解这道题并讲解。---```
**Question Stem:**
例: (2022 年全国甲 T5) 函数 y = (3ˣ - 3⁻ˣ) cos x 在区间 [-π/2, π/2] 的图像大致为
**Options:**
A.
B.
C.
D.
**Chart/Diagram Description:**
The image contains a question about the graph of a function and four options (A, B, C, D), each displaying a Cartesian coordinate system with a graph plotted.
**General Chart Elements:**
Each option shows a 2D Cartesian coordinate system with a horizontal x-axis and a vertical y-axis intersecting at the origin O.
The x-axis is labeled with points -π/2, O (origin), and π/2.
The y-axis is labeled with points O (origin) and 1.
**Graph Description for each Option:**
* **Option A:**
* Type: Line graph representing a function.
* Curve: Starts at x = -π/2, goes below the x-axis, passes through the origin O (0,0), goes above the x-axis, reaches a peak above y=1 between 0 and π/2, and ends at x = π/2 on the x-axis. The curve is smooth.
* Key Points: Passes through (-π/2, 0), (0, 0), and (π/2, 0).
* **Option B:**
* Type: Line graph representing a function.
* Curve: Starts at x = -π/2 on the x-axis, goes above the x-axis, reaches a peak above y=1 between -π/2 and 0, comes down through the origin O (0,0), goes above the x-axis again, reaches a peak above y=1 between 0 and π/2, and ends at x = π/2 on the x-axis. The curve is symmetric about the y-axis.
* Key Points: Passes through (-π/2, 0), (0, 0), and (π/2, 0).
* **Option C:**
* Type: Line graph representing a function.
* Curve: Starts at x = -π/2 on the x-axis, goes above the x-axis, reaches a peak between -π/2 and 0 (below y=1), comes down through the origin O (0,0), goes below the x-axis, reaches a minimum between 0 and π/2, and ends at x = π/2 on the x-axis.
* Key Points: Passes through (-π/2, 0), (0, 0), and (π/2, 0).
* **Option D:**
* Type: Line graph representing a function.
* Curve: Starts at x = -π/2 below the x-axis, goes up through the origin O (0,0), goes below the x-axis again, reaches a minimum between 0 and π/2, and ends at x = π/2 below the x-axis. The curve is symmetric about the y-axis.
* Key Points: Passes through (0, 0). Ends at points on the vertical lines x = -π/2 and x = π/2, but below the x-axis.
```
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答案文本
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这是一道关于函数图像的选择题。我们需要分析函数 y = (3的x次方 减去 3的负x次方) 乘以 cos x 在区间负二分之π到二分之π上的图像特征。首先让我们理解这个函数的结构，它是两个因子的乘积。
现在我们来判断函数的奇偶性。设f(x)等于括号3的x次方减去3的负x次方括号乘以cos x。计算f负x，我们得到括号3的负x次方减去3的x次方括号乘以cos负x。由于cos负x等于cos x，所以f负x等于括号3的负x次方减去3的x次方括号乘以cos x，这等于负的括号3的x次方减去3的负x次方括号乘以cos x，即负f(x)。因此这是一个奇函数，图像关于原点对称。
接下来计算函数在特殊点的值。当x等于0时，y等于括号3的0次方减去3的负0次方括号乘以cos 0，等于括号1减1括号乘以1，结果为0。当x等于二分之π时，cos二分之π等于0，所以无论前面的因子是什么，结果都是0。同样，当x等于负二分之π时，cos负二分之π也等于0，所以结果也是0。因此函数图像过三个关键点：负二分之π逗号0，原点0逗号0，和二分之π逗号0。
现在分析函数的符号变化。函数y等于g(x)乘以h(x)，其中g(x)等于3的x次方减去3的负x次方，h(x)等于cos x。对于g(x)，当x大于0时，3的x次方大于1，3的负x次方小于1，所以g(x)大于0。当x小于0时，3的x次方小于1，3的负x次方大于1，所以g(x)小于0。对于h(x)等于cos x，在区间负二分之π到二分之π内，除了端点外cos x都大于0。综合分析，当x在0到二分之π区间时，g(x)大于0且h(x)大于0，所以y大于0，图像在x轴上方。当x在负二分之π到0区间时，g(x)小于0且h(x)大于0，所以y小于0，图像在x轴下方。
根据我们的分析，正确的图像应该满足四个条件：过三个关键点，具有奇函数的对称性，在0到二分之π区间图像在x轴上方，在负二分之π到0区间图像在x轴下方。检查各个选项，选项A符合所有条件：图像过负二分之π逗号0、原点和二分之π逗号0三个点，关于原点对称，在右半区间图像在上方，在左半区间图像在下方。选项B在左半区间图像错误地在上方，选项C在右半区间图像错误地在下方，选项D不过端点且关于y轴对称。因此正确答案是A。