（江苏专用）新高考数学一轮复习 第二章 函数 强化训练 函数的性质-人教高三数学试题

强化训练函数的性质1下列函数中，既是偶函数又在区间(1,2)内单调递减的是()Af(x)Bf(x)Cf(x)2x2xDf(x)cosx答案B解析函数f(x)是偶函数，且在(1,2)内单调递减，符合题意2函数f(x)x(x0)是()A奇函数，且在(0,3)上是增函数B奇函数，且在(0,3)上是减函数C偶函数，且在(0,3)上是增函数D偶函数，且在(0,3)上是减函数答案B解析因为f(x)xf(x)，所以函数f(x)x为奇函数又f(x)1，在(0,3)上f(x)<0恒成立，所以f(x)在(0,3)上是减函数3若函数f(x)ax2bx8(a0)是偶函数，则g(x)2ax3bx29x是()A奇函数B偶函数C非奇非偶函数D既奇又偶函数答案A解析由f(x)是偶函数可得b0，g(x)2ax39x，g(x)是奇函数4(2019·湖北武汉重点中学联考)已知偶函数f(x)在0，)上单调递减，f(1)1，若f(2x1)1，则x的取值范围为()A(，1 B1，)C0,1D(，01，)答案C解析由题意，得f(x)在(，0上单调递增，且f(1)1，所以f(2x1)f(1)，则|2x1|1，解得0x1.故选C.5若定义在R上的奇函数f(x)满足对任意的xR，都有f(x2)f(x)成立，且f(1)8，则f(2019)，f(2020)，f(2021)的大小关系是()Af(2019)<f(2020)<f(2021)Bf(2019)>f(2020)>f(2021)Cf(2020)>f(2019)>f(2021)Df(2020)<f(2021)<f(2019)答案A解析因为定义在R上的奇函数f(x)满足对任意的xR，都有f(x2)f(x)成立，所以f(x4)f(x)，即函数f(x)的周期为4，且f(0)0，f(2)f(0)0，f(3)f(1)8，所以f(2019)f(4×5043)f(3)8，f(2020)f(4×505)f(0)0，f(2021)f(4×5051)f(1)8，即f(2019)<f(2020)<f(2021)6(2019·北京大兴区模拟)给出下列函数：f(x)sinx；f(x)tanx；f(x)f(x)则它们共同具有的性质是()A周期性B偶函数C奇函数D无最大值答案C解析f(x)sinx为奇函数，周期为2且有最大值；f(x)tanx为奇函数且周期为，但无最大值；作出f(x)的图象(图略)，由图象可知此函数为奇函数但无周期性和最大值；作出f(x)的图象(图略)，由图象可知此函数为奇函数但无周期性和最大值所以这些函数共同具有的性质是奇函数7(多选)定义在R上的奇函数f (x)为减函数，偶函数g(x)在区间0，)上的图象与f (x)的图象重合，设ab0，则下列不等式中成立的是( )Af(b)f(a)g(a)g(b)Bf(b)f(a)g(a)g(b)Cf(a)f(b)g(b)g(a)Df(a)f(b)g(b)g(a)答案AC解析函数f(x)为R上的奇函数，且为单调减函数，偶函数g(x)在区间0，)上的图象与f (x)的图象重合，由ab0，得f (a)f (b)0，f (a)g(a)，f (b)g(b)；对于A，f (b)f (a)g(a)g(b)f (b)f (a)g(a)g(b)2f (b)0(因为f (a)g(a)在a0上成立)，所以A正确；对于B，f (b)f (a)g(a)g(b)f (b)f (a)g(a)g(b)2f (b)0，这与f (b)0矛盾，所以B错误；对于C，f (a)f (b)g(b)g(a)f (a)f (b)g(b)g(a)2f (a)f (b)0，这与f (a)f (b)符合，所以C正确；对于D，f (a)f (b)g(b)g(a)f (a)f (b)g(b)g(a)2f (a)f (b)0，这与f (a)f (b)矛盾，所以D错误8(多选)(2020·济南模拟)函数f (x)的定义域为R，且f (x1)与f (x2)都为奇函数，则()Af (x)为奇函数B.f (x)为周期函数Cf (x3)为奇函数D.f (x4)为偶函数答案ABC解析由f (x1)与f (x2)都为奇函数知函数f (x)的图象关于点(1,0)，(2,0)对称，所以f (x)f (2x)0，f (x)f (4x)0，所以f (2x)f (4x)，即f (x)f (x2)，所以f (x)是以2为周期的函数所以函数f (x)的图象关于点(3,0)，(2,0)，(1,0), (0,0)对称9(2019·衡水中学调研)已知定义在R上的函数f (x)满足f (x)f ，且f (3)3，则f (2022)_.