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3.3函数的简单特性

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函数的基本性质，抽象函数

函数的基本性质

定义 1.3.1

函数的若干基本性质定义如下：

有界性：设函数$\displaystyle f(x)$的定义域为$\displaystyle D$，若存在两个常数$\displaystyle m,M$,对任意$\displaystyle x\in D$都满足
$\displaystyle m\leqslant f(x)\leqslant M$，

则称函数\(\displaystyle f(x)\)在\(\displaystyle D\)上有界,其中\(\displaystyle m\)是它的下界,\(\displaystyle M\)是它的上界.否则称函数\(\displaystyle f(x)\)在\(\displaystyle D\)上无界.

单调性：设函数\(\displaystyle f(x)\)的定义域为\(\displaystyle D\)，区间\(\displaystyle I\)是\(\displaystyle D\)的一个子集，若对任意\(\displaystyle x_1,x_2\in I\),当\(\displaystyle x_1<x_2\)时恒有\(\displaystyle f(x_1)\leqslant f(x_2)(f(x_1)<f(x_2))\),则称函数\(\displaystyle f\)在\(\displaystyle I\)上单调递增（严格单调递增）；若对任意\(\displaystyle x_1,x_2\in I\),当\(\displaystyle x_1<x_2\)时恒有\(\displaystyle f(x_1)\geqslant f(x_2)(f(x_1)>f(x_2))\),则称函数\(\displaystyle f\)在\(\displaystyle I\)上单调递减（严格单调递减）。

如果函数\(\displaystyle y=f(x)\)在某个区间\(\displaystyle I\)上单调递增（单调递减），那么称函数\(\displaystyle y=f(x)\)在区间\(\displaystyle I\)上是单调函数，并称区间\(\displaystyle I\)是函数\(\displaystyle y=f(x)\)的一个单调区间。

奇偶性：设函数\(\displaystyle y=f(x)\)的定义域为\(\displaystyle D\)，若对任意\(\displaystyle x\in D\)都有\(\displaystyle -x\in D\)，且\(\displaystyle f(x)=f(-x)\),则称函数\(\displaystyle f(x)\)是偶函数；若对任意\(\displaystyle x\in D\)都有\(\displaystyle -x\in D\)，且\(\displaystyle f(x)+f(-x)=0\),则称函数\(\displaystyle f(x)\)是奇函数.

周期性：若存在常数\(\displaystyle T>0\),使得对一切\(\displaystyle x\in D\)都有\(\displaystyle f(x)=f(x+T)\)成立,则称函数\(\displaystyle f(x)\)是周期函数,\(\displaystyle T\)称为它的一个周期.若存在满足上述条件的最小的\(\displaystyle T\),则称它为\(\displaystyle f(x)\)的最小正周期.

最值：设函数\(\displaystyle y=f(x)\)的定义域为\(\displaystyle D\)，已知\(\displaystyle x_0\in D\)，若对任意\(\displaystyle x\in D\)，恒有\(\displaystyle f(x)\geqslant f(x_0)\)（\(\displaystyle f(x)\leqslant f(x_0)\)），那么称\(\displaystyle f(x_0)\)为\(\displaystyle y=f(x)\)的最小值（最大值），\(\displaystyle x_0\)为\(\displaystyle y=f(x)\)的最小值点（最大值点）。

函数对称性的若干简单探究

对于定义域为\(\displaystyle \mathbb{R}\)函数\(\displaystyle y=f(x)\)，将其做平移与翻折等操作后得到新的图象，其对应的函数\(\displaystyle y=g(x)\)与原函数\(\displaystyle y=f(x)\)之间有何联系呢？

将\(\displaystyle y=f(x)\)的图象向上平移\(\displaystyle k\)个单位（\(\displaystyle k<0\)时相当于向下平移\(\displaystyle |k|\)个单位）,得到的新图象对应的函数为\(\displaystyle g(x)=f(x)+ k\)。这是显然的。

如果将\(\displaystyle y=f(x)\)的图象向左平移\(\displaystyle k\)个单位（\(\displaystyle k<0\)时相当于向右平移\(\displaystyle |k|\)个单位）呢？

假设点\(\displaystyle (x_0,f(x_0))\)是\(\displaystyle y=f(x)\)图象上的点，那么它平移后对应的点为\(\displaystyle (x_0-k,f(x_0))\)，设\(\displaystyle x_0-k=t\)，于是该点可写作\(\displaystyle (t,f(t+k))\)，于是新图象对应的函数为\(\displaystyle g(t)=f(t+k)\)，也就是\(\displaystyle g(x)=f(x+k)\)。

  1. 将\(\displaystyle y=f(x)\)的图象上所有的点保持纵坐标不变,横坐标变为原先的\(\displaystyle k(k\neq 0)\)倍,得到的新图象对应的函数为\(\displaystyle g(x)=f(\frac{x}{k})\)；
  2. 将\(\displaystyle y=f(x)\)的图象上纵坐标小于0的部分关于\(\displaystyle x\)轴翻折,其余部分保持不变,得到的新图象对应的函数为\(\displaystyle g(x)=|f(x)|\)；
  3. 将\(\displaystyle y=f(x)\)的图象上横坐标小于0的部分抹去,并将横坐标大于零的部分沿\(\displaystyle y\)轴向左翻折，并保留一份在原地，得到的新图象对应的函数为\(\displaystyle g(x)=f(|x|)\)；
  4. 试述一种操作，将\(\displaystyle y=f(x)\)的图象由变为\(\displaystyle y=f(ax+b),ab\neq 0\)的图象.

抽象函数

A 组习题

习\(\displaystyle \quad\) 题

A组

  1. 【2008辽宁理8】 将函数\(\displaystyle y = 2^{x} + 1\)的图象按向量\(\displaystyle a\)平移得到函数\(\displaystyle y = 2^{x+1}\)的图象，则

    • \(\displaystyle a = \left(-1, -1\right)\)
    • \(\displaystyle a = \left(1, -1\right)\)
    • \(\displaystyle a = \left(1, 1\right)\)
    • \(\displaystyle a = \left(-1, 1\right)\)

    ??? answer "答案"

    A.

    1. 【2008安徽理11】 若函数\(\displaystyle f\left(x\right), g\left(x\right)\)分别是\(\displaystyle \mathbb{R}\)上的奇函数、偶函数\(\displaystyle ,\)且满足\(\displaystyle f\left(x\right) - g\left(x\right) = \mathrm{e}^{x},\)则有

    2. \(\displaystyle f\left(2\right) < f\left(3\right) < g\left(0\right)\)

    3. \(\displaystyle g\left(0\right) < f\left(3\right) < f\left(2\right)\)
    4. \(\displaystyle f\left(2\right) < g\left(0\right) < f\left(3\right)\)
    5. \(\displaystyle g\left(0\right) < f\left(2\right) < f\left(3\right)\)
??? answer "答案"

D.

  1. 【2016浙江理5】 设函数\(\displaystyle f\left(x\right) = \sin^{2}x + b\sin x + c,\)则\(\displaystyle f\left(x\right)\)的最小正周期

  2. 与\(\displaystyle b\)有关，且与\(\displaystyle c\)有关

  3. 与\(\displaystyle b\)有关，但与\(\displaystyle c\)无关
  4. 与\(\displaystyle b\)无关，且与\(\displaystyle c\)无关
  5. 与\(\displaystyle b\)无关，但与\(\displaystyle c\)有关

??? answer "答案"

B.

  1. 【2011福建9】对于函数\(\displaystyle f(x)=a\sin x+bx+c\)（其中\(\displaystyle a,b\in\mathbb{R},c\in\mathbb{Z}\)）,选取\(\displaystyle a,b,c\)的一组值计算\(\displaystyle f(1)\)和\(\displaystyle f(-1)\),所得出的正确结果一定不可能是

  2. 4和6

  3. 3和1
  4. 2和4
  5. 1和2
答案

D.

\(\displaystyle g\left(x\right)=f\left(x\right)-c\)是奇函数，所以\(\displaystyle g\left(-1\right)=-g\left(1\right)\)，即\(\displaystyle f\left(-1\right)-c=-f\left(1\right)+c\)，所以\(\displaystyle f\left(1\right)+f\left(-1\right)=2c\)．由于\(\displaystyle c\)是整数，则\(\displaystyle f\left(1\right)+f\left(-1\right)\)是偶数．而\(\displaystyle D\)的选项中两数之和是奇数，不可能取到奇数．

  1. 【2019上海15】 已知\(\displaystyle \omega \in \mathbb{R},\)函数\(\displaystyle f\left(x\right) = \left(x - 6\right)^{2} \cdot \sin\left(\omega x\right),\)存在常数\(\displaystyle a \in R,\)使得\(\displaystyle f\left(x + a\right)\)为偶函数，则\(\displaystyle \omega\)可能的值为

  2. \(\displaystyle \frac{\pi}{2}\)

  3. \(\displaystyle \frac{\pi}{3}\)
  4. \(\displaystyle \frac{\pi}{4}\)
  5. \(\displaystyle \frac{\pi}{5}\)
答案

C. 根据\(\displaystyle f\left(x\right)\)的表达式，只能有\(\displaystyle a=6\)，所以\(\displaystyle f\left(x+6\right)=\sin\left(\omega\left(x+6\right)\right)\)是偶函数．由三角函数的奇偶性质，只能有\(\displaystyle \sin\left(\omega\left(x+6\right)\right)=\pm\cos\left(\omega x\right)\)，因此取\(\displaystyle x=0\)可得\(\displaystyle \sin\left(6\omega\right)=\pm1\)．检查四个答案，只能选C．

  1. 【2021浙江7】 已知函数\(\displaystyle f\left(x\right) = x^{2} + \frac{1}{4}, g\left(x\right) = \sin x,\)则图象为如图的函数可能是
答案

D.

  1. 【2021上海春考15】已知函数\(\displaystyle y=f(x)\)的定义域为\(\displaystyle \mathbb{R}\),下列是\(\displaystyle f(x)\)无最大值的充分条件是

  2. \(\displaystyle f(x)\)为偶函数且关于点\(\displaystyle (1,1)\)对称

  3. \(\displaystyle f(x)\)为偶函数且关于直线\(\displaystyle x=1\)对称
  4. \(\displaystyle f(x)\)为奇函数且关于点\(\displaystyle (1,1)\)对称
  5. \(\displaystyle f(x)\)为奇函数且关于直线\(\displaystyle x=1\)对称
??? answer "答案"

C.
  1. 【2007安徽理11】 定义在 \(\displaystyle \mathbb{R}\) 上的函数 \(\displaystyle f(x)\) 既是奇函数又是周期函数, \(\displaystyle T\) 是它的一个正周期. 将方程 \(\displaystyle f(x)=0\) 在闭区间 \(\displaystyle [-T,T]\) 上的根的个数记为 \(\displaystyle n\), 则\(\displaystyle n\)可能为
答案

D.

  1. 【2026上海春15】平移对称法在几何学中具有重要的应用.设平面直角坐标系\(\displaystyle xOy\)中有一图形\(\displaystyle \Omega\),过\(\displaystyle \Omega\)内任意一点\(\displaystyle P\)作垂直于\(\displaystyle x\)轴的直线\(\displaystyle l_P\),满足\(\displaystyle l_P\cap \Omega\)为一线段,现沿\(\displaystyle l_P\)方向平移这些线段,使得它们的中点均在\(\displaystyle x\)轴上,这样叫做平移对称法.对于\(\displaystyle -x^2+x+1,-x^2-x\),直线\(\displaystyle x=0\)和直线\(\displaystyle x=1\)围成的封闭图形\(\displaystyle \Omega\),对它进行一次平移对称,得到的图像大致为

    3.4.1

    答案

    A. 10. 【2024新高考I卷8】已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),\(\displaystyle f(x)>f(x-1)+f(x-2)\),且当\(\displaystyle x<3\)时,\(\displaystyle f(x)=x\),则下列结论中一定正确的是

  2. \(\displaystyle f(10)>100\)

  3. \(\displaystyle f(20)>1000\)
  4. \(\displaystyle f(10)<1000\)
  5. \(\displaystyle f(20)<10000\)
??? answer "答案"

B.
  1. 【2025“fiddie”模拟考9】（多选）设函数 \(\displaystyle f(x) = \mathrm{e}^x + ae^{-x} + bx^2 + (a-b)x\),则
答案

ABC.

对于 A 选项,当\(\displaystyle a=1, b=1\)时,\(\displaystyle f(x)=\mathrm{e}^x+\mathrm{e}^{-x}+x^2\)是偶函数,故 A 选项正确.

对于 B 选项,当\(\displaystyle a=-1, b=0\)时,\(\displaystyle f(x)=\mathrm{e}^x-\mathrm{e}^{-x}-x=0\),故 B 选项正确.

