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
EASYMEDIUMHARDPaper 1APaper 1BPaper 2Paper 3
Syllabus 2.6, 2.7, 2.12考纲 2.6、2.7、2.12AA HL
Name:姓名：Date:日期：
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 all algebraic working. State the discriminant condition explicitly whenever you appeal to it. For inequalities, give the answer set in interval or set-builder form, and indicate any excluded points. No calculator permitted.写出全部代数过程。使用判别式条件时必须显式写出该条件。对不等式，请以区间或集合记号给出解集，并标出任何被排除的点。不可使用计算器。
Q1EASYPaper 1A2.7 Discriminant[4 marks]
The equation $k x^{2} - 4 x + 1 = 0$ has two distinct real roots. Find the set of values of the real parameter $k$.方程 $k x^{2} - 4 x + 1 = 0$ 有两个不同的实数根。求实参数 $k$ 的取值集合。
(a)State the non-degeneracy condition on $k$ for the equation to be quadratic.写出使方程为真正二次方程的非退化条件（对 $k$）。[1]
(b)Write the discriminant $\Delta$ in terms of $k$ and apply the two-distinct-roots condition.将判别式 $\Delta$ 用 $k$ 表示，并应用两不同实根的条件。[2]
(c)State the final set of values of $k$.写出 $k$ 的最终取值集合。[1]
Q2MEDIUMPaper 1A2.6 Completing the Square[5 marks]
Consider the quadratic $y = 2 x^{2} - 12 x + 7$.考虑二次函数 $y = 2 x^{2} - 12 x + 7$。
(a)Write $y$ in vertex form $a(x - h)^{2} + k$ by completing the square.通过配方将 $y$ 写为顶点式 $a(x - h)^{2} + k$。[3]
(b)State the coordinates of the vertex and the equation of the axis of symmetry.写出顶点坐标与对称轴方程。[1]
(c)State the minimum value of $y$ on $\mathbb{R}$ and the $x$-value at which it occurs.写出 $y$ 在 $\mathbb{R}$ 上的最小值及取得最小值的 $x$。[1]
Q3MEDIUMPaper 1A2.7 Vieta · Symmetric Sum[5 marks]
Let $\alpha$ and $\beta$ be the (complex) roots of $x^{2} - 6 x + 11 = 0$.设 $\alpha$、$\beta$ 是 $x^{2} - 6 x + 11 = 0$ 的（复）根。
(a)Without solving the equation, state $\alpha + \beta$ and $\alpha \beta$ using Vieta's formulas.不求解方程，用韦达定理写出 $\alpha + \beta$ 与 $\alpha \beta$。[2]
(b)Hence compute $\alpha^{2} + \beta^{2}$ using a single symmetric identity (no solving for $\alpha, \beta$).由此用一个对称恒等式计算 $\alpha^{2} + \beta^{2}$（不要解出 $\alpha, \beta$）。[2]
(a)Factor the numerator and list all critical points (zeros of the numerator and of the denominator). Mark which value of $x$ is excluded from the domain.分解分子，列出全部临界点（分子的零点与分母的零点）。标出哪个 $x$ 值不属于定义域。[2]
(b)Build a sign chart for $\dfrac{(x - 2)(x + 2)}{x - 1}$ across the four intervals determined by the critical points.在临界点划分的四个区间上，给出 $\dfrac{(x - 2)(x + 2)}{x - 1}$ 的符号表。[3]
(c)State the full solution set in interval notation, treating each endpoint with care.用区间记号写出完整解集，仔细处理每个端点。[2]
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
Multi-part scaffolded problem. Each part feeds the next. State Vieta's identities explicitly when invoked, and write a sign chart for any polynomial inequality. No calculator permitted.分级支撑式长题，每一部分为下一部分提供基础。引用韦达定理时须显式写出；任何多项式不等式都要附符号表。不可使用计算器。
Q5HARDPaper 1B2.12 Cubic from Roots · Polynomial Inequality (HL)[11 marks]
The cubic $P(x) = x^{3} + a x^{2} + b x + c$ has roots $1$, $-2$, and $3$.三次式 $P(x) = x^{3} + a x^{2} + b x + c$ 的根为 $1$、$-2$、$3$。
(a)Write $P(x)$ in fully factored form.写出 $P(x)$ 的完全因式分解。[1]
(c)Build a sign chart for $P(x)$ across the four intervals determined by the three roots.在三根划分的四个区间上给出 $P(x)$ 的符号表。[3]
(d)Hence solve $P(x) > 0$. Give the answer in interval notation.由此求解 $P(x) > 0$，用区间记号给出答案。[2]
(e)Compute $\alpha^{2} + \beta^{2} + \gamma^{2}$ using Vieta only (do not square the roots individually).仅用韦达定理（不要逐根平方）计算 $\alpha^{2} + \beta^{2} + \gamma^{2}$。[1]
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 polynomial root-finding you may use the GDC's polyRoots or graph-and-trace tools directly, but state the input polynomial and read the roots to a sensible exactness (rational roots: give exact; irrational: 3 s.f.). Always cross-check with $P(\text{root}) = 0$.需要图形计算器（GDC）。多项式求根可直接调用 polyRoots 或图像追踪工具，但须写出输入的多项式，并以合理精度读出根（有理根给精确值；无理根 3 位有效数字）。务必用 $P(\text{root}) = 0$ 进行复核。
Q6MEDIUMPaper 22.12 Cubic Roots on GDC (HL)[7 marks]
