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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
Q9

What is the maximum value of ∣eiθ−2∣+∣eiθ+2∣|e^{i\theta}-2|+|e^{i\theta}+2|∣eiθ−2∣+∣eiθ+2∣ for 0≤θ≤2π0\leq\theta\leq2\pi0≤θ≤2π?

A. 5\sqrt55​
B. 444
C. 252\sqrt525 D.

1 mark
Multiple choice
2020HSC
Q10

Which of the following is equal to ∫02af(x) dx\displaystyle\int_0^{2a}f(x)\,dx∫02a​f(x)dx?

A. ∫0a B. C. D.

2020HSC
Q11 (a) (ii)

Consider the complex numbers w=−1+4iw=-1+4iw=−1+4i and z=2−iz=2-iz=2−i. Evaluate .

2020HSC
Q11 (a) (i)

Consider the complex numbers w=−1+4iw=-1+4iw=−1+4i and z=2−iz=2-iz=2−i. Evaluate .

2020HSC
Q11 (b)

Use integration by parts to evaluate
∫1exln⁡x dx.\int_1^e x\ln x\,dx.∫1e​xlnxdx.

3 marks
Short answer
2020HSC
Q11 (c)

A particle starts at the origin with velocity 111 and acceleration given by
a=v2+v,a=v^2+v,a=v2+v,
where vv is the velocity of the particle. Find an expression for , the displacement of the particle, in terms of .

2020HSC
Q11 (d)

Consider the two vectors u=−2i−j+3k\mathbf u=-2\mathbf i-\mathbf j+3\mathbf ku=−2i−j+3k and v=pi+j+2k\mathbf v=p\mathbf i+\mathbf j+2\mathbf kv=. For what values of are and perpendicular?

2020HSC
Q11 (e)

Solve z2+3z+(3−i)=0z^2+3z+(3-i)=0z2+3z+(3−i)=0, giving your answer(s) in the form , where and are real.

2020HSC
Q12 (b) (ii)

Using the projectile model from part (i), show that the Cartesian equation of the path of flight is
y=−gx22u2(tan⁡2θ−2u2gxtan⁡θ+1).y=\frac{-gx^2}{2u^2}\left(\tan^2\theta-\frac{2u^2}{gx}\tan\theta+1\right).y=2u

2020HSC
Q12 (b) (iii)

Given u2>gRu^2>gRu2>gR, prove that there are two distinct values of θ\thetaθ for which the particle will land at x=Rx=Rx=.

2020HSC
Q12 (a) (iii)

Find the velocity of the box after the first three seconds.

2 marks
Short answer
2020HSC
Q12 (a) (i)

A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200200200 newtons at an angle of 30∘30^\circ30∘ to the horizontal. The box experiences a resistive force of 0.3R0.3R0.3R newtons, where RRR is the normal force, as shown in the diagram. Take .

2020HSC
Q12 (b) (i)

A particle is projected from the origin with initial velocity u m/su\,\mathrm{m/s}um/s at an angle θ\thetaθ to the horizontal. The particle lands at x=Rx=Rx=R on the xx-axis. The acceleration vector is where is the acceleration due to gravity. (Do NOT prove this.)

2020HSC
Q12 (a) (ii)

Show that the net force horizontally is approximately 53.253.253.2 newtons.

2 marks
Short answer
2020HSC
Q13 (d) (ii)

By expanding (eiθ+e−iθ)4(e^{i\theta}+e^{-i\theta})^4(eiθ+e−iθ), show that

2020HSC
Q13 (d) (i)

Show that for any integer nnn,
einθ+e−inθ=2cos⁡(nθ).e^{in\theta}+e^{-in\theta}=2\cos(n\theta).einθ+e

2020HSC
Q13 (d) (iii)

Hence, or otherwise, find
∫0π/2cos⁡4θ dθ.\int_0^{\pi/2}\cos^4\theta\,d\theta.∫0π/2​cos4θ

2020HSC
Q13 (b)

Consider two lines in three dimensions given by:
r=(3−17)+λ1(121)andr=(3−62)+λ2(−2 By equating components, find the point of intersection of the two lines.

2020HSC
Q13 (a)

A particle is undergoing simple harmonic motion with period π3\dfrac{\pi}{3}3π​. The central point of motion of the particle is at x=3x=\sqrt3x=. When the particle has its maximum displacement of from the central point of motion. Find an equation for the displacement of the particle in terms of .

