2023~2024学年第一学期四校联考(二)参考答案题号123456789101112答案ABDDBCCBBCBCDBCDABD2413.614.15.k116.(2,).2ln21部分试题答案详解7.【答案】Cx23x1【解答】解:由题意知x23x1kex0有两个不同的解,即y与yk有exx23x1x2x2(x2)(x1)两个不同的交点,记g(x),则g(x),exexex当x2时,g(x)0,g(x)单调递增;当2x1时,g(x)0,g(x)单调递减;当x1时,g(x)0,g(x)单调递增.所以当x2时,函数g(x)有极大值e2,当x1时,函5数g(x)有极小值.e又因为x时,g(x)0;x时,g(x)0,且g(x)0,如下图:5数形结合可知k[0,e2)U{}时,函数f(x)恰有两个零e点.8.【答案】Bax1x0【解答】解:由题意知函数fx恰有两个“姊妹点对”,lnxx0等价于函数f(x)lnx,x0与函数g(x)ax1,x 0的图象恰好有两个交点,所以方程lnxax1,即lnxax10在(0,)上有两个不同的解.1构造函数h(x)lnxax1,则h(x)a,x当a0时,h(x)0,函数h(x)区间(0,)上单调递增,不符合题意;11当a0时,令h(x)0,解得0x,所以函数h(x)在区间(0,)上单调递增,aa第1页共9页{#{QQABIQaAoggoQAAAAAhCQwUQCAKQkBACCAoORBAIIAIAARFABAA=}#}11令h(x)0,解得x,所以函数h(x)在区间(,)上单调递减,aa1所以h()0,解得0ae2,a又h(e)lneae1ae0,1所以函数h(x)在(e,)上有且仅有一个零点,a112x令M(x)lnxx1,则M(x),x2x2x令M(x)0,解得0x4,所以函数M(x)在(0,4)上单调递增,令M(x)0,解得x4,所以函数M(x)在区间(4,)上单调递减.所以M(x)maxM(4)ln430,所以M(x)lnxx1M(4)0,即lnxx1.222222又h()lna11a1(12)0,a2a2a2a2a2a212所以函数h(x)在(,)上有且仅有一个零点.aa2综上可得0ae2.12.【答案】ABDx【解答】解:函数f(x)ex2的零点为x1,函数g(x)lnxx2的零点为x2,x1可得e2x1,lnx22x2,由yex与其反函数ylnx关于直线yx对称,xye与直线y2x的交点为(x1,2x1),ylnx与直线y2x的交点为(x2,2x2),可得x12x2,即x1x22,故A正确;直线与直线yx垂直,则点x1和也关于直线yx对称,则有y2x(x1,e)(x2,lnx2)第2页共9页{#{QQABIQaAoggoQAAAAAhCQwUQCAKQkBACCAoORBAIIAIAARFABAA=}#}x1x1x1lnx2,则有elnx2ex12,故B正确;3313又g(1)ln11210,glnlnlne0,22221113g(e)lnee2e222.2520,223所以xe,则xxx(2x)xlnx,221222223因为yxlnx,x,e,2y1lnx0,3所以yxlnx在,e上单调递增,2e所以xxxlnxelne,故C错误;1222233由上可知xxxlnxln,122222331127127因为lnln1ln0,2222828e33112所以ln,即xx,则x2x2xx2xx42xx3,2221221212121222所以x1x23,故D正确.15.【答案】解:(1)令xy0,得f(00)f(0)f(0),所以f(0)0.证明:令yx,得f(xx)f(x)f(x)f(0)0,所以f(x)f(x),所以f(x)为奇函数.由题知:f(k2x)f(4x18x2x)0f(0),即f(k2x4x18x2x)f(0),又yf(x)是定义在R上的增函数,所以k2x4x18x2x0对任意x[1,2]恒成立,所以k2x2x8x4x1,即k122x2x2,1令2xt,t[,4],22则g(t)t4t1,所以kg(t)max,第3页共9页{#{QQABIQaAoggoQAAAAAhCQwUQCAKQkBACCAoORBAIIAIAARFABAA=}#}当t4时,g(t)maxg(4)161611,所以k1.16.【解答】解:f(x)2xa(lnx1),2若函数f(x)x2axlnx在(,2)上不单调,e2则方程f(x)0在(,2)上有根e2x2即方程a在(,2)上有根且方程的根是函数f(x)的变号零点,lnx1e2x2lnx令g(x),则g(x),lnx1(lnx1)22x(,1)时,g(x)0,g(x)递减,x(1,2)时,g(x)0,g(x)递增,e244244又g(1)2,g(),g(2),由g(2)g()0,eeln2ln21eln21eln24得g(x)(2,),ln214故a(2,),ln214故答案为:(2,).ln212f(0)b3,17.【答案】解:(1)fxx2axb,结合题意可得................1分f(3)6ab90,a2解得,经检验符合题意,...............................................3分b31故fxx32x23x1.3所以在点0,f0处的切线方程为y3x1..............................................4分(2)由(1)知fxx24x3.令f(x)0,解得x3或x1,令f(x)0,解得1x3,故f(x)在,1,3,上单调递增,在1,3上单调递减,...................................6分7所以fxf1,fx极小值f31;...................................7分极大值3(3)f(x)在0,3上有极大值,无极小值,又因为f01,f31,.所以要使不等式能成立,则.............................8分fxm0fxminm第4页共9页{#{QQABIQaAoggoQAAAAAhCQwUQCAKQkBACCAoORBAIIAIAARFABAA=}#}所以m1............................................9分故m取值取值范围是是1,+............................................10分318.【答案】解:(1)角θ的终边上一点p1,y,且sin得2所以θ为第四象限角,则y<0,........................................1分y所以由sin,y3........................................3分1y2所以tanθ=-3.........................................4分(2)因为tanθ=-3,cos()-cos()sin+cos所以2=.......................................6分sin()+cos(+)sin-costanθ+1-3+1===2-3.........................................8分tanθ-1-3-110(3)因为,0,0,,且sin(+)得2210310+(0,),所以co(s+)1sin(2+),...........................10分210.....................11分所以coscos(+)-cos(+)cossin(+)sin3101103=+(-)102102=3103020...........................................................12分19.【答案】解:(1)当n1时,a1S11;..........................................................1分当时,22,分n 2anSnSn1n(n1)2n1...........................................................2经检验,当时,满足,因此分n1an2n1an2n1.......................................3当n1时,b1T13;.....................................4分n2n3Tn2nn当n2时,bn2(3)3,......................................5分 Tn1n1n13当时,满足n,因此n分n1bn3bn3.......................................6第5页共9页{#{QQABIQaAoggoQAAAAAhCQwUQCAKQkBACCAoORBAIIAIAARFABAA=}#}n(2)由(1)知anbn(2n1)3,23nMn133353(2n1)3,......................................7分234nn13Mn133353(2n3)3(2n1)3,......................................8分两式相减得234nn12Mn32(3333)(2n1)3......................................9分93n132(2n1)3n1......................................10分136(2n2)3n1,......................................11分n1故Mn3(n1)3..........................................................12分20.【答案】解:(1)f(x)(x22x)ex,求导得f(x)ex(x22)..........................................................1分因为ex0,令f(x)ex(x22)0,即x220,解得x2或x2,令f(x)ex(x22)0,即x220,解得2x2,............................
广东省四校联考2024届高三9月第一次联考数学参考答案(1)
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