对于 C 选项,当\(\displaystyle a=b=0\)时,\(\displaystyle f(x)=\mathrm{e}^x\)是\(\displaystyle (-\infty,+\infty)\)上的增函数,故 C 选项正确.

对于 D 选项,因为当\(\displaystyle x\to +\infty\)时,\(\displaystyle \mathrm{e}^x\)的增长速度比\(\displaystyle \mathrm{e}^{-x},x^2,x\)都快,所以\(\displaystyle \lim_{x\to +\infty}f(x)=+\infty\),故\(\displaystyle f(x)\)不可能是减函数.

（另解）证明存在\(\displaystyle x_0\)使得当\(\displaystyle x > x_0\)时\(\displaystyle f'(x) > 0\).

Fiddie评：本题改编自【2009浙江文8】.

  1. 【2023四省联考9】（多选）已知\(\displaystyle f(x)\)是定义在\(\displaystyle \mathbb{R}\)上的偶函数,是定义在\(\displaystyle \mathbb{R}\)上的奇函数,且\(\displaystyle f(x),g(x)\)在\(\displaystyle (-\infty,0]\)单调递减,则
答案

BD.

  1. 【2024九省联考11】（多选）已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),且\(\displaystyle f(\frac{1}{2})\neq 0\),若\(\displaystyle f(x+y)+f(x)f(y)=4xy\),则
答案

ABD.

  1. 【2025北京15改编】（多选）函数 \(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),则
答案

BC.

  1. 【2023新高考I卷11】（多选）已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),\(\displaystyle f(xy)=y^2f(x)+x^2f(y)\),则
答案

ABC.

  1. 【2014 全国 I卷文 5理 3】 设 \(\displaystyle f(x), g(x)\) 的定义域为 \(\displaystyle \mathbb{R}\),\(\displaystyle f(x)\) 是奇函数,\(\displaystyle g(x)\) 是偶函数,分别判断\(\displaystyle f(x)g(x),|f(x)|g(x),f(x)|g(x)|,|f(x)g(x)|\)的奇偶性.
  2. 【2021新高考II卷14】 写出一个同时具有下列性质的函数\(\displaystyle f\left(x\right).\),(1).\(\displaystyle f\left(x_1x_2\right) = f\left(x_1\right)f\left(x_2\right)\);(2) 当\(\displaystyle x \in \left(0, +\infty\right)\)时，\(\displaystyle f'\left(x\right) > 0\) ;(3)\(\displaystyle f'\left(x\right)\)是奇函数。
  3. 【2017全国III卷理11】已知函数\(\displaystyle f(x)=x^2-2x+a(\mathrm{e}^{x-1}+\mathrm{e}^{-x+1})\)有唯一零点,求\(\displaystyle a\)的值.
答案

\(\displaystyle \frac{1}{2}.\)

  1. 【2011上海理13】 设\(\displaystyle g\left(x\right)\)是定义在\(\displaystyle \mathbb{R}\)上，以\(\displaystyle 1\)为周期的函数，若函数\(\displaystyle f\left(x\right) = x + g\left(x\right)\)在区间\(\displaystyle \left[3, 4\right]\)上的值域为\(\displaystyle \left[-2, 5\right],\)求\(\displaystyle f\left(x\right)\)在区间\(\displaystyle \left[-10, 10\right]\)上的值域。
答案

\(\displaystyle [-15,11]\).

  1. 【2022新高考II卷8】 已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),满足\(\displaystyle f(x+y)+f(x-y)=f(x)f(y),f(1)=1\),求\(\displaystyle \sum_{k=1}^{22}f(k)\).
答案

\(\displaystyle -3\).

  1. 【2021全国甲卷理12】【同2026新高考II卷8】】已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),\(\displaystyle f(x+1)\)为奇函数,\(\displaystyle f(x+2)\)为偶函数,当\(\displaystyle x\in[1,2]\)时,\(\displaystyle f(x)=ax^{2}+b\),若\(\displaystyle f(0)+f(3)=6\),求\(\displaystyle f(\dfrac{9}{2})\)

??? answer "答案"

$\displaystyle \frac{5}{2}$.

    相似题： **【2021新高考II卷8】** 设函数$\displaystyle f\left(x\right)$的定义域为$\displaystyle R,$且$\displaystyle f\left(x + 2\right)$为偶函数，$\displaystyle f\left(2x + 1\right)$为奇函数，则 <div class="choices choices--4" markdown>

- $\displaystyle f\left(-\frac{1}{2}\right) = 0$
- $\displaystyle f\left(-1\right) = 0$
- $\displaystyle f\left(2\right) = 0$
- $\displaystyle f\left(4\right) = 0$

</div>

  1. 【2010湖南8】若函数\(\displaystyle f(x)=\min\left \{ |x|,|x+t| \right \}\)的图象关于直线\(\displaystyle x=-\frac{1}{2}\)对称,求\(\displaystyle t\). ??? answer "答案"

    \(\displaystyle 1\). 23. 【2015 全国 II卷文 12】 已知函数 \(\displaystyle f(x) = \ln(1+|x|) - \frac{1}{1+x^2}\),解不等式 \(\displaystyle f(x) > f(2x-1)\).

答案

\(\displaystyle (\frac{1}{3},1)\).

  1. 【2012 新课标I卷文 16】设函数 \(\displaystyle f(x) = \frac{(x+1)^2 + \sin x}{x^2 + 1}\) 的最大值为 \(\displaystyle M\),最小值为 \(\displaystyle m\),求\(\displaystyle M+m\). ??? answer "答案"

    2. 25. 【2022全国乙卷文16】若\(\displaystyle f(x)=\ln \left | a+\frac{1}{1-x} \right | +b\)是奇函数,求\(\displaystyle a,b\).

    答案

    \(\displaystyle a=-\frac{1}{2},b=\ln 2\). 26. 【2005江西文13】 若函数 \(\displaystyle f(x) = \log_a(x + \sqrt{x^2 + 2a^2})\) 是奇函数, 求 \(\displaystyle a\).

    答案

    \(\displaystyle a=\frac{\sqrt{2}}{2}\). 27. 【2013新课标I卷理16】若函数 \(\displaystyle f(x)=(1-x^2)(x^2+ax+b)\) 的图象关于直线 \(\displaystyle x=-2\) 对称, 求 \(\displaystyle f(x)\) 的最大值.

    答案

    16. 28. 已知函数\(\displaystyle y=f(x)\),解释： 29. 判断正误： 1. （A）函数\(\displaystyle f(x)\)在\(\displaystyle D\)上存在最大值、最小值当且仅当函数\(\displaystyle f(x)\)在\(\displaystyle D\)上有界； 2. （B）闭区间\(\displaystyle [a,b]\)上的连续函数\(\displaystyle f(x)\)一定存在最大值\(\displaystyle M\)与最小值\(\displaystyle m\),记\(\displaystyle f(x)\)在\(\displaystyle [a,b]\)上的极值点为\(\displaystyle x_1,x_2,\cdots,x_n,\cdots\),则一定有$\(\displaystyle M=\max\{f(a),f(b),f(x_1),f(x_2),\cdots\},\quad m=\min\{f(a),f(b),f(x_1),f(x_2),\cdots\}\)$ 3. （A）若函数 \(\displaystyle f(x)\) 在定义域 \(\displaystyle D\) 内无界,则对任意给定的正数 \(\displaystyle M\),在 \(\displaystyle D\) 内总存在一点 \(\displaystyle x_0\),使得 \(\displaystyle f(x_0) > M\)； 4. （A）函数 \(\displaystyle f(x)\) 在区间 \(\displaystyle I\) 上有界当且仅当\(\displaystyle |f(x)|\) 在 \(\displaystyle I\) 上有界. 30. 已知函数\(\displaystyle f(x)\)的定义域是\(\displaystyle \mathbb{R}\),则下列命题中可以作为“\(\displaystyle f(x)\)在\(\displaystyle \mathbb{R}\)上严格单调递增”的充分条件的是

    1. 对任意两个不相等实数\(\displaystyle x_1,x_2\),有\(\displaystyle (f(x_1)-f(x_2))(x_1-x_2)>0\)；
      1. 对任意实数\(\displaystyle x_0\),存在\(\displaystyle \delta>0\),使得当\(\displaystyle x\in(x_0,x_0+\delta)\)时有\(\displaystyle f(x)>f(x_0)\)；
      2. 存在\(\displaystyle \delta>0\),使得对任意两个实数\(\displaystyle x_1,x_2\),当\(\displaystyle 0<x_2-x_1<\delta\)时有\(\displaystyle f(x_1)<f(x_2)\)；
      3. 函数\(\displaystyle f(x)\)连续,且存在正实数\(\displaystyle p\),使得对任意实数\(\displaystyle x\),有\(\displaystyle f(x)<f(x+p)\)；
      4. 对任意实数\(\displaystyle x\),存在正实数\(\displaystyle p\),使得\(\displaystyle f(x)<f(x+p)\)；
      5. 对任意两个有理数\(\displaystyle r_1,r_2\),当\(\displaystyle r_1<r_2\)时有\(\displaystyle f(r_1)<f(r_2)\)；
      6. 对任意实数\(\displaystyle x\)和正整数\(\displaystyle n\),有\(\displaystyle f(x)<f(x+\frac{1}{n})\)；
      7. 对任意两个实数\(\displaystyle a,b\),当\(\displaystyle a<b\)时,函数\(\displaystyle f(x)\)在区间\(\displaystyle (a,b)\)上都不单调递减；
      8. 函数\(\displaystyle f(x)\)连续,对任意两个实数\(\displaystyle a,b\),当\(\displaystyle a<b\)时,函数\(\displaystyle f(x)\)在区间\(\displaystyle (a,b)\)上都不单调递减.

    2. 以下是关于函数图象的对称性与周期性关系的探究.设\(\displaystyle f(x)\)是定义在\(\displaystyle \mathbb{R}\)上的函数,证明或证伪：

    3. 若\(\displaystyle y=f(x)\)的图象既关于\(\displaystyle x=b\)对称,又关于\(\displaystyle x=a\)对称（\(\displaystyle a\neq b\)）,则\(\displaystyle f(x)\)是周期函数,它的一个周期是\(\displaystyle T=2|b-a|\),且\(\displaystyle y=f(x)\)的图象也关于\(\displaystyle x=b+\frac{1}{2}kT(k\in\mathbb{Z})\)对称；

      1. 若\(\displaystyle y=f(x)\)的图象既关于\(\displaystyle (a,y_a)\)对称,又关于\(\displaystyle (b,y_b)\)对称（\(\displaystyle a\neq b\)）,则\(\displaystyle f(x)\)是周期函数,它的一个周期是\(\displaystyle T=2|b-a|\),且\(\displaystyle y=f(x)\)的图象也关于\(\displaystyle (b+\frac{1}{2}kT,0)(k\in\mathbb{Z})\)对称；
      2. 若\(\displaystyle y=f(x)\)的图象既关于\(\displaystyle x=b\)对称,又关于\(\displaystyle (a,y_a)\)对称（\(\displaystyle a\neq b\)）,则\(\displaystyle y=f(x)\)是周期函数,它的一个周期是\(\displaystyle T=4|b-a|\),且\(\displaystyle y=f(x)\)的图象也关于\(\displaystyle x=b+\frac{1}{2}kT(k\in\mathbb{Z}),(a+\frac{1}{2}kT,0)\)对称；
    4. 证明或反驳：
    5. Dirichlet函数\(\displaystyle D(x)=\begin{cases} 1& x\in\mathbb{Q}\\ 0& x\notin \mathbb{Q} \end{cases}\) 是周期函数,且存在最小正周期；
      1. （B）若\(\displaystyle f(x),g(x)\)是定义在\(\displaystyle \mathbb{R}\)上的周期函数,则\(\displaystyle f(x)+g(x)\)一定是周期函数；
      2. （B）若\(\displaystyle f(x),g(x)\)是定义在\(\displaystyle \mathbb{R}\)上的周期函数,则\(\displaystyle f(x)g(x)\)一定是周期函数；
      3. （A）若\(\displaystyle f(x),g(x)\)是定义在\(\displaystyle \mathbb{R}\)上的周期函数,则\(\displaystyle f(g(x))\)一定是周期函数；
      4. （C）若\(\displaystyle y=f(x)\)的图象有一个对称中心和一条对称轴,那么\(\displaystyle f(x)\)是周期函数.
    6. 对于下列函数,若其图象为中心对称图形,则求出其对称中心；若为轴对称图形（此处仅考虑垂直于\(\displaystyle x\)轴的对称轴）,则求出其对称轴.

