IB Mathematics Analysis and Approaches Standard Level Paper 1 (November 2025) Study Guide

General Examination Information and Instructions

• Subject: Mathematics: analysis and approaches Standard level Paper 1.
• Session Date: 10 November 2025.
• Examination Zones: Zone A, Zone B, and Zone C afternoon sessions.
• Duration: 1 hour 30 minutes.
• Maximum Marks: $80\,marks$.
• Calculator Policy: Candidates are not permitted access to any calculator for this paper.
• Required Material: A clean copy of the mathematics: analysis and approaches SL formula booklet is required.
• General Rules:     * Show all working and/or explanations. Full marks are not necessarily awarded for a correct answer without working.     * Numerical answers must be given exactly or correct to three significant figures, unless otherwise stated.     * Section A questions must be answered within the provided boxes.     * Section B questions must be answered in the provided answer booklet.     * A candidate session number is required on both the paper and the answer booklet.

Section A: Short Answer Questions

Question 1: Graph Transformations

• Maximum Marks: $4\,marks$
• Function f Definition: The graph of $y = f(x)$ is provided for the domain $-3 \le x \le 3$.
• Task: Match the following transformed functions to their respective graphs (labeled A through F):     * Function 1: $y = f(x + 1)$. This represents a horizontal translation to the left by $1\,unit$.     * Function 2: $y = f(x) - 1$. This represents a vertical translation downward by $1\,unit$.     * Function 3: $y = f(-x)$. This represents a reflection across the $y$-axis.     * Function 4: $y = f(2x)$. This represents a horizontal compression by a scale factor of $\frac{1}{2}$.

Question 2: Arithmetic Sequences

• Maximum Marks: $6\,marks$
• Given Information:     * The $1st\,term$ ($u_1$) is $36$.     * The $5th\,term$ ($u_5$) is $12$.
• Part (a): Find the 13th term.     * Use the general formula: $u_n = u_1 + (n - 1)d$.     * Calculate common difference ($d$) using $u_5 = u_1 + 4d$: $12 = 36 + 4d$.     * Find $u_{13}$: $u_{13} = u_1 + 12d$.     * Mark Allocation: $[4]$
• Part (b): Sum of terms.     * The sum of the first $n\,terms$ ($S_n$) is zero.     * Use the formula: $S_n = \frac{n}{2}(2u_1 + (n - 1)d) = 0$.     * Solve for $n$.     * Mark Allocation: $[2]$

Question 3: Probability and Independence

• Maximum Marks: $6\,marks$
• Given Probabilities:     * $P(A) = 0.5$     * $P(B) = 0.6$     * $P(A \cap B) = 0.4$
• Part (a): Union of Two Events.     * Calculate $P(A \cup B)$ using the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.     * Mark Allocation: $[2]$
• Part (b): Independence Check.     * Show that events $A$ and $B$ are not independent.     * Requirement: For independence, $P(A \cap B) = P(A) \times P(B)$. Compare $0.4$ to $0.5 \times 0.6$.     * Mark Allocation: $[1]$
• Part (c): Conditional Probability.     * Find $P(A|B')$ where $B'$ is the complement of $B$.     * Formula: $P(A|B') = \frac{P(A \cap B')}{P(B')}$.     * Note: $P(A \cap B') = P(A) - P(A \cap B)$ and $P(B') = 1 - P(B)$.     * Mark Allocation: $[3]$

Question 4: Coordinate Geometry and Calculus

• Maximum Marks: $7\,marks$
• Given Information:     * Line equation: $y = -3x + 9$.     * Intersection points: $P(0, 9)$ and $Q(3, 0)$.     * Parabola equation: $y = ax^2 + c$, where $a, c \in \mathbb{Z}$.     * The parabola also passes through points $P$ and $Q$.
• Part (a): Parabola Coefficients.     * (i) Determine $c$ by substituting point $P(0, 9)$.     * (ii) Determine $a$ by substituting point $Q(3, 0)$ and the value of $c$.     * Mark Allocation: $[3]$
• Part (b): Area Under Curves.     * Find the area of the region enclosed by the line and the parabola.     * Calculated using integration: $\int_{0}^{3} (y_{parabola} - y_{line})\,dx$ or equivalent depending on which curve is upper.     * Mark Allocation: $[4]$

Question 5: Logarithmic Equations and Identity

• Maximum Marks: $6\,marks$
• Part (a): Algebraic Proof.     * Show that for $x > 7$: $\frac{x(x^2 - 1)}{(x^2 - 8x + 7)(x + 1)} = \frac{x}{x - 7}$.     * Steps involve factoring $x^2 - 1$ as $(x - 1)(x + 1)$ and $x^2 - 8x + 7$ as $(x - 7)(x - 1)$.     * Mark Allocation: $[2]$
• Part (b): Solve Logarithmic Equation.     * Equation: $\log_2[x(x^2 - 1)] - 1 = \log_2[(x^2 - 8x + 7)(x + 1)]$.     * Apply log properties: $\log_2[x(x^2 - 1)] - \log_2[(x^2 - 8x + 7)(x + 1)] = 1$.     * Simplify using parts from part (a): $\log_2(\frac{x}{x - 7}) = 1$.     * Convert to exponential form: $\frac{x}{x - 7} = 2^1$.     * Mark Allocation: $[4]$

