解:(Ⅰ)設(shè)M、N兩點(diǎn)的橫坐標(biāo)分別為x
1、x
2,
∵f′(x)=1-
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,
∴切線PM的方程為:y-(x
1+
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)=(1-
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)(x-x
1),
又∵切線PM過(guò)點(diǎn)P(1,0),∴有0-(x
1+
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)=(1-
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)(1-x
1),
即x
12+2tx
1-t=0,(1)
同理,由切線PN也過(guò)點(diǎn)P(1,0),得x
22+2tx
2-t=0.(2)
由(1)、(2),可得x
1,x
2是方程x
2+2tx-t=0的兩根,
∴
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(*)
|MN|=
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=
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,
把(*)式代入,得|MN|=
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,
因此,函數(shù)g(t)的表達(dá)式為g(t)=
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(t>0).
(Ⅱ)當(dāng)點(diǎn)M、N與A共線時(shí),k
MA=k
NA,
∴
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=
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,即
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=
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,
化簡(jiǎn),得(x
2-x
1)[t(x
2+x
1)-x
1x
2]=0
∵x
1≠x
2,∴t(x
2+x
1)=x
2x
1.(3)
把(*)式代入(3),解得t=
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.
∴存在t,使得點(diǎn)M、N與A三點(diǎn)共線,且t=
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.
分析:(I)設(shè)出M、N兩點(diǎn)的橫坐標(biāo)分別為x
1、x
2,對(duì)函數(shù)求導(dǎo)得到切線的斜率,寫出切線的方程,根據(jù)切線過(guò)一個(gè)點(diǎn),得到一個(gè)方程,根據(jù)根與系數(shù)的關(guān)系寫出兩點(diǎn)之間的長(zhǎng)度,得到函數(shù)的表示式.
(II)根據(jù)三點(diǎn)共線寫出其中兩點(diǎn)連線的斜率相等,整理出最簡(jiǎn)單形式,把上一問(wèn)做出的結(jié)果代入,求出t的值.
點(diǎn)評(píng):本題考查函數(shù)的綜合題目,主要應(yīng)用導(dǎo)函數(shù)求最值來(lái)解題,本題解題的關(guān)鍵是正確應(yīng)用導(dǎo)數(shù),本題是一個(gè)綜合題目,綜合性比較強(qiáng).