    \settasks{ label=(\arabic*), label-width=2em, label-offset=0.2em,

    column-sep=2em,

    after-item-skip=1.5ex }

(2)

    \task  $\displaystyle f(x) = \ln\frac{x+1}{x-1}$ \hfill
    \task  $\displaystyle f(x) = \ln\frac{3x+5}{3x-5}$ \hfill
    \task  $\displaystyle f(x) = \ln\frac{x-2}{x-4}$\hfill
    \task  $\displaystyle f(x) = \frac{2^x+1}{2^x-1}$ \hfill
    \task  $\displaystyle f(x) = \frac{2^x+5}{2^x-5}$ \hfill
    \task  $\displaystyle f(x) = \frac{5}{3^x+8}$\hfill
    \task  $\displaystyle f(x) = 4^x+5\times 4^{-x}$ \hfill
  1. 【2023全国乙卷21（2）】已知函数\(\displaystyle f(x)=(\frac{1}{x}+a)\ln(1+x)\).是否存在\(\displaystyle a,b\),使得曲线\(\displaystyle y=f(\frac{1}{x})\)关于直线\(\displaystyle x=b\)对称？若存在,求\(\displaystyle a,b\),否则说明理由.
答案

思路1：设\(\displaystyle g\left(x\right)=f\left(\frac{1}{x}\right)\)，则\(\displaystyle g\left(x\right)=\left(x+a\right)\ln\left(1+\frac{1}{x}\right)\)，\(\displaystyle g\left(x\right)\)的定义域为\(\displaystyle \left(-\infty,-1\right)\cup\left(0,+\infty\right)\)．

若存在\(\displaystyle a\),\(\displaystyle b\)，使得曲线\(\displaystyle y=g\left(x\right)\)关于直线\(\displaystyle x=b\)对称，则\(\displaystyle \left(-\infty,-1\right)\cup\left(0,+\infty\right)\)关于\(\displaystyle x=b\)对称，所以\(\displaystyle b=-\frac{1}{2}\)．

由\(\displaystyle g\left(x\right)=g\left(-1-x\right)\)得\(\displaystyle \left(x+a\right)\ln\left(1+\frac{1}{x}\right)=\left(-1-x+a\right)\ln\frac{x}{1+x}\)，整理得 $\(\displaystyle \left(x+a\right)\ln\left(1+\frac{1}{x}\right)=\left(x+1-a\right)\ln\left(1+\frac{1}{x}\right)\)$

所以\(\displaystyle a=\frac{1}{2}\)，故存在\(\displaystyle a=\frac{1}{2}\)，\(\displaystyle b=-\frac{1}{2}\)，使得曲线\(\displaystyle y=f\left(\frac{1}{x}\right)\)关于直线\(\displaystyle x=b\)对称．

思路2：设函数\(\displaystyle g\left(x\right)=f\left(\frac{1}{x}\right)\)，则\(\displaystyle g\left(x\right)=\left(x+a\right)\ln\left(1+\frac{1}{x}\right)\)．若曲线\(\displaystyle y=f\left(\frac{1}{x}\right)\)关于直线\(\displaystyle x=b\)对称，则\(\displaystyle g\left(x\right)=g\left(2b-x\right)\)，即 $\(\displaystyle \left(x+a\right)\ln\left(1+\frac{1}{x}\right)=\left(x-2b-a\right)\ln\frac{x-2b}{x-2b-1}\)$

由此可以得到，当 $\(\displaystyle \begin{cases} -2b=1,\\ -2b-1=0,\\ -2b-a=a, \end{cases}\)$

即\(\displaystyle a=\frac{1}{2}\)，\(\displaystyle b=-\frac{1}{2}\)时，曲线\(\displaystyle y=f\left(\frac{1}{x}\right)\)关于直线\(\displaystyle x=b\)对称．

B 组习题

B组

  1. 已知函数 \(\displaystyle f(x), g(x)\) 的定义域均为 \(\displaystyle \mathbb{R}\),且 \(\displaystyle f(x)\) 的图像关于点 \(\displaystyle (2, 0)\) 对称.设集合 \(\displaystyle A = \{a \mid \text{对任意 } x, f(x)+g(a) > 0\}\),\(\displaystyle B = \{a \mid \text{存在 } x, f(x)+g(a) > 0\}\),\(\displaystyle C = \{a \mid g(a) \geqslant f(1)\}\),则

    • \(\displaystyle A \cap C = A\)
    • \(\displaystyle A \cap C = C\)
    • \(\displaystyle B \cap C = B\)
    • \(\displaystyle B \cap C = C\)

    2. 【2009浙江10】对于正实数\(\displaystyle \alpha\),记\(\displaystyle M_{\alpha}\)为满足下述条件的函数\(\displaystyle f(x)\)构成的集合：对任意的\(\displaystyle x_1,x_2\in\mathbb{R}\),且\(\displaystyle x_2>x_1\),有\(\displaystyle -\alpha(x_2-x_1)<f(x_2)-f(x_1)<\alpha(x_2-x_1)\),下列结论中正确的是

    （A）若\(\displaystyle f(x)\in M_{\alpha_1},g(x)\in M_{\alpha_2}\),则\(\displaystyle f(x)g(x)\in M_{\alpha_1\alpha_2}\)

    （B）若\(\displaystyle f(x)\in M_{\alpha_1},g(x)\in M_{\alpha_2}\),且\(\displaystyle g(x)\neq0\),则\(\displaystyle \frac{f(x)}{g(x)}\in M_{\alpha_1/\alpha_2}\)

    （C）若\(\displaystyle f(x)\in M_{\alpha_1},g(x)\in M_{\alpha_2}\),则\(\displaystyle f(x)+g(x)\in M_{\alpha_1+\alpha_2}\)

    （D）若\(\displaystyle f(x)\in M_{\alpha_1},g(x)\in M_{\alpha_2}\),且\(\displaystyle \alpha_1>\alpha_2\),则\(\displaystyle f(x)-g(x)\in M_{\alpha_1-\alpha_2}\) 3. 【2026成都二诊11】（多选）函数\(\displaystyle f(x),g(x)\)的定义域为\(\displaystyle \mathbb{R}\),函数\(\displaystyle h(x)=\min\{f(x),g(x)\}\),则

    <div class="choices choices--4" markdown>

    4. 【2022新高考I卷12】（多选） 已知函数\(\displaystyle f(x)\)及其导函数\(\displaystyle f'(x)\)的定义域均为\(\displaystyle \mathbb{R}\),记\(\displaystyle g(x)=f'(x)\),若\(\displaystyle f(\frac{3}{2}-2x),g(2+x)\)均为偶函数,则

    <div class="choices choices--4" markdown>

    5. 已知函数\(\displaystyle f(x)=x^{3}-6x^{2}+12\),若函数\(\displaystyle g(x)=|f(x-m)+n|\)为偶函数,求\(\displaystyle m,n\)的值. 6. 已知曲线 \(\displaystyle C_1: y = x^3 - x\) 与 \(\displaystyle C_2\) 关于点 \(\displaystyle (p,q)\) 对称，且 \(\displaystyle C_1\) 与 \(\displaystyle C_2\) 有且仅有一个公共点，求\(\displaystyle pq\) 的最小值。 7. 【2005辽宁10】 已知 \(\displaystyle f(x)\) 是定义在 \(\displaystyle \mathbb{R}\) 上的单调函数, 实数 \(\displaystyle x_1 \neq x_2\), \(\displaystyle \lambda \neq -1\), \(\displaystyle \alpha = \frac{x_1 + \lambda x_2}{1+\lambda}\), \(\displaystyle \beta = \frac{x_2 + \lambda x_1}{1+\lambda}\), 若 \(\displaystyle |f(x_1)-f(x_2)| < |f(\alpha)-f(\beta)|\),求\(\displaystyle \lambda\)的取值范围. 8. 【2022全国乙卷12】 已知函数\(\displaystyle f(x),g(x)\)的定义域均为\(\displaystyle \mathbb{R}\),且\(\displaystyle f(x)+g(2-x)=5,g(x)-f(x-4)=7\).若\(\displaystyle y=g(x)\)的图象关于直线\(\displaystyle x=2\)对称,\(\displaystyle g(2)=4\),求\(\displaystyle \sum_{k=1}^{22}f(k)\). 9. 若存在函数\(\displaystyle f(x)\)满足：对任意实数\(\displaystyle x\in\mathbb{R}\),有\(\displaystyle f(x^2+ax)=|x+1|+|x+2|\),求\(\displaystyle a\). 10. 【2025高联一试B9】设 \(\displaystyle f(x)\) 是定义域为 \(\displaystyle \mathbb{R}\) 的函数,\(\displaystyle g(x)=(x-1)f(x),h(x)=f(x)+x\).若 \(\displaystyle g(x)\) 为奇函数,\(\displaystyle h(x)\) 为偶函数,求 \(\displaystyle \frac{f(1)f(3)\cdots f(99)}{f(2)f(4)\cdots f(100)}\) 的值. 11. 设非空集合\(\displaystyle S,T\)满足\(\displaystyle S\cap T=\varnothing,S\cup T=\mathbb{R}\),当\(\displaystyle x\in S\)时,\(\displaystyle f(x)=x+3\),当\(\displaystyle x\in T\)时,\(\displaystyle f(x)=2^x+2^{-x}\),若函数\(\displaystyle y=f(x)\)为偶函数,求\(\displaystyle |S|\)的最大值. 12. 【2016上海理18】设 \(\displaystyle f(x),g(x),h(x)\) 是定义域为 \(\displaystyle \mathbb{R}\) 的三个函数,证明或证伪：

    1. 若 $\displaystyle f(x)+g(x)、f(x)+h(x)、g(x)+h(x)$ 均为增函数, 则 $\displaystyle f(x)、g(x)、h(x)$ 中至少有一个为增函数；
    1. 若 \(\displaystyle f(x)+g(x)、f(x)+h(x)、g(x)+h(x)\) 均为以 \(\displaystyle T\) 为周期的函数, 则 \(\displaystyle f(x)、g(x)、h(x)\) 中至少有一个为以 \(\displaystyle T\) 为周期的函数.
    答案

    \par 新答案（来源：1.9 函数的周期性.md）： D

    【解题思路】本题需根据单个函数的和函数的性质，探讨单个函数的性质，不妨将单个函数用和函数来表示，例如

    \[\displaystyle f\left(x\right)=\frac{\left[f\left(x\right)+g\left(x\right)\right]+\left[f\left(x\right)+h\left(x\right)\right]-\left[g\left(x\right)+h\left(x\right)\right]}{2}\]

    若\(\displaystyle f\left(x\right)+g\left(x\right)\)、\(\displaystyle f\left(x\right)+h\left(x\right)\)、\(\displaystyle g\left(x\right)+h\left(x\right)\)均是以\(\displaystyle T\)为周期的函数，根据\(\displaystyle f\left(x\right)\)的上述表达式，易推得\(\displaystyle f\left(x\right)\)也是以\(\displaystyle T\)为周期的函数，\(\displaystyle g\left(x\right)\)与\(\displaystyle h\left(x\right)\)的情况同理，从而②为真命题．

    然而，若\(\displaystyle f\left(x\right)+g\left(x\right)\)、\(\displaystyle f\left(x\right)+h\left(x\right)\)、\(\displaystyle g\left(x\right)+h\left(x\right)\)均是增函数，根据\(\displaystyle f\left(x\right)\)的上述表达形式，却不能推得\(\displaystyle f\left(x\right)\)是增函数，也就是说\(\displaystyle f\left(x\right)\)可能不是增函数，例如\(\displaystyle f\left(x\right)=-x\)，\(\displaystyle g\left(x\right)=2x\)，\(\displaystyle h\left(x\right)=3x\)．同理，\(\displaystyle g\left(x\right)\)与\(\displaystyle h\left(x\right)\)也可能不是增函数．

    接下来要判断①的真假，还要考虑\(\displaystyle f\left(x\right)\)、\(\displaystyle g\left(x\right)\)、\(\displaystyle h\left(x\right)\)是否可能都不是增函数．在之前举例的基础上，可以找到\(\displaystyle f\left(x\right)\)、\(\displaystyle g\left(x\right)\)、\(\displaystyle h\left(x\right)\)都不是增函数的例子．

    例如，

    \[\displaystyle f\left(x\right)=\begin{cases}x,&x<0,\\ 0,&0\leqslant x\leqslant1,\\ 2x,&x>1,\end{cases}\]
    \[\displaystyle g\left(x\right)=\begin{cases}0,&x<0,\\ x,&0\leqslant x\leqslant1,\\ 2x,&x>1,\end{cases}\]
    \[\displaystyle h\left(x\right)=\begin{cases}x,&x<0,\\ x,&0\leqslant x\leqslant1,\\ 0,&x>1,\end{cases}\]

    所以①为假命题．

    【实测数据】本题理科难度为\(\displaystyle 0.215\)，区分度为\(\displaystyle 0.03\)，选\(\displaystyle A\)、\(\displaystyle B\)、\(\displaystyle C\)、\(\displaystyle D\)占比分别为\(\displaystyle 14.58\%\)、\(\displaystyle 15.15\%\)、\(\displaystyle 48.58\%\)、\(\displaystyle 21.49\%\)．

