题目

Problem

Figure 4 shows a sketch of part of the curve $C_1$ with equation

\begin{align*} y=12\sin x \end{align*}

Figure 4

where $x$ is measured in radians.

The point $P$ shown in Figure 4 is a maximum point on $C_1$ .

(a) Find the coordinates of $P$ .

(2)

The curve $C_2$ has equation

\begin{align*} y=12\sin x+k \end{align*}

where $k$ is a constant.

Given that the maximum value of $y$ on $C_2$ is $3$ ,

(b) find the coordinates of the minimum point on $C_2$ which has the smallest positive $x$ coordinate.

(2)

The curve $C_3$ has equation

\begin{align*} y=12\sin(x+B) \end{align*}

where $B$ is a positive constant.

Given that $\left(\dfrac{\pi}{4},A\right)$ , where $A$ is a constant, is the minimum point on $C_3$ which has the smallest positive $x$ coordinate,

(i) the value of $A$ ,

(ii) the smallest possible value of $B$ .

(2)

解答

(a)

解法一

思路

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$12\sin x$ 的最大值是 $12$ 。图中给出的最大点 $P$ 是右侧那一个，对应 $x=\frac{5\pi}{2}$ 。

答题过程

展开

For $y=12\sin x$ , the maximum value is $12$ .

The maximum point shown has

\begin{align*} x=\frac{5\pi}{2}. \end{align*}

Therefore

\begin{align*} P=\left(\frac{5\pi}{2},12\right). \end{align*}

(b)

解法一

思路

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$12\sin x$ 的最大值是 $12$ ，所以 $12\sin x+k$ 的最大值是 $12+k$ 。由最大值为 $3$ 可求 $k=-9$ 。最小值就是 $-12-9=-21$ 。

答题过程

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The maximum value of $12\sin x+k$ is

\begin{align*} 12+k. \end{align*}

Given that this maximum is $3$ ,

\begin{align*} 12+k=&\,3\\ k=&\,-9. \end{align*}

So the minimum value is

\begin{align*} -12-9=-21. \end{align*}

The smallest positive $x$ coordinate for a minimum of $\sin x$ is

\begin{align*} x=\frac{3\pi}{2}. \end{align*}

Therefore the point is

\begin{align*} \left(\frac{3\pi}{2},-21\right). \end{align*}

(c)

解法一

思路

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$12\sin(x+B)$ 的最小值一定是 $-12$ ，所以 $A=-12$ 。

最小点发生在

\begin{align*} x+B=\frac{3\pi}{2}+2n\pi. \end{align*}

题目给 $x=\frac{\pi}{4}$ ，求最小的正 $B$ 。

答题过程

展开

The minimum value of $12\sin(x+B)$ is

\begin{align*} -12. \end{align*}

Therefore

\begin{align*} A=-12. \end{align*}

For a minimum,

\begin{align*} x+B=\frac{3\pi}{2}+2n\pi. \end{align*}

Using $x=\frac{\pi}{4}$ and taking the smallest positive value of $B$ ,

\begin{align*} \frac{\pi}{4}+B=&\,\frac{3\pi}{2}\\ B=&\,\frac{3\pi}{2}-\frac{\pi}{4}\\ B=&\,\frac{5\pi}{4}. \end{align*}

\begin{align*} A=-12,\qquad B=\frac{5\pi}{4}. \end{align*}

IAL 2024 May Q11

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程