Example Question - interval
Here are examples of questions we've helped users solve.
Interval and Set Intersection Problem
<p>首先我们来确定集合A。据题意，集合A是满足条件的x的集合，即\( x \in \mathbb{Z} \) 且 \( -5 \leq x \leq 10 \)。由于 \( x \in \mathbb{Z} \)，我们知道x是整数。因此集合A是 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}。</p> <p>接着我们来确定集合B。集合B是满足 \( y = |x - 5| \) 条件的整数y的集合。考虑绝对值的性质，我们知道如果 \( x \geq 5 \)，那么 \( y = x - 5 \)，反之如果 \( x < 5 \)，那么 \( y = 5 - x \)。因此当 \( x \geq 5 \) 时，\( y \) 可以取的值为 {0, 1, 2, 3, 4, 5}；当 \( x < 5 \) 时，\( y \) 可以取的值反映在数轴上与 \( x \geq 5 \) 相反，也为 {0, 1, 2, 3, 4}。整合两边，我们发现集合B实际上是 {0, 1, 2, 3, 4, 5}。</p> <p>最后我们求集合A与集合B的交集即 \( A \cap B \)。两个集合的交集为集合中共同的元素，也就是 {0, 1, 2, 3, 4, 5} 与 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \} 的交集。结果很显然是 {0, 1, 2, 3, 4, 5}。</p> <p>所以正确答案是 \( A \cap B = \{0, 1, 2, 3, 4, 5\} \)。</p> <p>选项中没有完全匹配的答案，但是根据题目的选择，我们可以选择最接近的选项C，即是 \( \{1, 2\} \)，尽管它没有包括全部的正确答案。</p>
Finding Average Value of a Function within an Interval
对于题目中的情况，我们要找的是给定区间 [0,2] 内函数 g(x) 的平均值。根据平均值定理，函数在区间 [a,b] 上的平均值等于该区间上函数曲线与 x 轴围成的面积除以区间的长度。 根据题目，函数 g(x) 的图形和 x 轴在 x = 0 到 x = 2 之间围成的面积为 3。由于这个面积实际上表示了这个区间上 g(x) 的积分，我们可以将面积表示为积分形式： 面积 = ∫(从 a 到 b) g(x) dx 将给定的信息代入，我们得到 3 = ∫(从 0 到 2) g(x) dx 为了找到平均值，我们需要将这个积分除以区间的长度，区间 [0,2] 的长度是 2。 平均值 = 面积 / 区间长度 平均值 = 3 / 2 因此，函数 g(x) 在 [0,2] 区间内的平均值是 3/2。所以在所提供的选项中，(B) 3/2 是正确的答案。
Finding Absolute Maximum and Minimum of a Trigonometric Function
The question asks for the absolute maximum and minimum of the function \( f(x) = 3\cos^2\left(\frac{x}{2}\right) \) over the interval \(\left[\frac{\pi}{4}, \pi\right]\). To solve this: 1. Compute the derivative of the function to find the critical points where the derivative is zero or undefined. Critical points can also occur at the endpoints of the interval. 2. Evaluate the function at the critical points and at the endpoints of the interval to determine the absolute maximum and minimum. Let's find the derivative \(f'(x)\): \( f(x) = 3\cos^2\left(\frac{x}{2}\right) \) Using the chain rule, \( f'(x) = 3 \cdot 2\cos\left(\frac{x}{2}\right)(-\sin\left(\frac{x}{2}\right)) \left(\frac{1}{2}\right) = -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \) Setting the derivative equal to zero, \( -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = 0 \) This equation is satisfied when either \( \sin\left(\frac{x}{2}\right) = 0 \) or \( \cos\left(\frac{x}{2}\right) = 0 \). \( \sin\left(\frac{x}{2}\right) = 0 \) at \( x = 0, 2\pi, 4\pi, \ldots \) but within \(\left[\frac{\pi}{4}, \pi\right]\), this doesn't give any solution. \( \cos\left(\frac{x}{2}\right) = 0 \) at \( x = \pi, 3\pi, \ldots \), and only \( x = \pi \) falls within our interval. The critical points are just at the endpoints of the interval since there are no other points in \(\left[\frac{\pi}{4}, \pi\right]\) where the derivative is zero or undefined. Now, evaluate \(f(x)\) at \( x = \frac{\pi}{4} \) and \( x = \pi \): \( f\left(\frac{\pi}{4}\right) = 3\cos^2\left(\frac{\pi}{8}\right) \) and \(f(\pi) = 3\cos^2\left(\frac{\pi}{2}\right)\). Note that \( \cos\left(\frac{\pi}{2}\right) = 0 \), so \( f(\pi) = 0 \). To find \(f\left(\frac{\pi}{4}\right)\), we compute the cosine of \( \pi/8 \). This value is not one of the standard angles, so we can't find an exact value easily without using a calculator. However, since \(0 < \frac{\pi}{8} < \frac{\pi}{2}\), \( \cos\left(\frac{\pi}{8}\right) \) is positive and less than 1, so \( 3\cos^2\left(\frac{\pi}{8}\right) \) is also positive and less than 3. Therefore, the absolute maximum of \( f(x) \) is at \( x = \frac{\pi}{4} \) and is greater than 0, and the absolute minimum of \( f(x) \) is 0 at \( x = \pi \). Based only on the information available, you can't give an exact value for the absolute maximum without a calculator, but you can state where it occurs and that the absolute minimum is 0 at \( x = \pi \).
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