    【易错警示】三个错误选项中，选择率最高的是\(\displaystyle C\)．选择\(\displaystyle C\)的原因更多是猜测而不是推理：凭感觉认为命题①为真命题，且认为所给的两个命题一真一假，此外，还有选择题猜答案为\(\displaystyle C\)的必然性．选项\(\displaystyle A\)和选项\(\displaystyle B\)也较有迷惑性，前两个层次的考生选\(\displaystyle A\)多于选\(\displaystyle B\)，而后两个层次的考生选\(\displaystyle B\)多于选\(\displaystyle A\)．排除猜测的成分，命题②易于推理判断为真，而命题①则需举出反例说明为假，考生的选择从某种程度上也是他们对推理形式的掌握情况的反映．

    \par
 新答案（来源：4.2 多个量词的问题（1）函数.md）：

    （1）由已知，函数\(\displaystyle f\left(x\right)\)的定义域为\(\displaystyle \left(0,+\infty\right)\)，

    \(\displaystyle g\left(x\right)=f'\left(x\right)=2\left(x-a\right)-2\ln x-2\left(1+\frac{a}{x}\right)\)，

    所以，\(\displaystyle g'\left(x\right)=2-\frac{2}{x}+\frac{2a}{x^2}=\frac{2\left(x-\frac{1}{2}\right)^2+2\left(a-\frac{1}{4}\right)}{x^2}\)．

    当\(\displaystyle 0<a<\frac{1}{4}\)时，\(\displaystyle g\left(x\right)\)在区间\(\displaystyle \left(0,\frac{1-\sqrt{1-4a}}{2}\right)\)，\(\displaystyle \left(\frac{1+\sqrt{1-4a}}{2},+\infty\right)\)上单调递增，

    在区间\(\displaystyle \left(\frac{1-\sqrt{1-4a}}{2},\frac{1+\sqrt{1-4a}}{2}\right)\)上单调递减；

    当\(\displaystyle a\geqslant\frac{1}{4}\)时，\(\displaystyle g\left(x\right)\)在区间\(\displaystyle \left(0,+\infty\right)\)上单调递增．

    （2）由\(\displaystyle f'\left(x\right)=2\left(x-a\right)-2\ln x-2\left(1+\frac{a}{x}\right)=0\)，解得\(\displaystyle a=\frac{x-1-\ln x}{1+x^{-1}}\)．

    令\(\displaystyle \varphi\left(x\right)=-2\left(x+\frac{x-1-\ln x}{1+x^{-1}}\right)\ln x+x^2-2\left(\frac{x-1-\ln x}{1+x^{-1}}\right)x-2\left(\frac{x-1-\ln x}{1+x^{-1}}\right)^2+\frac{x-1-\ln x}{1+x^{-1}}\)，

    则\(\displaystyle \varphi\left(1\right)=1>0\)，\(\displaystyle \varphi\left(\mathrm{e}\right)=\frac{-\mathrm{e}\left(\mathrm{e}-2\right)}{1+\mathrm{e}^{-1}}-2\left(\frac{e-2}{1+\mathrm{e}^{-1}}\right)^2<0\)．

    故存在\(\displaystyle x_0\in\left(1,\mathrm{e}\right)\)，使得\(\displaystyle \varphi\left(x_0\right)=0\)．

    令\(\displaystyle a_0=\frac{x_0-1-\ln x_0}{1+x_0^{-1}}\)，\(\displaystyle u\left(x\right)=x-1-\ln x\left(x\geqslant1\right)\)．

    由\(\displaystyle u'\left(x\right)=1-\frac{1}{x}\geqslant0\)知，函数\(\displaystyle u\left(x\right)\)在区间\(\displaystyle \left(1,+\infty\right)\)上单调递增．

    所以\(\displaystyle 0=\frac{u\left(1\right)}{1+1}<\frac{u\left(x_0\right)}{1+x_0^{-1}}=a_0<\frac{u\left(\mathrm{e}\right)}{1+\mathrm{e}^{-1}}=\frac{e-2}{1+\mathrm{e}^{-1}}<1\)．

    即\(\displaystyle a_0\in\left(0,1\right)\)．

    当\(\displaystyle a=a_0\)时，有\(\displaystyle f'\left(x_0\right)=0\)，\(\displaystyle f\left(x_0\right)=\varphi\left(x_0\right)=0\)．

    由（1）知，\(\displaystyle f'\left(x\right)\)在区间\(\displaystyle \left(1,+\infty\right)\)上单调递增，

    故当\(\displaystyle x\in\left(1,x_0\right)\)时，\(\displaystyle f'\left(x\right)<0\)，从而\(\displaystyle f\left(x\right)>f\left(x_0\right)=0\)；

    当\(\displaystyle x\in\left(x_0,+\infty\right)\)时，\(\displaystyle f'\left(x\right)>0\)，从而\(\displaystyle f\left(x\right)>f\left(x_0\right)=0\)．

    所以，当\(\displaystyle x\in\left(1,+\infty\right)\)时，\(\displaystyle f\left(x\right)\geqslant0\)．

    综上所述，存在\(\displaystyle a\in\left(0,1\right)\)，使得\(\displaystyle f\left(x\right)\geqslant0\)在区间\(\displaystyle \left(1,+\infty\right)\)内恒成立，且\(\displaystyle f\left(x\right)=0\)在区间\(\displaystyle \left(1,+\infty\right)\)内有唯一解．

    【解题思路】当\(\displaystyle x\in\left(1,+\infty\right)\)且\(\displaystyle a\in\left(0,1\right)\)时，可以证明存在\(\displaystyle x_0\)，使得\(\displaystyle f'\left(x_0\right)=0\)，故\(\displaystyle f\left(x\right)\)在\(\displaystyle \left(1,x_0\right)\)递减，在\(\displaystyle \left(x_0,+\infty\right)\)递增．所以只需要保证\(\displaystyle f\left(x_0\right)=0\)即可得欲证结论．首先根据\(\displaystyle f'\left(x_0\right)=0\)得到关系式\(\displaystyle a=\frac{x_0-1-\ln x_0}{1+x_0^{-1}}\)（①），把此关系式代入到\(\displaystyle f\left(x_0\right)=0\)消去\(\displaystyle a\)，得到一个关于\(\displaystyle x_0\)的方程②．我们需要证明方程②有解，并把这个解代入到①，证明由①定义出来的\(\displaystyle a\in\left(0,1\right)\)．

    \par
 新答案（来源：081-100放缩方法与不等式证明.md）：
      1. \(\displaystyle f'\left(x\right)=\frac{x^2-bx+1}{x\left(x+1\right)^2}\left(x>1\right)\)，\(\displaystyle x>1\) 时 \(\displaystyle h\left(x\right)=\frac{1}{x\left(x+1\right)^2}>0\)，因此函数 \(\displaystyle f\left(x\right)\) 具有性质 \(\displaystyle P\left(b\right)\).

      2. \(\displaystyle b\leqslant2\) 时，\(\displaystyle x>1\Rightarrow x^2-bx+1\geqslant x^2-2x+1=\left(x-1\right)^2>0\)，即 \(\displaystyle f'\left(x\right)>0\)，\(\displaystyle f\left(x\right)\) 在 \(\displaystyle \left(1,+\infty\right)\) 上单调递增；

        \(\displaystyle b>2\) 时，对 \(\displaystyle f\left(x\right)\)，\(\displaystyle f'\left(x\right)\) 列表如下：

        \(\displaystyle x\) \(\displaystyle \left(1,\frac{b+\sqrt{b^2-4}}{2}\right)\) \(\displaystyle \frac{b+\sqrt{b^2-4}}{2}\) \(\displaystyle \left(\frac{b+\sqrt{b^2-4}}{2},+\infty\right)\)
        \(\displaystyle f'\left(x\right)\) \(\displaystyle -\) \(\displaystyle 0\) \(\displaystyle +\)
        \(\displaystyle f\left(x\right)\) \(\displaystyle \searrow\) 极小值 \(\displaystyle \nearrow\)

        由表可知 \(\displaystyle f\left(x\right)\) 在 \(\displaystyle \left(1,\frac{b+\sqrt{b^2-4}}{2}\right)\) 上单调递减，在 \(\displaystyle \left(\frac{b+\sqrt{b^2-4}}{2},+\infty\right)\) 上单调递增.

    2. 问其考察的重点实际上是头脑是否清醒，对 \(\displaystyle m\) 进行分类之后放缩一下即可讨论出结果，至于依据什么来分类，题干本身已经提示得非常明显了：\(\displaystyle m\leqslant0\)，\(\displaystyle 0<m<1\)，\(\displaystyle m>1\).

      依题意，\(\displaystyle g'\left(x\right)=h\left(x\right)\left(x^2-2x+1\right)=h\left(x\right)\left(x-1\right)^2>0\left(x>1\right)\)，

      因此 \(\displaystyle g\left(x\right)\) 在 \(\displaystyle \left(1,+\infty\right)\) 上单调递增.

      \(\displaystyle m\leqslant0\) 时，

      \[\displaystyle \alpha=mx_1+\left(1-m\right)x_2\geqslant mx_2+\left(1-m\right)x_2=x_2\]
      \[\displaystyle \beta=\left(1-m\right)x_1+mx_2\leqslant\left(1-m\right)x_1+mx_1=x_1\]

      于是

      \[\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right| =g\left(\alpha\right)-g\left(\beta\right) \geqslant g\left(x_2\right)-g\left(x_1\right) =\left|g\left(x_1\right)-g\left(x_2\right)\right|,\]

      不满足题意；

      \(\displaystyle 0<m<1\) 时，

      \[\displaystyle \alpha=mx_1+\left(1-m\right)x_2<mx_2+\left(1-m\right)x_2=x_2\]
      \[\displaystyle \alpha=mx_1+\left(1-m\right)x_2>mx_1+\left(1-m\right)x_1=x_1\]

      即 \(\displaystyle \alpha\in\left(x_1,x_2\right)\)，同理 \(\displaystyle \beta\in\left(x_1,x_2\right)\).

      由 \(\displaystyle g\left(x\right)\) 单调性可知 \(\displaystyle g\left(\alpha\right)\in\left(g\left(x_1\right),g\left(x_2\right)\right)\)，\(\displaystyle g\left(\beta\right)\in\left(g\left(x_1\right),g\left(x_2\right)\right)\)，因此

      \[\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right|<\left|g\left(x_1\right)-g\left(x_2\right)\right|,\]

      满足题意；

      \(\displaystyle m\geqslant1\) 时，

      \[\displaystyle \alpha=mx_1+\left(1-m\right)x_2\leqslant mx_1+\left(1-m\right)x_1=x_1\]
      \[\displaystyle \beta=\left(1-m\right)x_1+mx_2\geqslant\left(1-m\right)x_2+mx_2=x_2\]

      于是

      \[\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right| =g\left(\beta\right)-g\left(\alpha\right) \geqslant g\left(x_2\right)-g\left(x_1\right) =\left|g\left(x_1\right)-g\left(x_2\right)\right|,\]

      不满足题意；

      综上，\(\displaystyle m\in\left(0,1\right)\).