Consider the cubic equation $\;2 x^{3} - 5 x^{2} - 4 x + 3 \;=\; 0.$考虑三次方程 $\;2 x^{3} - 5 x^{2} - 4 x + 3 \;=\; 0$。
(a)Use your GDC to find all three real roots. Give each root in exact form (as a rational number).用 GDC 求出全部三个实根。以精确形式（有理数）给出每个根。[3]
(b)Verify the largest root by direct substitution into the original equation.将最大的根直接代回原方程进行验证。[2]
(c)Hence write the polynomial $2 x^{3} - 5 x^{2} - 4 x + 3$ as a product of three linear factors over $\mathbb{Z}$ (note that the leading coefficient $2$ must appear in one of the linear factors as $(2x - 1)$ rather than $\bigl(x - \tfrac{1}{2}\bigr)$).由此将多项式 $2 x^{3} - 5 x^{2} - 4 x + 3$ 在 $\mathbb{Z}$ 上写为三个一次因式之积（注意首项系数 $2$ 必须出现在某一个一次因式中，写成 $(2x - 1)$ 而非 $\bigl(x - \tfrac{1}{2}\bigr)$）。[2]
Q7HARDPaper 22.12 Remainder + Factor Theorems System (HL)[9 marks]
Let $P(x) = x^{3} + a x^{2} + b x + 6$ where $a, b \in \mathbb{R}$. When $P(x)$ is divided by $(x - 1)$ the remainder is $4$, and $(x + 3)$ is a factor of $P(x)$.设 $P(x) = x^{3} + a x^{2} + b x + 6$，其中 $a, b \in \mathbb{R}$。$P(x)$ 除以 $(x - 1)$ 的余数为 $4$，且 $(x + 3)$ 是 $P(x)$ 的因式。
(a)Apply the remainder theorem at $x = 1$ to obtain a linear equation in $a$ and $b$.在 $x = 1$ 处使用余数定理，得到一个关于 $a, b$ 的一次方程。[2]
(b)Apply the factor theorem at $x = -3$ to obtain a second linear equation in $a$ and $b$.在 $x = -3$ 处使用因式定理，得到第二个关于 $a, b$ 的一次方程。[2]
(c)Solve the $2 \times 2$ system to find $a$ and $b$.求解 $2 \times 2$ 方程组，得 $a$ 与 $b$。[2]
(d)Hence factor $P(x)$ completely over $\mathbb{Z}$ and list all real roots.由此将 $P(x)$ 在 $\mathbb{Z}$ 上完全因式分解，并列出全部实根。[3]
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. Method marks dominate. State each Vieta identity by name before invoking it; for symmetric-function computations, the chain $p_{1}, p_{2}, p_{3}$ in terms of the elementary symmetric polynomials $e_{1}, e_{2}, e_{3}$ is the canonical machinery (Newton's identities).需要图形计算器（GDC）。方法分占主导。引用韦达恒等式时先点出名称；对对称函数运算，幂和 $p_{1}, p_{2}, p_{3}$ 用初等对称多项式 $e_{1}, e_{2}, e_{3}$ 表示的链条是标准工具（Newton 恒等式）。
Q8HARDPaper 32.12 Vieta · Newton's Identities for a Cubic (HL)[15 marks]
Let $\alpha, \beta, \gamma$ be the (complex) roots of the monic cubic $\;x^{3} + p x^{2} + q x + r \;=\; 0,$ with $p, q, r \in \mathbb{R}$.设 $\alpha, \beta, \gamma$ 是首一三次 $\;x^{3} + p x^{2} + q x + r \;=\; 0$ 的（复）根，其中 $p, q, r \in \mathbb{R}$。
(a)State Vieta's identities for this cubic, giving $e_{1} = \alpha + \beta + \gamma$, $e_{2} = \alpha \beta + \beta \gamma + \gamma \alpha$, and $e_{3} = \alpha \beta \gamma$ each in terms of $p, q, r$.写出该三次的韦达恒等式，将 $e_{1} = \alpha + \beta + \gamma$、$e_{2} = \alpha \beta + \beta \gamma + \gamma \alpha$、$e_{3} = \alpha \beta \gamma$ 分别用 $p, q, r$ 表示。[3]
(c)Show that $\alpha^{2} \beta^{2} + \beta^{2} \gamma^{2} + \gamma^{2} \alpha^{2} = q^{2} - 2 p r$, by squaring $e_{2}$.通过对 $e_{2}$ 平方，证明 $\alpha^{2} \beta^{2} + \beta^{2} \gamma^{2} + \gamma^{2} \alpha^{2} = q^{2} - 2 p r$。[2]
(d)Show that $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = -\dfrac{q}{r}$, stating any non-degeneracy condition you impose on $r$.证明 $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = -\dfrac{q}{r}$，并指出对 $r$ 施加的非退化条件。[2]
(e)Use Newton's identity $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$ (with $p_{1} = e_{1}$) to express $\alpha^{3} + \beta^{3} + \gamma^{3}$ in terms of $p, q, r$.用 Newton 恒等式 $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$（其中 $p_{1} = e_{1}$）将 $\alpha^{3} + \beta^{3} + \gamma^{3}$ 用 $p, q, r$ 表示。[3]
(f)Apply the formulas of (b), (d), (e) to the specific cubic $x^{3} - 4 x^{2} + 5 x - 2 = 0$ to compute $p_{2}$, $\sum 1/\alpha$, and $p_{3}$. (You may check by spotting the root $x = 1$ and factoring; do this as a verification, not as the primary method.)将 (b)、(d)、(e) 的公式应用于具体三次 $x^{3} - 4 x^{2} + 5 x - 2 = 0$，计算 $p_{2}$、$\sum 1/\alpha$、$p_{3}$。（可通过观察根 $x = 1$ 并因式分解进行验证，但只作验证，不作主方法。）[3]