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IAL 2023 Jan Q8

A Level / Edexcel / P2

IAL 2023 Jan Paper · Question 8

题目

Problem

In this question you must show all stages of your working.

Solutions based entirely on calculator technology are not acceptable.

(i) Solve, for 23π<x<23π-\frac{2}{3}\pi < x < \frac{2}{3}\pi, the equation

5sin(3x+0.1)+2=05 \sin(3x + 0.1) + 2 = 0

giving your answers, in radians, to 2 decimal places.

(4)

(ii) Solve, for 0<θ<3600 < \theta < 360^{\circ}, the equation

2tanθsinθ=5+cosθ2 \tan\theta \sin\theta = 5 + \cos\theta

giving your answers, in degrees, to one decimal place.

(5)

解答

(i)

解法一

思路

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  1. 化簡方程5sin(3x+0.1)+2=0    sin(3x+0.1)=0.45 \sin(3x + 0.1) + 2 = 0 \implies \sin(3x + 0.1) = -0.4
  2. 換元與確定範圍: 令 y=3x+0.1y = 3x + 0.1。已知 23π<x<23π-\frac{2}{3}\pi < x < \frac{2}{3}\pi,乘以 3 再加 0.1,得到 yy 的取值範圍: 2π+0.1<y<2π+0.1-2\pi + 0.1 < y < 2\pi + 0.16.183<y<6.383-6.183 < y < 6.383
  3. yy 的解
    • siny=0.4\sin y = -0.4,其主值(principal value)為 α=arcsin(0.4)0.4115\alpha = \arcsin(-0.4) \approx -0.4115
    • 在一個週期 2π2\pi 內,另一解為 πα=π(0.4115)3.5531\pi - \alpha = \pi - (-0.4115) \approx 3.5531
    • 尋找在範圍 (2π+0.1,2π+0.1)(-2\pi + 0.1, 2\pi + 0.1) 內的所有解:
      • y1=α0.4115y_1 = \alpha \approx -0.4115
      • y2=πα3.5531y_2 = \pi - \alpha \approx 3.5531
      • y3=2π+α5.8717y_3 = 2\pi + \alpha \approx 5.8717
      • y4=πα2.7301y_4 = -\pi - \alpha \approx -2.7301
  4. xx 的解: 利用 x=y0.13x = \frac{y - 0.1}{3} 計算對應的 xx,並保留 2 位小數。

答题过程

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Simplify the equation:

5sin(3x+0.1)+2=0sin(3x+0.1)=0.4.\begin{align*} 5 \sin(3x + 0.1) + 2 =&\,\, 0 \\[2mm] \sin(3x + 0.1) =&\,\, -0.4. \end{align*}

Let y=3x+0.1y = 3x + 0.1. The range of xx is 23π<x<23π-\frac{2}{3}\pi < x < \frac{2}{3}\pi, which gives the range of yy:

3(23π)+0.1<y<3(23π)+0.12π+0.1<y<2π+0.1.\begin{align*} 3\left(-\frac{2}{3}\pi\right) + 0.1 <&\,\, y < 3\left(\frac{2}{3}\pi\right) + 0.1 \\[2mm] -2\pi + 0.1 <&\,\, y < 2\pi + 0.1. \end{align*}

Now solve siny=0.4\sin y = -0.4. The principal value of yy is:

y=arcsin(0.4)0.41152 rad.\begin{align*} y =&\,\, \arcsin(-0.4) \approx -0.41152\text{ rad}. \end{align*}

The other solution in the basic interval [π,π][-\pi, \pi] is:

y=π(0.41152)=π+0.411522.73007 rad.\begin{align*} y =&\,\, -\pi - (-0.41152) = -\pi + 0.41152 \approx -2.73007\text{ rad}. \end{align*}

Adding/subtracting multiples of 2π2\pi to find all solutions within the range 2π+0.1<y<2π+0.1-2\pi + 0.1 < y < 2\pi + 0.1:

  • For y=0.41152y = -0.41152:
    • y=0.41152y = -0.41152
    • y=2π0.411525.87167y = 2\pi - 0.41152 \approx 5.87167
  • For y=2.73007y = -2.73007:
    • y=2.73007y = -2.73007
    • y=2π2.73007y = 2\pi - 2.73007 (this is 3.553113.55311 which is within the range)

So the solutions for yy are:

y2.73007,0.41152,3.55311,5.87167.\begin{align*} y \approx&\,\, -2.73007, \,\, -0.41152, \,\, 3.55311, \,\, 5.87167. \end{align*}

Now calculate the corresponding values of xx using x=y0.13x = \frac{y - 0.1}{3}:

