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Найдите количество решений следующего уравнения в ...

作成日: 2025年11月3日

使用モデル GPT-5 Thinking by Chat01

質問

Найдите количество решений следующего уравнения в зависимости от значений параметра a (где [x] – целая часть числа x)a+[x]=-sqrt(9-x^2).

思考

Analyzing solutions for equation

The domain for x is [-3, 3], since we're taking the square root of (9 - x^2). The equation also has constraints: the right-hand side (RHS) is always less than or equal to 0. This means that [x], the floor function of x, has to be less than or equal to -a. I need to think about how the number of integer solutions varies with a.

Clarifying integer restrictions for equation

For x in the domain [-3, 3], I need to carefully check how the floor function operates on the boundaries. When x is in the range [2, 3), n = 2, but at x = 3, [3] = 3 and the equation simplifies to a + 3 = 0, so a = -3. Although I previously restricted n to {-3, ..., 2}, I need to include the endpoints, x = -3 and x = 3.

Refining solutions based on intervals

n can be -3 and 3 at the endpoints, but inside the intervals, n takes values from -3 to 2. The equation a + n = RHS becomes a + [x] = -sqrt(9 - x^2). For each fixed n, the values of x are restricted to specific intervals, e.g., n = -3 is satisfied at x = -3, and similarly for other values of n in the defined ranges.

Refining equation with intervals

On each interval where [x] = n, the equation simplifies to -sqrt(9 - x^2) = a + n. This implies that a + n must fall between [-3, 0]. Squaring both sides, we find x^2 = 9 - (a + n)^2. So for each fixed value of a + n (within [-3, 0]), x must lie within the interval corresponding to n.

Solving for valid x values

For each integer n, the possible solutions for x are determined by checking conditions where n ≤ x < n+1, with additional care for endpoint cases like x = ±3. For negative n, only -r may qualify, while for positive n, r is used. We must solve inequalities based on the range of r, which depends on the value of a + n.

Identifying value ranges for b

The value for b must satisfy specific constraints for each n. For n=0, b lies in [−3,−22)[-3, -2\sqrt{2}). For n=1, b shifts to [−1−22,−1−5)[-1 - 2\sqrt{2}, -1 - \sqrt{5}). For n=2, it further narrows to [−2−5,−2)[ -2 - \sqrt{5}, -2 ). Each case ensures b stays within the defined interval [−3,0][-3, 0]. Inputting the constraints for different n values ensures accurate solutions in each range.

Evaluating solution branches

For each n value, the equation yields intervals for "a" and corresponding solutions. For n=-3, a ∈ (3 - √5, 3]. For n=0, b=a, and solutions exist when a=-3. Other n ranges yield distinct x values, with at most one solution per n, except for special cases like r=0. Multiple values of n can produce different solutions, and each n condition needs careful scrutiny.

Verifying interval overlaps

I’m considering overlapping intervals for each n value and counting how many n-conditions each "a" belongs to. For example, between -4.236 and -3.828, only n=2 is active, while between -3.828 and -3.236, both n=2 and n=1 are active, so we have two solutions. I’m verifying each step carefully to count overlaps accurately.

Counting solutions across intervals

I’m checking the intervals for each "a" value. For example, between -3.236 and -3, only n=2 remains active. At a = -3, n=2 and n=0 are active, giving 2 solutions. At a = -2, no solution exists for n=2, but n=-1 is active between -2 and -1.828, yielding 1 solution there.

Finalizing piecewise solutions

I’m wrapping up the piecewise function, summarizing all of the solution counts across intervals. For example, between -2 and -√5, there's 1 solution; at a = -3, I found 3 solutions. At certain points like a = 0 and a = 2, there are no solutions, while for a = 1, there’s 1 solution. Some intervals, like (-∞, -2-√5), have no solutions.

Finalizing solution intervals

I've double-checked the intervals and made sure each boundary is treated correctly. The solution counts across intervals (like -4.236, -3.828, -0.236, 3, etc.) all match expectations. I'll present the piecewise solution clearly with intervals in increasing order, including endpoints correctly, and present the answer succinctly using radicals for clarity.

Finalizing solution mapping

I’ve carefully verified that for values of "a", no triple overlaps exist except at specific points like a = -3. I've summarized the result with exact intervals for zero solutions and mapped them to integer solutions. The solution count formula reflects conditions on "a" and includes the special case for a = -3. I'll present the final solution in a clear, concise manner for the user and include a table for better clarity.

