課時(shí)規(guī)范練12對(duì)數(shù)函數(shù)高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2026基礎(chǔ)鞏固練1234567891011121314151617

B

12345678910111213141516172.(2025·浙江開學(xué)考試)已知關(guān)于x的函數(shù)y=ln(x-a)在[1,2]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是(

)A.(-∞,1) B.(-∞,2)C.(1,+∞) D.(2,+∞)A解析

由于y=ln(x-a)在[1,2]上單調(diào)遞增,所以x-a>0在[1,2]上恒成立,即a<1.故選A.1234567891011121314151617

D1234567891011121314151617

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A

12345678910111213141516175.(2024·河南新鄉(xiāng)期末)若函數(shù)f(x)=log3ax(a>0,且a≠1)在[-1,2]上的值域?yàn)閇m,2],則m的值為(

)A.-4或-1 B.0或-2C.-2或-1 D.-4或-2A1234567891011121314151617

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ACD1234567891011121314151617

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f(x)=lnx

12345678910111213141516178.(2024·廣東汕頭模擬)不等式log2(x-1)+log2(x-2)>log26的解集為

.(4,+∞)

1234567891011121314151617

1234567891011121314151617綜合提升練

D1234567891011121314151617

123456789101112131415161711.(2024·安徽黃山模擬)“a<1”是“函數(shù)f(x)=log2[(1-a)x-1]在區(qū)間(1,+∞)上單調(diào)遞增”的(

)A.充分不必要條件B.充要條件C.必要不充分條件D.既不充分也不必要條件C解析

令u=(1-a)x-1,則y=log2u,若f(x)=log2[(1-a)x-1]在區(qū)間(1,+∞)上單調(diào)遞增,因?yàn)閥=log2u在(1,+∞)上單調(diào)遞增,則需使u=(1-a)x-1在區(qū)間(1,+∞)上單調(diào)遞增,且u>0,則1-a>0,且1-a-1≥0,解得a≤0,因?yàn)?-∞,0]?(-∞,1),故“a<1”是“a≤0”的必要不充分條件,故選C.1234567891011121314151617

B1234567891011121314151617

123456789101112131415161713.(2024·湖南岳陽(yáng)模擬)若函數(shù)f(x)=loga(x2-ax+1)有最小值,則a的取值范圍是

.(1,2)

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123456789101112131415161715.(13分)(2024·北京朝陽(yáng)期末)設(shè)函數(shù)f(x)=log2(4x+m)(m>-1).(1)當(dāng)m=0時(shí),求f(1)的值;(2)判斷f(x)在區(qū)間[0,+∞)內(nèi)的單調(diào)性,并用函數(shù)單調(diào)性的定義證明你的結(jié)論;(3)當(dāng)x∈[0,+∞)時(shí),f(x)的最小值為3,求m的值.1234567891011121314151617

1234567891011121314151617創(chuàng)新應(yīng)用練

A1234567891011121314151617

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ABD1