    1. 【2020上海16】若存在 \(\displaystyle a \in \mathbb{R}\) 且 \(\displaystyle a \neq 0\),对于任意的 \(\displaystyle x\),均有 \(\displaystyle f(x + a) < f(x) + f(a)\) 恒成立,则称函数 \(\displaystyle f(x)\) 具有性质 \(\displaystyle P\),已知 $\(\displaystyle q_1: f(x) \text{单调递减且} f(x) > 0 \text{恒成立}\)$ $\(\displaystyle q_2: f(x)\text{单调递增,存在} x_0 < 0\text{ 使得} f(x_0) = 0\)$

    2. 证明或证伪：\(\displaystyle q_1\)是“\(\displaystyle f(x)\) 具有性质 \(\displaystyle P\)”的充分条件；

    3. 证明或证伪：\(\displaystyle q_2\)是“\(\displaystyle f(x)\) 具有性质 \(\displaystyle P\)”的充分条件.
    4. 【2010江苏20】 设 \(\displaystyle f(x)\) 是定义在区间 \(\displaystyle (1,+\infty)\) 上的函数, 其导函数为 \(\displaystyle f'(x)\). 如果存在实数 \(\displaystyle a\) 和函数 \(\displaystyle h(x)\), 其中 \(\displaystyle h(x)\) 对任意的 \(\displaystyle x \in (1,+\infty)\) 都有 \(\displaystyle h(x)>0\), 使得 \(\displaystyle f'(x)=h(x)(x^2-ax+1)\), 则称函数 \(\displaystyle f(x)\) 具有性质 \(\displaystyle P(a)\).已知函数 \(\displaystyle g(x)\) 具有性质 \(\displaystyle P(2)\), 给定 \(\displaystyle x_1,x_2 \in (1,+\infty)\), \(\displaystyle x_1<x_2\), 设 \(\displaystyle m\) 为实数. \(\displaystyle \alpha=mx_1+(1-m)x_2\), \(\displaystyle \beta=(1-m)x_1+mx_2\), 且 \(\displaystyle \alpha>1\), \(\displaystyle \beta>1\), 若 \(\displaystyle |g(\alpha)-g(\beta)| < |g(x_1)-g(x_2)|\), 求 \(\displaystyle m\) 的取值范围.
    答案

    由题设知，\(\displaystyle g'\left(x\right)=h\left(x\right)\left(x^2-2x+1\right)\)，其中函数\(\displaystyle h\left(x\right)>0\)对于任意的\(\displaystyle x\in\left(1,+\infty\right)\)都成立．所以，当\(\displaystyle x>1\)时，\(\displaystyle g'\left(x\right)=h\left(x\right)\left(x-1\right)^2>0\)，从而\(\displaystyle g\left(x\right)\)在区间\(\displaystyle \left(1,+\infty\right)\)上单调递增．

    ①当\(\displaystyle m\in\left(0,1\right)\)时，有 $\(\displaystyle \alpha=mx_1+\left(1-m\right)x_2>mx_1+\left(1-m\right)x_1=x_1,\)$ $\(\displaystyle \alpha=mx_1+\left(1-m\right)x_2<mx_2+\left(1-m\right)x_2=x_2,\)$ 得\(\displaystyle \alpha\in\left(x_1,x_2\right)\)．同理可得\(\displaystyle \beta\in\left(x_1,x_2\right)\)．所以由\(\displaystyle g\left(x\right)\)单调性知，\(\displaystyle g\left(\alpha\right)\)，\(\displaystyle g\left(\beta\right)\in\left(g\left(x_1\right),g\left(x_2\right)\right)\)，从而有 $\(\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right|<\left|g\left(x_1\right)-g\left(x_2\right)\right|，\)$ 符合题设．

    ②当\(\displaystyle m\leqslant0\)时， $\(\displaystyle \alpha=mx_1+\left(1-m\right)x_2\geqslant mx_2+\left(1-m\right)x_1=x_2,\)$ $\(\displaystyle \beta=\left(1-m\right)x_1+mx_2\leqslant\left(1-m\right)x_1+mx_1=x_1,\)$ 于是由\(\displaystyle \alpha>1\)，\(\displaystyle \beta>1\)以及\(\displaystyle g\left(x\right)\)单调性， $\(\displaystyle g\left(\beta\right)\leqslant g\left(x_1\right)<g\left(x_2\right)\leqslant g\left(\alpha\right),\)$ 所以\(\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right|\geqslant\left|g\left(x_1\right)-g\left(x_2\right)\right|\)，与题设不符．

    ③当\(\displaystyle m\geqslant1\)时，同理可得\(\displaystyle \alpha\leqslant x_1\)，\(\displaystyle \beta\geqslant x_2\)，进而得 $\(\displaystyle \left|g\left(\alpha\right)-g\left(\beta\right)\right|\geqslant\left|g\left(x_1\right)-g\left(x_2\right)\right|,\)$ 与题设不符．

    因此，综合①、②、③得所求的\(\displaystyle m\)的取值范围是\(\displaystyle \left(0,1\right)\)．

    1. 【2026“漫游数海”（网络联考）预测卷17】设函数 \(\displaystyle f(x)\) 及其导函数 \(\displaystyle f'(x)\) 的定义域为 \(\displaystyle \mathbb{R}\),\(\displaystyle f(0) \neq 0\),且对任意 \(\displaystyle x_1, x_2 \in \mathbb{R}\),都有 \(\displaystyle f(2x_1) + f(2x_2) = f(x_1+x_2)f(x_1-x_2)\).

    2. 证明：\(\displaystyle f(x)\) 是偶函数；

    3. 若存在 \(\displaystyle a \in \mathbb{R}\),使 \(\displaystyle f(a) = 0\).证明\(\displaystyle f'(x)\) 是周期函数,并写出\(\displaystyle y=f'(x)\) 图象的2条对称轴.

    4. 【2003上海理22】已知集合 \(\displaystyle M\) 是满足下列性质的函数 \(\displaystyle f(x)\) 的全体: 存在非零常数 \(\displaystyle T\), 对任意 \(\displaystyle x \in \vv{R}\), 有 \(\displaystyle f(x+T) = Tf(x)\) 成立.

    5. 设函数 \(\displaystyle f(x) = a^x\) (\(\displaystyle a>0\), 且 \(\displaystyle a \neq 1\)) 的图象与 \(\displaystyle y = x\) 的图象有公共点, 证明: \(\displaystyle f(x) = a^x \in M\)；

    6. 若函数 \(\displaystyle f(x) = \sin kx \in M\), 求实数 \(\displaystyle k\) 的取值范围.
    答案

    （1）

        （2）由题意,若$\displaystyle f(x) = \sin kx \in M$,则存在非零常数$\displaystyle T$,对任意$\displaystyle x\in\mathbb{R}$有$\displaystyle \sin(kx + kT) = T \sin kx$
    ,猜测$\displaystyle T = \pm 1$,下证明：

    若 \(\displaystyle |T| < 1\),可知对任意\(\displaystyle T\),存在\(\displaystyle x_0\)使得\(\displaystyle kx_0+kT=\pi/2\),此时\(\displaystyle |T \sin kx_0| \leqslant |T| < 1=\sin(kx_0+kT)\),不符题意；若 \(\displaystyle |T| > 1\),可知,存在 \(\displaystyle x_0\) 满足 \(\displaystyle kx_0 = \pi/2\),此时\(\displaystyle x_0 = \pi/(2k)\),有 $\(\displaystyle |\sin(kx_0 + kT)| = \left| \sin\left(\frac{\pi}{2} + kT\right) \right| \leqslant 1 < |T \sin kx_0| = |T|\)$不符题意；

    当 \(\displaystyle T = 1\)时,等式化为\(\displaystyle \sin(kx+k) = \sin kx\),这对任意\(\displaystyle x\in\mathbb{R}\)成立,可得\(\displaystyle k = 2p\pi,p \in \mathbb{Z}\).同理,当\(\displaystyle T=-1\)时,\(\displaystyle k = (2p+1)\pi,p \in \mathbb{Z}\)

    综上\(\displaystyle k = p\pi, p \in \mathbb{Z}\)

    1. 【2005 上海 21】 对定义域分别是 \(\displaystyle D_f\)、 \(\displaystyle D_g\) 的函数 \(\displaystyle y = f(x)\)、 \(\displaystyle y = g(x)\), 规定：函数 $\(\displaystyle h(x) = \begin{cases} f(x) \cdot g(x) & \text{当 } x \in D_f \text{ 且 } x \in D_g \\ f(x) & \text{当 } x \in D_f \text{ 且 } x \notin D_g \\ g(x) & \text{当 } x \notin D_f \text{ 且 } x \in D_g \end{cases}\)$

    2. 若函数 \(\displaystyle f(x) = \frac{1}{x - 1}\), \(\displaystyle g(x) = x^2\),写出函数 \(\displaystyle h(x)\) 的解析式；

    3. 求问题 （1） 中函数 \(\displaystyle h(x)\) 的值域；
    4. 若 \(\displaystyle g(x) = f(x + \alpha)\),其中 \(\displaystyle \alpha\) 是常数,且 \(\displaystyle \alpha \in [0, \pi]\),请设计一个定义域为 \(\displaystyle \mathbb{R}\) 的函数 \(\displaystyle y = f(x)\),及一个 \(\displaystyle \alpha\) 的值,使得 \(\displaystyle h(x) = \cos 4x\),并予以证明.
    答案

    \par 新答案（来源：201-220零点定理与特殊点.md）： 1. \(\displaystyle x=x^2\left|x-2\right|\)，\(\displaystyle x=0\) 显然是该方程的解.

    若 $\displaystyle x\geqslant2$，$\displaystyle x\left(x-2\right)=1$，解得 $\displaystyle x=1+\sqrt{2}$.

若 $\displaystyle x\ne0$ 且 $\displaystyle x<2$，$\displaystyle x\left(2-x\right)=1$，解得 $\displaystyle x=1$.

综上，满足 $\displaystyle x=f\left(x\right)$ 的 $\displaystyle x$ 的集合为 $\displaystyle \left\{0,1,\sqrt{2}+1\right\}$.

    1. \(\displaystyle \left(2\right)\) 问考虑到 \(\displaystyle a-2\) 的符号问题，分成 \(\displaystyle a<1\)，\(\displaystyle 1\leqslant a\leqslant2\)，\(\displaystyle a>2\) 三种情形讨论：

      方便起见，以下只考虑 \(\displaystyle \left[1,2\right]\) 区间，记 \(\displaystyle f\left(x\right)\) 在 \(\displaystyle \left[1,2\right]\) 上的最小值为 \(\displaystyle f\left(x\right)_{\min}\).

      ① \(\displaystyle a<1\).

      \(\displaystyle f\left(x\right)=x^2\left(x-a\right)\)，\(\displaystyle f'\left(x\right)=3x^2-2ax=x\left(3x-2a\right)>0\)，\(\displaystyle f\left(x\right)\) 单调递增，

      \(\displaystyle f\left(x\right)_{\min}=f\left(1\right)=1-a\).

      ② \(\displaystyle 1\leqslant a\leqslant2\).

      \(\displaystyle f\left(x\right)\geqslant0\) 且 \(\displaystyle f\left(a\right)=0\)，\(\displaystyle f\left(x\right)_{\min}=0\).

      ③ \(\displaystyle a>2\).

      \(\displaystyle f\left(x\right)=x^2\left(a-x\right)\)，\(\displaystyle f'\left(x\right)=x\left(2a-3x\right)\).

      若 \(\displaystyle 2<a<3\)：

      \(\displaystyle 1<x<\frac{2}{3}a\) 时 \(\displaystyle f'\left(x\right)>0\)，\(\displaystyle f\left(x\right)\) 单调递增，\(\displaystyle \frac{2}{3}a<x<2\) 时 \(\displaystyle f'\left(x\right)<0\)，\(\displaystyle f\left(x\right)\) 单调递减，

      \(\displaystyle f\left(x\right)_{\min}=\min\left\{f\left(1\right),f\left(2\right)\right\}=\min\left\{a-1,4a-8\right\}\).

      即 \(\displaystyle 2<a\leqslant\frac{7}{3}\) 时，\(\displaystyle f\left(x\right)_{\min}=4a-8\)；\(\displaystyle \frac{7}{3}<a<3\) 时，\(\displaystyle f\left(x\right)_{\min}=a-1\).

      若 \(\displaystyle a\geqslant3\)：

      \(\displaystyle f'\left(x\right)\geqslant0\)，\(\displaystyle f\left(x\right)\) 单调递增，\(\displaystyle f\left(x\right)_{\min}=f\left(1\right)=a-1\).

      综上，

      \[\displaystyle f\left(x\right)_{\min}= \begin{cases} 1-a,&a<1,\\ 0,&1\leqslant a\leqslant2,\\ 4a-8,&2<a<\frac{7}{3},\\ a-1,&a\geqslant\frac{7}{3}. \end{cases}\]

      \par 新答案（来源：3.1 性质的证明（1）函数.md）：

    （1）易见\(\displaystyle h\left(x\right)=x+\sin\frac{x}{3}\)的定义域为\(\displaystyle \mathbb{R}\)，

    对任意\(\displaystyle x\in\mathbb{R}\)，\(\displaystyle h\left(x+6\pi\right)=x+6\pi+\sin\frac{x+6\pi}{3}=h\left(x\right)+6\pi\)，

    所以\(\displaystyle \cos h\left(x+6\pi\right)=\cos\left(h\left(x\right)+6\pi\right)=\cos h\left(x\right)\)，

    即\(\displaystyle h\left(x\right)\)是以\(\displaystyle 6\pi\)为余弦周期的余弦周期函数．

    （2）由于\(\displaystyle f\left(x\right)\)值域是\(\displaystyle \mathbb{R}\)，所以对任意\(\displaystyle c\in\left[f\left(a\right),f\left(b\right)\right]\)，\(\displaystyle c\)都是一个函数值，即有\(\displaystyle x_0\in\mathbb{R}\)，使得\(\displaystyle f\left(x_0\right)=c\)．

    若\(\displaystyle x_0<a\)，则由\(\displaystyle f\left(x\right)\)单调递增得到\(\displaystyle c=f\left(x_0\right)<f\left(a\right)\)，与\(\displaystyle c\in\left[f\left(a\right),f\left(b\right)\right]\)矛盾，因此\(\displaystyle x_0\geqslant a\)．同理可证\(\displaystyle x_0\leqslant b\)．故存在\(\displaystyle x_0\in\left[a,b\right]\)使得\(\displaystyle f\left(x_0\right)=c\)．

