**(2026年新高考1卷)** 已知函数 $f(x)$ 的定义域为 $\mathbf{R}$,且当 $x < 0$ 时, $f(x) = 2^x$.对任意 $x_0 \in \mathbf{R}$,定义集合 $D(x_0) = \{d \in \mathbf{R} \mid f(x_0 + d) > f(x_0)\}$.

（i）证明: $f(0) \geqslant 1$;

（ii）证明: $f(x)$ 在区间 $(0, +\infty)$ 单调递增.

**解.** (1)因为当 $x < 0$ 时, $f(x) = 2^x$;当 $x \geqslant 0$ 时, $f(x) = 1 - x$,则$f\left(-1\right)=f\left(\frac{1}{2}\right)
=\frac{1}{2}$,且
$f(x)$在$(-\infty,0)$上单调递增,
在$[0,+\infty)$上单调递减.

而$D\left( -1 \right) =\left\{ d\in \mathbf{R}\mid f\left( -1+d \right) >f\left( -1 \right) \right\} =\left\{ d\in \mathbf{R}\mid f\left( -1+d \right) >\frac{1}{2} \right\}$,
故$f\left( -1+d \right) >f\left(\frac{1}{2}\right)$
且$f\left( -1+d \right) >f\left( -1 \right)$,
则$-1<-1+d<\frac{1}{2}$,解得$0< d<\frac{3}{2}$.

故$D\left( -1 \right) =\left( 0,\frac{3}{2} \right)$.

(2) **证法一.** 若 $f(x)$ 是$\mathbf{R}$上的奇函数,则$f(-x)=-f(x)$, $f(0)=-f(0)$,故$f(0)=0$.

设$x>0$,则$-x<0$, $f(-x)=2^{-x}=-f(x)$,故$f(x)=-\frac{1}{2^x}$, $x>0$.

故$f(x)=\begin{cases}
-\frac{1}{2^x}, & \mbox{若} x>0,\\
0, & \mbox{若} x=0, \\
2^x, & \mbox{若} x<0.
\end{cases}$

因为$f(x_1) \leqslant f(x_2)$,且 $x_1 x_2 \neq 0$,则$x_1,x_2$均不为$0$.

(a)若$x_1\leqslant x_2<0$,则$D\left( x_2 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_2+d \right) >f\left( x_2 \right) \right\} =\left( 0,-x_2 \right)$; $D\left( x_1 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_1+d \right) >f\left( x_1 \right) \right\} =\left( 0,-x_1 \right)$.再由$x_1\leqslant x_2<0$, $-x_2\leqslant -x_1$可知$D\left( x_2 \right) \subseteq D\left( x_1 \right)$;

(b)若$0< x_1\leqslant x_2$,则$D\left( x_2 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_2+d \right) >f\left( x_2 \right) \right\} =\left( 0,+\infty \right) \cup \left( -\infty ,-x_2 \right]$; $D\left( x_1 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_1+d \right) >f\left( x_1 \right) \right\} =\left( 0,+\infty \right) \cup \left( -\infty ,-x_1 \right]$.再由$x_1\leqslant x_2<0$, $-x_2\leqslant -x_1$可知$D\left( x_2 \right) \subseteq D\left( x_1 \right)$.

(c)若$x_2<0< x_1$,则$D\left( x_2 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_2+d \right) >f\left( x_2 \right) \right\} =\left( 0,-x_2 \right)$; $D\left( x_1 \right) =\left\{ d\in \mathbf{R}\mid f\left( x_1+d \right) >f\left( x_1 \right) \right\} =\left( 0,+\infty \right) \cup \left( -\infty ,-x_1 \right]$.
再由$x_2<0< x_1$, $-x_2>0>-x_1$可知$D\left( x_2 \right) =\left( 0,-x_2 \right) \subseteq \left( 0,+\infty \right) \subseteq D\left( x_1 \right)$.

综上所述, $D\left( x_2 \right) \subseteq D\left( x_1 \right)$.