Thought for 4m 35s

回答

Решаем по “полочкам” (интервалам, где целая часть постоянна).
Область: x∈[−3,3]x\in[-3,3], правая часть −9−x2∈[−3,0]-\sqrt{9-x^2}\in[-3,0].

Пусть [x]=n[x]=n. Тогда на каждом интервале

In={[−3,−2),[−2,−1),[−1,0),[0,1),[1,2),[2,3)(n=−3,−2,−1,0,1,2),и отдельно x=3(n=3),I_n=\begin{cases} [-3,-2),[-2,-1),[-1,0),[0,1),[1,2),[2,3) & (n=-3,-2,-1,0,1,2),\\ \text{и отдельно }x=3 & (n=3), \end{cases}

уравнение сводится к

−9−x2=a+n,9−x2=−(a+n)≥0.-\sqrt{9-x^2}=a+n,\qquad \sqrt{9-x^2}=-(a+n)\ge 0.

Положим b=a+nb=a+n. Тогда b∈[−3,0]b\in[-3,0] и x=±9−b2x=\pm\sqrt{9-b^2}.
Из условия [x]=n[x]=n остается единственный знак: для n≥0n\ge0 нужен x=+9−b2∈[n,n+1)x=+\sqrt{9-b^2}\in[n,n+1); для n≤−1n\le-1 — x=−9−b2∈[n,n+1)x=-\sqrt{9-b^2}\in[n,n+1). Это дает такие условия на aa:

nинтервал для a2a∈[ −2−5, −2 )1a∈[ −1−22, −1−5 )0a∈[ −3, −22 )−1a∈( −2, 1−22 ]−2a∈( 2−22, 2−5 ]−3a∈( 3−5, 3 ]\begin{array}{c|c} n & \text{интервал для }a \\ \hline 2 & a\in\bigl[\,{-2-\sqrt5},\,-2\,\bigr) \\ 1 & a\in\bigl[\,{-1-2\sqrt2},\,{-1-\sqrt5}\,\bigr) \\ 0 & a\in\bigl[\,{-3},\,{-2\sqrt2}\,\bigr) \\ -1 & a\in\bigl(\,{-2},\,{1-2\sqrt2}\,\bigr] \\ -2 & a\in\bigl(\,{2-2\sqrt2},\,{2-\sqrt5}\,\bigr] \\ -3 & a\in\bigl(\,{3-\sqrt5},\,3\,\bigr] \\ \end{array}

(и отдельно x=3x=3 дает решение при a=−3a=-3).

Число решений N(a)N(a) — это количество интервалов из таблицы, в которых лежит данный aa, плюс 1 при a=−3a=-3 (за точку x=3x=3). Отсюда:

  • N(a)=0N(a)=0, если
    a∈(−∞, −2−5)  ∪  {−2}  ∪  (1−22, 2−22)  ∪  (2−5, 3−5)  ∪  (3,∞)a\in(-\infty,\,{-2-\sqrt5})\;\cup\;\{-2\}\;\cup\;(1-2\sqrt2,\,2-2\sqrt2)\;\cup\;(2-\sqrt5,\,3-\sqrt5)\;\cup\;(3,\infty).
  • N(a)=1N(a)=1, если
    a=−2−5a={-2-\sqrt5};; или a∈(−2−5, −1−22)a\in(-2-\sqrt5,\,-1-2\sqrt2);; или a∈(−1−5, −3)a\in(-1-\sqrt5,\,-3);; или a=−22a={-2\sqrt2};; или a∈(−22, −2)a\in(-2\sqrt2,\,-2);; или a∈(−2, 1−22]a\in(-2,\,1-2\sqrt2];; или a∈(2−22, 2−5]a\in(2-2\sqrt2,\,2-\sqrt5];; или a∈(3−5, 3]a\in(3-\sqrt5,\,3].
  • N(a)=2N(a)=2, если
    a∈[ −1−22, −1−5 )  ∪  (−3, −22)a\in[\,{-1-2\sqrt2},\,{-1-\sqrt5}\,)\;\cup\;(-3,\,-2\sqrt2).
  • N(a)=3N(a)=3 только при a=−3a=-3 (точки x=0,  x≈2.828,  x=3x=0,\;x\approx2.828,\;x=3).

Для ориентира численные значения границ:
5≈2.236,  22≈2.828,  1−22≈−1.828,  2−22≈−0.828,  2−5≈−0.236,  3−5≈0.764.\sqrt5\approx2.236,\;2\sqrt2\approx2.828,\;1-2\sqrt2\approx-1.828,\;2-2\sqrt2\approx-0.828,\;2-\sqrt5\approx-0.236,\;3-\sqrt5\approx0.764.

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