写出 2sin⁡3x+12\sin 3x + 1 的振幅和周期。 绘制 y=2sin⁡3x+1y = 2\sin 3x + 1 在 0≤x≤π0 \le x \le \pi 范围内的函数图像草图。 通过在草图上绘制一条合适的直线,确定方程 πsin⁡3x−x=0\pi\sin 3x - x = 0 的解的个数。

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完整解法

10(a)

1. Identify the amplitude from the coefficient of the sine term

The function is in the form y=asin⁡(bx)+cy = a\sin(bx) + c. The amplitude is given by ∣a∣|a|.
Amplitude = ∣2∣=2|2| = 2.

2. Identify the period from the coefficient of $x$

The period of sin⁡(bx)\sin(bx) is given by 2π∣b∣\frac{2\pi}{|b|}.
Period = 2π3\frac{2\pi}{3}.

答案

Amplitude = 22, Period = 2π3\frac{2\pi}{3}

10(b)

1. Determine the key features of the graph (maximum, minimum, and starting points)

The graph of y=2sin⁡3x+1y = 2\sin 3x + 1 has an amplitude of 22 and is shifted upwards by 11.
Maximum value =1+2=3= 1 + 2 = 3.
Minimum value =1−2=−1= 1 - 2 = -1.
At x=0x = 0, y=2sin⁡(0)+1=1y = 2\sin(0) + 1 = 1.

2. Determine the period and the number of cycles in the given domain

The period is 2π3\frac{2\pi}{3}.
The domain is 0≤x≤π0 \le x \le \pi.
Since π=1.5×2π3\pi = 1.5 \times \frac{2\pi}{3}, the graph will complete 1.51.5 cycles.

3. Sketch the curve and label key coordinates

Key points to plot:
(0,1)(0, 1)
(π6,3)(\frac{\pi}{6}, 3) (Maximum)
(π3,1)(\frac{\pi}{3}, 1)
(π2,−1)(\frac{\pi}{2}, -1) (Minimum)
(2π3,1)(\frac{2\pi}{3}, 1)
(5π6,3)(\frac{5\pi}{6}, 3) (Maximum)
(π,1)(\pi, 1)
Draw a smooth curve through these points.

答案

A smooth sine curve starting at (0,1)(0, 1), reaching a maximum of 33 at x=π6x = \frac{\pi}{6}, crossing y=1y=1 at x=π3x = \frac{\pi}{3}, reaching a minimum of −1-1 at x=π2x = \frac{\pi}{2}, crossing y=1y=1 at x=2π3x = \frac{2\pi}{3}, reaching a maximum of 33 at x=5π6x = \frac{5\pi}{6}, and ending at (π,1)(\pi, 1).

10(c)

1. Rearrange the given equation to match the function sketched in part (b).

πsin⁡3x−x=0\pi\sin 3x - x = 0
πsin⁡3x=x\pi\sin 3x = x
sin⁡3x=xπ\sin 3x = \frac{x}{\pi}
Multiply by 2 and add 1 to both sides:
2sin⁡3x+1=2xπ+12\sin 3x + 1 = \frac{2x}{\pi} + 1

2. Identify the straight line to be drawn.

The suitable straight line to draw is y=2xπ+1y = \frac{2x}{\pi} + 1.

3. Determine the number of intersections between the curve and the line in the given domain.

The line passes through (0,1)(0, 1) and (π,3)(\pi, 3).
Drawing this line on the sketch of y=2sin⁡3x+1y = 2\sin 3x + 1 for 0≤x≤π0 \le x \le \pi:

  • At x=0x = 0, both the curve and the line are at y=1y = 1 (1 intersection).
  • In the first cycle, the curve rises to y=3y = 3 at x=π6x = \frac{\pi}{6} while the line is at y=113y = 1\frac{1}{3}, so they intersect once as the curve descends (1 intersection).
  • In the third cycle, the curve rises to y=3y = 3 at x=5π6x = \frac{5\pi}{6} while the line is at y=223y = 2\frac{2}{3}. The curve crosses the line on the way up and on the way down (2 intersections).
    Total number of intersections = 1+1+2=41 + 1 + 2 = 4.
答案

4

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