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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
2023Trial
Q11 (b) (i)

The nnnth pentagonal number can be represented by the arithmetic series

P5(n)=1+4+7+⋯+(3n−2).P_5(n)=1+4+7+\cdots +(3n-2).P5
​
(
n
)
=
1+
4+
7+
⋯+
(3n−
2).

Hence show that

P5(n)=3n2−n2P_5(n)=\frac{3n^2-n}{2}P5​(n)=23n2−n​

for n∈Z+n\in\mathbb{Z}^+n∈Z+.

1 mark
Short answer
2023HSC
Q11 (a)

Solve the quadratic equation z2−3z+4=0z^2-3z+4=0z2−3z+4=0, where zzz is a complex number. Give your answers in Cartesian form.

2 marks
Short answer
2023Trial
Q11 (e) (i)

In the diagram below, OABCDOABCDOABCD is a solid figure where ∣OA→∣=∣OB→∣=4|\overrightarrow{OA}|=|\overrightarrow{OB}|=4∣OA∣=∣OB∣=4 units and ∣OD→∣=3|\overrightarrow{OD}|=3∣OD∣=3 units. The edge OD→\overrightarrow{OD}OD is vertical, DC→\overrightarrow{DC}DC is parallel to OB→\overrightarrow{OB}OB and ∣DC→∣=1|\overrightarrow{DC}|=1∣DC∣=1 unit. The base, OABOABOAB, is horizontal and ∠AOB=90∘\angle AOB=90^\circ∠AOB=90∘. Unit vectors i⃗\vec ii, j⃗\vec jj​ and k⃗\vec kk are parallel to OA→\overrightarrow{OA}OA, OB→\overrightarrow{OB}OB and OD→\overrightarrow{OD}OD respectively. The midpoint of AB‾\overline{AB}AB is MMM and the point NNN on BC‾\overline{BC}BC is such that NC→=2BN→\overrightarrow{NC}=2\overrightarrow{BN}NC=2BN.

Express vectors MD→\overrightarrow{MD}MD and ON→\overrightarrow{ON}ON in terms of , and .

2 marks
Short answer
2023HSC
Q11 (e)

A particle moves in simple harmonic motion described by x¨=−9(x−4)\ddot x=-9(x-4)x¨=−9(x−4). Find the period and the central point of motion.

2 marks
Short answer
2023Trial
Q11 (c)

Given a∈Za\in\mathbb{Z}a∈Z, prove that if 3a2−4a+53a^2-4a+53a2−4a+5 is even, then aaa is odd.

2 marks
Short answer
2023HSC
Q11 (b)

Find the angle between the vectors a=i+2j−3k\mathbf a=\mathbf i+2\mathbf j-3\mathbf ka=i+2j−3k and b=−i+4j+2k\mathbf b=-\mathbf i+4\mathbf j+2\mathbf kb=−i+4j+2k, giving your answer to the nearest degree.

3 marks
Short answer
2023Trial
Q11 (b) (ii)

A polygonal number is an integer which can be represented as a series of dots arranged
in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal
numbers are examples of polygonal numbers.

For an rrr-sided regular polygon, where r∈Z+r\in\mathbb{Z}^+r∈Z+, r≥3r\geq 3r≥3, the nnnth polygonal number Pr(n)P_r(n)Pr​(n)
is given by

Pr(n)=(r−2)n2−(r−4)n2P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}Pr​(n)=2(r−2)

where n∈Z+n\in\mathbb{Z}^+n∈Z+. Hence, the nnnth triangular number can be expressed as

P3(n)=n2+n2.P_3(n)=\frac{n^2+n}{2}.P3​(n)=2n2+n

The nnnth polygonal number, Pr(n)P_r(n)Pr​(n), can be represented by the series

Pr(n)=∑m=1n(1+(m−1)(r−2)),P_r(n)=\sum_{m=1}^{n}\left(1+(m-1)(r-2)\right),Pr​(n)=m=1∑

where r∈Z+r\in\mathbb{Z}^+r∈Z+, r≥3r\ge 3r≥3.

Use mathematical induction to prove that

Pr(n)=(r−2)n2−(r−4)n2P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}Pr​(n)=2(r−2)

where n∈Z+n\in\mathbb{Z}^+n∈Z+.

3 marks
Short answer
2023Trial
Q12 (b) (iii)

Hence or otherwise, evaluate

∫0π/2x1+cos⁡x+sin⁡x dx.\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx.∫0π/2​1+cosx+sinxx​dx.
2 marks
Short answer
2023Trial
Q12 (a) (iii)

Let α\alphaα be a real number and suppose that zzz is a complex number such that

z+1z=2cos⁡α.z+\frac{1}{z}=2\cos\alpha.z+z1​=2cosα.

You may assume that

zn+1zn=2cos⁡(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1​=

for all positive integer nnn. Let

ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1​.

Hence or otherwise, find all solutions of

16(cos⁡α)4+8(cos⁡α)3−12(cos⁡α)2−4cos⁡α+1=0.16(\cos\alpha)^4+8(\cos\alpha)^3-12(\cos\alpha)^2-4\cos\alpha+1=0.16(cosα)4+8(cosα)3−
2 marks
Short answer
2023HSC
Q12 (e) (i)

The complex number 2+i2+i2+i is a zero of P(z)=z4−3z3+cz2+dz−30P(z)=z^4-3z^3+cz^2+dz-30P(z)=z4−3z3+cz2+dz−30, where c,d∈Rc,d\in\mathbb Rc,d∈R. Explain why 2−i2-i2−i is also a zero of P(z)P(z)P(z).

