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Rational and Radical Expressions · Solutions有理表达式与根式表达式 · 参考答案

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EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 参考答案SAT MCQ + ON/BC short answer · 22 marksSAT 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详解

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Simplify Rationals化简有理表达式 · HSA-APR.D.6 [3 marks][3 分]

Simplified form of $\dfrac{x^{2} - 25}{x^{2} - 4 x - 5}$ with restrictions.$\dfrac{x^{2} - 25}{x^{2} - 4 x - 5}$ 的最简形式及限制条件。

Answer:答案：  (B)  $\dfrac{x + 5}{x + 1}$, $x \ne -1, 5$

(a) Factor numerator and denominator分子分母分别因式分解 M1·A1

Numerator: $x^{2} - 25 = (x - 5)(x + 5)$ (difference of squares). Denominator: $x^{2} - 4x - 5 = (x - 5)(x + 1)$ (integers with product $-5$ and sum $-4$ are $-5$ and $1$). The full restriction set comes from the original denominator zeros: $x - 5 = 0$ or $x + 1 = 0$, so $x \ne 5$ and $x \ne -1$.分子：$x^{2} - 25 = (x - 5)(x + 5)$（平方差公式）。分母：$x^{2} - 4x - 5 = (x - 5)(x + 1)$（满足乘积为 $-5$、和为 $-4$ 的整数是 $-5$ 与 $1$）。完整的限制条件集合来自原分母的零点：$x - 5 = 0$ 或 $x + 1 = 0$，即 $x \ne 5$ 且 $x \ne -1$。

(b) Cancel the common factor; reaffirm restrictions约去公因式，再次写出限制条件 A1

$$ \frac{(x - 5)(x + 5)}{(x - 5)(x + 1)} \;=\; \frac{x + 5}{x + 1}, \qquad x \ne -1, 5. $$ Match: option (B).对应选项 (B)。
Why the wrong choices fail.错误选项为何不成立。
  • (A) $\frac{x - 5}{x - 1}$, $x \ne 5$, denominator mis-factored as $(x-5)(x-1)$, which expands to $x^2-6x+5$, not $x^2-4x-5$. Also drops the $x \ne -1$ restriction.：将分母错误分解为 $(x-5)(x-1)$，展开后为 $x^2-6x+5$，并非 $x^2-4x-5$。同时遗漏限制条件 $x \ne -1$。
  • (C) $\frac{x + 5}{x - 1}$ (no restrictions)（无限制条件）, correct form of the simplified fraction would still need restrictions; "no restrictions" violates HSA-APR.D.6, which requires that the simplified rational be equivalent to the original on the original domain.：即便是最简形式仍需写出限制条件；"无限制条件"违反 HSA-APR.D.6，该标准要求最简有理表达式在原定义域上与原表达式等价。
  • (D) $\frac{x + 5}{x + 1}$, $x \ne -1$, correct simplified form but drops $x \ne 5$. After cancellation the $(x-5)$ factor is invisible, but the hole at $x = 5$ in the original function survives, restrictions must be stated before cancellation.：最简形式正确，但遗漏 $x \ne 5$。约分后因式 $(x-5)$ 不再显见，但原函数在 $x = 5$ 处的空洞依然存在——限制条件必须在约分前写出。
Restrictions are read off the original denominator, not the simplified one.限制条件须从原分母读出，而非化简后的分母。 Cancellation hides domain information; HSA-APR.D.6 requires that every zero of the original denominator be listed, even if the corresponding factor cancels. The diagnostic: state restrictions first, then simplify. A cancelled factor like $(x-5)$ in this question marks a point discontinuity (hole), not a vertical asymptote, the function value is undefined at $x = 5$ but the limit exists. (This vocabulary returns in Q9.) Distractor (D) is the classic trap: the simplified expression is correct but the restriction list is incomplete.约分会隐藏定义域信息；HSA-APR.D.6 要求列出原分母的每一个零点，即使对应因式被约去。诊断方法：先写限制条件，再化简。本题中被约去的 $(x-5)$ 对应一个可去间断点（空洞），而非垂直渐近线——在 $x = 5$ 处函数值无定义但极限存在。（这一概念将在 Q9 中再次出现。）干扰项 (D) 是典型陷阱：化简表达式正确但限制条件列表不完整。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §5 Radical Operations根式运算 · HSN-RN.A.2 [3 marks][3 分]

Simplify $\sqrt{50} + \sqrt{18} - \sqrt{8}$.化简 $\sqrt{50} + \sqrt{18} - \sqrt{8}$。

Answer:答案：  (A)  $6\sqrt{2}$

(a) Extract the largest perfect-square factor from each radicand从每个被开方数中提取最大的完全平方因数 M1·A1

$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$; $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$; $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$. Every term now carries the same radical $\sqrt{2}$, they are like radicals and combine just as $\sqrt{2}$ were a single variable.$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$；$\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$；$\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$。三项均含相同根号 $\sqrt{2}$，互为同类根式，可以像把 $\sqrt{2}$ 当作单一变量一样合并。

(b) Combine like radicals合并同类根式 A1

$$ 5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2} \;=\; (5 + 3 - 2)\sqrt{2} \;=\; 6\sqrt{2}. $$ Match: option (A).对应选项 (A)。
Why the wrong choices fail.错误选项为何不成立。
  • (B) $\sqrt{60}$, radicands cannot be added inside a single radical. $\sqrt{50}+\sqrt{18}-\sqrt{8} \ne \sqrt{50+18-8}$; the operation $\sqrt{a}+\sqrt{b} = \sqrt{a+b}$ is false in general.：被开方数不能在同一根号内相加。$\sqrt{50}+\sqrt{18}-\sqrt{8} \ne \sqrt{50+18-8}$；一般情况下 $\sqrt{a}+\sqrt{b} = \sqrt{a+b}$ 不成立。
  • (C) $10\sqrt{2}$, arithmetic slip: $5 + 3 + 2 = 10$ would arise from adding the third term instead of subtracting.：算术失误：$5 + 3 + 2 = 10$ 来自把第三项当成加号而非减号。
  • (D) $2\sqrt{10}$, algebraic-looking decoy with the wrong magnitude; $2\sqrt{10} \approx 6.32$ is close to the correct $6\sqrt{2} \approx 8.49$ but not equal.：看似合理的代数干扰项，但数值不符；$2\sqrt{10} \approx 6.32$ 与正确答案 $6\sqrt{2} \approx 8.49$ 相近但并不相等。
Like radicals combine like like terms, they must share both index and radicand.同类根式如同同类项才能合并——必须同时具有相同的根指数和相同的被开方数。 $\sqrt{2}$ and $\sqrt{3}$ do not combine; $\sqrt{2}$ and $\sqrt[3]{2}$ do not combine. The simplification habit: pull out the largest perfect-square factor first, then check whether the leftover radicals match. The SAT loves seeding choice (B) as the "add the radicands" trap, it tests whether the student has internalised that radicals are operators, not just labels.$\sqrt{2}$ 与 $\sqrt{3}$ 不能合并；$\sqrt{2}$ 与 $\sqrt[3]{2}$ 也不能合并。化简习惯：先提取最大的完全平方因数，然后再检查剩余根号是否相同。SAT 常用 (B)"把被开方数相加"作为陷阱，考察学生是否真正理解根号是运算符，而不仅是标签。
Q3EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §4 Horizontal Asymptote水平渐近线 · HSF-IF.B.4 [3 marks][3 分]

Horizontal asymptote of $f(x) = \dfrac{4 x^{2} + 3 x - 1}{2 x^{2} - 5}$.$f(x) = \dfrac{4 x^{2} + 3 x - 1}{2 x^{2} - 5}$ 的水平渐近线。

Answer:答案：  (B)  $y = 2$

(a) Compare degrees比较次数 M1

Numerator degree $= 2$; denominator degree $= 2$. Equal degrees. The horizontal-asymptote rule for rational functions:分子次数 $= 2$；分母次数 $= 2$，次数相等。有理函数水平渐近线规则如下：
  • If $\deg(\text{num}) < \deg(\text{den})$: HA at $y = 0$.若 $\deg(\text{分子}) < \deg(\text{分母})$：水平渐近线为 $y = 0$。
  • If $\deg(\text{num}) = \deg(\text{den})$: HA at $y = \dfrac{\text{leading coeff of num}}{\text{leading coeff of den}}$.若 $\deg(\text{分子}) = \deg(\text{分母})$：水平渐近线为 $y = \dfrac{\text{分子首项系数}}{\text{分母首项系数}}$。
  • If $\deg(\text{num}) = \deg(\text{den}) + 1$: no HA, but a slant asymptote (long division).若 $\deg(\text{分子}) = \deg(\text{分母}) + 1$：无水平渐近线，但有斜渐近线（用长除法求）。
  • If $\deg(\text{num}) > \deg(\text{den}) + 1$: no HA, no slant.若 $\deg(\text{分子}) > \deg(\text{分母}) + 1$：既无水平渐近线，也无斜渐近线。