2020HSC
Q13 (c) (i)

By considering the right-angled triangle below, or otherwise, prove that
a+b2≥ab,\frac{a+b}{2}\geq\sqrt{ab},2a+b​≥ab where .

2020HSC
Q13 (c) (ii)

Prove that
p2+4q2≥4pq.p^2+4q^2\geq4pq.p2+4q2≥4pq.

1 mark
Short answer
2020HSC
Q14 (c)

Prove by mathematical induction that, for n≥2n\geq2n≥2,
122+132+⋯+1n2<n−1n.\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{n^2}<\frac{n-1}{n}.2

2020HSC
Q14 (d)

Prove that for any integer n>1n>1n>1, log⁡n(n+1)\log_n(n+1)logn​(n+1) is irrational.

3 marks
2020HSC
Q14 (b)

A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2\mathrm{m/s^2}m/s2, is modelled by
a=10(1−(kv)2),a=10\bigl(1-(kv)^2\bigr),a=10(1 where is the velocity of the particle in and . Find the velocity of the particle after seconds.

2020HSC
Q14 (a) (ii)

Prove that z12+z22=z1z2.z_1^2+z_2^2=z_1z_2.z12​+z2

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​

101010
[f(x)−f(2a−x)] dx\displaystyle\int_0^a[f(x)-f(2a-x)]\,dx
∫0a​[f(x)−f(2a−x)]dx

∫0a[f(x)+f(2a−x)] dx\displaystyle\int_0^a[f(x)+f(2a-x)]\,dx∫0a​[f(x)+f(2a−x)]dx

2∫0af(x−a) dx\displaystyle2\int_0^a f(x-a)\,dx2∫0a​f(x−a)dx

∫0a12f(2x) dx\displaystyle\int_0^a\frac12f(2x)\,dx∫0a​21​f(2x)dx
1 mark
Multiple choice
wzˉw\bar{z}
wzˉ
2 marks
Short answer
∣w∣|w|
∣w∣
1 mark
Short answer
v
xxx
vvv
3 marks
Short answer
pi+
j+
2k
ppp
u−v\mathbf u-\mathbf vu−v
u+v\mathbf u+\mathbf vu+v
3 marks
Short answer
a+bia+bi
a+bi
aaa
bbb
4 marks
Short answer
2
−gx2
​
(tan2θ−gx2u2​tanθ+1)
.
3 marks
Short answer
R
2 marks
Short answer
g=10 m/s2g=10\,\mathrm{m/s^2}g=10m/s2

(i) By resolving the forces vertically, show that R=400R=400R=400.

2 marks
Short answer
x

a=(0−g),\mathbf a=\begin{pmatrix}0\\-g\end{pmatrix},a=(0−g​),

ggg

(i) Show that the position vector r(t)\mathbf r(t)r(t) of the particle is given by
r(t)=(utcos⁡θutsin⁡θ−12gt2).\mathbf r(t)=\begin{pmatrix}ut\cos\theta\\ut\sin\theta-\frac12gt^2\end{pmatrix}.r(t)=(utcosθutsinθ−21​gt2​).

3 marks
Short answer
4

cos⁡4θ=18(cos⁡(4θ)+4cos⁡(2θ)+3).\cos^4\theta=\frac18\bigl(\cos(4\theta)+4\cos(2\theta)+3\bigr).cos4θ=81​(cos(4θ)+4cos(2θ)+3).
3 marks
Short answer
−inθ
=
2cos(nθ).
1 mark
Short answer
d
θ
.
2 marks
Short answer
13).\mathbf r=\begin{pmatrix}3\\-1\\7\end{pmatrix}+\lambda_1\begin{pmatrix}1\\2\\1\end{pmatrix} \quad \text{and} \quad \mathbf r=\begin{pmatrix}3\\-6\\2\end{pmatrix}+\lambda_2\begin{pmatrix}-2\\1\\3\end{pmatrix}.
r=​3−17​​+λ1​​121​​3−62​λ2​​−213​

3 marks
Short answer
3
​
t=0t=0t=0
232\sqrt323​
xxx
ttt
3 marks
Short answer
​
,

a>b≥0a>b\geq0a>b≥0
2 marks
Short answer
2
1
​
+
321​+
⋯+
n21​<
nn−1​.
4 marks
Short answer
Short answer
−
(kv)2),

vvv
m/s\mathrm{m/s}m/s
k=0.01k=0.01k=0.01
555
4 marks
Short answer
2
​
=
z1​z2​.
3 marks
Short answer
​
and
r
=
​
+
​
.