logoChat01.ai

(多选)定义在 $(0,+\infty)$ 上的函数 $f(x)$ 满足 $f(x+1)=f(x)-...

생성일: 2025년 7월 10일

답변에 사용된 모델 O3 by Chat01

질문

(多选)定义在 (0,+∞)(0,+\infty) 上的函数 f(x)f(x) 满足 f(x+1)=f(x)−xf(x+1)=f(x)-x，当 0≤x≤10\leq x \leq 1 时，f(x)=x−xf(x)=\sqrt{x}-x，则（ ）

A. 当 2≤x≤32\leq x \leq 3 时，f(x)=x−2−2x+2f(x)=\sqrt{x-2}-2x+2

B. 当 nn 为正整数时，f(n)=n−n22f(n)=\frac{n-n^2}{2}

C. 对任意正实数 tt，f(x)f(x) 在区间 (t,t+1)(t,t+1) 内恰有一个极大值点

D. 若 f(x)f(x) 在区间 (0,k)(0,k) 内恰有 3 个极大值点，则 kk 的取值范围是 (7336,19364]\left(\frac{73}{36}, \frac{193}{64}\right] 注意是多选，请给出详细解答！！

사고

Solving the function problem

The user asked for a detailed solution to a function defined by a recurrence and initial condition. The task is to analyze statements about the function's behavior and determine which ones are correct. To do this, I’ll approach the problem using the functional equation and apply it to the given recurrence. Starting with the initial conditions on the interval [0,1], I’ll iterate the recurrence and apply it to values of x greater than 1, considering the process for all values in the domain (0,∞).

Evaluating the function

I'm working through an iterative process to compute the given recurrence relation. I'll define a sum S and derive the formula step by step, starting with f(t+n) = f(t) - S. I'll verify that the formula holds for n=1, and extending it to general n, confirming the structure. Then, I’ll check the first statement A. When substituting x = 2+t into the formula, I find that the expression does not match, as the coefficients of t and the constants differ. Therefore, the statement is incorrect.

Evaluating statements B and C

For option B, the derived formula matches for n=1 and n=2, so it's correct. For option C, I’m analyzing the derivative of f(x) and finding that there's exactly one local maximum inside each interval (n, n+1) for any t in (0,1). The second derivative shows it’s a maximum, confirming one per interval. However, the original statement about the function having one local maximum for any positive real t is true, confirming option C.

Thought for 1m 26s

답변

解题思路总览

  1. 把递推式展开到一般点
    设

    x=n+t,n∈{0,1,2,… },  t∈[0,1)x=n+t ,\qquad n\in\{0,1,2,\dots\},\;t\in[0,1)

    由 f(x+1)=f(x)−xf(x+1)=f(x)-x 连续迭代 nn 次可得

    f(n+t)=f(t)−∑k=0n−1(t+k)=f(t)−(nt+n(n−1)2).f(n+t)=f(t)-\sum_{k=0}^{n-1}(t+k)=f(t)-\bigl(nt+\tfrac{n(n-1)}2\bigr).

    又已知 f(t)=t−t  (0≤t≤1)f(t)=\sqrt t-t\;(0\le t\le1)，于是

     f(n+t)=t−(n+1)t−n(n−1)2 ,0≤t<1.\boxed{\,f(n+t)=\sqrt t-(n+1)t-\tfrac{n(n-1)}2\,},\qquad 0\le t<1.

  2. 逐项判断

    A 当 2≤x≤32\le x\le3 时
    取 x=2+t  (0≤t≤1)x=2+t\;(0\le t\le1)，此时 n=2n=2。

    f(2+t)=t−3t−1.f(2+t)=\sqrt t-3t-1.

    题中给出的是 t−2t−2\sqrt t-2t-2。二者在 t=0t=0 处即不相等（分别为 −1-1 与 −2-2），故 A 错。

    B nn 为正整数时
    令 t=0t=0，得

    f(n)=f(0)−n(n−1)2=n−n22(因 f(0)=0).f(n)=f(0)-\tfrac{n(n-1)}2=\frac{n-n^2}{2}\qquad(\text{因 }f(0)=0).

    与选项完全一致，故 B 对。

    C 任意正实数 tt，f(x)f(x) 在 (t,t+1)(t,t+1) 内恰有一个极大值点
    先求极值点：在区间 (n,n+1)(n,n+1) 内

    f′(x)=12t−(1+n)=0  ⟹  t=14(n+1)2,  xn=n+14(n+1)2.f'(x)=\frac1{2\sqrt t}-(1+n)=0\;\Longrightarrow\; t=\frac1{4(n+1)^2},\; x_n=n+\frac1{4(n+1)^2}.

    且 f′′(x)=−14t3/2<0f''(x)=-\dfrac1{4t^{3/2}}<0，故每个整段 (n,n+1)(n,n+1) 各有且仅有一个极大值点 xnx_n。
    但对一般长度为 1 的区间 (t,t+1)(t,t+1) 不一定只含一个 xnx_n。例：
    t=0.1t=0.1 时，(0.1,1.1)(0.1,1.1) 同时包含 x0=0.25x_0=0.25 与 x1=1.0625x_1=1.0625。
    因此 C 错。

    D (0,k)(0,k) 内恰有 3 个极大值点时 kk 的范围
    极大值点序列：

    x0=14,  x1=1+116=1716,  x2=2+136=7336,  x3=3+164=19364,  …x_0=\tfrac14,\; x_1=1+\tfrac1{16}= \tfrac{17}{16},\; x_2=2+\tfrac1{36}= \tfrac{73}{36},\; x_3=3+\tfrac1{64}= \tfrac{193}{64},\;\dots

    要使 (0,k)(0,k) 含 x0,x1,x2x_0,x_1,x_2 而不含 x3x_3，需

    x2<k≤x3  ⟹  k∈(7336,19364].x_2<k\le x_3\;\Longrightarrow\; k\in\Bigl(\tfrac{73}{36},\tfrac{193}{64}\Bigr].

    故 D 对。

结论

正确选项：B,  D\boxed{\text{正确选项：B,\;D}}

이 Q&A 공유하기