(b) Apply the equal-degree shortcut应用等次数捷径 A1·A1

Ratio of leading coefficients: $\dfrac{4}{2} = 2$. So $y = 2$ is the HA. Match: option (B).首项系数之比：$\dfrac{4}{2} = 2$。因此水平渐近线为 $y = 2$，对应选项 (B)。
Why the wrong choices fail.错误选项为何不成立。
  • (A) $y = 0$, applies only when the numerator's degree is strictly less than the denominator's. Here degrees are equal.：仅当分子次数严格小于分母次数时才成立。这里两次数相等。
  • (C) $y = -1/5$, wrong shortcut: ratio of the constant terms $-1/-5 = 1/5$ (with a sign error). The HA shortcut uses leading coefficients, not constants, as $x \to \infty$ the leading terms dominate.：使用了错误的捷径——常数项之比 $-1/-5 = 1/5$（且符号有误）。水平渐近线公式使用的是首项系数，而非常数项，因为 $x \to \infty$ 时由首项主导。
  • (D) No HA; slant asymptote instead无水平渐近线，为斜渐近线, applies only when $\deg(\text{num}) = \deg(\text{den}) + 1$ exactly. Here degrees are equal, not off by one.：仅当 $\deg(\text{分子}) = \deg(\text{分母}) + 1$ 时成立。这里两次数相等而非相差 $1$。
End-behaviour is a leading-coefficient race.末端行为是首项系数之间的角力。 As $|x| \to \infty$ every non-leading term becomes negligible: $\frac{4x^2 + 3x - 1}{2x^2 - 5} \approx \frac{4x^2}{2x^2} = 2$. The constant terms $-1$ and $-5$ never enter the asymptote, they affect transient behaviour for small $x$ but vanish at infinity. The four-case degree comparison above is worth memorising as a single block: AP/SAT/Provincial all test it identically and a student who reaches for long division on every problem wastes time on the easy half of the exam.当 $|x| \to \infty$ 时，所有非首项均可忽略：$\frac{4x^2 + 3x - 1}{2x^2 - 5} \approx \frac{4x^2}{2x^2} = 2$。常数项 $-1$ 与 $-5$ 不进入渐近线表达式——它们只影响 $x$ 较小时的瞬态行为，在无穷处消失。上述四种情形的次数比较值得作为整体记忆：AP/SAT/省考考察方式完全一致，凡题必长除法会浪费考试中容易得分的部分。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Add & Subtract Rationals有理表达式的加减 · MHF4U Strand C [6 marks][6 分]

Combine into single rational in lowest terms; state restrictions. (a) $\frac{3}{x-2}+\frac{5}{x+4}$. (b) $\frac{2x}{x^2-9}-\frac{1}{x-3}$.合并为最简形式的单一有理表达式，并写出限制条件。(a) $\frac{3}{x-2}+\frac{5}{x+4}$。 (b) $\frac{2x}{x^2-9}-\frac{1}{x-3}$。

Answer:答案：  (a) $\dfrac{2(4x + 1)}{(x - 2)(x + 4)}$, $x \ne 2, -4$  ·  (b) $\dfrac{1}{x + 3}$, $x \ne 3, -3$

(a) Form the LCD; combine; state restrictions通分求 LCD；合并；写出限制条件 M1·A1·A1

Denominators $(x-2)$ and $(x+4)$ share no factor, so LCD $= (x-2)(x+4)$. Restrictions: $x \ne 2$ and $x \ne -4$ (stated before clearing).分母 $(x-2)$ 与 $(x+4)$ 无公因式，所以 LCD $= (x-2)(x+4)$。限制条件：$x \ne 2$ 且 $x \ne -4$（在去分母之前写出）。 $$ \frac{3}{x-2} + \frac{5}{x+4} \;=\; \frac{3(x+4) + 5(x-2)}{(x-2)(x+4)} \;=\; \frac{3x + 12 + 5x - 10}{(x-2)(x+4)} \;=\; \frac{8x + 2}{(x-2)(x+4)} \;=\; \frac{2(4x + 1)}{(x-2)(x+4)}. $$ The numerator $8x + 2 = 2(4x + 1)$ shares no factor with the denominator, so the form is in lowest terms.分子 $8x + 2 = 2(4x + 1)$ 与分母无公因式，因此已是最简形式。

(b) Factor the first denominator; form LCD; combine先因式分解第一个分母；求 LCD；合并 M1·A1·A1

$x^2 - 9 = (x - 3)(x + 3)$, so denominators are $(x-3)(x+3)$ and $(x-3)$. LCD $= (x-3)(x+3)$. Restrictions: $x \ne 3$ and $x \ne -3$ (stated before clearing).$x^2 - 9 = (x - 3)(x + 3)$，所以分母分别是 $(x-3)(x+3)$ 与 $(x-3)$。LCD $= (x-3)(x+3)$。限制条件：$x \ne 3$ 且 $x \ne -3$（去分母之前写出）。 $$ \frac{2x}{(x-3)(x+3)} - \frac{1}{x-3} \;=\; \frac{2x - (x+3)}{(x-3)(x+3)} \;=\; \frac{2x - x - 3}{(x-3)(x+3)} \;=\; \frac{x - 3}{(x-3)(x+3)}. $$ The $(x-3)$ factor cancels, leaving因式 $(x-3)$ 约去后，剩下 $$ \frac{1}{x + 3}, \qquad x \ne 3, -3. $$ Reaffirm both restrictions:再次写出两个限制条件： the surviving denominator $x + 3$ forces $x \ne -3$, and the cancelled $(x - 3)$ leaves a point discontinuity at $x = 3$, both are required by HSA-APR.D.6.保留下的分母 $x + 3$ 要求 $x \ne -3$，而被约去的 $(x - 3)$ 在 $x = 3$ 处留下可去间断点——两者均为 HSA-APR.D.6 所要求。
State restrictions before finding the LCD, never after cancellation.限制条件须在通分之前写出，绝不能等到约分之后。 If you wait until the simplified form to read restrictions, the cancelled factor's zero becomes invisible and you lose marks. The professional protocol is a fixed three-step sequence: (1) factor every denominator, (2) read the restriction set off the union of zeros, (3) only then form the LCD and combine. Part (b) is the canonical trap: the answer looks like $\frac{1}{x+3}$ with restriction $x \ne -3$, but the original $\frac{2x}{x^2-9}$ also excluded $x = 3$. Provincial markers deduct an A1 for incomplete restrictions every time.若等到最简形式才写限制条件，被约去因式的零点就会"消失"，从而失分。专业流程是固定的三步序列：(1) 将每个分母因式分解，(2) 取所有零点的并集作为限制条件集合，(3) 然后才通分合并。(b) 是典型陷阱：答案 $\frac{1}{x+3}$ 看起来只需 $x \ne -3$，但原表达式 $\frac{2x}{x^2-9}$ 同样排除 $x = 3$。省考阅卷人遇到限制条件不完整每次都会扣 A1。
Q5MEDIUM中Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §7 Rational Exponents有理指数 · HSN-RN.A.1 / BC PC11 [7 marks][7 分]

$x > 0$ throughout. (a) Evaluate $27^{2/3}$ and $32^{-3/5}$. (b) Single-power form of $\frac{\sqrt{x}\cdot\sqrt[3]{x^2}}{\sqrt[6]{x^5}}$. (c) Cite HSN-RN.A.1.全程假设 $x > 0$。 (a) 求 $27^{2/3}$ 与 $32^{-3/5}$。 (b) 将 $\frac{\sqrt{x}\cdot\sqrt[3]{x^2}}{\sqrt[6]{x^5}}$ 写成单一幂形式。 (c) 引用 HSN-RN.A.1。

Answer:答案：  (a) $27^{2/3} = 9$, $32^{-3/5} = \dfrac{1}{8}$  ·  (b) $x^{1/3}$  ·  (c) see cited explanation见所引用的说明

(a) Evaluate using $a^{m/n} = (\sqrt[n]{a})^{m}$利用 $a^{m/n} = (\sqrt[n]{a})^{m}$ 求值 A1·A1

$27^{2/3} = (\sqrt[3]{27})^{2} = 3^{2} = 9$. (Take the cube root first to keep magnitudes small, then square, the inner-root-first order is computationally easier than $(27^{2})^{1/3} = 729^{1/3}$.)$27^{2/3} = (\sqrt[3]{27})^{2} = 3^{2} = 9$。（先开立方根以保持数值较小，再平方——先开根的顺序在计算上比 $(27^{2})^{1/3} = 729^{1/3}$ 简便。）
$32^{-3/5} = \dfrac{1}{32^{3/5}} = \dfrac{1}{(\sqrt[5]{32})^{3}} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$. (Negative exponent flips to reciprocal first; then root-then-power as above.)$32^{-3/5} = \dfrac{1}{32^{3/5}} = \dfrac{1}{(\sqrt[5]{32})^{3}} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$。（负指数先取倒数，再按"先开根后乘方"的顺序。）

