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Mathematics Extension 2

Stage 6 · HSC questions by topic

About this subject

Mathematics Extension 2

Atlas organises human-reviewed questions against the NSW Stage 6 Mathematics Extension 2syllabus. Every published question retains its exam source and page.

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

Proof

  • The Nature of Proof
  • Further Proof by Mathematical Induction

Vectors

  • Introduction to three-dimensional vectors
  • Further operations with three-dimensional vectors
  • Vectors and vector equations of lines

Complex Numbers

  • Arithmetic of complex numbers
  • Geometric representation of a complex number
  • Other representations of complex numbers
  • Solving equations with complex numbers
  • Geometrical implications of complex numbers

Calculus

  • Further Integration

Mechanics

  • Simple harmonic motion
  • Modelling motion without resistance
  • Resisted motion
  • Projectiles and resisted motion
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58 questions found

List
2022Trial
Q12 (a) (i)

Find AAA and BBB such that
15x2−2x−15=Ax+3+Bx−5.\frac{15}{x^2-2x-15}=\frac A{x+3}+\frac B{x-5}.

x2−2x−1515​
=
x+3A​+
x−5B​.
2 marks
Short answer
2022Trial
Q12 (a) (ii)

Hence, or otherwise, find
∫x2−2xx2−2x−15 dx.\int\frac{x^2-2x}{x^2-2x-15}\,dx.∫x2−2x−15x2−2x​dx.

2 marks
Short answer
2022Trial
Q12 (b)

Evaluate
∫0π/4sec⁡4xtan⁡2x dx.\int_0^{\pi/4}\sec^4x\tan^2x\,dx.∫0π/4​sec4xtan2xdx.

3 marks
Short answer
2022Trial
Q12 (e)

Use the substitution x=2sin⁡θx=2\sin\thetax=2sinθ to find
∫dx(4−x2)3/2.\int\frac{dx}{(4-x^2)^{3/2}}.∫(4−x2)3/2dx​.

3 marks
Short answer
2022Trial
Q13 (b)

Use the substitution t=tan⁡x2t=\tan\frac x2t=tan2x​, or otherwise, to evaluate
∫0π/213sin⁡x−4cos⁡x+5 dx.\int_0^{\pi/2}\frac{1}{3\sin x-4\cos x+5}\,dx.∫0π/2​3sinx−4cosx+51​

3 marks
Short answer
2022Trial
Q13 (a)

Find ∫cos⁡−1x dx\displaystyle\int\cos^{-1}x\,dx∫cos−1xdx.

3 marks
Short answer
2022HSC
Q14 (b) (v)

Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn​=∫01​xne−xdx. Using parts (ii) and (iv), prove that
e=lim⁡n→∞∑r=0n1r!.e=\lim_{n\to\infty}\sum_{r=0}^{n}\frac1{r!}.e=limn→∞​∑r=0n​

1 mark
Short answer
2022HSC
Q14 (b) (iv)

Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn​=∫01​xne−xdx, where nnn is a non-negative integer. Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all n≥0n\ge0n≥0,
Jn=n!−n!e∑r=0n1r!.J_n=n!-\frac{n!}{e}\sum_{r=0}^{n}\frac1{r!}.Jn​=n!−en!

2 marks
Short answer
2022HSC
Q14 (b) (i)

Let
Jn=∫01xne−x dx,J_n=\int_0^1x^ne^{-x}\,dx,Jn​=∫01​xne−xdx,
where nnn is a non-negative integer. Show that
J0=1−1e.J_0=1-\frac1e.J0​=1−e1​.

1 mark
Short answer
2022Trial
Q14 (d) (i)

Use a suitable substitution to show that
∫0af(x) dx=∫0af(a−x) dx.\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx.∫0a​f(x)dx=∫0a​f(a−x)dx.

1 mark
Short answer
2022HSC
Q14 (b) (iii)

Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn​=∫01​xne−xdx, where nnn is a non-negative integer. Show that
Jn=nJn−1−1e,n≥1.J_n=nJ_{n-1}-\frac1e,\qquad n\ge1.Jn​=nJn−1​−

2 marks
Short answer
2022HSC
Q14 (b) (ii)

Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn​=∫01​xne−xdx, where nnn is a non-negative integer. Show that
Jn≤1n+1.J_n\le\frac1{n+1}.Jn​≤n+11​.

2 marks
Short answer
2022Trial
Q14 (d) (ii)

Hence, or otherwise, evaluate
∫0π/4sin⁡xcos⁡x+sin⁡x dx.\int_0^{\pi/4}\frac{\sin x}{\cos x+\sin x}\,dx.∫0π/4​cosx+sinxsinx​dx.

