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CIE 9709 2024 June Paper 13 Q2

A Level / CIE / P1

CIE 9709 2024 June Paper 13 Paper · Question 2

题目

Problem

(a)

The diagram shows the curve y=kcos⁡(x−16π)y = k\cos\big(x - \frac{1}{6}\pi\big) where kk is a positive constant and xx is measured in radians. The curve crosses the xx-axis at point AA and BB is a minimum point.

Find the coordinates of AA and BB.

[3]

(b) Find the exact value of tt that satisfies the equation

3sin⁡−1(3t)+2cos⁡−1(122)=π.3\sin^{-1}(3t) + 2\cos^{-1}\bigg(\frac{1}{2}\sqrt{2}\bigg) = \pi.

[2]
题目中文翻译

(a)

图中显示曲线 y=kcos⁡(x−16π)y = k\cos\big(x - \frac{1}{6}\pi\big),其中 kk 是正常数,xx 以弧度为单位。曲线在点 AA 与 xx 轴相交,且 BB 是一个最低点。

求 AA 和 BB 的坐标。

[3]

(b) 求满足方程

3sin⁡−1(3t)+2cos⁡−1(122)=π3\sin^{-1}(3t) + 2\cos^{-1}\bigg(\frac{1}{2}\sqrt{2}\bigg) = \pi

的 tt 的精确值。

[2]

解答

(a)

解法一

思路

展开

按图判断是哪一次过零和哪一个谷底。AA 是从下往上穿过 xx 轴的点,BB 是它右边的第二个极值点。

答题过程

展开

At AA, the cosine argument is 3π/23\pi/2, the rising zero shown. Thus x=3π/2+π/6=5π/3x=3\pi/2+\pi/6=5\pi/3. The minimum BB shown after the next maximum has argument 3π3\pi, so x=3π+π/6=19π/6x=3\pi+\pi/6=19\pi/6 and y=−ky=-k. Therefore

A=(5π/3,0),B=(19π/6,−k).A=(5\pi/3,0),\qquad B=(19\pi/6,-k).

(b)

解法一

思路

展开

先求反余弦的精确角,再把反正弦单独留在等号一边。

答题过程

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The principal value cos⁡−1(2/2)\cos^{-1}(\sqrt2/2) is π/4\pi/4. Hence

3sin⁡−1(3t)= π−2(π/4)=π/2,sin⁡−1(3t)= π/6.\begin{align*} 3\sin^{-1}(3t)=&\,\pi-2(\pi/4)=\pi/2,\\ \qquad \sin^{-1}(3t)=&\,\pi/6. \end{align*}

Taking sine gives 3t=1/23t=1/2, so t=1/6t=1/6.