(b) Convert each radical to a rational exponent; combine via LCD of fractions将每个根式转化为有理指数；通过分数通分合并 M1·A1·A1

$\sqrt{x} = x^{1/2}$, $\sqrt[3]{x^{2}} = x^{2/3}$, $\sqrt[6]{x^{5}} = x^{5/6}$. So$\sqrt{x} = x^{1/2}$、$\sqrt[3]{x^{2}} = x^{2/3}$、$\sqrt[6]{x^{5}} = x^{5/6}$。因此 $$ \frac{\sqrt{x}\cdot\sqrt[3]{x^{2}}}{\sqrt[6]{x^{5}}} \;=\; \frac{x^{1/2} \cdot x^{2/3}}{x^{5/6}} \;=\; x^{1/2 + 2/3 - 5/6}. $$ LCD of $\{2, 3, 6\}$ is $6$: $\frac{1}{2} = \frac{3}{6}$, $\frac{2}{3} = \frac{4}{6}$, $\frac{5}{6} = \frac{5}{6}$. So the exponent is $\frac{3 + 4 - 5}{6} = \frac{2}{6} = \frac{1}{3}$. Final form: $\;x^{1/3}$ (equivalently $\sqrt[3]{x}$).$\{2, 3, 6\}$ 的最小公分母是 $6$：$\frac{1}{2} = \frac{3}{6}$、$\frac{2}{3} = \frac{4}{6}$、$\frac{5}{6} = \frac{5}{6}$。指数为 $\frac{3 + 4 - 5}{6} = \frac{2}{6} = \frac{1}{3}$。最终结果：$\;x^{1/3}$（等价于 $\sqrt[3]{x}$）。

(c) Justify the radical–exponent identification via利用 HSN-RN.A.1 说明根式与有理指数等同的合理性 R1·R1

HSN-RN.A.1: "Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents.解释有理指数的定义如何由把整数指数的运算律延伸至这些值而得到，从而允许用有理指数记号来表示根式。" Concretely: if the exponent law $(a^{p})^{q} = a^{pq}$ must continue to hold for rational $p, q$, then $(a^{1/n})^{n} = a^{(1/n) \cdot n} = a^{1} = a$. But the unique positive number whose $n$-th power is $a$ (for $a > 0$) is by definition $\sqrt[n]{a}$. Hence $a^{1/n} = \sqrt[n]{a}$ is forced by the requirement of preserving the integer-exponent law, it is the only definition that makes the algebra of exponents consistent across rationals.具体来说：若指数律 $(a^{p})^{q} = a^{pq}$ 在有理 $p, q$ 上仍要成立，则 $(a^{1/n})^{n} = a^{(1/n) \cdot n} = a^{1} = a$。但对 $a > 0$ 而言，满足 $n$ 次幂等于 $a$ 的唯一正数按定义就是 $\sqrt[n]{a}$。因此 $a^{1/n} = \sqrt[n]{a}$ 是保持整数指数运算律所被迫的定义——它是唯一能让指数代数在有理数上保持一致的定义。
Rational exponent and radical are two notations for one object, pick whichever simplifies the algebra in front of you.有理指数与根式是同一对象的两种记号——挑选当下能简化代数的那一种即可。 The rational-exponent form excels at multiplicative combination (Q5b: just add fractions in the exponent). The radical form excels at numerical evaluation when the inner root is clean (Q5a: $\sqrt[3]{27} = 3$ is a one-liner). The fluent move is to convert in the direction that shortens the work: rationals-in $\to$ exponent arithmetic; nice integer roots $\to$ stay radical. The HSN-RN.A.1 justification matters for AP graders because it underlies why the law $a^{m/n} \cdot a^{p/q} = a^{m/n + p/q}$ holds in the first place, it's not a definition imposed by convention but a theorem forced by consistency.有理指数形式擅长乘法合并（Q5b：在指数上做分数运算即可）。当被开方数能开出整数根时，根式形式擅长数值求值（Q5a：$\sqrt[3]{27} = 3$ 一步搞定）。熟练做法是按能缩短工作量的方向转换：含分数 $\to$ 指数运算；整齐的整数根 $\to$ 保留根式。HSN-RN.A.1 的说明对 AP 阅卷人尤为重要，因为它揭示了 $a^{m/n} \cdot a^{p/q} = a^{m/n + p/q}$ 这一律为何成立——它不是约定俗成的定义，而是由相容性所迫得出的定理。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  长答题 · 参考答案AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分

Section B · Worked SolutionsB 部分 · 详解

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Add & Subtract Rationals有理表达式的加减 · HSA-APR.D.6 [8 marks][8 分]

$R(x) = \frac{x+1}{x^2-4} + \frac{2}{x-2} - \frac{1}{x+2}$. (a) LCD + restrictions. (b) Combine; expand numerator. (c) Simplify; full restrictions.$R(x) = \frac{x+1}{x^2-4} + \frac{2}{x-2} - \frac{1}{x+2}$。 (a) LCD 与限制条件。 (b) 合并；展开分子。 (c) 化简；完整限制条件。

Answer:答案：  (a) LCD $= (x-2)(x+2)$, $x \ne 2, -2$  ·  (b) $R(x) = \dfrac{2x + 7}{(x-2)(x+2)}$  ·  (c) already in lowest terms; $x \ne 2, -2$已是最简形式；$x \ne 2, -2$

(a) Factor every denominator; state LCD and restrictions将每个分母因式分解；写出 LCD 与限制条件 M1·A1

$x^{2} - 4 = (x - 2)(x + 2)$; the other two denominators are $(x - 2)$ and $(x + 2)$. LCD $= (x - 2)(x + 2)$. Restriction set (stated before combining): $x \ne 2$ and $x \ne -2$.$x^{2} - 4 = (x - 2)(x + 2)$；另外两个分母分别为 $(x - 2)$ 与 $(x + 2)$。LCD $= (x - 2)(x + 2)$。限制条件集合（在合并之前写出）：$x \ne 2$ 且 $x \ne -2$。

(b) Rewrite each term over the LCD; expand and collect the numerator把每一项写到 LCD 之上；展开并合并分子 M1·A1·A1·A1

$$ R(x) \;=\; \frac{x+1}{(x-2)(x+2)} + \frac{2(x+2)}{(x-2)(x+2)} - \frac{1(x-2)}{(x-2)(x+2)} \;=\; \frac{(x+1) + 2(x+2) - (x-2)}{(x-2)(x+2)}. $$ Expand and collect:展开合并： $$ (x + 1) + (2x + 4) - (x - 2) \;=\; x + 1 + 2x + 4 - x + 2 \;=\; 2x + 7. $$ So $R(x) = \dfrac{2x + 7}{(x - 2)(x + 2)}$. The numerator $2x + 7$ is linear with root $x = -7/2$, which is neither $2$ nor $-2$, so no factor of the denominator divides it, already in lowest terms.因此 $R(x) = \dfrac{2x + 7}{(x - 2)(x + 2)}$。分子 $2x + 7$ 是一次式，其根 $x = -7/2$ 既非 $2$ 也非 $-2$，故分母中没有因式能整除它——已是最简形式。

(c) Final form; reaffirm full restriction set最终形式；再次写出完整限制条件集合 A1

$$ R(x) \;=\; \frac{2x + 7}{(x - 2)(x + 2)}, \qquad x \ne 2, -2. $$ No cancellation occurred, so the restriction set from part (a) carries through unchanged. The simplified form is equivalent to the original $R(x)$ on the original domain, satisfying HSA-APR.D.6.由于未发生约分，(a) 部分的限制条件原样保留。最简形式在原定义域上与原 $R(x)$ 等价，满足 HSA-APR.D.6。
Always do the sign of the subtraction before reaching for the LCD.先处理减号，再去通分。 The most common mark-losing error in Q6-style problems is dropping the minus sign on $\frac{1}{x+2}$: students write $(x+1) + 2(x+2) - 1 \cdot (x-2)$ but expand the last term as $(x - 2)$ instead of $-(x - 2) = -x + 2$. The bracket discipline: write the subtraction with its bracket, "$-(x - 2)$", before distributing, never as "$- x - 2$". When the numerator does not factor against the denominator (as here, $2x+7$ has root $-7/2$), the work is done; no cancellation step is needed beyond the audit.Q6 类题最常见的失分错误是把 $\frac{1}{x+2}$ 前的负号丢掉：学生写出 $(x+1) + 2(x+2) - 1 \cdot (x-2)$，却把最后一项展开成 $(x - 2)$ 而非 $-(x - 2) = -x + 2$。括号纪律：在分配律前把减号带括号写成 "$-(x - 2)$"，绝不可写成 "$- x - 2$"。当分子与分母不存在公因式（如本题 $2x+7$ 的根为 $-7/2$）时即告结束；除核对外不需要约分。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Rational Equations有理方程 · HSA-REI.A.2 [9 marks][9 分]