**证法二.** 任取$d\in D\left( x_2 \right)$,
则当$x_2<0$时, $0< d<-x_2$;当$x_2>0$时, $d>0$或$d\leqslant -x_2$.

当$x_2<0$时,因为$f(x_2)\geqslant f(x_1)$,则$x_1>0$或$x_1\leqslant x_2<0$.
若$x_1>0$,则$d\in(0,-x_2)\subseteq(0,+\infty)\subseteq D\left( x_1 \right)$,因此$d\in D\left( x_1 \right)$;

若$x_1\leqslant x_2<0$,则$d\in(0,-x_2)\subseteq(0,-x_1)\subseteq D\left( x_1 \right)$,因此$d\in D\left( x_1 \right)$;

当$x_2>0$时,因为$f(x_2)\geqslant f(x_1)$,则$0< x_1\leqslant x_2$.利用$d\in ((0,+\infty)\cup (-\infty,-x_2])\subseteq ((0,+\infty)\cup (-\infty,-x_1])$可知$d\in D\left( x_1 \right)$.

综上所述, $D\left( x_2 \right) \subseteq D\left( x_1 \right)$.

(3) (i)利用反证法,假设$f(0)<1$,则存在$\varepsilon\in (0,1)$,使得$-\varepsilon\in (-1,0)$且$f(0)< f(-\varepsilon)<1$,由①可知$D(-\varepsilon)\subseteq D(0)$.再由$(0,\varepsilon)\subseteq D(-\varepsilon)$可知$(0,\varepsilon)\subseteq D(0)$.

由②可知$D(0)\cap (0,1)=\varnothing$,矛盾!故假设不成立, $f(0)\geqslant 1$.

**证法二.** 对任意 $x_0 \in (-2,0)$,取 $d = -\frac{x_0}{2}$.则 $d \in (0,1)$,且 $x_0 + d = \frac{x_0}{2} < 0$.

故 $f(x_0 + d) = 2^{\frac{x_0}{2}} > 2^{x_0} = f(x_0)$,即 $d \in D(x_0)$.

由条件②, $d \in (0,1) \Rightarrow f(d) < f(0)$.而 $f(d) = f\left(-\frac{x_0}{2}\right) = 2^{-\frac{x_0}{2}}$.

故 $2^{-\frac{x_0}{2}} < f(0)$ 对任意 $x_0 \in (-2,0)$ 成立.

故 $f(0) \geqslant 2^0 = 1$.

(ii) **证法一.**
(a)利用反证法,假设存在$x_0\in (0,1)$,使得$f(x_0)>0$.
则存在$\varepsilon_0<0$,使得$f(\varepsilon_0)=2^{\varepsilon_0}< f(x_0)$.由②可知$f(x_0)< f(0)$,则$-x_0\in D(x_0)$.又$f(\varepsilon_0)< f(x_0)$,由①可知$D(x_0)\subseteq D(\varepsilon_0)$,所以$-x_0\in D(\varepsilon_0)$,
则$f(\varepsilon_0-x_0)> f(\varepsilon_0)$,
又$\varepsilon_0-x_0< \varepsilon_0<0$,函数$f(x)$在$(-\infty,0)$上单调递增,所以$f(\varepsilon_0-x_0)< f(\varepsilon_0)$,矛盾!

故对任意$0< x<1$,均有$f(x)\leqslant 0$.

(b)假设存在$m_0\geqslant 1$,使得$f(m_0)>0$.
则存在$\xi<0$,使得$f(\xi)=2^{\xi}< f(m_0)$,
则$m_0-\xi\in D(\xi)$.取$\xi_0=\frac{1}{2}-(m_0-\xi)
=\xi+\frac{1}{2}-m_0<\xi<0$,由函数$f(x)$在$(-\infty,0)$上单调递增可知$f(\xi_0)< f(\xi)$,由①可知$D(\xi)\subseteq D(\xi_0)$,则$m_0-\xi\in D(\xi_0)$,则$f(\xi_0+m_0-\xi)
=f\left(\frac{1}{2}\right)>f(\xi_0)>0$,
与(a)中$f\left(\frac{1}{2}\right)\leqslant 0$矛盾!

