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Unit B3 · Functions & EquationsUnit B3 · 函数与方程

Functions with Asymptotes含渐近线的函数

IB-Style Practice Questions · Paper 1A · Paper 1B · Paper 2 · Paper 3IB 风格练习题 · 第一卷 A 节 · 第一卷 B 节 · 第二卷 · 第三卷

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus 2.8, 2.13考纲 2.8、2.13AA HL

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PART I  ·  PAPER 1 SECTION A第一部分  ·  第一卷 A 节No calculator · short response · 21 marks不可使用计算器 · 简答题 · 21 分

Section A · Short ResponseA 节 · 简答题

Show every algebraic step. For asymptotes, state both kinds (vertical, horizontal, or oblique) explicitly with an equation of the form $x = c$ or $y = c$. For holes, give an ordered pair $(x_0, y_0)$, not just the $x$ value. No calculator permitted.写出每一步代数过程。求渐近线时，须明确写出种类（竖直、水平或斜），并以 $x = c$ 或 $y = c$ 的方程形式给出。可去间断点须以有序对 $(x_0, y_0)$ 表示，不能只给 $x$。不可使用计算器。

Q1EASY Paper 1A 2.8 Linear over Linear [4 marks]

Let $f(x) = \dfrac{3x - 1}{x + 2}$.设 $f(x) = \dfrac{3x - 1}{x + 2}$。

(a) State the equation of the vertical asymptote.写出竖直渐近线的方程。 [2]
(b) State the equation of the horizontal asymptote.写出水平渐近线的方程。 [2]
Q2MEDIUM Paper 1A 2.8 Removable Hole [5 marks]

Let $f(x) = \dfrac{x^{2} - 9}{x - 3}$.设 $f(x) = \dfrac{x^{2} - 9}{x - 3}$。

(a) Factor the numerator and simplify $f(x)$ for $x \ne 3$.将分子因式分解，并对 $x \ne 3$ 化简 $f(x)$。 [2]
(b) State the coordinates of the removable hole.写出可去间断点的坐标。 [2]
(c) State whether $f$ has a vertical asymptote, and justify in one line.说明 $f$ 是否有竖直渐近线，并用一句话给出理由。 [1]
Q3MEDIUM Paper 1A 2.8 Inverse of Mobius Map [6 marks]

Let $f(x) = \dfrac{2x - 1}{x + 3}$, with $x \ne -3$.设 $f(x) = \dfrac{2x - 1}{x + 3}$，$x \ne -3$。

(a) Find $f^{-1}(x)$ in the form $\dfrac{\alpha x + \beta}{\gamma + \delta x}$ with integer coefficients.求 $f^{-1}(x)$，写成 $\dfrac{\alpha x + \beta}{\gamma + \delta x}$（系数为整数）的形式。 [3]
(b) State the vertical asymptote of $f$.写出 $f$ 的竖直渐近线。 [1]
(c) State the vertical asymptote of $f^{-1}$ and comment on its relationship to the horizontal asymptote of $f$.写出 $f^{-1}$ 的竖直渐近线，并说明它与 $f$ 的水平渐近线之间的关系。 [2]
Q4HARD Paper 1A 2.13 Rational Inequality by Sign Chart [6 marks]

Solve $\dfrac{x - 2}{x + 1} \ge 0$, $x \in \mathbb{R}$.解 $\dfrac{x - 2}{x + 1} \ge 0$，$x \in \mathbb{R}$。

(a) State the critical points (where the numerator is zero and where the expression is undefined). Indicate which is to be included and which excluded.写出临界点（分子为零之处与表达式无定义之处），并标明哪个包含、哪个排除。 [2]
(b) Build a sign chart for $\dfrac{x - 2}{x + 1}$ on the three intervals determined by the critical points.在临界点划分出的三个区间上为 $\dfrac{x - 2}{x + 1}$ 作符号表。 [3]
(c) Write the solution set using interval notation.用区间记号写出解集。 [1]
PART II  ·  PAPER 1 SECTION B第二部分  ·  第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分

Section B · Extended ResponseB 节 · 长答题

For oblique (slant) asymptotes, perform polynomial long division and write the quotient explicitly: the slant asymptote is the polynomial part. Sketches must label both intercepts (when finite) and every asymptote with its equation.求斜渐近线时须做多项式长除法，并显式写出商：斜渐近线即为多项式部分。草图须标出两类截距（若有限）及每条渐近线的方程。

Q5HARD Paper 1B 2.13 Slant Asymptote & Full Analysis (HL) [11 marks]

Let $f(x) = \dfrac{x^{2} + 2x + 5}{x - 1}$, $x \ne 1$.设 $f(x) = \dfrac{x^{2} + 2x + 5}{x - 1}$，$x \ne 1$。

(a) Use polynomial long division to write $f(x)$ in the form $f(x) = (mx + c) + \dfrac{R}{x - 1}$, stating $m$, $c$, and $R$.用多项式长除法把 $f(x)$ 写成 $f(x) = (mx + c) + \dfrac{R}{x - 1}$ 的形式，并写出 $m$、$c$、$R$。 [4]
(b) Hence state the equations of the vertical and oblique (slant) asymptotes.由此写出竖直渐近线与斜渐近线的方程。 [2]
(c) Show that the graph of $f$ never crosses its slant asymptote, by solving $f(x) = mx + c$.通过求解 $f(x) = mx + c$，证明 $f$ 的图像不与其斜渐近线相交。 [2]
(d) Sketch the graph of $f$, marking the $y$-intercept, both asymptotes (with equations), and the side of each asymptote on which each branch lies.绘制 $f$ 的草图，标出 $y$ 截距、两条渐近线（含方程），以及每一支位于每条渐近线的哪一侧。 [3]
PART III  ·  PAPER 2第三部分  ·  第二卷Calculator · mixed response · 16 marks可使用计算器 · 混合题型 · 16 分