Solve, state restrictions before clearing, verify each candidate. (a) $\frac{1}{x-1}+\frac{1}{x+1}=\frac{4}{x^2-1}$. (b) $\frac{x}{x-3}-\frac{2}{x}=\frac{6}{x(x-3)}$. (c) Cite HSA-REI.A.2.求解，在去分母前写出限制条件，并对每个候选根进行验证。 (a) $\frac{1}{x-1}+\frac{1}{x+1}=\frac{4}{x^2-1}$。 (b) $\frac{x}{x-3}-\frac{2}{x}=\frac{6}{x(x-3)}$。 (c) 引用 HSA-REI.A.2。

Answer:答案：  (a) $x = 2$ (verified)$x = 2$（已验证）  ·  (b) $x = 2$; $x = 0$ extraneous$x = 2$；$x = 0$ 为增根  ·  (c) see cited explanation见所引用的说明

(a) Restrictions $\to$ LCD $\to$ solve $\to$ verify限制条件 $\to$ LCD $\to$ 求解 $\to$ 验证 M1·A1·A1·A1

$x^{2} - 1 = (x - 1)(x + 1)$. Restrictions: $x \ne 1, -1$. LCD $= (x - 1)(x + 1)$. Multiply through:$x^{2} - 1 = (x - 1)(x + 1)$。限制条件：$x \ne 1, -1$。LCD $= (x - 1)(x + 1)$。两边同乘 LCD： $$ (x + 1) + (x - 1) \;=\; 4 \;\Longrightarrow\; 2x \;=\; 4 \;\Longrightarrow\; x \;=\; 2. $$ Verification验证 (required for rational equations): $x = 2$ satisfies the restrictions ($2 \ne \pm 1$). LHS: $\frac{1}{1} + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3}$. RHS: $\frac{4}{4 - 1} = \frac{4}{3}$. LHS $=$ RHS. $\checkmark$ Solution: $x = 2$, with no extraneous candidates.（有理方程必需）：$x = 2$ 满足限制条件（$2 \ne \pm 1$）。左边：$\frac{1}{1} + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3}$。右边：$\frac{4}{4 - 1} = \frac{4}{3}$。左 $=$ 右。$\checkmark$ 解：$x = 2$，无增根。

(b) Restrictions $\to$ LCD $\to$ solve $\to$ verify; flag extraneous root限制条件 $\to$ LCD $\to$ 求解 $\to$ 验证；标出增根 M1·A1·A1·A1

Restrictions: $x \ne 0$ and $x \ne 3$. LCD $= x(x - 3)$. Multiply through:限制条件：$x \ne 0$ 且 $x \ne 3$。LCD $= x(x - 3)$。两边同乘 LCD： $$ x \cdot x \;-\; 2(x - 3) \;=\; 6 \;\Longrightarrow\; x^{2} - 2x + 6 \;=\; 6 \;\Longrightarrow\; x^{2} - 2x \;=\; 0 \;\Longrightarrow\; x(x - 2) \;=\; 0. $$ Candidate roots: $x = 0$ and $x = 2$.候选根：$x = 0$ 与 $x = 2$。
Verification of $x = 0$:验证 $x = 0$： violates the restriction $x \ne 0$ (would force division by zero in the original equation). Extraneous, reject.违反限制条件 $x \ne 0$（会导致原方程出现除以零）。增根，舍去。
Verification of $x = 2$:验证 $x = 2$： $2 \ne 0$ and $2 \ne 3$, so the restriction is satisfied. LHS: $\frac{2}{-1} - \frac{2}{2} = -2 - 1 = -3$. RHS: $\frac{6}{2 \cdot (-1)} = \frac{6}{-2} = -3$. LHS $=$ RHS. $\checkmark$ Solution: $x = 2$ only.$2 \ne 0$ 且 $2 \ne 3$，满足限制条件。左边：$\frac{2}{-1} - \frac{2}{2} = -2 - 1 = -3$。右边：$\frac{6}{2 \cdot (-1)} = \frac{6}{-2} = -3$。左 $=$ 右。$\checkmark$ 解：仅 $x = 2$。

(c) Cite引用 HSA-REI.A.2 R1

HSA-REI.A.2: "Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.求解一元简单有理方程与根式方程，并举例说明增根如何产生。" Multiplying both sides of an equation by an expression containing the variable (here, the LCD $x(x - 3)$) is not a reversible operation at the values that make the multiplier zero. Those zero-of-LCD values can become "solutions" of the cleared polynomial equation but are not solutions of the original rational equation, hence the verification step is structurally mandatory in part (b) to catch $x = 0$ as extraneous. (In part (a) the cleared equation happened to have no zero-of-LCD root, so verification turned up no extraneous solutions; the step is still required, not optional.)两边同乘含变量的表达式（此处为 LCD $x(x - 3)$）在使乘子为零的取值处不是可逆运算。那些使 LCD 为零的值可能成为去分母后多项式方程的"解"，但不是原有理方程的解——因此 (b) 的验证步骤在结构上是强制性的，用以识别 $x = 0$ 为增根。（在 (a) 中，去分母后的方程恰好没有使 LCD 为零的根，所以验证未发现增根；但该步骤仍是必需的，并非可选。）
Extraneous roots are an artefact of multiplying by a variable expression, not of arithmetic error.增根是同乘含变量表达式所产生的副产品，并非由算术错误造成。 Two distinct philosophies for solving rational equations: (i) Common denominator: multiply both sides by the LCD, reducing to a polynomial equation. Always works; produces extraneous roots when a candidate makes the LCD zero. (ii) Cross-multiplication: only valid when each side is a single fraction; faster on 2-term equations but error-prone when more terms are present (you must combine first). The provincial-style protocol locks in approach (i) with explicit restriction-then-verify because it scales to any number of terms. Approach (ii) is fine on $\frac{a}{b} = \frac{c}{d}$ but stops being a shortcut once the equation has three or more rational terms, by then you have to find the LCD anyway, so just start there.求解有理方程有两种思路：(i) 公分母法：两边同乘 LCD，化为多项式方程。永远有效；当某候选值使 LCD 为零时会产生增根。(ii) 交叉相乘法：仅在两边各为单一分式时成立；用于二项式方程较快，但当项数更多时容易出错（必须先合并）。省考流程锁定方法 (i)，并要求"先写限制条件、最后验证"——因为它对任意项数都适用。方法 (ii) 对 $\frac{a}{b} = \frac{c}{d}$ 没问题，但一旦出现三项或更多有理项就不再是捷径——届时仍需求 LCD，不如一开始就用方法 (i)。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Radical Equations根式方程 · BC PC11 radical equationsBC PC11 根式方程 [8 marks][8 分]

Isolate, square, verify; state radicand-non-negative condition. (a) $\sqrt{3x+1}=x-1$. (b) $\sqrt{x+4}-\sqrt{x-3}=1$. (c) Identify the extraneous-root step.分离根号、平方、验证；写出被开方数非负条件。 (a) $\sqrt{3x+1}=x-1$。 (b) $\sqrt{x+4}-\sqrt{x-3}=1$。 (c) 指出引入增根的步骤。

Answer:答案：  (a) $x = 5$; $x = 0$ extraneous$x = 5$；$x = 0$ 为增根  ·  (b) $x = 12$  ·  (c) squaring step introduces the extraneous root平方这一步引入增根

(a) Single squaring; reject any candidate that fails verification单次平方；舍去任何未通过验证的候选根 M1·A1·A1

Radicand-non-negative: $3x + 1 \ge 0 \Leftrightarrow x \ge -\tfrac{1}{3}$. RHS-non-negative (since the square root is non-negative): $x - 1 \ge 0 \Leftrightarrow x \ge 1$. Combined constraint: $x \ge 1$.被开方数非负：$3x + 1 \ge 0 \Leftrightarrow x \ge -\tfrac{1}{3}$。右边非负（因为根号本身非负）：$x - 1 \ge 0 \Leftrightarrow x \ge 1$。合并约束：$x \ge 1$。
Square both sides:两边平方： $$ 3x + 1 \;=\; (x - 1)^{2} \;=\; x^{2} - 2x + 1 \;\Longrightarrow\; x^{2} - 5x \;=\; 0 \;\Longrightarrow\; x(x - 5) \;=\; 0. $$ Candidates: $x = 0$ and $x = 5$.候选根：$x = 0$ 与 $x = 5$。
Verify $x = 0$:验证 $x = 0$： LHS $= \sqrt{1} = 1$; RHS $= -1$. LHS $\ne$ RHS. Also fails the $x \ge 1$ constraint. Extraneous.左边 $= \sqrt{1} = 1$；右边 $= -1$。左 $\ne$ 右。同时不满足 $x \ge 1$。增根。
Verify $x = 5$:验证 $x = 5$： LHS $= \sqrt{16} = 4$; RHS $= 4$. $\checkmark$ Solution: $x = 5$.左边 $= \sqrt{16} = 4$；右边 $= 4$。$\checkmark$ 解：$x = 5$。

