高考数学二轮复习小题专练作业（十四）函数的图象与性质理.docx

小题专练作业(十四)函数的图象与性质1已知集合M是函数y的定义域，集合N是函数yx24的值域，则MN()解析由题意得M，N4，)，所以MN。故选B。答案B2已知函数f (x)则f (f (1)()A B2C4 D11解析因为f (1)1223，所以f (f (1)f (3)34。故选C。答案C3已知函数f (x)若f (a)3，则f (a2)()A B3C或3 D或3解析当a0时，若f (a)3，则log2aa3，解得a2(满足a0)；当a0时，若f (a)3，则4a213，解得a3，不满足a0，所以舍去。于是，可得a2。故f (a2)f (0)421。故选A。答案A4函数yln(2|x|)的大致图象为()解析令f (x)ln(2|x|)，易知函数f (x)的定义域为x|2x2，且f (x)ln(2|x|)ln(2|x|)f (x)，所以函数f (x)为偶函数，排除C、D。当x时，f lnba Bbca Cacb Dabc解析因为a2 0172 01701,0blog2 017log2 0172 0171，clog2 018bc。故选D。答案D6(2018贵阳摸底)已知函数f (x)a(aR)是奇函数，则函数f (x)的值域为()A(1,1) B(2,2)C(3,3) D(4,4)解析解法一：由f (x)是奇函数知f (x)f (x)，所以aa，得2a，所以a1，所以f (x)1。因为ex11，所以01，111，所以01，111，所以函数f (x)的值域为(1,1)。故选A。答案A7(2018开封考试)已知定义在R上的函数f (x)满足f (x)f (x2)，当x(0,2时，f (x)2xlog2x，则f (2 015)()A5B C2D2解析由f (x)f (x2)，得f (x4)f (x)，所以函数f (x)是周期为4的周期函数，所以f (2 015)f (50343)f (3)f (12)f (1)(20)2。故选D。答案D8(2018祁阳二模)已知偶函数f ，当x时，f (x)xsinx，设af (1)，bf (2)，cf (3)，则()Aabc BbcaCcba Dcab解析因为当x时，ysinx单调递增，yx也为增函数，所以函数f (x)xsinx也为增函数。因为函数f 为偶函数，所以f f ，f (x)的图象关于x对称，所以f (2)f (2)，f (3)f (3)，因为0312，所以f (3)f (1)f (2)，即caf (0)对任意的x(0,2都成立，则f (x)在0,2上是增函数”为假命题的一个函数是_。解析这是一道开放性试题，答案不唯一。只要满足f (x)f (0)对任意的x(0,2都成立，且函数f (x)在0,2上不是增函数即可。如f (x)sinx，答案不唯一。答案f (x)sinx(答案不唯一)10(2018蚌埠质检)已知函数f (x)e|x|lg(ax)图象关于原点对称，则实数a的值为_。解析依题意有f (x)f (x)0，f (x)e|x|lg(ax)，f (x)f (x)e|x|lg(14x2a2x2)0，故a24，a2。答案211(2018洛阳统考)若函数f (x)同时满足下列两个条件，则称该函数为“优美函数”：(1)xR，都有f (x)f (x)0；(2)x1，x2R，且x1x2，都有0。f (x)sinx；f (x)2x3；f (x)1x；f (x)ln(x)。以上四个函数中，“优美函数”的个数是_。解析由条件(1)，得f (x)是R上的奇函数，由条件(2)，得f (x)是R上的单调减函数。对于，f (x)sinx在R上不单调，故不是“优美函数”；对于，f (x)2x3既是奇函数，又在R上单调递减，故是“优美函数”；对于，f (x)1x不是奇函数，故不是“优美函数”；对于，易知f (x)在R上单调递增，故不是“优美函数”。答案112(2018浙江高考)函数y2|x|sin2x的图象可能是()解析设f (x)2|x|sin2x，其定义域关于坐标原点对称，又f (x)2|x|sin(2x)f (x)，所以yf (x)是奇函数，故排除A、B；令f (x)0，所以sin2x0，所以2xk(kZ)，所以x(kZ)，故排除C。故选D。答案D13(2018河北名校联考)已知奇函数f (x)满足f (x1)f (1x)，若当x(1,1)时，f (x)lg，且f (2 018a)1，则实数a的值可以是()A BC D解析因为f (x1)f (1x)，所以f (x)f (2x)，又函数f (x)为奇函数，所以f (x)f (x)，所以f (x)f (2x)，所以f (2x)f (x)，所以f (x4)f (x2)f (x)，所以函数f (x)为周期函数，周期为4。当x(1,1)时，令f (x)lg1，得x，又f (2 018a)f (2a)f (a)，所以a可以是。故选A。答案A14(2018成都检测)已知定义在R上的奇函数f (x)满足f (x2)f (x)0，且当x0,1时，f (x)log2(x1)，则下列不等式正确的是()Af (log27)f (5)f (6)Bf (log27)f (6)f (5)Cf (5)f (log27)f (6)Df (5)f (6)f (log27)解析f (x2)f (x)0f (x2)f (x)f (x4)f (x2)f (x)，所以f (x)是周期为4的周期函数。又f (x)f (x)，且有f (2)f (0)0，所以f (5)f (5)f (1)log221，f (6)f (2)0。又2log273，所以0log2721，即0log21，f (log27)f (log272)0f (log27)f (log272)f log2log2，又1log22，所以0log21，所以1log20，所以f (5)f (log27)f (6)。故选C。答案C15(2018衡水中学第十五次模拟)定义在R上的函数f (x)若满足：对任意x1，x2(x1x2)，都有(x1x2)f (x1)f (x2)0；对任意x，都有f (ax)f (ax)2b，则称函数f (x)为“关于点(a，b)的斜函数”。已知函数yf (x1)是关于点(1,0)的斜函数，若m，n满足不等式f (m22n)f (n22m)，则当m1,4时，的取值范围为_。解析由可知函数f (x)为R上的减函数，由及函数yf (x1)是关于点(1,0)的斜函数，可知函数yf (x)是关于点(