  1. For y=2.73007y = -2.73007: x=2.730070.130.943360.94x = \frac{-2.73007 - 0.1}{3} \approx -0.94336 \approx -0.94
  2. For y=0.41152y = -0.41152: x=0.411520.130.170510.17x = \frac{-0.41152 - 0.1}{3} \approx -0.17051 \approx -0.17
  3. For y=3.55311y = 3.55311: x=3.553110.131.151041.15x = \frac{3.55311 - 0.1}{3} \approx 1.15104 \approx 1.15
  4. For y=5.87167y = 5.87167: x=5.871670.131.923891.92x = \frac{5.87167 - 0.1}{3} \approx 1.92389 \approx 1.92

Therefore, the solutions are:

x=0.94,0.17,1.15,1.92(to 2 d.p.).\begin{align*} x =&\,\, -0.94, \,\, -0.17, \,\, 1.15, \,\, 1.92 \quad \text{(to 2 d.p.)}. \end{align*}

(ii)

解法一

思路

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  1. 替換切切:利用關係式 tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}2(sinθcosθ)sinθ=5+cosθ    2sin2θcosθ=5+cosθ2 \left( \frac{\sin\theta}{\cos\theta} \right) \sin\theta = 5 + \cos\theta \implies \frac{2\sin^2\theta}{\cos\theta} = 5 + \cos\theta
  2. 去分母與統一三角函數: 方程兩邊同乘以 cosθ\cos\theta(需注意 cosθ0\cos\theta \neq 0): 2sin2θ=5cosθ+cos2θ2 \sin^2\theta = 5 \cos\theta + \cos^2\theta 利用 sin2θ=1cos2θ\sin^2\theta = 1 - \cos^2\theta 將正弦轉換為餘弦: 2(1cos2θ)=5cosθ+cos2θ    3cos2θ+5cosθ2=02(1 - \cos^2\theta) = 5 \cos\theta + \cos^2\theta \implies 3\cos^2\theta + 5\cos\theta - 2 = 0
  3. 因式分解求解: 將二次三項式因式分解為 (3cosθ1)(cosθ+2)=0(3\cos\theta - 1)(\cos\theta + 2) = 0。 由於餘弦值的範圍為 [1,1][-1, 1],所以 cosθ=2\cos\theta = -2 無解。 故 cosθ=13\cos\theta = \frac{1}{3}
  4. θ\theta 的解: 在範圍 0<θ<3600 < \theta < 360^{\circ} 內,餘弦值為正,有兩個解:第一象限角和第四象限角。

答题过程

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Using tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}:

2tanθsinθ=5+cosθ2(sinθcosθ)sinθ=5+cosθ2sin2θcosθ=5+cosθ.\begin{align*} 2 \tan\theta \sin\theta =&\,\, 5 + \cos\theta \\[2mm] 2 \left(\frac{\sin\theta}{\cos\theta}\right) \sin\theta =&\,\, 5 + \cos\theta \\[3mm] \frac{2\sin^2\theta}{\cos\theta} =&\,\, 5 + \cos\theta. \end{align*}

Multiply by cosθ\cos\theta:

2sin2θ=5cosθ+cos2θ.\begin{align*} 2\sin^2\theta =&\,\, 5\cos\theta + \cos^2\theta. \end{align*}

Use the identity sin2θ=1cos2θ\sin^2\theta = 1 - \cos^2\theta:

2(1cos2θ)=5cosθ+cos2θ22cos2θ=5cosθ+cos2θ3cos2θ+5cosθ2=0(3cosθ1)(cosθ+2)=0.\begin{align*} 2(1 - \cos^2\theta) =&\,\, 5\cos\theta + \cos^2\theta \\[2mm] 2 - 2\cos^2\theta =&\,\, 5\cos\theta + \cos^2\theta \\[2mm] 3\cos^2\theta + 5\cos\theta - 2 =&\,\, 0 \\[2mm] (3\cos\theta - 1)(\cos\theta + 2) =&\,\, 0. \end{align*}

Since 1cosθ1-1 \le \cos\theta \le 1, cosθ=2\cos\theta = -2 has no real solutions.

Thus, we have:

3cosθ1=0cosθ=13.\begin{align*} 3\cos\theta - 1 =&\,\, 0 \\[2mm] \cos\theta =&\,\, \frac{1}{3}. \end{align*}

For 0<θ<3600 < \theta < 360^{\circ}:

θ=arccos(13)70.528870.5,\begin{align*} \theta =&\,\, \arccos\left(\frac{1}{3}\right) \approx 70.5288^{\circ} \approx 70.5^{\circ}, \end{align*}

and the other solution is:

θ=36070.5288289.4712289.5.\begin{align*} \theta =&\,\, 360^{\circ} - 70.5288^{\circ} \approx 289.4712^{\circ} \approx 289.5^{\circ}. \end{align*}

Therefore, the solutions are:

θ=70.5,289.5(to 1 d.p.).\begin{align*} \theta =&\,\, 70.5^{\circ}, \,\, 289.5^{\circ} \quad \text{(to 1 d.p.)}. \end{align*}