故对任意$x>0$,均有$f(x)\leqslant 0$.

假设存在$0< y_1< y_2$,使得$f(y_2)\leqslant f(y_1)$,则$f(y_1-y_2)>0\geqslant f(y_1)$,
由①可知$D(y_1-y_2)\subseteq D(y_1)$.

又$y_2-y_1\in D(y_1-y_2)$,所以$y_2-y_1\in D(y_1)$,与$f(y_2)=f(y_1+(y_2-y_1))\leqslant f(y_1)$, $y_2-y_1\notin D(y_1)$矛盾!
故对任意$y_1,y_2\in (0,+\infty)$,且$y_1< y_2$,均有$f(y_1)< f(y_2)$, $f(x)$ 在区间 $(0, +\infty)$ 单调递增.

**证法二.** 由①可知,若$f(x_1)=f(x_2)$,则$f(x_1) \leqslant f(x_2)$, $D(x_2) \subseteq D(x_1)$;并且有$f(x_2) \leqslant f(x_1)$, $D(x_1) \subseteq D(x_2)$,
于是$D(x_1)=D(x_2)$.

若存在$x_1<0< x_2$,使得$f(x_1)=f(x_2)$,则由②可知$f(x_2)= f(x_1)<1\leqslant f(x_2-x_2)=f(0)$,则$-x_2\in D(x_2)$, $-x_2\in D(x_1)$,又$x_1-x_2< x_1<0$,函数$f(x)$在$(-\infty,0)$上单调递增,故
$f(x_1-x_2)< f(x_1)$, $-x_2\notin D(x_1)$,矛盾!

于是对任意$x>0$,均有$f(x)\leqslant 0$或$f(x)\geqslant 1$.

当$0< x< 1$时,若$f(x)\geqslant 1$,由②可知$1\leqslant f(x)< f(0)$,取$x_1<0< x_2<1$,则$f(x_1)< f(x_2)< f(0)$,
由①可知$D(x_2) \subseteq D(x_1)$.此时$-x_2\in D(x_2)$,则$-x_2\in D(x_1)$.因为$f(x_1)>f(x_1-x_2)$,
故$-x_2\notin D(x_1)$,矛盾!
故当$0< x<1$时, $f(x)\leqslant 0$.

当$x\geqslant 1$时, 若$f(x)\geqslant 1$,取$m_0\geqslant 1$.

(a)若$1\leqslant f(m_0)< f(0)$,则任取$m_1<0$,有$f(m_1)< f(m_0)$,由①可知$D(m_0)\subseteq D(m_1)$,由$f(0)=f(-m_0+m_0)> f(m_0)$可知$-m_0\in D(m_0)$,则$-m_0\in D(m_1)$.而由$f(-m_0+m_1)< f(m_1)$可知$-m_0\notin D(m_1)$,矛盾!

(b)若$f(m_0)\geqslant f(0)$,则任取$m_1<0$,
$f(m_0-m_1+m_1)=f(m_0)\geqslant f(0)> f(m_1)$,
则$m_0-m_1\in D(m_1)$.取$m_2=-m_0+m_1+\varepsilon< m_1<0$,其中$\varepsilon\in (0,1)$,则$f(m_2)< f(m_1)$,由①可知$D(m_1)\subseteq D(m_2)$,于是$m_0-m_1\in D(m_2)$.而由$f(m_0-m_1+m_2)=f(\varepsilon)<0< f(m_2)$可知$m_0-m_1\notin D(m_2)$,矛盾!

因此对任意$x>0$,均有$f(x)\leqslant 0$.

任取$y_1,y_2\in (0,+\infty)$,且$y_1< y_2$,则$y_1-y_2<0$.