(b) Two squarings; isolate one radical at a time平方两次；每次只分离一个根号 M1·A1·A1·A1

Radicand-non-negative conditions: $x + 4 \ge 0$ and $x - 3 \ge 0$, so $x \ge 3$.被开方数非负：$x + 4 \ge 0$ 且 $x - 3 \ge 0$，所以 $x \ge 3$。
Isolate one radical: $\sqrt{x + 4} = 1 + \sqrt{x - 3}$. Square both sides:先分离一个根号：$\sqrt{x + 4} = 1 + \sqrt{x - 3}$。两边平方： $$ x + 4 \;=\; \bigl(1 + \sqrt{x - 3}\bigr)^{2} \;=\; 1 + 2\sqrt{x - 3} + (x - 3) \;=\; x - 2 + 2\sqrt{x - 3}. $$ Isolate the remaining radical: $6 = 2\sqrt{x - 3}$, so $\sqrt{x - 3} = 3$. Square again: $x - 3 = 9 \Rightarrow x = 12$.再分离剩余根号：$6 = 2\sqrt{x - 3}$，所以 $\sqrt{x - 3} = 3$。再次平方：$x - 3 = 9 \Rightarrow x = 12$。
Verify $x = 12$:验证 $x = 12$： $\sqrt{16} - \sqrt{9} = 4 - 3 = 1$. $\checkmark$ Also $12 \ge 3$. Solution: $x = 12$ (no extraneous candidate emerged this time).$\sqrt{16} - \sqrt{9} = 4 - 3 = 1$。$\checkmark$ 且 $12 \ge 3$。解：$x = 12$（本题未出现增根）。

(c) BC PC11 Big Idea + extraneous-root sourceBC PC11 大概念 + 增根来源 R1

BC PC11 Big Idea:BC PC11 大概念： "The meanings of, and connections between, operations extend to powers, radicals, and polynomials.运算的意义及其相互联系延伸至幂、根式与多项式。" The single step in part (b) that introduces the possibility of an extraneous root is the squaring: $\sqrt{x + 4} = 1 + \sqrt{x - 3}$ implies $x + 4 = (1 + \sqrt{x - 3})^{2}$, but the converse fails because $(\cdot)^{2}$ is not injective, both $+u$ and $-u$ square to $u^{2}$. (Each subsequent squaring is another chance for an extraneous root, which is why two-radical equations need verification at the end.)(b) 中引入增根可能性的那一步就是平方：由 $\sqrt{x + 4} = 1 + \sqrt{x - 3}$ 可推出 $x + 4 = (1 + \sqrt{x - 3})^{2}$，但反向并不成立，因为 $(\cdot)^{2}$ 不是单射——$+u$ 与 $-u$ 平方后都等于 $u^{2}$。（每次再平方都可能再引入增根，所以含两个根号的方程末了必须验证。）
Squaring is a non-injective map; verification is not optional.平方是非单射映射；验证不可省略。 The function $u \mapsto u^{2}$ collapses $+u$ and $-u$ to the same image, so the equation after squaring may have more solutions than the original. The "candidate set $\supseteq$ true solution set" inclusion can only be checked by substitution into the original equation. Two radical-specific protocol moves: (1) state the radicand-non-negative conditions before squaring, they prune candidates before you compute; (2) when the RHS of a $\sqrt{\cdot} = (\text{expression})$ equation has a sign you can determine (e.g., RHS $= x - 1$ in (a) must be $\ge 0$), add that as a constraint too. Both habits compress the work and catch extraneous roots structurally rather than by accident.映射 $u \mapsto u^{2}$ 将 $+u$ 与 $-u$ 压缩到同一像点，因此平方后的方程可能比原方程多出解。"候选集 $\supseteq$ 真解集"的包含关系只能通过代回原方程来核实。根式方程两条专用纪律：(1) 在平方之前写出被开方数非负条件，可在计算前就剔除部分候选；(2) 若 $\sqrt{\cdot} = (\text{表达式})$ 的右边正负可判（如 (a) 中右边 $= x - 1$ 必须 $\ge 0$），也把它加入约束。两个习惯都能压缩工作量，并在结构上而非偶然中捕获增根。
Q9HARD难Honors荣誉级 🇨🇦 BC卑 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §4 Rational Function有理函数 · BC PC12 / MHF4U Strand C [11 marks][11 分]

$f(x)=\frac{x^2-x-6}{x^2-4x+3}$. (a) Factor. (b) Hole. (c) VA. (d) HA, $x$-int, $y$-int. (e) Domain.$f(x)=\frac{x^2-x-6}{x^2-4x+3}$。 (a) 因式分解。 (b) 空洞。 (c) 垂直渐近线。 (d) 水平渐近线、$x$ 截距、$y$ 截距。 (e) 定义域。

Answer:答案：  (a) $\dfrac{(x-3)(x+2)}{(x-3)(x-1)}$, common $(x-3)$$\dfrac{(x-3)(x+2)}{(x-3)(x-1)}$，公因式为 $(x-3)$  ·  (b) hole at $\bigl(3, \tfrac{5}{2}\bigr)$空洞位于 $\bigl(3, \tfrac{5}{2}\bigr)$  ·  (c) VA $x = 1$垂直渐近线 $x = 1$  ·  (d) HA $y = 1$; $x$-int $(-2, 0)$; $y$-int $(0, -2)$水平渐近线 $y = 1$；$x$ 截距 $(-2, 0)$；$y$ 截距 $(0, -2)$  ·  (e) $\{x \in \mathbb{R} \,:\, x \ne 1, 3\}$

(a) Factor numerator and denominator分子分母分别因式分解 M1·A1

Numerator $x^{2} - x - 6$: integers with product $-6$ and sum $-1$ are $-3$ and $2$, so $(x - 3)(x + 2)$. Denominator $x^{2} - 4x + 3$: integers with product $3$ and sum $-4$ are $-3$ and $-1$, so $(x - 3)(x - 1)$. Common factor: $(x - 3)$. So分子 $x^{2} - x - 6$：乘积为 $-6$、和为 $-1$ 的整数对是 $-3$ 与 $2$，因此为 $(x - 3)(x + 2)$。分母 $x^{2} - 4x + 3$：乘积为 $3$、和为 $-4$ 的整数对是 $-3$ 与 $-1$，因此为 $(x - 3)(x - 1)$。公因式为 $(x - 3)$。因此 $$ f(x) \;=\; \frac{(x - 3)(x + 2)}{(x - 3)(x - 1)}. $$

(b) Point discontinuity (hole)可去间断点（空洞） M1·A1

The cancelled factor $(x - 3)$ produces a hole at $x = 3$. After cancellation, $f^{*}(x) = \dfrac{x + 2}{x - 1}$, and $f^{*}(3) = \dfrac{5}{2}$. So the hole is at $\bigl(3, \tfrac{5}{2}\bigr)$.被约去的因式 $(x - 3)$ 在 $x = 3$ 处产生空洞。约分后 $f^{*}(x) = \dfrac{x + 2}{x - 1}$，且 $f^{*}(3) = \dfrac{5}{2}$。因此空洞位于 $\bigl(3, \tfrac{5}{2}\bigr)$。

(c) Vertical asymptote(s)垂直渐近线 M1·A1

Apply the rule "denominator-only zero $\Rightarrow$ VA; common-factor zero $\Rightarrow$ hole." Denominator zeros are $x = 3$ (common factor) and $x = 1$ (denominator-only). So the only VA is at $x = 1$. Sign analysis confirms: as $x \to 1^{-}$, $f^{*}(x) = \frac{x+2}{x-1} \to \frac{3}{0^{-}} = -\infty$; as $x \to 1^{+}$, $f^{*}(x) \to +\infty$. Standard VA behaviour.应用规则"仅分母为零 $\Rightarrow$ 垂直渐近线；公因式为零 $\Rightarrow$ 空洞"。分母零点为 $x = 3$（公因式）与 $x = 1$（仅分母）。因此唯一垂直渐近线在 $x = 1$。符号分析确认：当 $x \to 1^{-}$，$f^{*}(x) = \frac{x+2}{x-1} \to \frac{3}{0^{-}} = -\infty$；当 $x \to 1^{+}$，$f^{*}(x) \to +\infty$。典型垂直渐近线行为。

