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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.

Current syllabus: NESA Stage 6 (2017)
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64 questions found

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2020HSC
Q14 (a) (i)

Let z1z_1z1​ be a complex number and let
z2=eiπ/3z1.z_2=e^{i\pi/3}z_1.z2​=eiπ/3z1​.
The diagram shows points AAA and BBB, which represent z1z_1z1​ and z2z_2z2​, respectively, in the Argand plane.

(i) Explain why triangle OABOABOAB is an equilateral triangle.

2 marks
Short answer
2020HSC
Q15 (a) (ii)

Using the proposition PPP from part (i), write down the contrapositive of PPP.

1 mark
Short answer
2020HSC
Q15 (b) (iv)

Using parts (ii) and (iii), or otherwise, prove that TTT is the point that divides the interval PRPRPR in the ratio 2:12:12:1.

1 mark
Short answer
2020HSC
Q15 (b) (iii)

Let OPQROPQROPQR be a parallelogram with OP→=p\overrightarrow{OP}=\mathbf pOP and . The point is the midpoint of and is the intersection of and , as shown in the diagram.

2020HSC
Q15 (b) (ii)

Using the same definitions as in part (i), prove that
OC→=mm+na+nm+nb.\overrightarrow{OC}=\frac m{m+n}\mathbf a+\frac n{m+n}\mathbf b.OC=

2020HSC
Q15 (b) (i)

The point CCC divides the interval ABABAB so that
CBAC=mn.\frac{CB}{AC}=\frac mn.AC The position vectors of and are and , respectively, as shown in the diagram.

2020HSC
Q15 (a) (iii)

Using the proposition PPP from part (i), write down the converse of PPP and state, with reasons, whether this converse is true or false.

3 marks
Short answer
2020HSC
Q15 (a) (i)

In the set of integers, let PPP be the proposition: “If k+1k+1k+1 is divisible by 333, then k3+1k^3+1k is divisible by $3”.

2020HSC
Q16 (b) (iii)

Let
Jn=∫01xn(1−x)n dx,n=0,1,2,….J_n=\int_0^1x^n(1-x)^n\,dx,\qquad n=0,1,2,\ldots.Jn​=∫

2020HSC
Q16 (b) (ii)

Using the result of part (i), deduce that
In=22n(n!)2(2n+1)!.I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}.In​=(2

2020HSC
Q16 (b) (i)

Let
In=∫0π/2sin⁡2n+1(2θ) dθ,n=0,1,….I_n=\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta,\qquad n=0,1,\ldots.In​=∫

2020HSC
Q16 (a) (ii)

Given that v<gmkv<\dfrac{gm}{k}v<kgm​, show that when
t=3mkln⁡2,t=\frac{3m}{k}\ln2, the velocity of the larger mass is

2020HSC
Q16 (a) (i)

Two masses, 2m2m2m kg and 4m4m4m kg, are attached by a light string. The string is placed over a smooth pulley as shown. The two masses are at rest before being released and vvv is the velocity of the larger mass at time ttt seconds after they are released. The force due to air resistance on each mass has magnitude kvkv, where is a positive constant.

2020HSC
Q16 (b) (iv)

Prove that
(2nn!)2≤(2n+1)!.\bigl(2^n n!\bigr)^2\leq(2n+1)!.(2nn!)2≤(2n+

← PreviousPage 3 of 3
=
p
OR→=r\overrightarrow{OR}=\mathbf rOR=r
SSS
QRQRQR
TTT
PRPRPR
OSOSOS

(iii) Show that
OT→=23r+13p.\overrightarrow{OT}=\frac23\mathbf r+\frac13\mathbf p.OT=32​r+31​p.

3 marks
Short answer
m+nm​
a
+
m+nn​b.
1 mark
Short answer
CB
​
=
nm​.

AAA
BBB
a\mathbf aa
b\mathbf bb

(i) Show that
AC→=nm+n(b−a).\overrightarrow{AC}=\frac n{m+n}(\mathbf b-\mathbf a).AC=m+nn​(b−a).

2 marks
Short answer
3
+
1

(i) Prove that the proposition PPP is true.

2 marks
Short answer
0
1
​
xn
(
1
−
x)ndx,n=
0,1,2,….

Using the result of part (ii), or otherwise, show that
Jn=(n!)2(2n+1)!.J_n=\frac{(n!)^2}{(2n+1)!}.Jn​=(2n+1)!(n!)2​.

3 marks
Short answer
n
+
1
)!
22n(n!)2
​
.
3 marks
Short answer
0
π/2
​
sin2n+1
(
2
θ
)
d
θ
,
n
=
0,1,….

(i) Prove that
In=2n2n+1In−1,n≥1.I_n=\frac{2n}{2n+1}I_{n-1},\qquad n\geq1.In​=2n+12n​In−1​,n≥1.

3 marks
Short answer
t=
k3m​ln2,


gm2k.\frac{gm}{2k}.2kgm​.
3 marks
Short answer
kv
kkk

(i) Show that
dvdt=gm−kv3m.\frac{dv}{dt}=\frac{gm-kv}{3m}.dtdv​=3mgm−kv​.

2 marks
Short answer
1)!.
2 marks
Short answer