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IAL 2024 Jan Q6

A Level / Edexcel / P1

IAL 2024 Jan Paper · Question 6

题目

Problem

Figure 2 shows a plot of part of the curve C1C_1 with equation

y=5cos⁡x\begin{align*} y=5\cos x \end{align*}

Figure 2

with xx being measured in degrees.

The point PP, shown in Figure 2, is a minimum point on C1C_1.

(a) State the coordinates of PP.

(2)

The point QQ lies on a different curve C2C_2.

Given that point QQ

  • is a maximum point on the curve
  • is the maximum point with the smallest xx coordinate, x>0x>0

(b) find the coordinates of QQ when

(i) C2C_2 has equation y=5cos⁡x−2y=5\cos x-2

(ii) C2C_2 has equation y=−5cos⁡xy=-5\cos x.

(4)

解答

(a)

解法一

思路

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5cos⁡x5\cos x 的最小值是 −5-5，发生在 x=180∘,540∘,…x=180^\circ,540^\circ,\ldots。图中标出的 PP 是右侧那个最小点，所以取 x=540∘x=540^\circ。

答题过程

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For y=5cos⁡xy=5\cos x, the minimum value is −5-5.

The minimum point shown has

x=540∘.\begin{align*} x=540^\circ. \end{align*}

Therefore

P=(540∘,−5).\begin{align*} P=(540^\circ,-5). \end{align*}

(b)(i)

解法一

思路

展开

5cos⁡x5\cos x 的最大值是 55，所以 5cos⁡x−25\cos x-2 的最大值是 33。最小正 xx 的最大点在 x=360∘x=360^\circ，因为题目要求 x>0x>0。

答题过程

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For

y=5cos⁡x−2,\begin{align*} y=5\cos x-2, \end{align*}

the maximum value is

5−2=3.\begin{align*} 5-2=3. \end{align*}

The maximum point with the smallest positive xx coordinate occurs at

x=360∘.\begin{align*} x=360^\circ. \end{align*}

Therefore

Q=(360∘,3).\begin{align*} Q=(360^\circ,3). \end{align*}

(b)(ii)

解法一

思路

展开

−5cos⁡x-5\cos x 的最大值是 55，这发生在 cos⁡x=−1\cos x=-1 的位置。最小正解是 x=180∘x=180^\circ。

答题过程

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For

y=−5cos⁡x,\begin{align*} y=-5\cos x, \end{align*}

the maximum value is 55, when

cos⁡x=−1.\begin{align*} \cos x=-1. \end{align*}

The smallest positive value of xx is

x=180∘.\begin{align*} x=180^\circ. \end{align*}

Therefore

Q=(180∘,5).\begin{align*} Q=(180^\circ,5). \end{align*}