速解 2026 高考一卷数学压轴 3-2

Question. $f (x) = 2^x $ for \(x < 0\), let \(D (x_0) = \{ d \in \mathbb{R} | f (x_0 + d) > f (x_0) \}\), \(f (x) < f (0)\) for \(0 < x < 1\).

And \(f (x_1) \leqslant f (x_2) \Rightarrow D (x_2) \subseteq D (x_1) .\)

Prove (1) \(f (0) \geqslant 1 ;\) (2) \(f (u) < f (v)\) for any \(0 < u < v.\)

Proof.

We first find that \(\forall x_1, x_2, x_3, s.t. f (x_1) \leqslant f (x_2) < f (x_3)\) then \(f (x_1) < f (x_1 - x_2 + x_3) .\)

(1) If \(f (0) \leqslant 0\), choose \(x_1 = 0, x_2 = - 1, x_3 = - 0.5, \Rightarrow f (0) < f (0.5)\) contradiction!

If \(0 < f (0) < 1\), choose $x_1 \rightarrow - \infty, x_2 = 0, x_3 \nearrow 0, \Rightarrow f (x_1) < f
(x_1 + x_3) $ contradiction!

Hence \(f (0) \geqslant 1\).

(2) We first claim \(f (x) \leqslant 0\) for \(x \in (0, + \infty)\):

• If \(0 < f (x) < f (0)\), choose \(x_1 \rightarrow - \infty, x_2 = x, x_3 = 0, \Rightarrow f (x_1) < f (x_1 - x)\) contradiction!

Hence we get \(f (x) \leqslant 0\) for \(x \in (0, 1)\). for \(x \geqslant 1\),

• If \(f (x) \geqslant f (0)\), choose \(x_1 \searrow - x, x_2 = 0, x_3 = x, \Rightarrow f (x_1) < f (x_1 + x) < 0\) contradiction!

So \(f (x) \leqslant 0\) for \(x \in (0, + \infty)\). for any \(0 < u < v\),

choose \(x_1 = u, x_2 = - v, x_3 = - u,\) then \(f (u) = f (x_1) < f (x_1 - x_2 + x_3) = f (v)\).