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Mathematics Extension 1

Stage 6 · HSC questions by topic

About this subject

Mathematics Extension 1

Atlas organises human-reviewed questions against the NSW Stage 6 Mathematics Extension 1syllabus. Every published question retains its exam source and page.

Syllabus: NESA Stage 6 (2017)
Explore the NESA Stage 6 (2017) syllabus topics

Functions

  • Further Work with Functions
  • Polynomials

Trigonometric Functions

  • Inverse Trigonometric Functions
  • Further Trigonometric Identities
  • Trigonometric Equations

Calculus

  • Rates of Change
  • Further Calculus Skills
  • Applications of Calculus

Combinatorics

  • Working with Combinatorics

Proof

  • Proof by Mathematical Induction

Vectors

  • Introduction to Vectors

Statistical Analysis

  • The Binomial Distribution
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5 questions found

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2023HSC
Q8

The diagram shows the graph of a function.

[Graph of the function]

Which of the following is the equation of the function?
A. y=∣1−∣∣x∣−2∣∣y=\left|1-\left||x|-2\right|\right|y=∣1−∣∣x∣−2∣ B. C. D.

∣

y=∣2−∣∣x∣−1∣∣y=\left|2-\left||x|-1\right|\right|y=∣2−∣∣x∣−1∣∣

y=∣1−∣x−2∣∣y=\left|1-\left|x-2\right|\right|y=∣1−∣x−2∣∣

y=∣2−∣x−1∣∣y=\left|2-\left|x-1\right|\right|y=∣2−∣x−1∣∣
1 mark
Multiple choice
2023HSC
Q9

The graph of a cubic function, y=f(x)y=f(x)y=f(x), is given below.

[Graph of the cubic function y=f(x)y=f(x)y=f(x)]

Which of the following functions has an inverse relation whose graph has more than 3 points with an xxx-coordinate of 111?
A. y=f(x)y=\sqrt{f(x)}y=f(x)​
B. y=1f(x)y=\dfrac{1}{f(x)}y=f(x)1​
C. y=f(∣x∣)y=f(|x|)y=f(∣x∣)
D. y=∣f(x)∣y=|f(x)|y=∣f(x)∣

1 mark
Multiple choice
2023HSC
Q11 (a)

The parametric equations of a line are given below.

x=1+3ty=4t\begin{aligned} x&=1+3t\\ y&=4t \end{aligned}xy​=1+3t=4t​

Find the Cartesian equation of this line in the form y=mx+cy=mx+cy=mx+c.

2 marks
Short answer
2023HSC
Q14 (a) (i)

Let f(x)=2x+ln⁡xf(x)=2x+\ln xf(x)=2x+lnx, for x>0x>0x>0.

(i) Explain why the inverse of f(x)f(x)f(x) is a function.

1 mark
Short answer
2023HSC
Q14 (a) (ii)

Let f(x)=2x+ln⁡xf(x)=2x+\ln xf(x)=2x+lnx, for x>0x>0x>0.

(ii) Let g(x)=f−1(x)g(x)=f^{-1}(x)g(x)=f−1(x). By considering the value of f(1)f(1)f(1), or otherwise, evaluate g′(2)g'(2)g′(2).

2 marks
Short answer