2023年赣州高三一诊考试数学试卷含解析

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2023
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2+V=4,ᵫQsin(e+?ᐔ)=/᜛sin8+pcos^=1,*ÌÒ1ḄÌÍᙶ᪗ÇÈ\x+y=1*42É2
ßÌÒ/ḄÓᦪÇÈ\a\ÓᦪÊ,[6y-Id-----12éᐭ(x—2y+V=4,//+3"+1=0,ᑣA=18—4=14>0,

15ßAஹ8«âãÓᦪᑖä\᜜ஹt,r,+z=-3>/2<0»=l>0,22"<0,f<0,/.IA^4|+|MB|=I?!I+|z1=+Z|=3V2.222ூ«¬௃⚪ὃ¯³ÓᦪÇÈஹ᩽ᙶ᪗ÇÈÁÅÆÇÈìsḄíᓄ±²ὃ¯³ÌÒÓᦪÇÈîïwZḄãᵨὃ¯ðÿÀĹÔ⚪.20.(1)a=2,b=l(2)!"ñU᪆ூU᪆௃(1)°K//'(1)=a,f(1)=—be,òᔠóxᑗÒÇÈᓽK°/“஺ḄQ*õ22(2)ᑭᵨᦪK//(ᵨ)=2஺--=2/“,-------xe(l,2),ö᪀⌼»ᦪa)=2--=-,l0,g'(1)=2-e<0,ᦑᙠ᳝e(0,1),gG)=0,xe(0,xJg'(x)>(),g(x)ᓫ⌴+8)g'(x)<0,g(x)ᓫ⌴ᵫg(0)=l>0,g(1)=2>0,8(2)=4-e2<0,ᦑᙠ(1,2),g(x°)=0,xe(0,x0)g(x)>0,/'(x)>0,!"ᓫ⌴xe(%+8)g(x)<0,fXx)<0,!"ᓫ⌴,

16ᦑ¹ᦪÂᙠÄ
Ḅ᩽ᜧ%,£8(%)=2%-%0*+*=0,ᓽe%=Tx°£(l,2),/T*2È!J/Uo)=2É%----=2?0-------r,%%-1222@h(x)=2bvc-------,l0,x—1x(%—1)ᦑv஺)ᙠ(1,2)ÌᓫÎ⌴Ð!2ᵫ/£(1,2),ᦑ/?(/)

17⚪ὃᔠḄùú.ᦪḄẆÔ!`ᔠᦪÛÜÝ«Ḅqᵨ!X/û⚪.22.(1)0=2cose+4sin஺!(x-1)2+(y-2)2=5'(2)x+y-l=O.ூ"᪆௃(Dëp,Q᩽ᙶ᪗ᑖ3,e),38)ᵫþ=2ÿ
᜛='2°=2cos஺+4sin஺,᦮ᳮᓽ
ᑮ᩽ᙶ᪗
ᙶ᪗ফA,3Ḅᦪᑖ!"#$ᑣ|4|=()ṹ+21,-/Ḅᦪ/ᐭGḄᙶ᪗12ᑭᵨ567ᳮ
A+8=2(cosa+sina),tt=-3,ᑣ1A/ᵨ=(+(=:fI="(4+G:;4<
x2+|MB]=ᨬ#?@aAᔠḄᩩD,
-IḄEF.ூIJ௃(1)P,Q᩽ᙶ᪗ᑖ!"(A஺)(஺,஺)O"9=2ᑣ0=g2o=2cos6+4sin஺ᡠQR-GḄ᩽ᙶ᪗"P=2cos6+4sin0,ST(U᜛,
p~=2pcos஺+4spinQ,ᡠQC2Ḅᙶ᪗"/+V=2x+4y,ᓽ(x—1:+(y—2)2=5.JQ—fCOS(XW45Ḅᦪᑖ!"3,X!||Z4|=|,|[ὡ=,2|,,-/Ḅᦪ];Qᦪ)/[y=1+rsinaᐭ஺2Ḅᙶ᪗(*-1)2+(V—2)2=51᦮ᳮ
a;2(cosa+sina"-3=0.ᵫ567ᳮ
c+t=2(cose+sine),tt=-3,2t2ᡠQ+=1]|+/21=d1;+=+.)2-4e="(cosa+sinaf+12=J4sin2a+16>2G,fghfsin2a=—1@ijᡂlXmltana=-l,ᡠQf|M4|+|n8|=
ᨬ#?@-/ḄEF"x+y-1=0.ூo௃p⚪ὃs᩽ᙶ᪗tᙶ᪗Ḅuᓄὃsᑭᵨ-ḄᦪẆx-tᙊḄz{ᐵ}.