(d) Horizontal asymptote; intercepts水平渐近线与截距 M1·A1·A1

HA by degree comparison (Q3 shortcut): both num and den have degree $2$, ratio of leading coefficients $\frac{1}{1} = 1$. So $y = 1$.用次数比较法求水平渐近线（Q3 中的捷径）：分子分母同为 2 次，首项系数之比为 $\frac{1}{1} = 1$。故 $y = 1$。
$x$-intercept: solve $f^{*}(x) = 0 \Leftrightarrow x + 2 = 0 \Leftrightarrow x = -2$. (Numerator zero, denominator non-zero: $-2 - 1 = -3 \ne 0$.) Point: $(-2, 0)$.$x$ 截距：解 $f^{*}(x) = 0 \Leftrightarrow x + 2 = 0 \Leftrightarrow x = -2$。（分子为零、分母非零：$-2 - 1 = -3 \ne 0$。）点：$(-2, 0)$。
$y$-intercept: $f^{*}(0) = \frac{2}{-1} = -2$. (Also satisfies the original: $0$ is not in the excluded set.) Point: $(0, -2)$.$y$ 截距：$f^{*}(0) = \frac{2}{-1} = -2$。（在原函数中亦成立：$0$ 不在排除集合中。）点：$(0, -2)$。

(e) Domain in set-builder notation用集合构造记号写出定义域 A1·A1

The original denominator zeros are $x = 1$ and $x = 3$. Both are excluded, even though the $x = 3$ exclusion is invisible in the simplified $f^{*}$, BC PC12's "domain" elaboration explicitly requires that cancelled-factor zeros remain excluded, since the original $f$ is undefined there. Domain:原分母的零点为 $x = 1$ 与 $x = 3$。两者均排除——即使 $x = 3$ 在化简后的 $f^{*}$ 中"看不见"。BC PC12 的"定义域"细化明确要求被约去因式的零点仍需排除，因为原 $f$ 在此无定义。定义域： $$ \{\,x \in \mathbb{R} \,:\, x \ne 1 \text{ and且 } x \ne 3\,\}, \qquad \text{or equivalently或等价地} \qquad \mathbb{R} \setminus \{1, 3\}. $$
Hole vs. vertical asymptote: read the factorisation, not the equation.空洞 vs. 垂直渐近线：看因式分解，不看方程。 The rule is structural and never about substitution: (denominator factor that cancels) $\Rightarrow$ hole at that $x$; (denominator factor that does not cancel) $\Rightarrow$ vertical asymptote at that $x$. Sign analysis (the $\pm \infty$ behaviour as $x$ approaches the VA from each side) is a consequence, not the test. The hole's $y$-coordinate is read off the simplified $f^{*}$ at the cancelled $x$, not off the original (which is undefined there). Domain follows the original: every original-denominator zero is excluded, full stop. BC PC12 markers test this distinction every year, the simplified function and the original function have the same VA and HA but different domains, and the gap is the cancelled-factor hole.这条规则是结构性的，与代入无关：（被约去的分母因式） $\Rightarrow$ 在该 $x$ 处为空洞；（未被约去的分母因式） $\Rightarrow$ 在该 $x$ 处为垂直渐近线。符号分析（$x$ 从两侧趋近垂直渐近线时趋向 $\pm \infty$ 的行为）只是结果，并非判据。空洞的 $y$ 坐标须用化简后的 $f^{*}$ 在被约去的 $x$ 处计算，而非原函数（原函数在此无定义）。定义域则依据原函数：原分母的每个零点都被排除，无一例外。BC PC12 阅卷人每年都会考这一区别——化简后的函数与原函数有相同的垂直渐近线与水平渐近线，但定义域不同，差距就是那个被约去因式所对应的空洞。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 参考答案Universal · 32 marks通用 · 共 32 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Radical Equation Modeling根式方程建模 · HSA-REI.A.2 [10 marks][10 分]

Pendulum: $T = 2\pi\sqrt{L/g}$, $g = 9.8$. (a) $T$ for $L = 0.994$. (b) Solve for $L(T)$. (c) $L$ when $T = 3.5$. (d) Verify; cite HSA-REI.A.2.单摆：$T = 2\pi\sqrt{L/g}$，$g = 9.8$。 (a) 求 $L = 0.994$ 时的 $T$。 (b) 解出 $L(T)$。 (c) 求 $T = 3.5$ 时的 $L$。 (d) 验证；引用 HSA-REI.A.2。

Answer:答案：  (a) $T \approx 2.001$ s秒  ·  (b) $L = \dfrac{g T^{2}}{4\pi^{2}}$  ·  (c) $L \approx 3.04$ m米  ·  (d) back-substitution gives $T \approx 3.500$ s. $\checkmark$回代得 $T \approx 3.500$ 秒。$\checkmark$

(a) Substitute and compute代入并计算 M1·A1

$$ T \;=\; 2\pi\sqrt{\dfrac{0.994}{9.8}} \;=\; 2\pi\sqrt{0.10143} \;=\; 2\pi \cdot 0.31848 \;\approx\; 2.001 \;\text{s}. $$ (To three decimal places. The $L = 0.994$ m specification is chosen so that the period is essentially exactly $2$ seconds, the standard grandfather-clock "seconds pendulum" calibration.)（保留三位小数。$L = 0.994$ 米的取值是为了让周期几乎正好为 $2$ 秒——这是标准落地老式钟"秒摆"的校准值。）

(b) Solve $T = 2\pi\sqrt{L/g}$ for $L$由 $T = 2\pi\sqrt{L/g}$ 解出 $L$ M1·A1·A1

Square both sides (legal because $T > 0$ and $L > 0$, so both sides are non-negative and squaring is reversible on the situational domain):两边平方（因为 $T > 0$、$L > 0$，两边均非负，所以在情境定义域上平方运算是可逆的）： $$ T^{2} \;=\; 4\pi^{2} \cdot \dfrac{L}{g} \;\Longrightarrow\; L \;=\; \dfrac{g T^{2}}{4 \pi^{2}}, \qquad L > 0, \;\; T > 0. $$

(c) Compute $L$ for $T = 3.5$ s求 $T = 3.5$ 秒时的 $L$ M1·A1·A1

$$ L \;=\; \dfrac{9.8 \cdot (3.5)^{2}}{4\pi^{2}} \;=\; \dfrac{9.8 \cdot 12.25}{4\pi^{2}} \;=\; \dfrac{120.05}{39.478} \;\approx\; 3.04 \;\text{m}. $$

(d) Verify by back-substitution; cite回代验证；引用 HSA-REI.A.2 A1·R1

Back-substitute $L = 3.04$ into the original: $T = 2\pi\sqrt{3.04/9.8} = 2\pi\sqrt{0.31020} = 2\pi \cdot 0.55696 \approx 3.500$ s. $\checkmark$把 $L = 3.04$ 代回原模型：$T = 2\pi\sqrt{3.04/9.8} = 2\pi\sqrt{0.31020} = 2\pi \cdot 0.55696 \approx 3.500$ 秒。$\checkmark$
HSA-REI.A.2: "Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.求解一元简单有理方程与根式方程，并举例说明增根如何产生。" Squaring is a non-injective operation ($u^{2} = v^{2}$ does not imply $u = v$, only $u = \pm v$), so the algebra in part (b) could in principle introduce extraneous solutions, explicit verification in part (d) confirms that the $L$ found really does produce the target $T$ in the original radical equation. (In this question the situational constraint $T > 0$, $L > 0$ already rules out the spurious negative branch, but the verification step is the formal closure of the radical-equation protocol.)平方是非单射运算（$u^{2} = v^{2}$ 并不意味着 $u = v$，仅能得 $u = \pm v$），因此 (b) 的代数操作原则上可能引入增根；(d) 的明确验证确认所求 $L$ 确实在原根式方程中产生目标 $T$。（本题情境约束 $T > 0$、$L > 0$ 已排除负分支，但验证步骤仍是根式方程流程的正式收尾。）
Squaring is reversible on the situational domain $T, L > 0$, but you must say so.在情境定义域 $T, L > 0$ 上平方是可逆的，但必须明确写出这一点。 Models with physical quantities (period, length, mass, time) almost always restrict to positives, and on the positive ray squaring is injective. Once you cite the restriction, the algebra is bidirectional and no extraneous root can creep in. AP markers reward students who name this, the citation of HSA-REI.A.2 plus the back-substitution check is what distinguishes a 4-mark answer from a 3-mark answer. The pendulum model is also useful as a sanity check: the period scales like $\sqrt{L}$, so quadrupling $L$ doubles $T$. Compare (a) ($L \approx 1$ m, $T \approx 2$ s) with (c) ($L \approx 3$ m, $T \approx 3.5$ s), the ratio $\sqrt{3.04/0.994} = \sqrt{3.057} \approx 1.749$ matches $3.500/2.001 \approx 1.749$. $\checkmark$含物理量（周期、长度、质量、时间）的模型几乎总限制为正值，而在正半轴上平方是单射。一旦引用了该限制，代数操作就是双向的，增根无从产生。AP 阅卷人会奖励能点出这一点的学生：引用 HSA-REI.A.2 加上回代验证就是 4 分答案与 3 分答案的分水岭。单摆模型也可用作合理性检验：周期与 $\sqrt{L}$ 成正比，所以 $L$ 翻四倍则 $T$ 翻倍。比较 (a)（$L \approx 1$ 米，$T \approx 2$ 秒）与 (c)（$L \approx 3$ 米，$T \approx 3.5$ 秒）：比值 $\sqrt{3.04/0.994} = \sqrt{3.057} \approx 1.749$ 与 $3.500/2.001 \approx 1.749$ 相符。$\checkmark$
Q11MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Rational Equation Modeling有理方程建模 · MHF4U Strand C [11 marks][11 分]

