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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

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2022Trial
Q15 (b) (i)

ABCDABCDABCD is a parallelogram. NNN is on ADADAD so that . is on the diagonal . Letting and , express in terms of and .

AN→=13AD→\overrightarrow{AN}=\frac13\overrightarrow{AD}
AN=31​AD
MMM
ACACAC
AB→=p⃗\overrightarrow{AB}=\vec pAB=p​
AD→=q⃗\overrightarrow{AD}=\vec qAD=q​
NB→\overrightarrow{NB}NB
p⃗\vec pp​
q⃗\vec qq​
1 mark
Short answer
2022Trial
Q15 (b) (ii)

ABCDABCDABCD is a parallelogram. NNN is on ADADAD so that AN→=13AD→\overrightarrow{AN}=\frac13\overrightarrow{AD}AN=31​AD. MMM is on the diagonal ACACAC. By letting AM→=λAC→\overrightarrow{AM}=\lambda\overrightarrow{AC}AM=λAC and MB→=δNB→\overrightarrow{MB}=\delta\overrightarrow{NB}MB=δNB, prove that
AM→=14AC→.\overrightarrow{AM}=\frac14\overrightarrow{AC}.AM=41​

3 marks
Short answer
2022Trial
Q15 (a) (i)

Show that a2+9b2≥6aba^2+9b^2\ge6aba2+9b2≥6ab, where aaa and bbb are real numbers.

1 mark
Short answer
2022Trial
Q15 (a) (iii)

Hence, if a>b>c>0a>b>c>0a>b>c>0, show that
a2+5b2+9c2>9bc.a^2+5b^2+9c^2>9bc.a2+5b2+9c2>9bc.

2 marks
Short answer
2022HSC
Q15 (a) (ii)

Hence, or otherwise, show that the point PPP cannot be lifted to a position 2h3\frac{2h}{3}32h​ metres above the floor.

2 marks
Short answer
2022HSC
Q15 (b)

The diagrams show two positions of a single piston in the cylinder chamber of a motorcycle. The piston moves vertically, in simple harmonic motion, between a maximum height of 0.170.170.17 metres and a minimum height of 0.050.050.05 metres.

The mass of the piston is 0.8 kg0.8\,\mathrm{kg}0.8kg. The piston completes 404040 cycles per second.

What is the resultant force on the piston, in newtons, that produces the maximum acceleration of the piston? Give your answer correct to the nearest newton.

3 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
2022HSC
Q15 (d)

The complex number zzz satisfies
∣z−4z∣=2.\left|z-\frac4z\right|=2.​z−z4​​=2.

Using the triangle inequality, or otherwise, show that
∣z∣≤5+1.|z|\le\sqrt5+1.∣z∣≤5​+1.

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
2022Trial
Q15 (c) (i)

The line has vector equation
r⃗=λ[131].\vec r=\lambda\begin{bmatrix}1\\3\\1\end{bmatrix}.r=λ​131​​.
For P(x1,y1,z1)P(x_1,y_1,z_1)P(x1​,y1​,z1​), the point QQQ on the line is such that PQPQPQ is a minimum. Using vector projection methods, show that
OQ→=x1+3y1+z111[131].\overrightarrow{OQ}=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}.OQ​=

2 marks
Short answer
2022Trial
Q15 (c) (ii)

It is given that
c⃗=[140]+t[−115]\vec c=\begin{bmatrix}1\\4\\0\end{bmatrix}+t\begin{bmatrix}-1\\1\\5\end{bmatrix}c=​140​​+t​−115​​
does not intersect r⃗\vec rr. Point TTT is on c⃗\vec cc and SSS is a point on r⃗\vec rr such that TSTSTS is a minimum. By first expressing c⃗\vec cc in parametric form, find an expression for TS→\overrightarrow{TS}TS in terms of ttt.

3 marks
Short answer
2022Trial
Q15 (a) (ii)

Hence, or otherwise, show that
a2+5b2+9c2≥3(ab+bc+ac).a^2+5b^2+9c^2\ge3(ab+bc+ac).a2+5b2+9c2≥3(ab+bc+ac).