Question 6: Quadratic Functions and Their Vertices

• Maximum Marks: $8\,marks$
• Given Functions:     * $f(x) = -x^2 + 4x + p$     * $g(x) = x^2 + qx - 1$     * $p$ and $q$ are non-zero constants.     * Vertex of $f(x)$ is at point $A$. Vertex of $g(x)$ is at point $B$.     * Points $A$ and $B$ lie on the line $y = 2x - 1$.
• Part (a): Solve for p.     * Find the x-coordinate of vertex $A$: $x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2$.     * Since $A$ is on $y = 2x - 1$, find the y-coordinate: $y = 2(2) - 1 = 3$.     * Substitute $(2, 3)$ into $f(x)$ to show $p = -1$.     * Mark Allocation: $[3]$
• Part (b): Solve for q.     * Find the x-coordinate of vertex $B$ in terms of $q$: $x = -\frac{q}{2(1)} = -\frac{q}{2}$.     * Since $B$ is on $y = 2x - 1$, its y-coordinate is $y = 2(-\frac{q}{2}) - 1 = -q - 1$.     * Substitute vertex coordinates $B(-\frac{q}{2}, -q - 1)$ into $g(x) = x^2 + qx - 1$.     * Solve the resulting equation for $q$.     * Mark Allocation: $[5]$

Section B: Long Answer Questions

Question 7: Composite and Inverse Functions

• Maximum Marks: $14\,marks$
• Given Functions:     * $f(x) = 2x + 3$.     * $h(x)$ is linear with $x-intercept = 2$ and $y-intercept = -1$.
• Part (a): Basic Function Operations.     * (i) Evaluate $f(2)$.     * (ii) Find the inverse function $f^{-1}(x)$.     * Mark Allocation: $[3]$
• Part (b): Finding Linear Expression.     * Write an expression for $h(x)$ using the given intercepts.     * Use slope-intercept form $y = mx + c$ or similar.     * Mark Allocation: $[2]$
• Part (c): Inverse Linear Equation.     * Solve $h^{-1}(x) = -2$.     * Alternatively, this is equivalent to solving $h(-2) = x$.     * Mark Allocation: $[2]$
• Part (d): Composite Function Parameters.     * Defined: $g(x) = mx + c$ where $m, c \in \mathbb{Q}$.     * Given $h(x) = (f^{-1} \circ g)(x)$.     * Equate and solve for $m$ and $c$.     * Mark Allocation: $[4]$
• Part (e): Finding k(x).     * If $h(k(x)) = x$, then $k(x)$ is the inverse function $h^{-1}(x)$.     * Find the expression for $k(x)$.     * Mark Allocation: $[2]$
• Part (f): Identifying Transformation.     * State the single transformation mapping $y = k(x)$ onto $y = h(x)$.     * This usually involves a reflection in the line $y = x$ as they are inverses.     * Mark Allocation: $[1]$

Question 8: Derivatives and Integration

• Maximum Marks: $14\,marks$
• Derivative Given: $f'(x) = 3x^2 + 12x - 15$.
• Part (a): Stationary Points.     * Find $a$ and $b$ ($a < b$) where horizontal tangents exist ($f'(x) = 0$).     * Solve $3x^2 + 12x - 15 = 0$.     * Mark Allocation: $[3]$
• Part (b): Local Extrema.     * A sign diagram is provided: $+$ for $x < a$, $-$ for $a < x < b$, and $+$ for $x > b$.     * State whether $x = a$ is a local maximum or minimum by referencing the sign change of $f'(x)$.     * Mark Allocation: $[2]$
• Part (c): Second Derivative.     * Find the second derivative $f''(x)$.     * Solve for $c$ where $f''(x) = 0$.     * Mark Allocation: $[3]$
• Part (d): Point of Inflexion.     * A sign diagram for $f''(x)$ is provided: $-$ for $x < c$ and $+$ for $x > c$.     * State if a point of inflexion exists at $x = c$ with a reason (change in concavity).     * Mark Allocation: $[2]$
• Part (e): Finding f(x).     * Determine the original function $f(x)$ by integrating $f'(x)$.     * Use the initial condition $f(-2) = 36$ to find the constant of integration.     * Mark Allocation: $[4]$

Question 9: Trigonometry Identity and Application

• Maximum Marks: $15\,marks$
• Part (a): Trigonometric Proof.     * Show that $\frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \sin(2\theta)$.     * Hint: use equalities $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ and $1 + \tan^2(\theta) = \sec^2(\theta) = \frac{1}{\cos^2(\theta)}$.     * Mark Allocation: $[3]$
• Part (b): Solving Equations.     * Solve $\frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \frac{1}{2}$ for $0^{\circ} \le \theta \le 180^{\circ}$.     * Replace the left side with $\sin(2\theta)$ and solve for $2\theta$.     * Mark Allocation: $[5]$
• Part (c): Exact Values - Part 1.     * Let $x = \tan(22.5^{\circ})$.     * Show that $x^2 + 2x - 1 = 0$ using the result from part (a).     * Note: For $\theta = 22.5^{\circ}$, $2\theta = 45^{\circ}$, and $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$.     * Mark Allocation: $[4]$
• Part (d): Exact Values - Part 2.     * Find the exact value of $\tan(22.5^{\circ})$ by solving the quadratic equation from part (c).     * Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.     * Select the positive root since $22.5^{\circ}$ is in the first quadrant.     * Mark Allocation: $[3]$

Documentation and Metadata

• Copyright: © International Baccalaureate Organization 2025. All rights reserved. Use by third parties is prohibited.
• Exam Code: 16EP01 through 16EP16 (page identifiers).
• Paper ID: 8825–7109.
• Session Type: afternoon.