3 marks
Short answer
2022Trial
Q15 (b)

Consider
In=∫01xn1−x dx.I_n=\int_0^1\frac{x^n}{\sqrt{1-x}}\,dx.In​=∫01​1−x​xn​dx.
Use integration by parts to show that
In=2n2n+1In−1.I_n=\frac{2n}{2n+1}I_{n-1}.In​=2n+12n​I

4 marks
Short answer
2022HSC
Q15 (c)

Using the substitution x=tan⁡2θx=\tan^2\thetax=tan2θ, evaluate
∫01sin⁡−1x1+x dx.\int_0^1\sin^{-1}\sqrt{\frac{x}{1+x}}\,dx.∫01​sin−11+xx​​dx.

4 marks
Short answer
2022Trial
Q16 (a) (iii)

Hence find
∫0π/2cos⁡7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2​cos7θdθ.

2 marks
Short answer
2022Trial
Q16 (c)

Find
∫ln⁡x−1(x+ln⁡x)2 dx.\int\frac{\ln x-1}{(x+\ln x)^2}\,dx.∫(x+lnx)2lnx−1​dx.

3 marks
Short answer
2022Trial
Q16 (b) (i)

If
In=∫0π/2cos⁡nθ dθ,I_n=\int_0^{\pi/2}\cos^n\theta\,d\theta,In​=∫0π/2​cosnθdθ,
show that
In=n−1nIn−2.I_n=\frac{n-1}{n}I_{n-2}.In​=nn−1​I

2 marks
Short answer
2022Trial
Q16 (b) (ii)

If
In=∫0π/2cos⁡nθ dθ,I_n=\int_0^{\pi/2}\cos^n\theta\,d\theta,In​=∫0π/2​cosnθdθ,
hence find
∫0π/2cos⁡7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2​cos7θdθ.

2 marks
Short answer
2021HSC
Q2

Which expression is equal to ∫x5e7x dx\int x^5e^{7x}\,dx∫x5e7xdx?

A. 17x5e7x−57∫x4e7x dx\frac17x^5e^{7x}-\frac57\int x^4e^{7x}\,dx71​x5e7x−7
B. 17x5e7x−57∫x5e7x dx\frac17x^5e^{7x}-\frac57\int x^5e^{7x}\,dx71​x5e7x−
C. 57x4e7x−57∫x4e7x dx\frac57x^4e^{7x}-\frac57\int x^4e^{7x}\,dx75​x4e7x−
D. 57x4e7x−57∫x5e7x dx\frac57x^4e^{7x}-\frac57\int x^5e^{7x}\,dx75​x4e7x−

1 mark
Multiple choice
2021HSC
Q11 (f)

Express 3x2−5(x−2)(x2+x+1)\frac{3x^2-5}{(x-2)(x^2+x+1)}(x−2)(x2+x+1)3x2−5​ as a sum of partial fractions over R\mathbb RR.

3 marks
Short answer
2021HSC
Q12 (a)

Find ∫2x+3x2+2x+2 dx\int\frac{2x+3}{x^2+2x+2}\,dx∫x2+2x+22x+3​dx.

3 marks
Short answer
2021HSC
Q13 (c) (ii)

Region AAA is bounded by y=1y=1y=1 and x2+y2=1x^2+y^2=1x2+y2=1 between x=0x=0x=0 and x=1x=1x=1. Region BBB is bounded by y=1y=1y=1 and y=ln⁡xy=\ln xy=lnx between x=1x=1x=1 and x=ex=ex=e. Using part (i), or otherwise, show that the ratio of the volumes formed by rotating regions AAA and BBB about the xxx-axis is VA:VB=1:3V_A:V_B=1:3VA​:VB​=1:3.

4 marks
Short answer
2021HSC
Q15 (c) (i)

An object of mass 1 kg1\,\mathrm{kg}1kg is projected vertically upwards with initial velocity u m s−1u\,\mathrm{m\,s^{-1}}ums−1. It experiences air resistance of magnitude kv2kv^2kv2 newtons, where vvv is velocity in m s−1\mathrm{m\,s^{-1}}ms−1 and k>0k>0k>0. The height above its starting point is xxx metres, time is ttt seconds, and acceleration due to gravity is g m s−2g\,\mathrm{m\,s^{-2}}gms−2. Show that the time to reach maximum height is
1gkarctan⁡(ukg) seconds.\frac1{\sqrt{gk}}\arctan\left(u\sqrt{\frac{k}{g}}\right)\text{ seconds}.gk​1​

3 marks
Short answer
2021HSC
Q15 (c) (ii)

For the resisted vertical motion described in part (i), find an expression for the maximum height reached by the object, in terms of kkk, ggg and uuu.

3 marks
Short answer
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