题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

Figure 5 shows the curve with equation

y=\frac{1+2\cos x}{1+\sin x}\qquad \frac{\pi}{2}<x<\frac{3\pi}{2}

The point $M$ , shown in Figure 5, is the minimum point on the curve.

(a) Show that the $x$ coordinate of $M$ is a solution of the equation

2\sin x+\cos x=-2

(4)

(b) Hence find, to 3 significant figures, the $x$ coordinate of $M$ .

(5)

题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

图 5 给出了曲线

y=\frac{1+2\cos x}{1+\sin x}\qquad \frac{\pi}{2}<x<\frac{3\pi}{2}

的图像。

图中的点 $M$ 是曲线上的最低点。

(a) 证明点 $M$ 的 $x$ 坐标是方程

2\sin x+\cos x=-2

的一个解。

(b) 由此求点 $M$ 的 $x$ 坐标，答案保留 3 位有效数字。

解答

We have

y=\frac{1+2\cos x}{1+\sin x}

Using the quotient rule,

\frac{dy}{dx} =\frac{(1+\sin x)(-2\sin x)-(1+2\cos x)\cos x}{(1+\sin x)^2}

At the minimum point $M$ ,

\frac{dy}{dx}=0

Since

(1+\sin x)^2\ne0

inside the interval, the numerator must be zero:

(1+\sin x)(-2\sin x)-(1+2\cos x)\cos x=0

Expand:

-2\sin x-2\sin^2x-\cos x-2\cos^2x=0

Using

\sin^2x+\cos^2x=1

we get

-2\sin x-\cos x-2=0

Therefore

2\sin x+\cos x=-2

So the $x$ coordinate of $M$ is a solution of

\boxed{2\sin x+\cos x=-2}

as required.

Write

2\sin x+\cos x=R\sin(x+\alpha)

Then

R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha

Comparing coefficients,

R\cos\alpha=2, \qquad R\sin\alpha=1

R=\sqrt5

and

\tan\alpha=\frac12

Thus

\alpha=\tan^{-1}\frac12

The equation becomes

\sqrt5\sin(x+\alpha)=-2

\sin(x+\alpha)=-\frac{2}{\sqrt5}

Now

\alpha=0.4636\ldots

For

\frac{\pi}{2}<x<\frac{3\pi}{2}

the solution giving the minimum point is

x+\alpha=\pi+\sin^{-1}\frac{2}{\sqrt5}

Therefore

x=\pi+\sin^{-1}\frac{2}{\sqrt5}-\tan^{-1}\frac12

x=3.785\ldots

Hence, to 3 significant figures,

\boxed{x=3.79}