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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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290 questions found

List
2022Trial
Q16 (b) (ii)

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

0π/2​
cosn
θ
d
θ
,


∫0π/2cos⁡7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2​cos7θdθ.
2 marks
Short answer
2022Trial
Q16 (c) (i)

A projectile is launched from OOO at V m/sV\,\mathrm{m/s}Vm/s so that it will hit the point A(a,b)A(a,b)A(a,b) on the slope. Air resistance is neglected and the acceleration due to gravity is g m/s2g\,\mathrm{m/s^2}gm/s2. Given that
y=−gx22V2(1+tan⁡2α)+xtan⁡α,y=-\frac{gx^2}{2V^2}(1+\tan^2\alpha)+x\tan\alpha,y=−2V2gx
and that α1<α2\alpha_1<\alpha_2α1​<α2​, show that
tan⁡α1=V2−V4−g2a2−2V2gbga\tan\alpha_1=\frac{V^2-\sqrt{V^4-g^2a^2-2V^2gb}}{ga}tanα1​=gaV
and
tan⁡α2=V2+V4−g2a2−2V2gbga.\tan\alpha_2=\frac{V^2+\sqrt{V^4-g^2a^2-2V^2gb}}{ga}.tanα2​=gaV

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

By showing that
tan⁡(α1+α2)=tan⁡(90∘+θ),\tan(\alpha_1+\alpha_2)=\tan(90^\circ+\theta),tan(α1​+α2​)=tan(90∘+θ),
show that
∠COY=∠BOA.\angle COY=\angle BOA.∠COY=∠BOA.

4 marks
Short answer
2022HSC
Q16 (d)

Find all the complex numbers z1,z2,z3z_1,z_2,z_3z1​,z2​,z3​ that satisfy the following three conditions simultaneously:
∣z1∣=∣z2∣=∣z3∣,|z_1|=|z_2|=|z_3|,∣z1​∣=∣z2​∣=∣z
z1+z2+z3=1,z_1+z_2+z_3=1,z1​+z2​+z3​
z1z2z3=1.z_1z_2z_3=1.z1​z2​z3​=1.

3 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) (ii)

Given
Sn=133+143+153+⋯+1n3,n≥3,S_n=\frac1{3^3}+\frac1{4^3}+\frac1{5^3}+\cdots+\frac1{n^3},\qquad n\ge3,Sn​=331​+431​+531​+⋯+n31​,n≥3,
use part (i), or otherwise, to prove that
Sn<112.S_n<\frac1{12}.Sn​<121​.

3 marks
Short answer
2021HSC
Q1

Four cubes are placed in a line as shown in the diagram. Which vector is equal to AB→+CQ→\overrightarrow{AB}+\overrightarrow{CQ}AB+CQ​?

A. AQ→\overrightarrow{AQ}AQ​
B. CP→\overrightarrow{CP}CP C. D.

1 mark
Multiple choice
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
Q3

Which of the following is a vector equation of the line joining A(4,2,5)A(4,2,5)A(4,2,5) and B(−2,2,1)B(-2,2,1)B(−2,2,1)?

A. r⃗=(425)+λ(123)\vec r=\begin{pmatrix}4\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\3\end{pmatrix}r=​4
B. r⃗=(425)+λ(302)\vec r=\begin{pmatrix}4\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\2\end{pmatrix}r=
C. r⃗=(123)+λ(425)\vec r=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}4\\2\\5\end{pmatrix}r=
D. r⃗=(302)+λ(425)\vec r=\begin{pmatrix}3\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}4\\2\\5\end{pmatrix}r=

1 mark
Multiple choice
2021HSC
Q4

Consider the statement: “For all integers nnn, if nnn is a multiple of 666, then nnn is a multiple of 222.” Which is its contrapositive?

A. There exists an integer nnn such that nnn is a multiple of 666 and not a multiple of 222.
B. There exists an integer nnn such that nnn is a multiple of 222 and not a multiple of 666.
C. For all integers nnn, if nnn is not a multiple of 222, then nnn is not a multiple of 666.
D. For all integers nnn, if nnn is not a multiple of 666, then nnn is not a multiple of 222.

1 mark
Multiple choice
2021HSC
Q5

Which of the following statements is FALSE?

A. ∀a,b∈R\forall a,b\in\mathbb R∀a,b∈R, a<b⇒a3<b3a<b\Rightarrow a^3<b^3a<b⇒a3<b3.
B. ∀a,b∈R\forall a,b\in\mathbb R∀a,b∈R, a<b⇒e−a>e−ba<b\Rightarrow e^{-a}>e^{-b}a<b⇒e−a>e−b.
C. ∀a,b∈(0,+∞)\forall a,b\in(0,+\infty)∀a,b∈(0,+∞), a<b⇒ln⁡a<ln⁡ba<b\Rightarrow\ln a<\ln ba<b⇒lna<lnb.
D. ∀a,b∈R\forall a,b\in\mathbb R∀a,b∈R, with a,b≠0a,b\ne0a,b=0, a<b⇒1a>1ba<b\Rightarrow\frac1a>\frac1ba<b⇒a1​>b.

