2022届高考数学统考一轮复习 课后限时集训11 函数性质的综合问题（理含解析）新人教版.doc

1、课后限时集训(十一)函数性质的综合问题建议用时：40分钟一、选择题1定义在R上的函数f (x)满足f (x1)f (x1)，且f (x)其中aR，若f (5)f (4.5)，则a()A0.5 B1.5 C2.5 D3.5C由f (x1)f (x1)，得f (x)是周期为2的周期函数，又f (5)f (4.5)，所以f (1)f (0.5)，即1a1.5，所以a2.5，故选C2定义在R上的奇函数f (x)满足f (x2)f (x)，且在0,1上是减函数，则有()Af f f Bf f f Cf f f Df f f C由f (x2)f (x)及f (x)是奇函数得f f f f ，又函数f (x

2、)在1,1上是减函数，所以f f f ，即f f f ，故选C3设f (x)是定义在2b,3b上的偶函数，且在2b,0上为增函数，则f (x1)f (3)的解集为()A3,3 B2,4C1,5 D0,6B因为f (x)是定义在2b,3b上的偶函数，所以有2b3b0，解得b3，由函数f (x)在6,0上为增函数，得f (x)在(0,6上为减函数，故f (x1)f (3)f (|x1|)f (3)|x1|3，故2x4.4设奇函数f (x)定义在(，0)(0，)上，f (x)在(0，)上为增函数，且f (1)0，则不等式0的解集为()A(1,0)(1，)B(，1)(0,1)C(，1)(1，)D(1,

3、0)(0,1)D奇函数f (x)定义在(，0)(0，)上，在(0，)上为增函数，且f (1)0，函数f (x)的图象关于原点对称，且过点(1,0)和(1,0)，且f (x)在(，0)上也是增函数函数f (x)的大致图象如图所示f (x)f (x)，不等式0可化为0，即xf (x)0.不等式的解集即为自变量与对应的函数值异号的x的范围，据图象可知x(1,0)(0,1)5(2020全国卷)设函数f (x)ln|2x1|ln|2x1|，则f (x)()A是偶函数，且在单调递增B是奇函数，且在单调递减C是偶函数，且在单调递增D是奇函数，且在单调递减D由得函数f (x)的定义域为，其关于原点对称，因为f

4、 (x)ln|2(x)1|ln|2(x)1|ln|2x1|ln|2x1|f (x)，所以函数f (x)为奇函数，排除A，C当x时，f (x)ln(2x1)ln(12x)，易知函数f (x)单调递增，排除B当x时，f (x)ln(2x1)ln(12x)lnln，易知函数f (x)单调递减，故选D6(2020全国卷)已知函数f (x)sin x，则()Af (x)的最小值为2Bf (x)的图象关于y轴对称Cf (x)的图象关于直线x对称Df (x)的图象关于直线x对称D由题意得sin x1,0)(0,1对于A，当sin x(0,1时，f (x)sin x22，当且仅当sin x1时取等号；当sin

5、 x1,0)时，f (x)sin x22，当且仅当sin x1时取等号，所以A错误对于B，f (x)sin(x)f (x)，所以f (x)是奇函数，图象关于原点对称，所以B错误对于C，f (x)sin(x)，f (x)sin(x)sin x，则f (x)f (x)，f (x)的图象不关于直线x对称，所以C错误对于D，f sincos x，f sincos x，所以f f ，f (x)的图象关于直线x对称，所以D正确故选D二、填空题7已知f (x)是定义在R上的偶函数，且f (x4)f (x2)若当x3,0时，f (x)6x，则f (919)_.6f (x4)f (x2)，f (x6)f (x)

6、，f (x)的周期为6，91915361，f (919)f (1)又f (x)为偶函数，f (919)f (1)f (1)6.8定义在实数集R上的函数f (x)满足f (x)f (x2)0，且f (4x)f (x)现有以下三个命题：8是函数f (x)的一个周期；f (x)的图象关于直线x2对称；f (x)是偶函数其中正确命题的序号是_f (x)f (x2)0，f (x2)f (x)，f (x4)f (x2)f (x)，f (x)的周期为4，故正确；又f (4x)f (x)，所以f (2x)f (2x)，即f (x)的图象关于直线x2对称，故正确；由f (x)f (4x)得f (x)f (4x)

7、f (x)，故正确9定义在R上的奇函数f (x)满足f (x)f (3x)，f (2 020)2，则f (1)_.2由f (x)f (3x)得f (3x)f (x)，从而f (6x)f (x)，即函数f (x)是周期为6的周期函数，所以f (2 020)f (4)f (1)f (1)2.所以f (1)2.三、解答题10设f (x)是定义域为R的周期函数，最小正周期为2，且f (1x)f (1x)，当1x0时，f (x)x.(1)判断f (x)的奇偶性；(2)试求出函数f (x)在区间1,2上的表达式解(1)f (1x)f (1x)，f (x)f (2x)又f (x2)f (x)，f (x)f