    （3）若\(\displaystyle u_0\)是\(\displaystyle \cos f\left(x\right)=1\)在\(\displaystyle \left[0,T\right]\)的解，则\(\displaystyle \cos f\left(u_0\right)=1\)，且\(\displaystyle u_0+T\in\left[T,2T\right]\)，\(\displaystyle \cos f\left(u_0+T\right)=\cos f\left(u_0\right)=1\)，即\(\displaystyle u_0+T\)为方程\(\displaystyle \cos f\left(x\right)=1\)在\(\displaystyle \left[T,2T\right]\)上的解．

    同理，若\(\displaystyle u_0+T\)是方程\(\displaystyle \cos f\left(x\right)=1\)在\(\displaystyle \left[T,2T\right]\)上的解，则\(\displaystyle u_0\)是方程在\(\displaystyle \left[0,T\right]\)上的解．

    以下证明最后一部分结论．

    由（2）所证可知存在\(\displaystyle 0=x_0<x_1<x_2<x_3<x_4=T\)，使得\(\displaystyle f\left(x_i\right)=i\pi\)，\(\displaystyle i=0,1,2,3,4\)．而\(\displaystyle \left[x_i,x_{i+1}\right]\)是函数\(\displaystyle \cos f\left(x\right)\)的单调区间，\(\displaystyle i=0,1,2,3\)．

    与之前类似地可以证明：\(\displaystyle u_0\)是\(\displaystyle \cos f\left(x\right)=-1\)在\(\displaystyle \left[0,T\right]\)上的解当且仅当\(\displaystyle u_0+T\)是\(\displaystyle \cos f\left(x\right)=-1\)在\(\displaystyle \left[T,2T\right]\)上的解．从而\(\displaystyle \cos f\left(x\right)=\pm1\)在\(\displaystyle \left[0,T\right]\)与\(\displaystyle \left[T,2T\right]\)上的解个数相同．

    故\(\displaystyle f\left(x_i+T\right)=f\left(x_i\right)+4\pi\)，\(\displaystyle i=0,1,2,3,4\)．

    对于\(\displaystyle x\in\left[0,x_1\right]\)，\(\displaystyle f\left(x\right)\in\left[0,\pi\right]\)，\(\displaystyle f\left(x+T\right)\in\left[4\pi,5\pi\right]\)，

    而\(\displaystyle \cos f\left(x+T\right)=\cos f\left(x\right)\)，故\(\displaystyle f\left(x+T\right)=f\left(x\right)+4\pi=f\left(x\right)+f\left(T\right)\)．

    类似地，当\(\displaystyle x\in\left[x_i,x_{i+1}\right]\)，\(\displaystyle i=1,2,3\)时，有\(\displaystyle f\left(x+T\right)=f\left(x\right)+f\left(T\right)\)．

    结论成立．

    【实测数据】本题第（1）问难度为\(\displaystyle 0.571\)，区分度为\(\displaystyle 0.61\)；第（2）问难度为\(\displaystyle 0.182\)，区分度为\(\displaystyle 0.55\)；第（3）问难度为\(\displaystyle 0.076\)，区分度为\(\displaystyle 0.55\)．

    【易错警示】（1）不理解新定义；

    （2）本题的证明分两个环节：\(\displaystyle x_0\)的存在性以及其所属的区间．在作答的考生中，失分的主要原因是忽略了存在性，直接证明\(\displaystyle x_0\in\left[a,b\right]\)，也有一些考生说理不够清楚．

    （3）全体得分率很低，第一层次考生明显好于后面层次考生．本题需证明两个结论，虽然前一个结论的证明并不难，但不少考生整个小题都没有作答．后一个结论的证明有难度，在作答的考生中，没能突破难点的原因在于没有充分利用前面的结论、从中找到解题策略和思路，有不少考生想到从\(\displaystyle \cos f\left(x+T\right)=\cos f\left(x\right)\)着手得到\(\displaystyle f\left(x+T\right)=2k\pi\pm f\left(x\right)\)但可惜的是想当然地把其中的\(\displaystyle k\)认为是常数，反映出对函数概念的理解不够深刻．

    1. 【2025上海春21】已知函数\(\displaystyle y=f(x)\)的定义域是\(\displaystyle D\),对于\(\displaystyle t\in D\),定义集合\(\displaystyle S_{f(t)}=\{x\mid f(x)\geqslant f(t)\}\).

    （1）已知\(\displaystyle f(x)=\log_2 x\),求\(\displaystyle S_{f(16)}\)；

    （2）对于集合\(\displaystyle A\),若对任意\(\displaystyle x\in A\)都有\(\displaystyle -x\in A\),则称\(\displaystyle A\)是对称集.若\(\displaystyle D\)是对称集,证明：“函数\(\displaystyle y=f(x)\)是偶函数”的充要条件是“对任意\(\displaystyle t\in D\),是\(\displaystyle S_{f(t)}\)对称集”；

    （3）若\(\displaystyle x\in\mathbb{R},f(x)=\mathrm{e}^x-\frac{1}{2}mx^2\),求\(\displaystyle m\)的取值范围,使得对于任意\(\displaystyle t_1<t_2\in D\),都有\(\displaystyle S_{f(t_2)}\subseteq S_{f(t_1)}\).

    答案

    （1）\(\displaystyle \{x|x\geqslant 16\}\)

        （2） 充分性：任取 $\displaystyle x \in S_{f(t)}$,因为 $\displaystyle f(-x) = f(x) \geqslant f(t)$,所以 $\displaystyle -x \in S_{f(t)}$,从而$\displaystyle S_{f(t)}$是对称集.

    必要性： 充分性：设对任意 \(\displaystyle t\in D\),\(\displaystyle S_{f\left(t\right)}\) 是对称集.下证 \(\displaystyle f\left(x\right)\) 是偶函数.

    对任意 \(\displaystyle x\in D\),因为 \(\displaystyle f\left(x\right)\geqslant f\left(x\right)\),所以 \(\displaystyle x\in S_{f\left(x\right)}\).\(\displaystyle \left(*\right)\)

    而 \(\displaystyle S_{f\left(x\right)}\) 是对称集,所以 \(\displaystyle -x\in S_{f\left(x\right)}\),从而 \(\displaystyle f\left(-x\right)\geqslant f\left(x\right)\)①.

    因为 \(\displaystyle f\left(-x\right)\geqslant f\left(-x\right)\),所以 \(\displaystyle -x\in S_{f\left(-x\right)}\).而 \(\displaystyle S_{f\left(-x\right)}\) 是对称集,所以 \(\displaystyle x\in S_{f\left(-x\right)}\),从而 \(\displaystyle f\left(x\right)\geqslant f\left(-x\right)\)②.

    由①②可得 \(\displaystyle f\left(-x\right)=f\left(x\right)\).所以 \(\displaystyle f\left(x\right)\) 是偶函数.

    或使用反证法：假设 \(\displaystyle f(x)\) 不是偶函数,即存在 \(\displaystyle x_0\) 使得 \(\displaystyle f(x_0) \neq f(-x_0)\).

    若 \(\displaystyle f(x_0) > f(-x_0)\),取 \(\displaystyle t = x_0\),有 \(\displaystyle x_0 \in S_{f(t)}\).但 \(\displaystyle f(-x_0) < f(x_0) = f(t)\),可得 \(\displaystyle -x_0 \notin S_{f(t)}\),这与 \(\displaystyle S_{f(t)}\) 是对称集矛盾.

    若 \(\displaystyle f(x_0) < f(-x_0)\),取 \(\displaystyle t = -x_0\),有 \(\displaystyle -x_0 \in S_{f(t)}\),但 \(\displaystyle f(x_0) < f(-x_0) = f(t)\),可得 \(\displaystyle x_0=-(-x_0) \notin S_{f(t)}\),这与 \(\displaystyle S_{f(t)}\) 是对称集矛盾.

    故 \(\displaystyle f(x)\) 必为偶函数.

    （3）\(\displaystyle S_{f(t_2)} \subseteq S_{f(t_1)}\)等价于\(\displaystyle \{x \mid f(x) \geqslant f(t_2) \} \subseteq \{x \mid f(x) \geqslant f(t_1) \}\),即对任意\(\displaystyle x\),若\(\displaystyle f(x)\geqslant f(t_2)\),则\(\displaystyle f(x)\geqslant f(t_1)\),当\(\displaystyle f(t_2) \geqslant f(t_1)\)时显然满足上述条件.当\(\displaystyle f(t_2)<f(t_1)\)时,由于\(\displaystyle f(x)\)是连续函数,故存在\(\displaystyle x_0\)使得\(\displaystyle f(x)=(f(t_1)+f(t_2))/2\),此时\(\displaystyle f(x)> f(t_2),f(x)< f(t_1)\),与上述条件矛盾,于是\(\displaystyle S_{f(t_2)}\subseteq S_{f(t_1)}\)当且仅当\(\displaystyle f(t_2)\geqslant f(t_1)\),这对任意的\(\displaystyle t_1<t_2\in D\)成立,于是\(\displaystyle f(x)\)在定义域上单调递增,据此可以求出\(\displaystyle m \in [0, \mathrm{e}]\).

    1. 【2021上海21】已知 \(\displaystyle x_1, x_2 \in \mathbb{R}\),若对任意 \(\displaystyle x_2 - x_1 \in S\),\(\displaystyle f(x_2) - f(x_1) \in S\),则称\(\displaystyle f(x)\)在 \(\displaystyle S\) 上关联．
    2. 判断\(\displaystyle f(x) = 2x - 1\) 是否在 \(\displaystyle [0, +\infty)\) 上关联？是否在 \(\displaystyle [0, 1]\) 上关联？
    3. 若 \(\displaystyle f(x)\) 在 \(\displaystyle \{3\}\) 上关联,且当 \(\displaystyle x \in [0, 3)\) 时,\(\displaystyle f(x) = x^2 - 2x\),求解不等式：\(\displaystyle 2 \leqslant f(x) \leqslant 3\)．
    4. 证明：“\(\displaystyle f(x)\) 在 \(\displaystyle \{1\}\) 上关联且在 \(\displaystyle [0, +\infty)\) 上关联”当且仅当 “\(\displaystyle f(x)\) 在 \(\displaystyle [1, 2]\) 上关联”．
    答案

    （1） ① 是.当 \(\displaystyle x_1-x_2\geqslant0\) 时,

    $\(\displaystyle f\left(x_1\right)-f\left(x_2\right)=\left(2x_1-1\right)-\left(2x_2-1\right)=2\left(x_1-x_2\right)\geqslant0.\)$

    所以 \(\displaystyle f\left(x\right)=2x-1\) 是 \(\displaystyle \left[0,+\infty\right)\) 关联的.

    ② 不是.当 \(\displaystyle x_1=1\),\(\displaystyle x_2=0\) 时,\(\displaystyle x_1-x_2=1\in\left[0,1\right]\),但是 \(\displaystyle f\left(x_1\right)-f\left(x_2\right)=2\notin\left[0,1\right]\),所以 \(\displaystyle f\left(x\right)=2x-1\) 不是 \(\displaystyle \left[0,1\right]\) 关联的.

    （2）当 \(\displaystyle x\in\left[0,3\right)\) 时,由二次函数性质可知 \(\displaystyle f\left(x\right)\in\left[-1,3\right)\).由条件,当 \(\displaystyle x_1-x_2=3\) 时,\(\displaystyle f\left(x_1\right)-f\left(x_2\right)=3\).

    所以当 \(\displaystyle x\in\left[3k,3k+3\right)\)（\(\displaystyle k\in\mathbb{Z}\)）时,\(\displaystyle f\left(x\right)\in\left[-1+3k,3+3k\right)\).

    因此当 \(\displaystyle x\geqslant6\) 时,\(\displaystyle f\left(x\right)\geqslant5>3\),不等式 \(\displaystyle 2\leqslant f\left(x\right)\leqslant3\) 无解；

    当 \(\displaystyle x<0\) 时,\(\displaystyle f\left(x\right)<0<2\),不等式 \(\displaystyle 2\leqslant f\left(x\right)\leqslant3\) 无解.

    当 \(\displaystyle x\in\left[0,3\right)\) 时,不等式 \(\displaystyle 2\leqslant f\left(x\right)\leqslant3\) 的解集为 \(\displaystyle \left[1+\sqrt3,3\right)\)；

    当 \(\displaystyle x\in\left[3,6\right)\) 时,\(\displaystyle f\left(x\right)=f\left(x-3\right)+3=\left(x-3\right)^2-2\left(x-3\right)+3=x^2-8x+18\),

    所以不等式 \(\displaystyle 2\leqslant f\left(x\right)\leqslant3\) 的解集为 \(\displaystyle \left[3,5\right]\).

    综上,不等式 \(\displaystyle 2\leqslant f\left(x\right)\leqslant3\) 的解集为 \(\displaystyle \left[1+\sqrt3,5\right]\).