Paper 2 · Calculator Permitted第二卷 · 允许使用计算器

A graphing calculator is required. For range from a graph, find any turning points exactly when possible and state explicitly which endpoints of the range are attained (closed bracket) versus only approached (open bracket). Quotient-rule derivatives must be presented in fully simplified form.需要图形计算器（GDC）。由图象求值域时，能精确求出转折点的尽量精确求出，并明确说明值域端点是取到（闭区间）还是仅趋近（开区间）。商法则求导后须化为最简形式。

Q6MEDIUM Paper 2 2.13 Range from GDC (HL) [7 marks]

Let $f(x) = \dfrac{x^{2} - 1}{x^{2} + 1}$, $x \in \mathbb{R}$.设 $f(x) = \dfrac{x^{2} - 1}{x^{2} + 1}$，$x \in \mathbb{R}$。

(a) State the equation of the horizontal asymptote, and use a limit argument to justify it.写出水平渐近线的方程，并用极限论证。 [2]
(b) Sketch $f$ on your GDC. Identify the global minimum value, with the $x$ at which it occurs.在 GDC 上绘制 $f$。指出全局最小值及其取值点 $x$。 [3]
(c) State the range of $f$ using interval notation. Use brackets carefully to mark attained versus unattained endpoints.用区间记号写出 $f$ 的值域。注意区分取到与未取到的端点。 [2]
Q7HARD Paper 2 2.13 Quotient Rule + Asymptote Analysis (HL) [9 marks]

Let $f(x) = \dfrac{x}{x^{2} + 1}$, $x \in \mathbb{R}$.设 $f(x) = \dfrac{x}{x^{2} + 1}$，$x \in \mathbb{R}$。

(a) Show that $f$ has no vertical asymptote, and state the horizontal asymptote.证明 $f$ 无竖直渐近线，并写出水平渐近线。 [2]
(b) Use the quotient rule to show that $f'(x) = \dfrac{1 - x^{2}}{(x^{2} + 1)^{2}}$.用商法则证明 $f'(x) = \dfrac{1 - x^{2}}{(x^{2} + 1)^{2}}$。 [3]
(c) Solve $f'(x) = 0$ to find the $x$ coordinates of the turning points, classify each as a maximum or minimum (justify briefly using the sign of $f'$), and compute the corresponding values of $f$.解 $f'(x) = 0$ 求出转折点的 $x$ 坐标，根据 $f'$ 的符号简要判定每点为极大或极小，并求对应的 $f$ 值。 [3]
(d) Hence write down the range of $f$.由此写出 $f$ 的值域。 [1]
PART IV  ·  PAPER 3第四部分  ·  第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 · HL Extended Problem第三卷 · HL 长题探究

A graphing calculator is required. This problem investigates the family of bilinear maps $f(x) = \dfrac{ax + b}{cx + d}$ with $ad - bc \ne 0$ and $c \ne 0$. Method marks dominate. State every algebraic condition you impose on $a, b, c, d$ before invoking it.需要图形计算器（GDC）。本题探究双线性映射族 $f(x) = \dfrac{ax + b}{cx + d}$，其中 $ad - bc \ne 0$ 且 $c \ne 0$。方法分占主导。在使用 $a, b, c, d$ 的任何代数条件前，须先把该条件写出。

Q8HARD Paper 3 2.8 Bilinear Maps · Self-Inverse (HL) [15 marks]

Throughout this problem, $f(x) = \dfrac{ax + b}{cx + d}$ where $a, b, c, d \in \mathbb{R}$, $c \ne 0$, and $ad - bc \ne 0$.本题中，$f(x) = \dfrac{ax + b}{cx + d}$，其中 $a, b, c, d \in \mathbb{R}$，$c \ne 0$，$ad - bc \ne 0$。

(a) State, in terms of $a, b, c, d$, the equations of the vertical asymptote and the horizontal asymptote of $f$.用 $a, b, c, d$ 表示 $f$ 的竖直渐近线和水平渐近线的方程。 [2]
(b) Show that $f^{-1}(x) = \dfrac{-dx + b}{cx - a}$.证明 $f^{-1}(x) = \dfrac{-dx + b}{cx - a}$。 [3]
(c) Hence state the vertical asymptote of $f^{-1}$, and show that it equals the horizontal asymptote of $f$. (This is the geometric meaning of "inverting a function reflects its graph in $y = x$".)由此写出 $f^{-1}$ 的竖直渐近线，并证明它等于 $f$ 的水平渐近线。（这正是"函数取逆使图象关于 $y = x$ 反射"的几何含义。） [2]
(d) Show that $f$ is self-inverse (that is, $f^{-1}(x) = f(x)$ for all $x$ in the domain) if and only if $a + d = 0$. (You may compare numerators and denominators of $f(x)$ and $f^{-1}(x)$ after expressing both in lowest terms.)证明 $f$ 是自逆函数（即对定义域内一切 $x$，$f^{-1}(x) = f(x)$）当且仅当 $a + d = 0$。（提示：将 $f(x)$ 与 $f^{-1}(x)$ 化为最简后比较分子和分母。） [5]
(e) Verify the result of (d) on $f(x) = \dfrac{2x + 3}{x - 2}$ by computing $f(f(x))$ and simplifying.在 $f(x) = \dfrac{2x + 3}{x - 2}$ 上验证 (d) 的结论：计算 $f(f(x))$ 并化简。 [3]