题目

Problem

Figure 4 shows a sketch of the curve with equation

\begin{align*} y=\tan x,\qquad -2\pi\le x\le2\pi. \end{align*}

Figure 4

The line $l$ , shown in Figure 4, is an asymptote to $y=\tan x$ .

(a) State an equation for $l$ .

(1)

A copy of Figure 4, labelled Diagram 1, is shown on the next page.

Diagram 1

(b)

(i) On Diagram 1, sketch the curve with equation

\begin{align*} y=\frac1x+1,\qquad -2\pi\le x\le2\pi, \end{align*}

stating the equation of the horizontal asymptote of this curve.

(ii) Hence, giving a reason, state the number of solutions of the equation

\begin{align*} \tan x=\frac1x+1 \end{align*}

in the region $-2\pi\le x\le2\pi$ .

(4)

(i) $0\le x\le40\pi$ ,

(ii) $-10\pi\le x\le\dfrac52\pi$ .

(2)

解答

(a)

解法一

思路

展开

$\tan x$ 的竖直渐近线在 $x=\frac\pi2+n\pi$ 。图中标出的右侧渐近线是 $x=\frac{3\pi}{2}$ 。

答题过程

展开

An equation for $l$ is

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

(b)

解法一

思路

展开

$y=\frac1x+1$ 是 reciprocal 图像上移 $1$ 。所以竖直渐近线是 $x=0$ ，水平渐近线是 $y=1$ 。方程的解就是两张图像的交点个数。

答题过程

展开

For

\begin{align*} y=\frac1x+1, \end{align*}

the horizontal asymptote is

\begin{align*} y=1. \end{align*}

The sketch should show two reciprocal branches with vertical asymptote $x=0$ and horizontal asymptote $y=1$ .

The number of solutions of

\begin{align*} \tan x=\frac1x+1 \end{align*}

is the number of intersections of the two graphs. From the sketch on $-2\pi\le x\le2\pi$ , there are

\begin{align*} 5 \end{align*}

solutions.

(c)

解法一

思路

展开

在正半轴上，每个长度为 $\pi$ 的 tangent 分支与 $y=\frac1x+1$ 有一个交点。负半轴区间要结合图像分支数来数。

答题过程

展开

For $0\le x\le40\pi$ , there are

\begin{align*} 40 \end{align*}

solutions.

For $-10\pi\le x\le\frac52\pi$ , there are

\begin{align*} 14 \end{align*}

solutions.

IAL 2021 June Q9

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