    （3）充分性：假设 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left\{1\right\}\) 关联且是 \(\displaystyle \left[0,+\infty\right)\) 关联的.

    当 \(\displaystyle x_1-x_2\in\left[0,+\infty\right)\) 时,\(\displaystyle f\left(x_1\right)-f\left(x_2\right)\in\left[0,+\infty\right)\),

    所以 \(\displaystyle f\left(x\right)\) 关于 \(\displaystyle x\) 单调递增.

    当 \(\displaystyle y-x=1\) 时,\(\displaystyle f\left(y\right)-f\left(x\right)=1\),所以 \(\displaystyle f\left(x\right)\) 满足等式 \(\displaystyle f\left(x+1\right)-f\left(x\right)=1\).

    注意到命题“\(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[1,2\right]\) 关联的”等价于“当 \(\displaystyle y-x\in\left[1,2\right]\) 时,\(\displaystyle f\left(y\right)-f\left(x\right)\in\left[1,2\right]\)”.

    当 \(\displaystyle x_1-x_2\in\left[1,2\right]\) 即 \(\displaystyle 1\leqslant x_1-x_2\leqslant2\) 时,由 \(\displaystyle f\left(x\right)\) 的单调性,

    $\(\displaystyle f\left(x_1\right)-f\left(x_2\right)\leqslant f\left(x_2+2\right)-f\left(x_2\right)\leqslant\left[f\left(x_2+2\right)-f\left(x_2+1\right)\right]+\left[f\left(x_2+1\right)-f\left(x_2\right)\right]=2,\)$

    $\(\displaystyle f\left(x_1\right)-f\left(x_2\right)\geqslant f\left(x_2+1\right)-f\left(x_2\right)=1.\)$

    所以 \(\displaystyle f\left(x_1\right)-f\left(x_2\right)\in\left[1,2\right]\),即 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[1,2\right]\) 关联的,因此充分性成立.

    必要性：假设 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[1,2\right]\) 关联的.则当 \(\displaystyle x_1-x_2\in\left[1,2\right]\) 时,\(\displaystyle f\left(x_1\right)-f\left(x_2\right)\in\left[1,2\right]\),取 \(\displaystyle x_1,x_2,x_3\in\mathbb{R}\) 满足 \(\displaystyle x_1=x_2+1\),\(\displaystyle x_3=x_2-1\),则 \(\displaystyle x_1-x_2=1\in\left[1,2\right]\),\(\displaystyle x_1-x_3=2\in\left[1,2\right]\).因此 \(\displaystyle 1\leqslant f\left(x_1\right)-f\left(x_3\right)\leqslant2\),\(\displaystyle 1\leqslant f\left(x_2\right)-f\left(x_3\right)\leqslant2\),故

    $\(\displaystyle f\left(x_1\right)-f\left(x_2\right)=\left[f\left(x_1\right)-f\left(x_3\right)\right]-\left[f\left(x_2\right)-f\left(x_3\right)\right]\leqslant1.\)$

    另一方面,由于 \(\displaystyle x_1-x_2=1\in\left[1,2\right]\),所以 \(\displaystyle 1\leqslant f\left(x_1\right)-f\left(x_2\right)\leqslant2\),因此必有 \(\displaystyle f\left(x_1\right)-f\left(x_2\right)=1\),所以 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left\{1\right\}\) 关联的.

    下面证明 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[0,+\infty\right)\) 关联的.取 \(\displaystyle x_1,x_2\) 使得 \(\displaystyle x_1-x_2\geqslant0\).则 \(\displaystyle x_1+1-x_2\geqslant1\),构造一列 \(\displaystyle \left\{y_n\right\}\)（\(\displaystyle n\geqslant2\)）使得

    $\(\displaystyle x_2=y_1<y_2<\cdots<y_n=x_1+1,\)$

    其中 \(\displaystyle y_i-y_{i-1}\in\left[1,2\right]\)（\(\displaystyle i=1,2,\cdots,n\)）.则由 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[1,2\right]\) 关联可知

    $\(\displaystyle f\left(x_1+1\right)-f\left(x_2\right)=\sum_{k=1}^{n-1}\left[f\left(y_{k+1}\right)-f\left(y_k\right)\right]\geqslant n-1\geqslant1.\)$

    由 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left\{1\right\}\) 关联的,则 \(\displaystyle f\left(x_1+1\right)=f\left(x_1\right)+1\),所以 \(\displaystyle f\left(x_1\right)-f\left(x_2\right)\geqslant0\),所以 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left[0,+\infty\right)\) 关联的.

    根据上述证明过程,不难得到如下结论：如果 \(\displaystyle f\left(x\right)\) 是 \(\displaystyle \left\{k\right\}\) 关联的,则对于任意正整数 \(\displaystyle n\),\(\displaystyle f\left(x\right)\) 也是 \(\displaystyle \left\{nk\right\}\) 关联的.

    1. 【2004江苏22】已知函数 \(\displaystyle f(x)\) (\(\displaystyle x \in \mathbb{R}\)) 满足下列条件: 对任意的实数 \(\displaystyle x_1\), \(\displaystyle x_2\) 都有 \(\displaystyle \lambda(x_1 - x_2)^2 \leqslant (x_1 - x_2)[f(x_1) - f(x_2)]\) 和 \(\displaystyle |f(x_1) - f(x_2)| \leqslant |x_1 - x_2|\), 其中 \(\displaystyle \lambda\) 是大于 \(\displaystyle 0\) 的常数. 设实数 \(\displaystyle a_0\), \(\displaystyle a\), \(\displaystyle b\) 满足 \(\displaystyle f(a_0) = 0\) 和 \(\displaystyle b = a - \lambda f(a)\)。
    2. 证明: \(\displaystyle \lambda \leqslant 1\), 并且不存在 \(\displaystyle b_0 \neq a_0\), 使得 \(\displaystyle f(b_0) = 0\)；
    3. 证明: \(\displaystyle (b - a_0)^2 \leqslant (1 - \lambda^2)(a - a_0)^2\)；
    4. 证明: \(\displaystyle [f(b)]^2 \leqslant (1 - \lambda^2)[f(a)]^2\)。
    5. 【2015上海理23】对于定义域为 \(\displaystyle \mathbb{R}\) 的函数 \(\displaystyle g(x)\),若存在正常数 \(\displaystyle T\),使得 \(\displaystyle \cos g(x)\) 是以 \(\displaystyle T\) 为周期的函数,则称 \(\displaystyle g(x)\) 为余弦周期函数,且称 \(\displaystyle T\) 为其余弦周期．已知 \(\displaystyle f(x)\) 是以 \(\displaystyle T\) 为余弦周期的余弦周期函数,其值域为 \(\displaystyle \mathbb{R}\)．设 \(\displaystyle f(x)\) 单调递增,\(\displaystyle f(0) = 0, f(T) = 4\pi\)．
    6. 验证 \(\displaystyle g(x) = x + \sin \frac{x}{3}\) 是以 \(\displaystyle 6\pi\) 为周期的余弦周期函数；
    7. 设 \(\displaystyle a < b\),证明对任意 \(\displaystyle c \in [f(a), f(b)]\),存在 \(\displaystyle x_0 \in [a, b]\),使得 \(\displaystyle f(x_0) = c\)；
    8. 证明：“\(\displaystyle u_0\) 为方程 \(\displaystyle \cos f(x) = 1\) 在 \(\displaystyle [0, T]\) 上得解,”的充要条件是“\(\displaystyle u_0 + T\) 为方程 \(\displaystyle \cos f(x) = 1\) 在区间 \(\displaystyle [T, 2T]\) 上的解”,并证明对于任意 \(\displaystyle x \in [0, T]\),都有 \(\displaystyle f(x+T) = f(x) + f(T)\)．
    9. 【2025上海21】已知函数 \(\displaystyle y = f(x)\) 的定义域为 \(\displaystyle \mathbb{R}\)．对于正实数 \(\displaystyle a\),定义集合 \(\displaystyle M_a = \{x \mid f(x+a) = f(x)\}\)．
    10. 若 \(\displaystyle f(x) = \sin x\),判断 \(\displaystyle \frac{\pi}{3}\) 是否是 \(\displaystyle M_\pi\) 中的元素,请说明理由；
    11. 若 \(\displaystyle f(x) = \begin{cases} x+2, & x < 0 \\ \sqrt{x}, & x \geqslant 0 \end{cases}\),\(\displaystyle M_a \neq \varnothing\),求 \(\displaystyle a\) 的取值范围；
    12. 若 \(\displaystyle y = f(x)\) 是偶函数,当 \(\displaystyle x \in (0, 1]\) 时,\(\displaystyle f(x) = 1 - x\),且对任意 \(\displaystyle a \in (0, 2)\),均有 \(\displaystyle M_a \subseteq M_2\)．写出 \(\displaystyle y = f(x), x \in (1, 2)\) 解析式,并证明：对任意实数 \(\displaystyle c\),函数 \(\displaystyle y = f(x) - c\) 在 \(\displaystyle [-3, 3]\) 上至多有 \(\displaystyle 9\) 个零点．

    13. 【2025 “Fiddie”模拟考19】 设函数 \(\displaystyle f\left(x\right)=x^3+ax^2+bx+c\),其中 \(\displaystyle a,b,c\in\mathbb{Z}\).对于 \(\displaystyle n\in\mathbb{N}^{*}\),定义 \(\displaystyle f_1\left(x\right)=f\left(x\right)\),\(\displaystyle f_{n+1}\left(x\right)=f\left(f_n\left(x\right)\right)\).记集合 \(\displaystyle S_f=\left\{x\in\mathbb{Q}\mid \exists n\in\mathbb{N}^{*},f_n\left(x\right)=x\right\}\),函数 \(\displaystyle g\left(x\right)=f\left(x\right)-x\).

    14. 若 \(\displaystyle a=-2\),\(\displaystyle b=4\),\(\displaystyle c=0\),求 \(\displaystyle g\left(x\right)\) 的单调区间与\(\displaystyle S_f\) 的元素个数；

    15. 记 \(\displaystyle x_0=1+\left\lvert a\right\rvert+\left\lvert b\right\rvert+\left\lvert c\right\rvert\).证明：当 \(\displaystyle x>x_0\) 时,\(\displaystyle g\left(x\right)>0\)；当 \(\displaystyle x<-x_0\) 时,\(\displaystyle g\left(x\right)<0\)；
    16. 证明：\(\displaystyle S_f\subseteq\mathbb{Z}\),且 \(\displaystyle S_f\) 只有有限个元素.
    答案

    （1）当 \(\displaystyle a=-2,b=4,c=0\) 时

    \[\displaystyle f(x)=x^{3}-2x^{2}+4x,g(x)=x^{3}-2x^{2}+3x,g'(x)=3x^{2}-4x+3=3(x-\frac{2}{3})^{2}+\frac{5}{3}>0\]

    所以 \(\displaystyle g(x)\) 在 \(\displaystyle (-\infty, +\infty)\) 单调递增.

    因为 \(\displaystyle g(0)=0\),即 \(\displaystyle f_1(0)=f(0)=0\),所以 \(\displaystyle 0\in S_f\).

    当 \(\displaystyle x>0\) 时,\(\displaystyle g(x)>g(0)=0\), 即 \(\displaystyle f(x)>x>0(*)\)

    将\(\displaystyle (*)\)式的 \(\displaystyle x\) 替换为 \(\displaystyle f(x)\),得 \(\displaystyle f(f(x))>f(x)>0\),即 \(\displaystyle f_2(x)>f_1(x)>x>0\),一直下去有 $\(\displaystyle f_n(x)>f_{n-1}(x)>\cdots>f(x)>x>0\)$

    故 \(\displaystyle x\notin S_f\).

    当 \(\displaystyle x<0\) 时,同理可证\(\displaystyle x\notin S_f\)

    综上,\(\displaystyle S_f=\{0\}\),\(\displaystyle S_f\) 的元素个数是 \(\displaystyle 1\).

    （2）!!! tip "证明"

    当 $\displaystyle x>x_{0}=1+|a|+|b|+|c|$ 时,由 $\displaystyle x_{0}\geqslant 1$ 得 $\displaystyle x>1$,所以
$$\displaystyle g(x)=x^{3}+ax^{2}+bx+c-x&>(1+|a|+|b|+|c|)x^{2}-|a|x^{2}-|b|x-|c|-x
&=(|b|+1)(x^2-x)+|c|(x^2-1)>0$$
所以 $\displaystyle g(x)>0$

当 $\displaystyle x<-x_{0}=-(1+|a|+|b|+|c|)\leqslant -1$ 时,

$$\displaystyle g(x)=x^{3}+ax^{2}+bx+c-x&<-(1+|a|+|b|+|c|)x^{2}+|a|x^{2}-|b|x+|c|-x
&=-(|b|+1)(x^2+x)+|c|(1-x^2)<0$$

所以$\displaystyle g(x)<0$

    （3）!!! tip "证明"

    使用反证法,若 $\displaystyle S_f$ 不包含于 $\displaystyle \mathbb{Z}$,则有一个有理数 $\displaystyle r=\frac{p}{q}\in S_f$,其中 $\displaystyle p\neq 0,q\geqslant 2,\frac{p}{q}$ 是既约分数.