1 mark
Short answer
2023Trial
Q12 (c) (i)

Let

In=∫e−11(1+log⁡ex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn​=∫e−11​(1+loge​x)ndx

and

Jn=∫e−11(log⁡ex)(1+log⁡ex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn​=∫e

for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….

Show that In=1−nIn−1I_n=1-nI_{n-1}In​=1−nIn−1​ for n=1,2,3,….

2 marks
Short answer
2023Trial
Q12 (c) (ii)

Let

In=∫e−11(1+log⁡ex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn​=∫e−11​(1+loge​x)ndx

and

Jn=∫e−11(log⁡ex)(1+log⁡ex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn​=∫e

for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….

Show that

Jn=1−(n+2)InJ_n=1-(n+2)I_nJn​=1−(n+2)In​

for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….

2 marks
Short answer
2023HSC
Q12 (c) (ii)

Using the same inclined-plane model, given that the object is initially at rest, find its velocity v(t)\mathbf v(t)v(t) in terms of ggg, θ\thetaθ, ttt and i\mathbf ii.

2 marks
Short answer
2023Trial
Q12 (a) (i)

Let α\alphaα be a real number and suppose that zzz is a complex number such that

z+1z=2cos⁡α.z+\frac{1}{z}=2\cos\alpha.z+z1​=2cosα.

You may assume that

zn+1zn=2cos⁡(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1​=

for all positive integer nnn. Let

ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1​.

Show that

ω4+ω3−3ω2−2ω=z+1z+z2+1z2+z3+1z3+z4+1z4.\omega^4+\omega^3-3\omega^2-2\omega=z+\frac{1}{z}+z^2+\frac{1}{z^2}+z^3+\frac{1}{z^3}+z^4+\frac{1}{z^4}.ω4+ω3−3
2 marks
Short answer
2023HSC
Q12 (a)

Prove that 23\sqrt{23}23​ is irrational.

3 marks
Short answer
2023HSC
Q12 (b)

Prove that for all real numbers xxx and yyy, where x2+y2≠0x^2+y^2\ne0x2+y2=0, (x+y)2x2+y2≤2.\frac{(x+y)^2}{x^2+y^2}\le2.x2+y2(x+y)

2 marks
Short answer
2023Trial
Q12 (a) (ii)

Let α\alphaα be a real number and suppose that zzz is a complex number such that

z+1z=2cos⁡α.z+\frac{1}{z}=2\cos\alpha.z+z1​=2cosα.

You may assume that

zn+1zn=2cos⁡(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1​=

for all positive integer nnn. Let

ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1​.

Find the ninth roots of unity.

2 marks
Short answer
2023HSC
Q12 (d)

Find the cube roots of 2−2i2-2i2−2i. Give your answer in exponential form.

3 marks
Short answer
2023HSC
Q12 (c) (i)

An object of mass mmm slides down a smooth inclined plane. The plane makes an angle θ\thetaθ with the horizontal. The normal reaction is R\mathbf RR, gravity is g\mathbf gg with magnitude ggg, and i,j\mathbf i,\mathbf ji,j are unit vectors parallel and perpendicular to the plane. Show that the resultant force is F=−(mgsin⁡θ)i.\mathbf F=-(mg\sin\theta)\mathbf i.F=−(mgsinθ)i.

2 marks
Short answer
2023Trial
Q12 (b) (i)

Show that

∫0π/2dx1+cos⁡x+sin⁡x=ln⁡2.\int_0^{\pi/2}\frac{dx}{1+\cos x+\sin x}=\ln 2.∫0π/2​1+cosx+sinxdx​=ln2.
3 marks
Short answer
2023Trial
Q12 (b) (ii)

By making the substitution u=a−xu=a-xu=a−x, 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
2023Trial
Q12 (c) (iii)

Let

In=∫e−11(1+log⁡ex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn​=∫e−11​(1+loge​x)ndx

and

Jn=∫e−11(log⁡ex)(1+log⁡ex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn​=∫e

for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….

Hence find the value of J3J_3J3​ in simplest exact form.

2 marks
Short answer
2023Trial
Q13 (d) (ii)

Another line, ℓ2\ell_2ℓ2​, is defined by the vector equation

r⃗=(012)+λ(a1−1),λ,a∈R.\vec r=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}a\\1\\-1\end{pmatrix},\qquad \lambda,a\in\mathbb{R}.r=​0

Find the possible values of aaa when the angle between ℓ1\ell_1ℓ1​ and ℓ2\ell_2ℓ2​ is π4\frac{\pi}{4}.

2 marks
Short answer
2023Trial
Q13 (c) (i)

Find real numbers aaa, bbb and ccc such that

10(x+1)(x2+4)≡ax+1+bx+cx2+4.\frac{10}{(x+1)(x^2+4)}\equiv\frac{a}{x+1}+\frac{bx+c}{x^2+4}.(x+1)(x2+4)10​≡x+1a
3 marks
Short answer
2023Trial
Q13 (a) (i)

Use De Moivre’s Theorem to solve the equation

z3=42(1+i).z^3=4\sqrt{2}(1+i).z3=42​(1+i).
2 marks
Short answer
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