Work-rate. Aanya alone $= x$ h, Ben alone $= x+3$ h, together $= 2$ h. (a) Rates. (b) Rational equation + restrictions. (c) Quadratic + formula. (d) Reject negative; state times. (e) Verify $1/x + 1/(x+3) = 1/2$.工作效率。Aanya 单独 $= x$ 小时，Ben 单独 $= x+3$ 小时，合作 $= 2$ 小时。 (a) 工作效率。 (b) 有理方程 + 限制条件。 (c) 二次方程 + 求根公式。 (d) 舍去负根；写出时间。 (e) 验证 $1/x + 1/(x+3) = 1/2$。

Answer:答案：  (a) Aanya $\frac{1}{x}$, Ben $\frac{1}{x+3}$ (jobs/h)Aanya $\frac{1}{x}$，Ben $\frac{1}{x+3}$（件/小时）  ·  (b) $\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}$, $x > 0$, $x \ne 0, -3$  ·  (c) $x^{2} - x - 6 = 0$, $x = 3$ or $x = -2$$x^{2} - x - 6 = 0$，$x = 3$ 或 $x = -2$  ·  (d) Aanya $3.0$ h, Ben $6.0$ hAanya $3.0$ 小时，Ben $6.0$ 小时  ·  (e) $\frac{1}{3} + \frac{1}{6} = \frac{1}{2} = 0.500$. $\checkmark$

(a) Rates of work工作效率 A1·A1

"Rate of work" $=$ fraction of one job completed per hour. Aanya finishes one full job in $x$ hours, so her rate is $\frac{1}{x}$ jobs per hour. Ben's rate is $\frac{1}{x + 3}$."工作效率" $=$ 每小时完成一件工作的分数。Aanya 用 $x$ 小时完成一件，故效率为 $\frac{1}{x}$ 件/小时。Ben 的效率为 $\frac{1}{x + 3}$。

(b) Combined-rate equation; restrictions合作效率方程；限制条件 M1·A1

Together for $2$ hours they complete one full job, so $2 \cdot (\text{Aanya} + \text{Ben}) = 1$:合作 $2$ 小时完成一件工作，故 $2 \cdot (\text{Aanya} + \text{Ben}) = 1$： $$ 2\left(\frac{1}{x} + \frac{1}{x + 3}\right) \;=\; 1 \;\Longrightarrow\; \frac{1}{x} + \frac{1}{x + 3} \;=\; \frac{1}{2}. $$ Algebraic restrictions: $x \ne 0$ and $x \ne -3$. Situational restriction: $x > 0$ (an alone-time of zero or negative hours is meaningless).代数限制：$x \ne 0$ 且 $x \ne -3$。情境限制：$x > 0$（单独完成时间为零或负数没有意义）。

(c) Multiply by LCD; reduce to quadratic; quadratic formula两边同乘 LCD；化为二次方程；使用求根公式 M1·A1·A1·A1

LCD $= 2x(x + 3)$. Multiply through:LCD $= 2x(x + 3)$。两边同乘： $$ 2(x + 3) + 2x \;=\; x(x + 3) \;\Longrightarrow\; 4x + 6 \;=\; x^{2} + 3x \;\Longrightarrow\; x^{2} - x - 6 \;=\; 0. $$ Quadratic formula (Unit 2):求根公式（第 2 单元）： $$ x \;=\; \dfrac{1 \pm \sqrt{1 + 24}}{2} \;=\; \dfrac{1 \pm 5}{2} \;\Longrightarrow\; x \;=\; 3 \;\text{ or或 }\; x \;=\; -2. $$ (The roots are exact integers; equivalently $x^{2} - x - 6 = (x - 3)(x + 2) = 0$.)（两根恰为整数；亦可由 $x^{2} - x - 6 = (x - 3)(x + 2) = 0$ 得出。）

(d) Reject negative root; state alone-times舍去负根；写出单独完成时间 A1·A1

The negative root $x = -2$ violates the situational domain $x > 0$ (a worker cannot have a negative completion time, even though the algebra produces it as a candidate). The positive root $x = 3$ is the unique physical solution. Aanya: $x = 3.0$ h. Ben: $x + 3 = 6.0$ h.负根 $x = -2$ 违反情境定义域 $x > 0$（工人不可能有负的完成时间，尽管代数会把它当作候选）。正根 $x = 3$ 是唯一的物理解。Aanya：$x = 3.0$ 小时。Ben：$x + 3 = 6.0$ 小时。
(Note: the Practice prompt asks for one-decimal rounding, but the roots here are exact integers, see the Practice inconsistency flagged at the foot of this solution. Stating $3.0$ h and $6.0$ h satisfies the rounding instruction trivially.)（注：练习题要求保留一位小数，但这里两根恰为整数——参见本答案末尾标出的练习题不一致说明。写成 $3.0$ 小时与 $6.0$ 小时即满足取整要求。）

(e) Verification验证 A1

$$ \frac{1}{3} + \frac{1}{6} \;=\; \frac{2}{6} + \frac{1}{6} \;=\; \frac{3}{6} \;=\; \frac{1}{2} \;=\; 0.500. \;\checkmark $$ The combined hourly rate is exactly $\frac{1}{2}$ jobs/h, so two hours of joint work complete exactly one job, as required.合作每小时效率正好为 $\frac{1}{2}$ 件，因此合作 $2$ 小时恰好完成 $1$ 件工作，符合题意。
Modeling restrictions can be stronger than algebraic ones.建模限制可能比代数限制更严格。 The algebra forbids only $x = 0$ and $x = -3$ (zeros of the LCD), but the situational context forbids the entire interval $x \le 0$, a negative alone-time is meaningless. The negative root $x = -2$ passes the algebraic restriction (it's not $0$ or $-3$) and even satisfies the original equation as a number ($\frac{1}{-2} + \frac{1}{1} = -\frac{1}{2} + 1 = \frac{1}{2}$ $\checkmark$), but it's rejected on situational grounds, not algebraic ones. The ON Provincial marker awards the A1 for the justification, not the rejection alone, saying "$x > 0$ is required because an alone-completion-time must be positive" is the load-bearing sentence.代数只禁止 $x = 0$ 与 $x = -3$（LCD 的零点），但情境背景禁止整个区间 $x \le 0$——负数单独完成时间没有意义。负根 $x = -2$ 通过了代数限制（既非 $0$ 也非 $-3$），甚至作为数值能满足原方程（$\frac{1}{-2} + \frac{1}{1} = -\frac{1}{2} + 1 = \frac{1}{2}$ $\checkmark$），但仍因情境原因被舍去，而非代数原因。安大略省考阅卷人奖励的是理由说明，不只是舍去本身——"$x > 0$ 是必需的，因为单独完成时间必须为正"才是关键句。
Q12HARD难 🇨🇦 BC卑 🇺🇸 US美 BC Provincial-style卑诗省考风格 §2 + §4 Rational Function App有理函数应用 · BC PC12 / HSA-CED.A.1 [11 marks][11 分]

Bottling plant. $\bar{C}(x) = (240 + 0.60x)/x$, $x > 0$. (a) Split into sum. (b) $\bar{C}(100), \bar{C}(400), \bar{C}(1200)$. (c) HA + interpretation. (d) Break-even at $0.85. (e) Situational domain.瓶装厂。$\bar{C}(x) = (240 + 0.60x)/x$，$x > 0$。 (a) 拆分为和。 (b) 求 $\bar{C}(100), \bar{C}(400), \bar{C}(1200)$。 (c) 水平渐近线 + 解释。 (d) 在 $0.85 处的盈亏平衡。 (e) 情境定义域。

Answer:答案：  (a) $\bar{C}(x) = \dfrac{240}{x} + 0.60$  ·  (b) $2.50, $1.20, $0.80  ·  (c) HA $\bar{C} = 0.60$水平渐近线 $\bar{C} = 0.60$  ·  (d) $x = 960$ bottles/day$x = 960$ 瓶/日  ·  (e) $x > 0$

