Year 11 and 10 Accelerated Mathematics Advanced Assessment Task 2 Study Guide

Assessment Task Framework: Year 11 and 10 Accelerated Mathematics Advanced

• Assessment Identification: Validation Test (Task Number 2).
• Subject: Mathematics Advanced.
• Target Student Cohort: Year 11 Advanced and Year 10 Accelerated.
• Teaching Staff: Ms Ghougassian, Mrs Thill, Mr Mikhaeel, and Ms Tran.
• Time Allocation: $50\,\text{minutes}$.
• Weighting: $30\%$.
• Exam Date: Tuesday, 3 June 2025 (Week 6, Day 2).
• Scheduled Periods: Period 4 (11 ADV) and Period 2 (10 ACC).
• Assessed Topics:
  • Functions.
  • Trigonometric Functions (Topic T1.1, excluding 3D trigonometry).
• Structural Composition:
  • Section 1: Multiple Choice ($5$ marks, recommended time: $7\,\text{minutes}$).
  • Section 2: Short Answer Responses ($35$ marks, recommended time: $43\,\text{minutes}$).
  • Total Marks: $40\,\text{marks}$.
• General Materials & Requirements:
  • Writing implementation: Black pen.
  • Calculators: NESA-approved devices only.
  • Standard Reference Sheet provided.
  • Working: Explicit mathematical reasoning or calculations required for Section II.

Functions: Polynomial and Rational Analysis

• Graphic Representation of Polynomials:
  • Function analysis: $f(x) = x(x + 2)^2(x - 3)^3$.
  • Key components:
    • Root at $x = 0$ (multiplicity 1, straight crossing).
    • Root at $x = -2$ (multiplicity 2, bounce/touching point).
    • Root at $x = 3$ (multiplicity 3, horizontal inflection/point of inflexion crossing).
• Sketching Specific Function Types:
  • Hyperbolas: Sketching functions in the form $y = \frac{1}{x + 1} + 2$.
    • Vertical Asymptote: $x = -1$.
    • Horizontal Asymptote: $y = 2$.
    • Translation: Shifted $1$ unit left and $2$ units up from the parent function $y = \frac{1}{x}$.
  • Absolute Value Functions: Sketching $y = |x - 3|$.
    • Vertex point: $(3, 0)$.
    • Shape: "V" shape passing through the y-intercept at $(0, 3)$.
• Even and Odd Functions:
  • Definition of an Even Function: A function where $f(x) = f(-x)$, resulting in symmetry across the y-axis.
  • Verification: For $f(x) = \sqrt{4 - x^2}$, let $x \rightarrow -x$:
    • $f(-x) = \sqrt{4 - (-x)^2}$
    • $f(-x) = \sqrt{4 - x^2} = f(x)$
  • Geometry: The function $f(x) = \sqrt{4 - x^2}$ represents a semi-circle with a radius of $2$ and center $(0, 0)$, located above the x-axis.
  • Domain: $x \in [-2, 2]$.
  • Range: $y \in [0, 2]$.

Quadratic Equations and Parabolic Models

• Nature of the Discriminant ($\Delta$):
  • Formula: $\Delta = b^2 - 4ac$.
  • If $\Delta$ is not a perfect square number, the roots are categorised as irrational (provided $a, b, c$ are rational).
  • Discriminant conditions for real roots: $\Delta \geq 0$.
  • Application: For the equation $2x^2 - 8x + m = 0$, real roots exist when:
    • $(-8)^2 - 4(2)(m) \geq 0$
    • $64 - 8m \geq 0$
    • $64 \geq 8m \rightarrow m \leq 8$.
• Parabola Construction from Geometric Features:
  • Information required: Passing through $(1, 4)$, $(2, 8)$, and having an axis of symmetry at $x = -\frac{1}{2}$.
  • Vertex form utilization: $y = a(x - h)^2 + k$ where the vertex is at $(-\frac{1}{2}, k)$, or general form $y = ax^2 + bx + c$ where $-\frac{b}{2a} = -\frac{1}{2}$.

Linear Relationships and Coordinate Geometry

• Points and Lines:
  • Locating missing coordinates: If $A(1, k)$ lies on $12x - 5y + 3 = 0$, then $12(1) - 5k + 3 = 0 \rightarrow 15 = 5k \rightarrow k = 3$.
• Perpendicular Lines:
  • The gradient ($m_1$) of $12x - 5y + 3 = 0$ is $\frac{12}{5}$.
  • The gradient of the perpendicular line ($m_2$) is $-\frac{1}{m_1}$, which is $-\frac{5}{12}$.
  • General form requirement: $Ax + By + C = 0$ format.
• Economic Modeling (Linear Systems):
  • Cost Function: $C(x) = 200 + 10x$.
  • Revenue Function: $R(x) = 50x$.
  • Unit Price: $\$50$.
  • Profit Analysis: Profit occurs when $R(x) > C(x)$.
    • $50x > 200 + 10x$
    • $40x > 200$
    • $x > 5$ (Sale of more than $5$ units is required for profit).

Circle Geometry and Intersections

• General Form of a Circle: $x^2 + y^2 - 6x + 8y + 9 = 0$.
  • Conversion to Standard Form: Completing the square is necessary to find the center and radius.
  • $(x^2 - 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16$
  • $(x - 3)^2 + (y + 4)^2 = 4^2$
  • Center: $(3, -4)$, Radius: $4$.
• Simultaneous Solution (Intersections):
  • Determining points where the circle and the parabola $y = \frac{(x+1)^2}{4} - 4$ meet via algebraic or graphical methods.

Trigonometric Principles and Applications

• Trigonometric Ratios and Identities:
  • Given $\sin(\theta) = \frac{6}{\sqrt{54}}$ and $\tan(\theta) = \frac{6}{\sqrt{18}}$.
  • Relationship: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \rightarrow \cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}$.
  • Calculation: $\cos(\theta) = \frac{6}{\sqrt{54}} \div \frac{6}{\sqrt{18}} = \frac{\sqrt{18}}{\sqrt{54}} = \sqrt{\frac{18}{54}} = \sqrt{\frac{1}{3}} = \frac{\sqrt{3}}{3}$.
• Triangle Measurements:
  • Area of Triangle ($\Delta CDE$): $103.4\,cm^2$.
  • Cosine Rule: Used to find the angle opposite the longest side in a triangle with sides $9$, $12$, and $16\,cm$.
    • $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$.
• Bearings and Navigation:
  • Leg 1: Point $M$ to $P$, $300\,km$ on a bearing of $N65^\circ E$.
  • Leg 2: Point $P$ to $Q$, $400\,km$ on a bearing of $S25^\circ E$.
  • Internal Angle Measurement: Calculation of $\angle MPQ$.
  • Displacement: Distance from the original position $M$ to $Q$.
  • Bearings Calculation: Finding $\angle PQM$ and the true bearing of $Q$ from $M$.
• Elevation Analysis:
  • Scene: Two points ($A$ and $B$) on either side of a hill ($DC$).
  • Angles of Elevation: $47^\circ$ from point $A$ and $68^\circ$ from point $B$.
  • Distance $AC = x$ metres. Total baseline distance $AB = 850\,m$.
  • Height derived via trigonometry: $DC$ in terms of $x$ from $\Delta ADC$, then solving for the total height of the hill to $3$ significant figures.