解:(1)∵點D的橫坐標(biāo)為1,點D在y=x+1的圖象上,∴D(1,2),
∴直線BD的解析式為y=3x-1,∴A(0,1),C(
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,0),
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∴S
四邊形AOCD=S
△AOD+S
△COD=
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×1×1+
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×
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×2=
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;
(2)①當(dāng)DP=DB時,設(shè)P(0,y),
∵B(0,-1),D(1,2),
∴DP
2=1
2+(y-2)
2=DB
2=1
2+(2+1)
2,
∴P(0,5);
②當(dāng)BP=DB時,DB=
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,∴P(0,-1-
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)或P(0,
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-1);
③當(dāng)PB=PD時,設(shè)P(0,a),則(a+1)
2=1+(2-a)
2,解得a=
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,
∴P(0,
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);
(3)若一次函數(shù)y=kx+b的圖象與函數(shù)y=x+1的圖象的交點D始終在第一象限,則系數(shù)k的取值范圍是:k>1.
分析:(1)先求出點D的坐標(biāo),再求出BD的解析式,然后根據(jù)S
四邊形AOCD=S
△AOD+S
△COD即可求解;
(2)分三種情況討論:①當(dāng)DP=DB時,②當(dāng)BP=DB時,③當(dāng)PB=PD時;
(3)根據(jù)圖象即可得出答案.
點評:本題考查了一次函數(shù)綜合知識,難度適中,關(guān)鍵是掌握分類討論思想的運用.