https://file.aiquickdraw.com/mathbot/file/17266658178234murtsqb.jpg

The problem in the image reads:

"実数 \(x, y\) が \(x^2 \leq y \leq 2x \sin \theta + \theta\) ( \(\theta > 0\)) を満たしているとき \((x - 1)^2 + y\) の範囲を求めよ。"

Translation:
Let \(x\) and \(y\) be real numbers that satisfy the inequality \(x^2 \leq y \leq 2x \sin \theta + \theta\) (with \(\theta > 0\)). Find the range of \((x - 1)^2 + y\).

### Solution Outline:

1. **Inequality Setup:**
   We are given that \(x^2 \leq y \leq 2x \sin \theta + \theta\). This implies that the value of \(y\) is constrained between the parabola \(x^2\) and the expression \(2x \sin \theta + \theta\), which varies depending on \(\theta\).

2. **Expression to Maximize/Minimize:**
   We need to find the range of \((x - 1)^2 + y\). This expression consists of two parts:
   - \((x - 1)^2\) is always non-negative and reaches its minimum value of \(0\) when \(x = 1\).
   - \(y\) is bounded as described by the inequality.

3. **Analyzing the Boundaries:**
   We will first consider the bounds of \(y\), then analyze how \((x - 1)^2\) affects the overall expression.

Let me proceed with the detailed calculation.The expression we are trying to analyze is:

\[
(x - 1)^2 + y \leq \theta + 2x \sin(\theta) + (x - 1)^2
\]

This gives us the upper bound for the expression \((x - 1)^2 + y\). To find the range, we need to determine the possible values this expression can take, depending on \(x\) and \(\theta\), while considering the lower bound \(y \geq x^2\).

I will now calculate the overall range of this expression.The lower bound for the expression \((x - 1)^2 + y\) is:

\[
(x - 1)^2 + x^2
\]

This simplifies to:

\[
x^2 + (x - 1)^2 = x^2 + (x^2 - 2x + 1) = 2x^2 - 2x + 1
\]



実数 x, y が x^2 ≤ y ≤ 2x sinθ + θ (θ > 0) を満たしているとき (x - 1)^2 + y の範囲を求めよ。

Find the range of the expression (x - 1)^2 + y, where x and y are real numbers that satisfy the inequality x^2 ≤ y ≤ 2x sin(θ) + θ for θ > 0. This problem combines concepts of quadratic functions and trigonometry.

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