8、(x)又f (x)的定义域为R，f (x)是偶函数(2)当x0,1时，x1,0，则f (x)f (x)x；从而当1x2时，1x20，f (x)f (x2)(x2)x2.故f (x)11设函数f (x)是(，)上的奇函数，f (x2)f (x)，当0x1时，f (x)x.(1)求f ()的值；(2)当4x4时，求函数f (x)的图象与x轴所围成图形的面积解(1)由f (x2)f (x)得，f (x4)f (x2)2f (x2)f (x)，所以f (x)是以4为周期的周期函数，所以f ()f (14)f (4)f (4)(4)4.(2)由f (x)是奇函数且f (x2)f (x)，得f (x1)2

9、f (x1)f (x1)，即f (1x)f (1x)故函数yf (x)的图象关于直线x1对称又当0x1时，f (x)x，且f (x)的图象关于原点成中心对称，则f (x)的图象如图所示当4x4时，设f (x)的图象与x轴围成的图形面积为S，则S4SOAB44.1已知定义在R上的奇函数f (x)满足f (x1)f (1x)，且当x0,1时，f (x)2xm，则f (2 019)()A1 B1 C2 D2Bf (x)是定义在R上的奇函数，且f (x1)f (1x)，f (x2)f (x)f (x)，f (x4)f (x)，f (x)的周期为4.x0,1时，f (x)2xm，f (0)1m0，m1，

10、x0,1时，f (x)2x1，f (2 019)f (15054)f (1)f (1)1.故选B2定义在R上的函数f (x)满足：对任意xR有f (x4)f (x)；f (x)在0,2上是增函数；f (x2)的图象关于y轴对称则下列结论正确的是()Af (7)f (6.5)f (4.5)Bf (7)f (4.5)f (6.5)Cf (4.5)f (6.5)f (7)Df (4.5)f (7)f (6.5)D由知函数f (x)的周期为4，由知f (x2)是偶函数，则有f (x2)f (x2)，即函数f (x)图象的一条对称轴是x2，由知函数f (x)在0,2上单调递增，则在2,4上单调递减，且在

11、0,4上越靠近x2，对应的函数值越大，又f (7)f (3)，f (6.5)f (2.5)，f (4.5)f (0.5)，由以上分析可得f (0.5)f (3)f (2.5)，即f (4.5)f (7)f (6.5)故选D3已知函数yf (x)在定义域1,1上既是奇函数又是减函数(1)求证：对任意x1，x21,1，有f (x1)f (x2)(x1x2)0；(2)若f (1a)f (1a2)0，求实数a的取值范围解(1)证明：若x1x20，显然不等式成立若x1x20，则1x1x21，因为f (x)在1,1上是减函数且为奇函数，所以f (x1)f (x2)f (x2)，所以f (x1)f (x2)

12、0.所以f (x1)f (x2)(x1x2)0成立若x1x20，则1x1x21，同理可证f (x1)f (x2)0.所以f (x1)f (x2)(x1x2)0成立综上得证，对任意x1，x21,1，有f (x1)f (x2)(x1x2)0恒成立(2)因为f (1a)f (1a2)0f (1a2)f (1a)f (a1)，所以由f (x)在定义域1,1上是减函数，得即解得0a1.故所求实数a的取值范围是0,1)1定义在R上的函数f (x)满足f (xy)f (x)f (y)，f (x2)f (x)且f (x)在1,0上是增函数，给出下列几个命题：f (x)是周期函数；f (x)的图象关于x1对称；

13、f (x)在1,2上是减函数；f (2)f (0)其中正确命题的序号是_(请把正确命题的序号全部写出来)因为f (xy)f (x)f (y)对任意x，yR恒成立令xy0，所以f (0)0.令xy0，所以yx，所以f (0)f (x)f (x)所以f (x)f (x)，所以f (x)为奇函数因为f (x)在1,0上为增函数，又f (x)为奇函数，所以f (x)在0,1上为增函数由f (x2)f (x)f (x4)f (x2)f (x4)f (x)，所以周期T4，即f (x)为周期函数f (x2)f (x)f (x2)f (x)又因为f (x)为奇函数所以f (2x)f (x)，所以函数图象关于x

14、1对称由f (x)在0,1上为增函数，又关于x1对称，所以f (x)在1,2上为减函数由f (x2)f (x)，令x0得f (2)f (0)f (0)2函数f (x)的定义域为Dx|x0，且满足对于任意x1，x2D，有f (x1x2)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 (x1x2)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：f (x)定义域关于原点对称，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)，所以f (x)为偶函数(3)依题设有f (44)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15x17且x1，所以x的取值范围是(15,1)(1,17)