State the range of principal values of tan⁡−1x\tan^{-1} x. The curve y=a+bcos⁡cxy = a + b \cos c x is shown below for 0≤x≤3π20 \le x \le \tfrac{3\pi}{2} radians. The curve has maximum points at (π2,7)\left(\tfrac{\pi}{2},7\right) and (3π2,7)\left(\tfrac{3\pi}{2},7\right). The curve also has minimum points at (0,1)(0,1) and (π,1).(\pi,1). The curve y=a+bcos⁡cxy = a + b \cos c x is shown below for 0≤x≤3π20 \le x \le \tfrac{3\pi}{2} radians. The curve has maximum points at (π2,7)\left(\tfrac{\pi}{2},7\right) and (3π2,7)\left(\tfrac{3\pi}{2},7\right). The curve also has minimum points at (0,1)(0,1) and (π,1).(\pi,1). Find the values of aa, bb and cc. Find the values of aa, bb and cc. On the same axes, draw a straight line to determine the number of solutions for the equation aπ+bπcos⁡cx=5π−2xa\pi + b\pi \cos c x = 5\pi - 2x for 0≤x≤3π20 \le x \le \tfrac{3\pi}{2}.

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Worked solution

4(a)

1. State the range of principal values for the inverse tangent function

The principal values of tan⁡−1x\tan^{-1} x lie strictly between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}.

−π2<tan⁡−1x<π2-\frac{\pi}{2} < \tan^{-1} x < \frac{\pi}{2}

Answer

−π2<tan⁡−1x<π2-\frac{\pi}{2} < \tan^{-1} x < \frac{\pi}{2}

4(b)(i)

1. Find the value of $a$ (the vertical shift).

The maximum value is 77 and the minimum value is 11.
a=max+min2=7+12=4a = \frac{\text{max} + \text{min}}{2} = \frac{7 + 1}{2} = 4

2. Find the value of $b$ (related to amplitude).

The amplitude is max−min2=7−12=3\frac{\text{max} - \text{min}}{2} = \frac{7 - 1}{2} = 3.
Since the curve has a minimum at x=0x = 0 instead of a maximum, the cosine curve is reflected about its principal axis.
Therefore, b=−3b = -3.

3. Find the value of $c$ (related to the period).

The curve completes one full cycle between the minimum points at x=0x = 0 and x=πx = \pi.
Period =π−0=π= \pi - 0 = \pi
We know that Period =2πc= \frac{2\pi}{c}.
2πc=π  ⟹  c=2\frac{2\pi}{c} = \pi \implies c = 2

Answer

a=4a = 4, b=−3b = -3, c=2c = 2

4(b)(ii)

1. Find the values of $a$, $b$, and $c$ from the given graph properties

From the graph, the maximum value is 77 and the minimum is 11.
Amplitude =7−12=3= \frac{7 - 1}{2} = 3.
Since the curve starts at a minimum at x=0x=0, b=−3b = -3.
Vertical shift a=7+12=4a = \frac{7 + 1}{2} = 4.
The period is π\pi (from minimum at 00 to minimum at π\pi), so c=2ππ=2c = \frac{2\pi}{\pi} = 2.
Thus, the curve is y=4−3cos⁡2xy = 4 - 3\cos 2x.

2. Rearrange the given equation to match the curve's equation on one side

The given equation is aπ+bπcos⁡cx=5π−2xa\pi + b\pi \cos cx = 5\pi - 2x.
Substitute a=4a=4, b=−3b=-3, and c=2c=2:
4π−3πcos⁡2x=5π−2x4\pi - 3\pi \cos 2x = 5\pi - 2x
Divide the entire equation by π\pi:
4−3cos⁡2x=5−2πx4 - 3\cos 2x = 5 - \frac{2}{\pi}x
The left side is exactly the equation of the curve y=4−3cos⁡2xy = 4 - 3\cos 2x.
Therefore, we need to draw the straight line y=5−2πxy = 5 - \frac{2}{\pi}x.

3. Determine the number of intersection points between the curve and the line

Plot the line y=5−2πxy = 5 - \frac{2}{\pi}x on the same axes:

  • At x=0x = 0, y=5y = 5
  • At x=π2x = \frac{\pi}{2}, y=5−1=4y = 5 - 1 = 4
  • At x=πx = \pi, y=5−2=3y = 5 - 2 = 3
  • At x=3π2x = \frac{3\pi}{2}, y=5−3=2y = 5 - 3 = 2

Tracing the intersections:

  1. Between x=0x=0 and x=π2x=\frac{\pi}{2}, the curve goes from 11 to 77, while the line goes from 55 to 44. (1 intersection)
  2. Between x=π2x=\frac{\pi}{2} and x=πx=\pi, the curve goes from 77 to 11, while the line goes from 44 to 33. (1 intersection)
  3. Between x=πx=\pi and x=3π2x=\frac{3\pi}{2}, the curve goes from 11 to 77, while the line goes from 33 to 22. (1 intersection)

Total number of intersections = 3.

Answer

Equation of line: y=5−2πxy = 5 - \frac{2}{\pi}x
Number of solutions: 3

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