因为 $\displaystyle f(\frac{p}{q})=\frac{p^{3}+aqp^{2}+bq^{2}p+cq^{3}}{q^{3}}$,而 $\displaystyle q\mid aqp^{2}+bq^{2}p+cq^{3},q\nmid p$

故 $\displaystyle \frac{p^{3}+aqp^{2}+bq^{2}p+cq^{3}}{q^{3}}$ 是既约分数.

对于既约分数 $\displaystyle \frac{p}{q}$（$\displaystyle p\in \mathbb{Z},p\neq 0,q\in \mathbb{N}^{*},q\geqslant 2$）,记 $\displaystyle R(\frac{p}{q})=\frac{1}{q}$.
所以 $\displaystyle R(f(\frac{p}{q}))=\frac{1}{q^{3}}=[R(\frac{p}{q})]^{3}$.

即对任意 $\displaystyle r\in \mathbb{Q}$,当 $\displaystyle r$ 不是整数时,都有 $\displaystyle R(f(r))=[R(r)]^{3}$.

将 $\displaystyle r$ 替换为 $\displaystyle f_n(\frac{p}{q})$ 可得 $\displaystyle R(f_{n+1}(\frac{p}{q}))=[R(f_n(\frac{p}{q}))]^{3}$.
从而 $\displaystyle R(f_n(\frac{p}{q}))=\frac{1}{q^{3^{n}}}$.

则对任意 $\displaystyle n\in \mathbb{N}^{*},R(f_n(r))<R(r)$,不可能有 $\displaystyle f_n(r)=r$,矛盾.

因此,$\displaystyle S_f\subseteq \mathbb{Z}$.

由第（2）问结论,当 $\displaystyle |x|>x_{0}$ 时,$\displaystyle |f(x)|>|x|>x_{0}$,所以

$$\displaystyle |f_n(x)|>|f_{n-1}(x)|>\cdots >|f_2(x)|>|f_1(x)|>|x|>x_{0}$$

所以当 $\displaystyle |x|>x_{0}$ 时,对任意 $\displaystyle n\in \mathbb{N}^{*}$ 都有 $\displaystyle |f_n(x)|>|x|$,不可能有 $\displaystyle f_n(x)=x$,即 $\displaystyle x\notin S_f$.

因此,$\displaystyle S_f\subseteq (-x_{0}, x_{0})$.

而 $\displaystyle [-x_{0}, x_{0}]\cap \mathbb{Z}$ 只有有限个元素,故 $\displaystyle S_f$ 也只有有限个元素.

    Fiddie评：第（3）问解答中引入 \(\displaystyle R(x)\) 是为了获取一个有理数的“分母部分”

    $\(\displaystyle R(x)=\begin{cases}\frac{1}{q}, & x\in \mathbb{Q}, x=\frac{p}{q}, p, q \text{互质}, q\neq 1 \\ 0, & x\in \mathbb{Z} \text{或} x\notin \mathbb{Q} \end{cases}\)$ 叫做 Riemann 函数.证明 \(\displaystyle S_f\subseteq \mathbb{Z}\) 时,用到的思想与“有理根定理”比较类似.

    本题改编自 GTM241, Exercise 0.6,原题引入了诸多记号,比较复杂.本题设计时为照顾考生,第（1）问尽可能的亲民,同时整题以初等的形式呈现.通过几个小问的铺垫引导考生解决最终的命题：三次函数 \(\displaystyle f(x)\) 的各阶有理不动点必为整数,并且个数有限.通过提升最后一题的分值（原卷为19分,本题仍遵照高考要求定为17分）,希望能够鼓励学生勇于尝试压轴题.在设计过程中,通过解答第（1）问的第二步,可以初步了解通过对数列 \(\displaystyle \{f_n(x)\}\) 进行不等式迭代的解决方法,为第（3）问的解决打下基础.

    第（2）问是一个比较经典的问题,对于多项式函数 \(\displaystyle p(x)\),一定有 \(\displaystyle \lim_{|x|\to +\infty} |p(x)| = +\infty\),即对任意 \(\displaystyle M>0\),存在 \(\displaystyle x_{0}\) 使得当 \(\displaystyle |x|>x_{0}\),有 \(\displaystyle |f(x)|>M\).本题给出了这里的 \(\displaystyle x_{0}\) 的显式表达式,所以解决第（2）问只需用不等式作差和绝对值不等式的知识来证明,不需要求导.

    第（3）问需要观察从有理数出发的迭代数列 \(\displaystyle \{f_n(\frac{p}{q})\}\) 的特征.

C 组习题

C组

    1. 分别写出函数$\displaystyle y=\cos x+\cos(x+\alpha),y=\cos x\cos(x+\alpha),y=\frac{\cos x}{\cos(x+\alpha)}(\alpha \neq k\pi ,k\in\mathbb{Z})$的图象的一个对称中心.
  1. 函数\(\displaystyle y=(x+a)(\cos2x-\sin x)\)的图象是否可能存在对称中心？是否可能存在垂直于\(\displaystyle x\)轴的对称轴？
  2. 【2025集英苑5月（网络联考）14】已知\(\displaystyle a>0\),若函数\(\displaystyle f(x)=\frac{b}{e^x+a}\)与\(\displaystyle g(x)=\ln(\frac{b}{x}-a)\)图像的对称中心重合,则\(\displaystyle b\)的最小值是\(\displaystyle (\triangle)\).
  3. 若函数\(\displaystyle f(x)=x^4+4x^3+ax(a\in\mathbb{R})\)的图象存在与\(\displaystyle x\)轴垂直的对称轴,求\(\displaystyle f(x)\)的最小值.
  4. 【2014辽宁理12】 已知定义在\(\displaystyle \left[0, 1\right]\)上的函数\(\displaystyle f\left(x\right)\)满足：①\(\displaystyle f\left(0\right) = f\left(1\right) = 0;\)②对所有\(\displaystyle x, y \in \left[0, 1\right],\)且\(\displaystyle x \neq y,\)有\(\displaystyle |f\left(x\right) - f\left(y\right)| < \frac{1}{2}|x - y|.\)若对所有\(\displaystyle x, y \in \left[0, 1\right], |f\left(x\right) - f\left(y\right)| < k\)恒成立，求\(\displaystyle k\)的最小值。

杂\(\displaystyle \quad\)题

  1. 定义在\(\displaystyle \mathbb{R}\)上的函数\(\displaystyle y=f(x)\)和\(\displaystyle y=g(x)\)的最小正周期分别是\(\displaystyle T_{1}\)和\(\displaystyle T_{2}\),已知\(\displaystyle y=f(x)+g(x)\)的最小正周期为\(\displaystyle 1\),则下列选项中可能成立的是

    • \(\displaystyle T_{1}=1\),\(\displaystyle T_{2}=2\)
    • \(\displaystyle T_{1}=\dfrac{1}{2},T_{2}=\dfrac{3}{4}\)
    • \(\displaystyle T_{1}=\dfrac{3}{4},T_{2}=\dfrac{5}{4}\)
    • \(\displaystyle T_{1}=\dfrac{3}{2},T_{2}=3\)

    2. 【2025新高考I卷4改编】若点\(\displaystyle (a,0)\)是函数\(\displaystyle y=2\tan (x-\frac{2\pi}{3})\)的图象的一个对称中心,求正数\(\displaystyle a\)的最小值. 3. 【2024新高考II卷6】设函数\(\displaystyle f(x)=a(x-1)^2-1,g(x)=\cos x+2ax\),当\(\displaystyle x\in (-1,1)\)时,曲线\(\displaystyle y=f(x)\)与\(\displaystyle y=g(x)\)恰有一个交点,求\(\displaystyle a\)的值. 4. 【2026“集英苑”5月模拟考8】若 \(\displaystyle f(x) = \frac{1}{x+1} + \frac{a}{x^2 + bx + c}\) 是奇函数,求\(\displaystyle a\). 5. 【2026长沙适应性考试10】（多选）已知函数 \(\displaystyle f(x)\) 的定义域为 \(\displaystyle (-\infty,0)\cup(0,+\infty)\),且 \(\displaystyle f(xy)=\frac{f(x)}{y}+\frac{f(y)}{x}\)．当 \(\displaystyle x>1\) 时,\(\displaystyle f(x)>0\),则

    • \(\displaystyle f(1)=0\)
    • \(\displaystyle f(x)\) 是偶函数
    • 当 \(\displaystyle -1<x<0\) 时,\(\displaystyle f(x)>0\)
    • \(\displaystyle x=1\) 为 \(\displaystyle f(x)\) 的极值点

    6. 【2025浙江Z20联考11】（多选）已知函数\(\displaystyle f(x)\)的定义域为\(\displaystyle \mathbb{R}\),且满足\(\displaystyle f(\cos x)=1-\cos nx\),则下列说法正确的是

    • 若\(\displaystyle n=2\),则函数\(\displaystyle f(x)\)的最大值为\(\displaystyle 2\)
    • 若\(\displaystyle n=3\),则函数\(\displaystyle f(x)\)为奇函数
    • 存在\(\displaystyle n\in\mathbb{Z}\),使得\(\displaystyle f(\sin x)=1-\sin nx\)
    • 若\(\displaystyle f(\sin x)+f(\cos x)=2\),则\(\displaystyle n=4k+2,k\in\mathbb{Z}\)

    7. 已知函数 \(\displaystyle f(x)\) 的定义域为 \(\displaystyle \mathbb{R}\),若存在常数 \(\displaystyle T>0\) 与 \(\displaystyle H\),使得对任意 \(\displaystyle x\in\mathbb{R}\),有 \(\displaystyle f(x+T)=f(x)+H\),则称函数 \(\displaystyle f(x)\) 是广义周期函数,证明或反驳： 1. 若 \(\displaystyle f(x)\) 是广义周期函数,则存在实数 \(\displaystyle k\),使得 \(\displaystyle f(x)-kx\) 是周期函数 2. 若 \(\displaystyle f(x)\) 有两个不同的对称中心,则 \(\displaystyle f(x)\) 是广义周期函数 3. 若 \(\displaystyle f(x)\) 与 \(\displaystyle g(x)\) 都是广义周期函数,则 \(\displaystyle f(x)+g(x)\) 也是广义周期函数 8. （多选）已知\(\displaystyle f(x)\)是定义在\(\displaystyle \mathbb{R}\)上的连续函数,设函数\(\displaystyle g_a(x)=\frac{f(x)-f(a)}{x-a} (a \in \mathbb{R})\),则下列说法正确的是

    （A）若\(\displaystyle f(x)\)在\(\displaystyle \mathbb{R}\)上单调递增,则存在实数\(\displaystyle a\),使得\(\displaystyle g_a(x)\)在\(\displaystyle (a, +\infty)\)上单调递增

    （B）对任意实数\(\displaystyle a\),存在\(\displaystyle \mathbb{R}\)上的单调递减函数\(\displaystyle f(x)\),使得\(\displaystyle g_a(x)\)在\(\displaystyle (a, +\infty)\)上单调递增

    （C）对于任意实数\(\displaystyle a\),若存在实数\(\displaystyle M_1 > 0\),使得\(\displaystyle |f(x)| < M_1\),则存在实数\(\displaystyle M_2 > 0\),使得\(\displaystyle |g_a(x)| < M_2\)

    （D）若函数\(\displaystyle g_a(x)\)满足：当\(\displaystyle x \in (a, +\infty)\)时,\(\displaystyle g_a(x) \geqslant 0\),当\(\displaystyle x \in (-\infty, a)\)时,\(\displaystyle g_a(x) \leqslant 0\),则\(\displaystyle f(a)\)为\(\displaystyle f(x)\)的最小值. 9. 已知函数\(\displaystyle f(x),g(x)\)定义域为\(\displaystyle \mathbb{R}\),若\(\displaystyle f(x+2)-g(1-x)=2,f'(x)=g'(x+1)\),且\(\displaystyle g(x+1)\)为奇函数,求\(\displaystyle \sum_{k=1}^{2021}f(k)g(k)\). 10. 【2023青岛一模16】设函数 \(\displaystyle f(x)\) 是定义在整数集\(\displaystyle \mathbb{Z}\)上的函数,且满足 \(\displaystyle f(0)=1\),\(\displaystyle f(1)=0\),对任意的 \(\displaystyle x, y \in \mathbb{Z}\) 都有 \(\displaystyle f(x+y)+f(x-y)=2f(x)f(y)\),求 \(\displaystyle \frac{f(1^2+2^2+\cdots+2023^2)}{f(1^2)+f(2^2)+\cdots+f(2023^2)}\).

D 组习题