·n 3 槡30 /0cos〈m,n〉= = = ,„„„„„„„„„„„„„„„„„„„„„„„„ 8¶ m n 槡6×槡5 10 (槡30 )2 槡70 /0[jACFz[jBCEˆk66@»s+槡1- = .„„„„„„„„„„„„„ 9¶ 10 10 → → → (3)oQ(a,b,c),}QwxDQ=λDF+μDB, O(a-2,b,c)=λ(-2,0,2)+μ (-2,2,0)a=2-2λ-2μ ,b=2μ ,c=2λ,„„„„„„„„„„„ 11¶ → OQ(2-2λ-2μ ,2μ ,2λ),/0AQ=(2-2λ-2μ ,2μ ,2λ).„„„„„„„„„„„„„„„„„„ 12¶ → → *+AQ⊥[jBCE,/0AQ;[jBCE6(cid:160)]`Y(cid:156),/0AQ∥m, O 2-2λ-2μ = 2μ = 2λ ,‚Sλ= 1 , μ= 1 ,„„„„„„„„„„„„„„„„„„„„„„„„ 14¶ 1 2 1 4 2 → (1 1) → 1 1 槡6 2AQ= ,1, ,AQ =槡+1+ = .„„„„„„„„„„„„„„„„„„„„„„ 15¶ 2 2 4 4 2 【“!"”#$%&·!"#$%&’( ’ 4(()6() W】18.‚:(1)*+.˜{a n }+¡¯.˜,:˘oa n =xn+y(x,y∈R), }a 2n =2a n +1wS2xn+y=2(xn+y)+1=2xn+2y+1,2y=2y+1,‚Sy=-1, /0a=xn-1,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 1¶ n S=4S,Oa+a=4a,Oa=3a, 2 1 2 1 1 2 1 /02x-1=3(x-1),‚Sx=2,„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 2¶ 2a=2n-1,S= n(a 1 +a n ) = n(1+2n-1) =n2.„„„„„„„„„„„„„„„„„„„„„ 4¶ n n 2 2 (2)M`(cid:160):}(1)S:a=2n-1, n ∴Cn≥21n∈ND, b n = (2n+1)(2n-3) , b (2n-1)2 n-1 b b b b b ∴b= n · n-1· n-2·…· 3· 2·b „„„„„„„„„„„„„„„„„„„„„„„„„ 6¶ n b b b b b 1 n-1 n-2 n-3 2 1 (2n+1)(2n-3) (2n-1)(2n-5) (2n-3)(2n-7) 7×3 5×1 2n+1 1 2n+1 = × × ×…× × ×1= × = , (2n-1)2 (2n-3)2 (2n-5)2 52 32 2n-1 3 6n-3 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 9¶ 2n+1 Cn=1D,b=1˙¨b= , 1 n 6n-3 2n+1 (cid:201)(cid:153)/˚:b= (n∈N).„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 10¶ n 6n-3 M`%:}(1)S:a=2n-1, n ∵b=1,a>0,ba2=b a a (n≥2,n∈N), 1 n n n n-1 n-1 n+1 a a ∴b· n =b · n-1, „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 6¶ n a n-1 a n+1 n a uc=b· n ,4.˜{c}+¸.˜, „„„„„„„„„„„„„„„„„„„„„„„„„„ 7¶ n n a n n+1 a a a 1 1 c=b· n =b · n-1=…=b· 1=1× = , „„„„„„„„„„„„„„„„„„„„ 9¶ n n a n-1 a 1 a 3 3 n+1 n 2 1 a 2n+1 ∴b= × n+1= .„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 10¶ n 3 a 6n-3 n 1 1 1 ( 1) (3)}(1)x, = ,(cid:204)j˝˛ >ln1+ ,„„„„„„„„„„„„„„„„„„„„„„ 11¶ 槡S n n n n of(x)=x-ln(x+1),x>0, x 4f′(x)= ,Cx>0D,f′(x)>0,f(x)^ˇ—(cid:209), x+1 /0f(x)>f(0)=0,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 13¶ (1) 1 ( 1) 1 ( 1) /0f = -ln1+ >0,O >ln1+ ,„„„„„„„„„„„„„„„„„„„„ 15¶ n n n n n 1 1 1 ( 1) ( 1) ( 1) /0T=1+ + +…+ >ln(1+1)+ln1+ +ln1+ +…+ln1+ n 2 3 n 2 3 n ( 3 4 n+1) =ln2× × ×…× =ln(n+1), 2 3 n /0T>ln(n+1).„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 17¶ n 19.