2 marks
Short answer
2022Trial
Q15 (a) (i)

Show by finding it that the only point of intersection between
(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 the sphere
(x−5)2+(y+2)2+(z−3)2=38(x-5)^2+(y+2)^2+(z-3)^2=38(x−5)2+(y+2)2+(
is (718)\begin{pmatrix}7\\1\\8\end{pmatrix}​718​.

3 marks
Short answer
2022Trial
Q15 (a) (iii)

Show that the line
(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}​xyz​​=​15−16​δ​4−1−1​
is also tangent to the sphere
(x−5)2+(y+2)2+(z−3)2=38(x-5)^2+(y+2)^2+(z-3)^2=38(x−5)2+(y+2)2+(
by showing that it is perpendicular to the radius of the sphere at (718)\begin{pmatrix}7\\1\\8\end{pmatrix}​718​.

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

Hence show that
cos⁡7θ=164(cos⁡7θ+7cos⁡5θ+21cos⁡3θ+35cos⁡θ).\cos^7\theta=\frac1{64}\left(\cos7\theta+7\cos5\theta+21\cos3\theta+35\cos\theta\right).cos7θ=641​(cos7θ+7cos5θ+21cos3θ+35cosθ).

2 marks
Short answer
2022Trial
Q16 (a) (i)

Prove that
cos⁡θ+isin⁡θ−1cos⁡θ+isin⁡θ+1=itan⁡θ2.\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=i\tan\frac\theta2.cosθ+isinθ+1cosθ+isinθ−1​=itan2θ​.

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

Find the roots of w5=1w^5=1w5=1. Express your answers in the form cos⁡θ+isin⁡θ\cos\theta+i\sin\thetacosθ+isinθ, where −π<θ≤π-\pi<\theta\le\pi−π<θ≤π.

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

Using parts (i) and (ii), find the roots of
(2+z2−z)5=1.\left(\frac{2+z}{2-z}\right)^5=1.(2−z2+z​)5=1.

3 marks
Short answer
2022HSC
Q16 (a)

A square in the Argand plane has vertices 5+5i5+5i5+5i, 5−5i5-5i5−5i, −5−5i-5-5i−5−5i and −5+5i-5+5i−5+5i. The complex numbers zA=5+iz_A=5+izA​=5+i, zBz_BzB​ and zCz_CzC​ lie on the square and form the vertices of an equilateral triangle, as shown in the diagram.

Find the exact value of the complex number zBz_BzB​.

4 marks
Short answer
2022HSC
Q16 (b)

A projectile of mass M kgM\,\mathrm{kg}Mkg is launched vertically upwards from a horizontal plane with initial speed v0 m s−1v_0\,\mathrm{m\,s^{-1}}v0​ms−1, where v0<100 m s−1v_0<100\,\mathrm{m\,s^{-1}}v0​<100ms−1. The projectile experiences a resistive force of magnitude 0.1Mv0.1Mv0.1Mv newtons, where v m s−1v\,\mathrm{m\,s^{-1}}vms−1 is the speed of the projectile. The acceleration due to gravity is 10 m s−210\,\mathrm{m\,s^{-2}}10ms−2.

The projectile lands on the horizontal plane 777 seconds after launch.

Find the value of v0v_0v0​, correct to 111 decimal place.

4 marks
Short answer
2022HSC
Q16 (c) (i)

It is given that for positive numbers x1,x2,x3,…,xnx_1,x_2,x_3,\ldots,x_nx1​,x2​,x3​,…,xn​ with arithmetic mean AAA,
x1x2x3⋯xnAn≤1.\frac{x_1x_2x_3\cdots x_n}{A^n}\le1.Anx1​x
(Do NOT prove this.)

Suppose a rectangular prism has dimensions a,b,ca,b,ca,b,c and surface area SSS. Show that
abc≤(S6)3/2.abc\le\left(\frac S6\right)^{3/2}.abc≤(

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

Using part (i), show that when the rectangular prism with surface area SSS is a cube, it has maximum volume.

2 marks
Short answer
2022Trial
Q16 (a) (i)

Using Euler’s theorem, show that if z=eiθ=cos⁡θ+isin⁡θz=e^{i\theta}=\cos\theta+i\sin\thetaz=eiθ=cosθ+isinθ, then
zn+z−n=2cos⁡nθ.z^n+z^{-n}=2\cos n\theta.zn+z−n=2cosnθ.

1 mark
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 (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
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