1、高考资源网（） 您身边的高考专家学业分层测评(十)函数的奇偶性(建议用时：45分钟)学业达标一、选择题1.(2016广州高一检测)函数f (x)x的图象关于()A.y轴对称 B.直线yx对称C.坐标原点对称 D.直线yx对称【解析】f (x)xf (x)，f (x)x是奇函数，所以f (x)的图象关于原点对称，故选C.【答案】C2.(2016洛阳高一检测)设函数f (x)，g(x)的定义域都为R，且f (x)是奇函数，g(x)是偶函数，则下列结论中正确的是()A.f (x)g(x)是偶函数 B.|f (x)|g(x)是奇函数C.f (x)|g(x)|是奇函数 D.|f (x)g(x)|是奇函数
2、【解析】f (x)是奇函数，g(x)是偶函数，|f (x)|为偶函数，|g(x)|为偶函数.再根据两个奇函数的积是偶函数、两个偶函数的积还是偶函数、一个奇函数与一个偶函数的积是奇函数，可得f (x)|g(x)|为奇函数，故选C.【答案】C3.(2016济南高一检测)已知f (x)是偶函数，且在区间(0，)上是增函数，则f (0.5)，f (1)，f (0)的大小关系是() 【导学号：60210044】A.f (0.5)f (0)f (1)B.f (1)f (0.5)f (0)C.f (0)f (0.5)f (1)D.f (1)f (0)f (0.5)【解析】函数f (x)为偶函数，f (0.5
3、)f (0.5)，f (1)f (1)，又f (x)在区间(0，)上是增函数，f (0)f (0.5)f (1)，即f (0)f (0.5)f (1)，故选C.【答案】C4.一个偶函数定义在区间7,7上，它在0,7上的图象如图217，下列说法正确的是()图217A.这个函数仅有一个单调增区间B.这个函数有两个单调减区间C.这个函数在其定义域内有最大值是7D.这个函数在其定义域内有最小值是7【解析】根据偶函数在0,7上的图象及其对称性，作出在7,7上的图象，如图所示，可知这个函数有三个单调增区间；有三个单调减区间；在其定义域内有最大值是7；在其定义域内最小值不是7.故选C.【答案】C5.设f (
4、x)是(，)上的奇函数，且f (x2)f (x)，当0x1时，f (x)x，则f (7.5)等于()A.0.5 B.0.5 C.1.5 D.1.5【解析】由f (x2)f (x)，则f (7.5)f (5.52)f (5.5)f (3.52)f (3.5)f (1.52)f (1.5)f (0.52)f (0.5)f (0.5)0.5.【答案】B二、填空题6.(2016沈阳高一检测)函数f (x)在R上为偶函数，且x0时，f (x)1，则当x0时，f (x)_.【解析】f (x)为偶函数，x0时，f (x)1，当x0时，x0，f (x)f (x)1，即x0时，f (x)1.【答案】17.若函数
5、f (x)是定义在R上的偶函数，在(，0)上是增函数，且f (2)0，则使得f (x)0的x的取值范围是_.【解析】函数f (x)是定义在R上的偶函数，且在(，0)上是增函数，又f (2)0，f (x)在(0，)上是减函数，且f (2)f (2)0，当x2或x2时，f (x)0，如图，即f (x)0的解为x2或x2，即不等式的解集为x|x2或x2.【答案】x|x2或x28.已知函数yf (x)是奇函数，若g(x)f (x)2，且g(1)1，则g(1)_.【解析】由g(1)1，且g(x)f (x)2，f (1)g(1)21，又yf (x)是奇函数.f (1)f (1)1，从而g(1)f (1)2
6、3.【答案】3三、解答题9.若函数f (x)当a为何值时，f (x)是奇函数？并证明.【解】假设f (x)是奇函数，则有f (x)f (x).当x0时，即x0时，f (x)x2x，f (x)x2x.f (x)f (x)，即ax2xx2x，a1.下面证明f (x)是奇函数.证明：当x0时，即x0，则f (x)(x)2(x)x2x(x2x)f (x)；当x0时，f (0)0f (0)；当x0，则f (x)(x)2(x)x2x(x2x)f (x)，于是f (x)f (x)f (x).假设成立，即a1时，f (x)是奇函数.10.设定义在2,2上的偶函数f (x)在区间0,2上单调递减，若f (1m)
7、f (m)，求实数m的取值范围.【解】f (x)是偶函数，f (x)f (x)f (|x|)，不等式f (1m)f (m)等价于f (|1m|)f (|m|).又当x0,2时，f (x)是减函数.解得1m.故实数m的取值范围为.能力提升1.若xR，nN*，规定Hx(x1)(x2)(xn1)，例如：H(4)(3)(2)(1)24，则f (x)xH的奇偶性为()A.是奇函数不是偶函数B.是偶函数不是奇函数C.既是奇函数又是偶函数D.既不是奇函数又不是偶函数【解析】由定义可知，f (x)xHx(x2)(x1)x(x1)(x2)x2(x21)(x24)，因为f (x)x2(x21)(x24)f (x)
8、，所以函数f (x)是偶函数不是奇函数.故选B.【答案】B2.(2016四平高一检测)若x，yR，且f (xy)f (x)f (y)，则() 【导学号：60210045】A.f (0)0且f (x)为奇函数B.f (0)0且f (x)为偶函数C.f (x)为增函数且为奇函数D.f (x)为增函数且为偶函数【解析】对任意的x，yR，都有f (xy)f (x)f (y)，令xy0，得f (0)f (0)f (0)2f (0)，f (0)0，令yx，得f (xx)f (x)f (x)f (0)0，f (x)f (x)，函数f (x)为奇函数.【答案】A3.(2016德阳高一检测)定义在R上的奇函数f
9、 (x)，满足f 0，且在(0，)上单调递减，则xf (x)0的解集为()A.B.C.D.【解析】函数f (x)是奇函数，在(0，)上单调递减，且f 0，f 0，且在区间(，0)上单调递减，当x0时，f (x)0，此时xf (x)0，当0x时，f (x)0，此时xf (x)0，综上，xf (x)0的解集为x或x0.【答案】B4.已知函数f (x)，当x，yR时，恒有f (xy)f (x)f (y).当x0时，f (x)0.(1)求证：f (x)是奇函数；(2)若f (1)，试求f (x)在区间2,6上的最值.【解】(1)证明：令x0，y0，则f (0)2f (0)，f (0)0.令yx，则f (0)f (x)f (x)，f (x)f (x)，即f (x)为奇函数.(2)任取x1，x2R，且x1x2，f (xy)f (x)f (y)，f (x2)f (x1)f (x2x1)，当x0时，f (x)0，且x1x2，f (x2x1)0，即f (x2)f (x1)，f (x)为增函数，当x2时，函数有最小值，f (x)minf (2)f (2)2f (1)1.当x6时，函数有最大值，f (x)maxf (6)6f (1)3.高考资源网版权所有，侵权必究！