‚:(1)Cm=1D,f(x)=excosx-x+1,f′(x)=ex(cosx-sinx)-1, uh(x)=f′(x),4h′(x)=[(cosx-sinx)+(-sinx-cosx)]ex=(-2sinx)ex, „„„„„„ 2¶ Cx∈[0,π]D,h′(x)≤0, /0h(x)=f′(x)J[0,π](cid:153)^ˇ—(cid:210).„„„„„„„„„„„„„„„„„„„„„„„„„„ 4¶ (2)˝˛:g(x)=f(x)+mx=emxcosx+1,g′(x)=emx(mcosx-sinx)=-槡m2+1sin(x+θ)emx, ( π ) ⁄(cid:134)θ˙¨tanθ=-m,m>0,θ∈ - ,0 ,„„„„„„„„„„„„„„„„„„„„„„„„ 6¶ 2 ug′(x)=0,Sx=-θ,Cx∈(0,-θ)D,g′(x)>0,g(x)^ˇ—(cid:209); 0 0 【“!"”#$%&·!"#$%&’( ’ 5(()6() W】( π) Cx∈ -θ, D,g′(x)<0,g(x)^ˇ—(cid:210). „„„„„„„„„„„„„„„„„„„„„„„ 8¶ 2 [ π] /0g(x)J 0, 2 (cid:153)(cid:211)J(cid:212)rsex 0 , 1tanx 0 =tan(-θ)=m. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 10¶ [ π] (3)}(2)xg(x)J 0, (cid:153)6qrs+g(-θ)=e-mθcos(-θ)+1=e-mθcosθ+1.„„„„„„„ 11¶ 2 U˝x∈ [ 0, π] ,(cid:213)Sg(x)>e1-m 1 +1~(cid:214)Rm>0(cid:215)(cid:216)i, 2 O˝e-mθcosθ+1>e1-m 1 +1~(cid:214)Rm>0(cid:215)(cid:216)i, O˝cosθ>emθ+1-m 1 ~(cid:214)Rm>0(cid:216)i,Gm=-tanθ, /0O˝cosθ>e-θtanθ+1+ta 1 nθ~(cid:214)Rθ∈ ( - π ,0 ) (cid:215)(cid:216)i, 2 [ 1 ] ( π ) O˝ ln(cosθ)- +θtanθ-1 >0,⁄(cid:134)θ∈ - ,0 .„„„„„„„„„„„„„„„„ 12¶ tanθ 2 min 1 ( π ) u φ (x)=ln(cosx)- +xtanx-1,x∈ - ,0 , tanx 2 ( 1 ) (cosx) 1 (sinx) 1 -sinx *+ ′= ′=- ,(tanx′)= ′= ,[ln(cosx)′]= =-tanx, tanx sinx sin2x cosx cos2x cosx 1 x cos2x+xsin2x /0 φ′(x)=-tanx+ sin2x +tanx+ cos2x = sin2xcos2x . ( π ) un(x)=cos2x+xsin2x,x∈ - ,0 , 2 4n′(x)=-2cosxsinx+sin2x+2xsinxcosx=sinx[sinx+2(x-1)cosx]>0, ( π ) ( π) 1 π 4n(x)J - ,0 (cid:153)^ˇ—(cid:209),Gn(-1)=cos2<0,n - = - >0, 2 4 2 8 ( π) 4x∈ -1,- ,(cid:213)n(x)=cos2x+xsin2x=0, 1 4 1 1 1 1 1 x ‚Stanx=-槡- ,/0cosx=槡 1 . 1 x 1 x-1 1 1 ( π ) Cx∈ - 2 ,x 1 D,n(x)<0,O φ′(x)<0, φ (x)^ˇ—(cid:210); Cx∈(x 1 ,0)D,n(x)>0,O φ′(x)>0, φ (x)^ˇ—(cid:209). /0 φ (x)Jx=x 1 Dp(cid:217)(cid:212)(cid:218)s,(cid:219);q(cid:218)s,„„„„„„„„„„„„„„„„„„„„„„ 14¶ 1 2 [ φ (x)] min =φ (x 1 )=ln(cosx 1 )- tanx +x 1 tanx 1 -1=- tanx -1+ln(cosx 1 ) 1 1 1 ( 1) =2槡-x -1- ln1- . „„„„„„„„„„„„„„„„„„„„„„„„„„„„„ 15¶ 1 2 x 1 1 ( 1) ( π) ut(x)=2槡-x-1- ln1- ,x∈ -1,- , 2 x 4 1 1 1 1 4t′(x)=- - × · <0, 槡-x 2 1 x2 1- x ( π) ( π) Ot(x)J -1,- 4 (cid:153)^ˇ—(cid:210), φ (x) min =φ (x 1 )=t(x 1 )>t- 4 ,„„„„„„„„„„„„ 16¶ ( π) π 1 ( 4) 1 7 7 1 7 Gt- =2槡-1- ln1+ >槡3-1- ln > - ln >0, 4 4 2 π 2 3 10 2 3 ( π ) [ 1 ] OCθ∈ - ,0 D,ln(cosθ)- +θtanθ-1 >0, 2 tanθ min /0x∈ [ 0, π] ,(cid:213)Sg(x)>e1-m 1 +1~(cid:214)Rm>0(cid:215)(cid:216)i,(cid:220)QS˝. „„„„„„„„„„„ 17¶ 2 【“!"”#$%&·!"#$%&’( ’ 6(()6() W】