答案3解析f (x)f ，f (x3)f f f (x)f (x)是以3为周期的周期函数则f (2022)f (673×33)f (3)3.10已知f (x)是定义在R上的奇函数，f (x1)是偶函数，当x(2,4)时，f (x)|x3|，则f (1)f (2)f (3)f (4)f (2020)_.答案0解析因为f (x)为奇函数，f (x1)为偶函数，所以f (x1)f (x1)f (x1)，所以f (x2)f (x)，所以f (x4)f (x2)f (x)，所以函数f (x)的周期为4，所以f (4)f (0)0，f (3)f (1)f (1)在f (x1)f (x1)中，令x1，可得f (2)f (0)0，所以f (1)f (2)f (3)f (4)0.所以f (1)f (2)f (3)f (4)f (2020)505f (1)f (2)f (3)f (4)0.11已知函数f (x)是(，)上的偶函数，若对于x0，都有f (x2)f (x)，且当x0,2)时，f (x)log2(x1)，求：(1)f (0)，f (2)，f (3)的值；(2)f (2021)f (2022)的值解(1)f (0)log210，f (2)f (0)0，f (3)f (12)f (1)log2(11)1.(2)依题意得，当x0时，f (x4)f (x2)f (x)，即当x0时，f (x)是以4为周期的函数因此，f (2 021)f (2 022)f (2 021)f (2 022)f (1)f (2)而f (2)0，f (1)log2(11)1，故f (2 021)f (2 022)1.12已知g(x)为偶函数，h(x)为奇函数，且满足g(x)h(x)2x，若存在x1,1，使得不等式m·g(x)h(x)0有解，求实数m的最大值解因为g(x)h(x)2x，所以g(x)h(x)2x.又g(x)为偶函数，h(x)为奇函数，所以g(x)h(x)2x，联立，得g(x)，h(x).由m·g(x)h(x)0，得m1.因为y1为增函数，所以当x1,1时，max1，所以m，即实数m的最大值为.13(2020·福州模拟)已知函数f (x)(xR)满足f (x)2f (x)，若函数y与yf (x)图象的交点为(x1，y1)，(x2，y2)，(xm，ym)，则(xiyi)等于()A0BmC2mD4m答案B解析因为f (x)f (x)2，y1.所以函数yf (x)与y的图象都关于点(0,1)对称，所以i0，i×2m，故选B.14已知函数f (x)则f (2019)_.答案1010解析当x>0时，f (x)f (x2)1，则f (2 019)f (2 017)1f (2 015)2f (1)1 009f (1)1 010, 而f (1)0，故f (2 019)1 010.15已知定义在R上的奇函数f (x)满足f (x4)f (x)，且在区间0,2上是增函数若方程f (x)m(m>0)在区间8,8上有四个不同的根x1，x2，x3，x4，则x1x2x3x4_.答案8解析因为定义在R上的奇函数满足f (x4)f (x)，所以f (x4)f (x)由f (x)为奇函数，所以函数图象关于直线x2对称，且f (0)0.由f (x4)f (x)知f (x8)f (x)，所以函数的周期为8.又因为f (x)在区间0,2上是增函数，所以函数在区间2,0上也是增函数，作出函数f (x)的大致图象如图所示，那么方程f (x)m(m>0)在区间8,8上有四个不同的根x1，x2，x3，x4，不妨设x1<x2<x3<x4，由对称性可知x1x212，x3x44，所以x1x2x3x48.16函数f (x)的定义域为Dx|x0，且满足对于任意x1，x2D，有f (x1·x2)f (x1)f (x2)(1)求f (1)的值；(2)判断f (x)的奇偶性并证明你的结论；(3)如果f (4)1，f (x1)<2，且f (x)在(0，)上是增函数，求x的取值范围解(1)因为对于任意x1，x2D，有f (x1·x2)f (x1)f (x2)，所以令x1x21，得f (1)2f (1)，所以f (1)0.(2)f (x)为偶函数，证明如下：令x1x21，有f (1)f (1)f (1)，所以f (1)f (1)0.令x11，x2x，有f (x)f (1)f (x)，所以f (x)f (x)，又f (x)的定义域关于原点对称，所以f (x)为偶函数(3)依题设有f (4×4)f (4)f (4)2，由(2)知，f (x)是偶函数，所以f (x1)<2，等价于f (|x1|)<f (16)又f (x)在(0，)上是增函数所以0<|x1|<16，解得15<x<17且x1.所以x的取值范围是x|15<x<17且x1