所以,a7b27
(21)【解析】(1)
fxx33x2ax2∴f′x3x26xa,f′0a设切点A02,切线与x轴交点为B20则kABf′0即所以a120a02
2当k1时,令fxkx2x33x2xkx40则x23x1令gxx23x1
4kx≠0x
442x33x24则gx2x32xxx2令hx2x33x24,则hx6x26x6xx1∴当x∈01时,hx0hx递减当x∈∞,0,或1,∞时,hx0hx递增;且h00h20∴当x2时,hx0gx0gx在∞,002上递减;当x2时,hx0gx0gx在0,∞上递增;∴当x∈02∪0,∞时,gx≥g21当x∈∞,0时,单调递减,且gx∈∞∞∴当k1时,gxk仅有一个根点图像如图所示所以当k1时,yfx与ykx2仅有一个交点
(22)【解析】(1)
PC2PAPDDC∴PAPD,ΔPAD为等腰三角形。连接AB则∠PAB∠DEBβ∠BCE∠BAEα∠PAB∠BCE∠PAB∠BAD∠PAD∠PDA∠DEB∠DBE∴βαβ∠DBE即α∠DBE,即∠BCE∠DBE,所以BEEC
(23)解析:(Ⅰ)设点Mxy是曲线C上任意一点,
2cos,∴∵x2y22x,即:x12y21
C的参数方程为∴
x1cos,为参数.ysi
C在D处的切线与直线l:y3x2垂直(Ⅱ)设点D1cossi
,∵
∴ta
3,
3
,,∴点D的坐标为:1cos
3
si
33,即322
文科数学试题第8页(共9页)
f(24)解析:(Ⅰ)∵fxx∴fx
111xaxxaa,且a0aaa
1a2,当且仅当a1时,取“”a故:fx211(Ⅱ)∵f35,∴f333a3a35aa1即:3a35aa30a3∴或113a3533a5aa
解之:
15521a22
文科数学试题第9页(共9页)
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