(a) Split single rational into a sum把单一有理表达式拆成和 M1·A1

$$ \bar{C}(x) \;=\; \dfrac{240 + 0.60 x}{x} \;=\; \dfrac{240}{x} + \dfrac{0.60 x}{x} \;=\; \dfrac{240}{x} + 0.60. $$ The $\frac{240}{x}$ term came from the fixed cost ($240/day) spread across $x$ bottles, it's a true rational term that shrinks as production rises. The constant $0.60$ came from the variable cost ($0.60/bottle) and is independent of $x$.$\frac{240}{x}$ 项来自固定成本（$240/日）摊到 $x$ 瓶上——这是真正的有理项，会随产量上升而减小。常数 $0.60$ 来自可变成本（$0.60/瓶），与 $x$ 无关。

(b) Three point values三个点处的取值 A1·A1·A1

$\bar{C}(100) = \frac{240}{100} + 0.60 = 2.40 + 0.60 = 2.50$, i.e. $2.50/bottle.，即 $2.50/瓶。
$\bar{C}(400) = \frac{240}{400} + 0.60 = 0.60 + 0.60 = 1.20$, i.e. $1.20/bottle.，即 $1.20/瓶。
$\bar{C}(1200) = \frac{240}{1200} + 0.60 = 0.20 + 0.60 = 0.80$, i.e. $0.80/bottle.，即 $0.80/瓶。
As $x$ increases, the $\frac{240}{x}$ term shrinks toward $0$, so the per-bottle cost decreases, the fixed cost gets amortised over more units (economies of scale).随 $x$ 增大，$\frac{240}{x}$ 趋向 $0$，因此每瓶成本下降——固定成本被分摊到更多的瓶子上（规模经济）。

(c) Horizontal asymptote and interpretation水平渐近线与解释 M1·A1·A1

Two ways to see the HA:两种方法判断水平渐近线：
Algebraic:代数法： degree of numerator $= 1$, degree of denominator $= 1$, so ratio of leading coefficients $= \frac{0.60}{1} = 0.60$.分子次数 $= 1$、分母次数 $= 1$，故首项系数之比 $= \frac{0.60}{1} = 0.60$。
Split form:拆分形式： as $x \to \infty$, $\frac{240}{x} \to 0$ and $\bar{C}(x) \to 0.60$ directly from the sum decomposition.当 $x \to \infty$ 时，$\frac{240}{x} \to 0$，所以由拆分式直接得 $\bar{C}(x) \to 0.60$。
Either way, the HA is $\bar{C} = 0.60$ (or, in context, $0.60 per bottle). Interpretation: as the plant scales up, the average cost per bottle approaches $0.60, the variable cost, but never quite reaches it, because the fixed cost is always being amortised over some finite number of bottles. The HA is the asymptotic floor.两法结果相同：水平渐近线为 $\bar{C} = 0.60$（结合情境即 $0.60/瓶）。解释：工厂规模扩大时，每瓶平均成本趋近 $0.60（即可变成本），但永远达不到——因为固定成本始终被分摊到有限数量的瓶子上。水平渐近线就是这条渐近下界。

(d) Break-even production盈亏平衡产量 M1·A1

Set $\bar{C}(x) = 0.85$:令 $\bar{C}(x) = 0.85$： $$ \dfrac{240}{x} + 0.60 \;=\; 0.85 \;\Longrightarrow\; \dfrac{240}{x} \;=\; 0.25 \;\Longrightarrow\; x \;=\; \dfrac{240}{0.25} \;=\; 960. $$ So break-even occurs at $x = 960$ bottles/day. Below this, $\bar{C}(x) > 0.85$ and the plant runs at a loss; above this, $\bar{C}(x) < 0.85$ and the plant is profitable.所以盈亏平衡产量为 $x = 960$ 瓶/日。低于此值时 $\bar{C}(x) > 0.85$，工厂亏损；高于此值时 $\bar{C}(x) < 0.85$，工厂盈利。

(e) Situational domain; algebraic + contextual exclusion of $x = 0$情境定义域；代数与情境两方面均排除 $x = 0$ R1

$x > 0$. The exclusion of $x = 0$ is justified twice: algebraically, $x = 0$ is a denominator-only zero (no common-factor cancellation), so $\bar{C}$ has a vertical asymptote at $x = 0$, the average cost diverges as production tends to zero. Contextually, a plant producing zero bottles has no per-bottle cost defined (any fixed cost divided by zero output is undefined). BC PC12's rational functions elaboration names exactly this checklist, asymptotes, intercepts, point discontinuities, domain, end-behaviour, and Q12 hits each item: VA $x = 0$, no point discontinuities (no common factor), HA $\bar{C} = 0.60$, domain $x > 0$, end-behaviour $\bar{C} \to 0.60$ as $x \to \infty$ and $\bar{C} \to \infty$ as $x \to 0^{+}$.$x > 0$。$x = 0$ 的排除有双重依据：代数上，$x = 0$ 是仅分母为零的零点（无公因式可约），因此 $\bar{C}$ 在 $x = 0$ 处有垂直渐近线——当产量趋于零时平均成本发散。情境上，生产 $0$ 瓶的工厂"每瓶成本"无定义（固定成本除以零产量未定义）。BC PC12 的有理函数细化恰好给出此清单：渐近线、截距、可去间断点、定义域、末端行为，本题逐项命中：垂直渐近线 $x = 0$、无可去间断点（无公因式）、水平渐近线 $\bar{C} = 0.60$、定义域 $x > 0$、末端行为 $x \to \infty$ 时 $\bar{C} \to 0.60$ 且 $x \to 0^{+}$ 时 $\bar{C} \to \infty$。
Split-form $\frac{240}{x} + 0.60$ shows the asymptote, the economies of scale, and the interpretation all at once.拆分形式 $\frac{240}{x} + 0.60$ 一次性呈现渐近线、规模经济与情境解释。 The unified rational form $\frac{240 + 0.60x}{x}$ is correct but hides the structure. The split decomposes the rational into a "diminishing" piece ($\frac{240}{x}$, the fixed-cost-per-unit term) and a "constant" piece ($0.60$, the variable cost). The HA $y = 0.60$ then reads directly as the constant piece, with no degree-comparison shortcut needed. This is the reverse of the §4.2 "combine into single fraction" move and is the load-bearing technique for asymptotic interpretation in BC PC12. Whenever the numerator decomposes cleanly against the denominator, splitting is strictly better than degree comparison, you get the interpretation for free.统一有理形式 $\frac{240 + 0.60x}{x}$ 虽然正确，却隐藏了结构。拆分把有理表达式分解为"递减项"（$\frac{240}{x}$，每单位固定成本）与"常数项"（$0.60$，可变成本）。水平渐近线 $y = 0.60$ 直接对应常数项，无需用次数比较捷径。这是 §4.2"合并为单一分式"操作的逆向，也是 BC PC12 中渐近解释的核心技巧。每当分子能干净地拆向分母时，拆分严格优于次数比较——情境解释也一并得到。
Drafter's notes · Practice inconsistencies flagged作者注 · 标出的练习题不一致

Q11 (Work-rate):（工作效率）： the Practice prompt instructs the student to round Aanya's and Ben's times "to one decimal place," but the quadratic $x^{2} - x - 6 = 0$ factors over the integers as $(x - 3)(x + 2)$, yielding the exact roots $x = 3$ and $x = -2$. The intended pedagogical effect (round-from-irrational) does not arise; stating $3.0$ h and $6.0$ h satisfies the rounding instruction trivially. Recommendation for v1.1: either change the combined-time from $2$ h to a value that produces an irrational root (e.g. $2.5$ h gives $x \approx 3.85$ h), or remove the "one decimal place" instruction in (d). Either edit closes the inconsistency.练习题要求把 Aanya 与 Ben 的时间"保留一位小数"，但二次方程 $x^{2} - x - 6 = 0$ 在整数范围内可因式分解为 $(x - 3)(x + 2)$，得到精确的根 $x = 3$ 与 $x = -2$。原本意图的教学效果（从无理数取整）并不会出现；写成 $3.0$ 小时与 $6.0$ 小时即满足要求。v1.1 建议：要么把合作时间从 $2$ 小时改为产生无理根的值（例如 $2.5$ 小时给出 $x \approx 3.85$ 小时），要么删除 (d) 中"保留一位小数"的指令。任一改动都能消除不一致。

All other answers spot-checked and consistent其余答案均已抽查并与练习题一致 with the Practice prompts. Mark sums per question reconcile to 90 marks total: Part I $3+3+3+6+7 = 22$; Part II $8+9+8+11 = 36$; Part III $10+11+11 = 32$; grand total $22 + 36 + 32 = 90$. $\checkmark$。各题分值总和为 90 分：第一部分 $3+3+3+6+7 = 22$；第二部分 $8+9+8+11 = 36$；第三部分 $10+11+11 = 32$；合计 $22 + 36 + 32 = 90$。$\checkmark$