因为$f(0)\geqslant 1>f(y_1-y_2)>0\geqslant f(y_1)$,故$y_2-y_1\in D(y_1-y_2)$,由①可知$D(y_1-y_2)\subseteq D(y_1)$,则$y_2-y_1\in D(y_1)$, $f(y_1+(y_2-y_1))=f(y_2)>f(y_1)$.

**溯源1 (2014上海春)**
如果存在非零常数$c$,对于函数$y = f(x)$定义域$\mathbf{R}$上的任意$x$,都有$f(x + c) > f(x)$成立,那么称函数$f(x)$为“$Z$函数”.

(1)求证:若$y = f(x)\ (x \in \mathbf{R})$是单调函数,则它是“$Z$函数”;

(2) 若函数$g(x) = ax^3 + bx^2$是“$Z$函数”,求实数$a, b$满足的条件.

**溯源2 (2025上海春)**
已知函数$y = f(x)$的定义域是$D$,对于$x \in D$,定义集合$S_{f(t)} = \{ x \in D \mid f(x) \geqslant f(t) \}$.

(1)设$f(x) = \log_2 x$,求$S_{f(16)}$;

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

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

**溯源3 (2025上海秋)**
已知函数$y = f(x)$的定义域为$\mathbf{R}$,对于正实数$a$,定义集合$M_a = \{ x \mid f(x + a) = f(x) \}$.

(1) 若$f(x) = \sin x$,判断$\frac{\pi}{3}$是否是$M_{\pi}$中的元素,请说明理由;

(2) 若$f(x) = \begin{cases}
x + 2, & x < 0, \\
\sqrt{x}, & x \geqslant 0, \end{cases}$
$M_a \neq \varnothing$,求$a$的取值范围;

(3) 若$y = f(x)$是偶函数,当$x \in (0, 1]$时, $f(x) = 1 - x$,且对任意$a \in (0, 2)$,均有$M_a \subseteq M_2$.写出$y = f(x)$, $x \in (1, 2)$解析式,并证明:对任意实数$c$,函数$y = f(x) - c$在$[-3, 3]$上至多有9个零点.

**溯源4 (2026年中山一模)** 已知函数$y = f(x)$的定义域为$D$,对于给定实数$t$,定义集合$V_{f(t)}=\{x \mid f(x+t) \geqslant f(x)\}$.

(1) 若$f(x) = x^3 - 13x$,求$V_{f(2)}$;

(2) 若$D = \mathbf{R}$,求证: “$y = f(x)$为周期函数”的充要条件是“存在非零常数$t$,使得$V_{f(t)} = V_{f(-t)} = D$”;

(3) 若$f(x) = \dfrac{1}{2}a\left(x - \dfrac{\mathrm{e}}{\mathrm{e}-1}\right)^2 - x\left(\ln x - \dfrac{\mathrm{e}}{\mathrm{e}-1}\right)$, $D = (0,+\infty)$,且对于任意的$t \in D$,都有$V_{f(t)} = D$,求实数$a$的取值范围.

**溯源5 (2021上海)** 对于实数集$\mathbf{R}$的子集$S$,如果对任意$x_1, x_2 \in \mathbf{R}$满足:当$x_1 - x_2 \in S$时,都有$f(x_1) - f(x_2) \in S$,则称$f(x)$是$S$关联的.

(1) 判断并证明: ① $f(x) = 2x - 1$是否为$[0, +\infty)$关联的; ② $f(x) = 2x - 1$是否为$[0, 1]$关联的;

(2) 已知$f(x)$是$\{3\}$关联的,且当$x \in [0, 3)$时,有$f(x) = x^2 - 2x$,求不等式$2 \leqslant f(x) \leqslant 3$的解集;

(3)证明: “$f(x)$是$\{1\}$关联的,且是$[0, +\infty)$关联的”当且仅当“$f(x)$是$[1, 2]$关联的”.