ATLASBeta

Subjects

Ancient HistoryComing soon
BiologyBusiness Studies
ChemistryComing soon
CAFSEconomics
EngineeringComing soon
English AdvancedEnglish Ext 1English Standard
English StudiesComing soon
Enterprise ComputingComing soon
Geography
Health & MovementComing soon
Legal StudiesMaths AdvancedMathematics Ext 1Mathematics Ext 2
Maths Standard 1Coming soon
Maths Standard 2Coming soon
Modern HistoryComing soon
PhysicsReligion IReligion II
Visual ArtsComing soon
Save your favourites

Accounts and progress tracking are coming later.

Coming later

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
Filters

More filters

290 questions found

List
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​dx.
3 marks
Short answer
2022Trial
Q13 (c) (ii)

The equation of line l2l_2l2​ is
x+3k=4−y6=z−1.\frac{x+3}{k}=\frac{4-y}{6}=z-1.kx+3​=64−y​=z−1.
Find the value of kkk for which l1l_1l1​ and l2l_2l2​ intersect.

3 marks
Short answer
2022Trial
Q13 (d)

The points XXX, YYY and ZZZ bisect ABABAB, BCBCBC and CACACA respectively. Show that
AY→+BZ→+CX→=0⃗.\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX}=\vec0.AY+BZ

3 marks
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) (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) (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
2022Trial
Q14 (c) (i)

On the Argand diagram, zzz and zwzwzw are represented by AAA and BBB. Given z=reiθz=re^{i\theta}z=reiθ and w=eiπ/3w=e^{i\pi/3}w=eiπ/3, where r>0r>0r>0, explain why OABOABOAB is an equilateral triangle.

2 marks
Short answer
2022Trial
Q14 (b)

Simplify fully
11+ω+11+ω2,\frac1{1+\omega}+\frac1{1+\omega^2},1+ω1​+1+ω21​,
given that ω\omegaω is a non-real cube root of unity.

2 marks
Short answer
2022Trial
Q14 (a) (i)

A particle is moving so that v2=n2(a2−x2)v^2=n^2(a^2-x^2)v2=n2(a2−x2). Show that the particle is moving in simple harmonic motion.

1 mark
Short answer
2022Trial
Q14 (a) (ii)

If the particle is moving at 15 m/s\sqrt{15}\,\mathrm{m/s}15​m/s when it is 2 m2\,\mathrm m2m to the right of the centre of motion and at 23 m/s2\sqrt3\,\mathrm{m/s}23​m/s when it is 4 m4\,\mathrm m4m to the right of the centre of motion, find the period and amplitude of the motion.

3 marks
Short answer
2022Trial
Q14 (b) (i)

A particle is moving so that x=2sin⁡3t−23cos⁡3tx=2\sin3t-2\sqrt3\cos3tx=2sin3t−23​cos3t. By expressing the motion in the form Asin⁡(3t−α)A\sin(3t-\alpha)Asin(3t−α), find the period and amplitude.

3 marks
Short answer
2022Trial
Q14 (b) (ii)

A particle is moving so that x=2sin⁡3t−23cos⁡3tx=2\sin3t-2\sqrt3\cos3tx=2sin3t−23​cos3t. Find the initial position and velocity of the particle.

2 marks
Short answer
2022Trial
Q14 (c) (i)

A 20 kg20\,\mathrm{kg}20kg projectile is launched upwards from the ground at 500 m/s500\,\mathrm{m/s}500m/s. It experiences a gravitational force of 200 N200\,\mathrm N200N and air resistance, opposite to its motion, of v218 N\frac{v^2}{18}\,\mathrm N18v2​N. Letting the ground be x=0x=0x=0, show that the acceleration is
x¨=−10−v2360.\ddot{x}=-10-\frac{v^2}{360}.x¨=−10−360v2​

1 mark
Short answer
2022Trial
Q14 (c) (ii)

Show that
x=180ln⁡(2536003600+v2)x=180\ln\left(\frac{253600}{3600+v^2}\right)x=180ln(3600+v2253600​)
and find the maximum height the projectile reaches.

3 marks
Short answer
2022Trial
Q14 (c) (iii)

Show that the time taken for the projectile to reach velocity v m/sv\,\mathrm{m/s}vm/s is
t=6tan⁡−1(253)−6tan⁡−1(v60),t=6\tan^{-1}\left(\frac{25}{3}\right)-6\tan^{-1}\left(\frac{v}{60}\right),t=6tan−1(325​)−6tan−1(60v​),
and find the time taken for the projectile to reach its maximum height.

2 marks
Short answer
2022Trial
Q14 (a)

If x≤1x\le1x≤1 and y≥1y\ge1y≥1, show that x+y≥1+xyx+y\ge1+xyx+y≥1+xy.

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
2022HSC
Q14 (a) (iv)

The point PPP is defined by
AP→=−6AB→−8AC→.\overrightarrow{AP}=-6\overrightarrow{AB}-8\overrightarrow{AC}.AP=−6AB−8AC.

Does PPP lie on the line ALALAL? Justify your answer.

2 marks
Short answer
2022HSC
Q14 (a) (iii)

Using part (ii), or otherwise, determine the position of LLL by showing that
BL→=47BC→.\overrightarrow{BL}=\frac47\overrightarrow{BC}.BL=74​BC.

2 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 (a) (i)

The two non-parallel vectors u⃗\vec uu and v⃗\vec vv satisfy λu⃗+μv⃗=0⃗\lambda\vec u+\mu\vec v=\vec0λu+μv for some real numbers λ\lambdaλ and μ\muμ. Show that λ=μ=0\lambda=\mu=0λ=μ=0.

2 marks
Short answer
2022Trial
Q14 (e)

A recurrence relation is defined by
un+1=3un−14un−1,u1=1.u_{n+1}=\frac{3u_n-1}{4u_n-1},\qquad u_1=1.un+1​=4un​−13un​−1​,u1​=1.
Use mathematical induction to prove that
un=n2n−1for n≥1.u_n=\frac n{2n-1}\quad\text{for }n\ge1.un​=2n−1n​for n≥

3 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 (a) (ii)

Show by finding it that the point of intersection between the lines
(xyz)=(1105)+λ(2−31)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}​xyz​​=​1105​λ​2−31​
and
(xyz)=(15−16)+δ(4−1−1)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}
is (718)\begin{pmatrix}7\\1\\8\end{pmatrix}​718​.

3 marks
Short answer
← PreviousPage 7 of 12Next →
+
CX=
0.
​
∑r=0n​
r!1​
.
e
1
​
,
n
≥
1.
.
r!1​
.
=
0
1.
​
+
​
​
xyz​
​
=
​15−16​​+
δ​4−1−1​​
​