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(-∞,0)∪(4,+∞)
分析:A 曲線即 (x-2)
2+y
2=1,表示以C(2,0)為圓心,以1為半徑的圓,
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表示圓上的點(diǎn)與原點(diǎn)連線的斜率,由r=1=
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,可得 k 的值,由此求得
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的取值范圍.
B 由于x-2在[0,2]上小于或等于0,故應(yīng)有|a-2x|在[0,2]上恒正,2x≠a,故
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<0,或
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>2,由此求得a的取值范圍.
解答:A 曲線
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(θ為參數(shù))即 (x-2)
2+y
2=1,表示以C(2,0)為圓心,以1為半徑的圓.
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表示圓上的點(diǎn)與原點(diǎn)連線的斜率,如圖所示,設(shè)切線的斜率為k,則切線的方程為y=kx,
由r=1=
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,可得 k=±
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,故
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的取值范圍為
故答案為:
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.
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B 關(guān)于x的不等式|a-2x|>x-2在[0,2]上恒成立,由于x-2在[0,2]小于或等于0,
故應(yīng)有|a-2x|恒正,∴2x≠a,即 x≠
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,∴
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<0,或
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>2,
∴a<0,或a>4,則a的取值范圍為 (-∞,0)∪(4,+∞),
故答案為:(-∞,0)∪(4,+∞).
點(diǎn)評(píng):本題考查斜率公式,圓的切線性質(zhì),參數(shù)方程與普通方程之間的轉(zhuǎn)化,圓的參數(shù)方程,絕對(duì)值不等式的解法,得到
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<0,或
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>2,是解題的難點(diǎn)和關(guān)鍵.