2025優(yōu)化設(shè)計一輪課時規(guī)范練35　三角函數(shù)的圖象與性質(zhì)

課時規(guī)范練35三角函數(shù)的圖象與性質(zhì)高考總復(fù)習(xí)優(yōu)化設(shè)計GAOKAOZONGFUXIYOUHUASHEJI202512345678910111213141516171819基礎(chǔ)鞏固練D12345678910111213141516171819A123456789101112131415161718193.(2024·廣西玉林模擬)若函數(shù)f(x)=asinx+1的最大值為4,則函數(shù)g(x)=cos(ax+1)的最小正周期為(

)D解析

由-1≤sin

x≤1,函數(shù)f(x)=asin

x+1的最大值為4,則a=±3,故函數(shù)g(x)=cos(ax+1)的最小正周期為123456789101112131415161718194.(2024·貴州貴陽模擬)已知a=sin1,b=sin,c=sin2,則(

)A.a<b<c B.c<b<aC.c<a<b D.a<c<bD123456789101112131415161718195.(多選題)(2024·福建三明模擬)已知函數(shù)f(x)=1-2sin2(x+),則下列結(jié)論錯誤的有(

)A.函數(shù)f(x)是偶函數(shù)B.函數(shù)f(x)的最小正周期為2πC.函數(shù)f(x)的圖象關(guān)于直線x=-對稱D.f(1)>f(2)ABD12345678910111213141516171819B12345678910111213141516171819D12345678910111213141516171819BD1234567891011121314151617181912345678910111213141516171819A123456789101112131415161718191234567891011121314151617181911.(2024·浙江金麗衢十二校聯(lián)考)寫出一個滿足下列條件的正弦型函數(shù),f(x)=

.

①最小正周期為π;②f(x)在區(qū)間[0,]上單調(diào)遞增;③?x∈R,|f(x)|≤2成立.123456789101112131415161718191234567891011121314151617181912345678910111213141516171819綜合提升練B解析

函數(shù)定義域為{x|x≠kπ,k∈Z},且f(x)=2sin

x,所以f(x)為奇函數(shù),其圖象關(guān)于點(kπ,0),k∈Z對稱,關(guān)于直線x=+kπ,k∈Z對稱,故選B.12345678910111213141516171819A1234567891011121314151617181915.(2024·山東沂水模擬)函數(shù)f(x)=tan(ωx+φ)(ω>0,|φ|<)的圖象如圖所示,圖中陰影部分的面積為6π,則φ=(

)A12345678910111213141516171819解析

如圖所示,區(qū)域①和區(qū)域③面積相等,故陰影部分的面積即為矩形ABCD的面積,可得AB=3.設(shè)函數(shù)f(x)的最小正周期為T,則AD=T,由題意可得3T=6π,解得T=2π,12345678910111213141516171819BC12345678910111213141516171819解析

對于A,f(x+π)=cos[2(x+π)]+|sin(x+π)|=cos(2x+2π)+|-sin

x|=cos

2x+|sin

x|=f(x),所以函數(shù)f(x)為周期函數(shù),且該函數(shù)的最小正周期不是2π,故A錯誤;對于B,對任意的x∈R,f(-x)=cos(-2x)+|sin(-x)|=cos

2x+|sin

x|=f(x),所以函數(shù)f(x)為偶函數(shù),故B正確;1234567891011121314151617181912345678910111213141516171819(答案不唯一)12345678910111213141516171819123456789101112131415161718191234567891011121314151617181919.(2024·北京房山模擬)已知函數(shù)f(x)=sin(ωx+φ)(ω>0,0<φ<π)的最小正周期為π.(1)求ω的值;(2)再從條件①、條件②、條件③三個條件中選擇一個作為已知,確定f(x)的解析式.設(shè)函數(shù)g(x)=f