1 mark
Multiple choice
2021HSC
Q6

Which polynomial could have 2+i2+i2+i as a zero, given that kkk is real?

A. x3−4x2+kxx^3-4x^2+kxx3−4x2+kx
B. x3−4x2+kx+5x^3-4x^2+kx+5x3−4x2+kx+5
C. x3−5x2+kxx^3-5x^2+kxx3−5x2+kx
D. x3−5x2+kx+5x^3-5x^2+kx+5x3−5x2+kx+5

1 mark
Multiple choice
2021HSC
Q7

Which diagram best shows the curve described by r⃗(t)=−5cos⁡(t) i⃗+5sin⁡(t) j⃗+t k⃗\vec r(t)=-5\cos(t)\,\vec i+5\sin(t)\,\vec j+t\,\vec kr(t)=−5cos(t)i+5sin(t)j​+tk for 0≤t≤4π0\le t\le4\pi0≤t≤4π?

1 mark
Multiple choice
2021HSC
Q8

A particle is travelling from AAA on the curve joining AAA to BBB. At a particular time, the particle is at PPP and has velocity v⃗\vec vv, as shown. The speed is increasing. Which diagram shows an acceleration a⃗\vec aa that would allow the particle to follow the curve to BBB?

1 mark
Multiple choice
2021HSC
Q9

Four cards have either RED or BLACK on one side and either WIN or TRY AGAIN on the other side. The visible cards are: Card 1 WIN; Card 2 BLACK; Card 3 TRY AGAIN; Card 4 RED.
For the statement “If a card is RED, then it has WIN written on the other side.”

What is the minimum number of cards Sam should turn over to check if the statement is true and which cards should be turned over?

A. 1 and 4
B. 3 and 4
C. 1, 2 and 4
D. 1, 3 and 4

1 mark
Multiple choice
2021HSC
Q10

Consider the two non-zero complex numbers zzz and www as vectors. Which expression is the projection of zzz onto www?

A. Re⁡(zw)∣w∣w\frac{\operatorname{Re}(zw)}{|w|}w∣w∣Re(zw)​w
B. ∣zw∣w\left|\frac zw\right|w​wz​​
C. Re⁡(zw)w\operatorname{Re}\left(\frac zw\right)wRe(wz​)w
D. Re⁡(z)∣w∣w\frac{\operatorname{Re}(z)}{|w|}w∣w∣Re(z)​w

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
Q11 (e)

The complex numbers z=5+iz=5+iz=5+i and w=2−4iw=2-4iw=2−4i are given. Find zˉw\frac{\bar z}{w}wzˉ​, giving your answer in Cartesian form.

2 marks
Short answer
2021HSC
Q11 (d) (ii)

Hence, or otherwise, solve z2+2z+1+i=0z^2+2z+1+i=0z2+2z+1+i=0, giving your solutions in the form a+iba+iba+ib where aaa and bbb are real numbers.

2 marks
Short answer
2021HSC
Q11 (d) (i)

Find the two square roots of −i-i−i, giving the answers in the form x+iyx+iyx+iy, where xxx and yyy are real numbers.

2 marks
Short answer
2021HSC
Q11 (c)

Find the angle between the vectors a⃗=(204)\vec a=\begin{pmatrix}2\\0\\4\end{pmatrix}a=​204​​ and b⃗=(−312)\vec b=\begin{pmatrix}-3\\1\\2\end{pmatrix}b=, giving the angle in degrees correct to 1 decimal place.

3 marks
Short answer
2021HSC
Q11 (a)

The complex numbers z=2eiπ/2z=2e^{i\pi/2}z=2eiπ/2 and w=6eiπ/6w=6e^{i\pi/6}w=6eiπ/6 are given. Find zwzwzw, giving the answer in the form reiθre^{i\theta}reiθ.

2 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
Q12 (b) (i)

Consider Statement A: “If n2n^2n2 is even, then nnn is even.” What is the converse of Statement A?

1 mark
Short answer
2021HSC
Q12 (e) (i)

The diagram shows pyramid ABCDSABCDSABCDS, where ABCDABCDABCD is a square. The diagonals of the square bisect each other at HHH. Show that
HA→+HB→+HC→+HD→=0⃗.\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD}=\vec0.HA+HB+

1 mark
Short answer
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2
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(
1
+
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