2014 hsc mathematics ext2 solutions

SOLUTIONS

- 2 -

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Section I - Multiple ChoiceQuestion #

1 (a) (b) (b) (d)

2 (a) (b) (c) (d)

3 (a) (b) (c) (d)

4 (a) (b) (c) (d)

5 (a) (b) (c) (d)

6 (a) (b) (c) (d)

7 (a) (b) (c) (d)

8 (a) (b) (c) (d)

9 (a) (b) (c) (d)

10 (a) (b) (c) (d)

Question 11 (15 marks)

(a)

i) z + w = 1 − i

= 2cis (− π4 )

ii)zw

= −2 − 2i3 + i

× 3 − i3 − i

= −8 − 4i10

= −4 − 2i5

= − 45 − 2

5 i

(b) I = ∫12

0(3x − 1)cos (π x) d x

We use integration by parts:

I = [ 3x − 1π

sin(π x)]12

0− 3

π ∫12

0sin(π x)d x

= 12π

+ 3π2 (0 − 1)

= 12π

− 3π2

(c) See diagram below

(d) See diagram below

Notes.

We can make use of the fact that the function is even!

It also has as an asymptote y = x2 for x → ± ∞, and x = 0

as a vertical asymptote.


Section II

π4π4

u 3x − 1

cos (πx)3

1π

sin (πx)

u′�

v

v′�

(e)

Vcylinder = π (R2 − r2)h

= π ((y + d y)2 − y2) x

= π (y2 + 2yd y + (d y)2 − y2) x

= 2π x yd y as (d y)2 is negligible

∴ V ≈6

∑y=0

= 2π x yd y

= limdy→0

6

∑y=0

2π x yd y

= 2π∫6

0y2(6 − y)d y

= 2π∫6

0(6y2 − y3)d y

= 2π [2y3 − y4

4 ]6

0

= 216π u3

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(a)

i) See diagram below

ii) See diagram below

- 5 -


(b)

i) x = 2 cos θ is a solution to x3 − 3x = 3

∴ 8 cos3 θ − 6 cos θ = 3

∴ 4 cos3 θ − 3 cos θ =3

2

∴ cos 3θ =3

2

ii) cos 3θ =3

2

∴ 3θ = π6 , 11π

6 , 12π6

∴ θ = π18 , 11π

18 , 13π18

Hence, the three real solutions are

x = 2 cos ( π18 ),

x = 2 cos ( 11π18 ), and

x = 2 cos ( 13π18 )

(c) L : x2 − y2 = 5 −(1)

Z : x y = 6 −(2)

Differentiating L with respect to x:

2x − 2yd yd x

= 0

d yd x

= xy

At (x0, y0)

d yd x

= mL = x0y0

Secondly, we repeat for Z:

y + xd yd x

= 0

d yd x

= − yx

At (x0, y0)

mZ = − y0x0

∴ At (x0, y0),

mL × mZ = x0y0

× − y0x0

= − 1

Hence the tangents to these curves are

perpendicular to one another.

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(d)

i) I0 = ∫1

0

1x2 + 1 d x

= [tan−1 x]10

= π4

ii) In + In−1

= ∫1

0

x2n + x2n−2

x2 + 1 d x

= ∫1

0x2n−2 ( x2 + 1

x2 + 1 ) d x

= ∫1

0x2n−2d x

= [ x2n − 12n − 1 ]

1

0

= 12n − 1

iii) ∫1

0

x4

x2 + 1 = I2

= 13 − I1 (by part (ii))

= 13 − (1 − I0)

= 13 − 1 + π

4

= π4 − 2

3

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(a) Let t = tan x2

x = 2 tan-1 t

d xdt

= 21 + t2

d x = 21 + t2 dt

When x = π2 , t = tan π

4 = 1

When x = π3 , t = tan π

6 = 13

∫π2

π3

d x3 sin x − 4 cos x + 5

= ∫1

13

13. 2t

1 + t2− 4. 1 − t2

1 + t2+ 5

. 21 + t2 dt

= ∫1

13

26t − 4 + 4t2 + 5 + 5t2 dt

= ∫1

13

29t2 + 6t + 1 dt

= ∫1

13

2(3t + 1)2 dt

= − 23 [ 1

3t + 1 ]1

13

, simplifying yields

=2 3 − 3

6

(b)

Using Pythagoras’s theorem we can express the perpendicular

height of the trapezium in the equation

h2 + ( 12 x2)2 = (x2)2

h2 = 34 x4

h =3

2 x2

The area of the trapezium is hence:

h2 (a + b) =

34 x2(2x2 + x2)

=3 3

4 x4

The volume of the solid is then:

V = limdx→0

Σ20

3 34 x4 d x

= ∫2

0

3 34 x4 d x

=3 3

4 [ x5

5 ]2

0

=3 3

4 . 325

=24 3

5 u3

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x2

x2

x2 x2

x2

2x2

2

h

(c)

i) We sub point M into the equation of the hyperbola:

L HS =a2(t2 + 1)2

4t2

a2 −b2(t2 − 1)2

4t2

b2

= t4 + 2t2 + 14t2 − t4 − 2t2 + 1

4t2

= 4t2

4t2

= 1

= R HS

Hence, M lies on the hyperbola.

ii) From part i), we know M lies on the hyperbola.

Thus it will be sufficient to show that mPQ = mtangM.

mPQ = bt + b

tat − a

t

= bt2 + bat2 − a

mtangM = d yd x

= d ydt

. dtd x

=ddt ( bt

2 − b2t )

ddt ( at

2 − a2t )

= bt2 + bat2 − a

= mPQ

Hence, PQ is a tangent to the hyperbola at M.

iii) Consider OS2 = a2e2 = a2 + b2, as e2 = a2 + b2

a2 .

Then OP × OQ = t2 (a2 + b2) × 1t2 (a2 + b2)

= a2 + b2

= OS2, as required.

iv) P and S share the same abscissa.

Hence, at = ae, and t = e.

The gradient of MS is then:

mMS =b(t2 − 1)

2ta(t2 + 1)

2t − ae

=b(e2 − 1)

2ea(e2 + 1)

2e − ae

= b(e2 − 1)a(e2 + 1) − 2ae2

= b(e2 − 1)ae2 + a − 2ae2

= b(e2 − 1)a − ae2

= b(e2 − 1)a(1 − e2)

= − ba

,

which is the gradient of an asymptote.

Hence, MS is parallel to one of the asymptotes of the

hyperbola.

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(a) P(x) = x5 − 10x2 + 15x − 6

i) P(1) = 1 − 10 + 15 − 6 = 0

∴ x = 1 is a root

P′�(x) = 5x4 − 20x + 15

P′�(1) = 5 − 20 + 15 = 0

P′�′�(x) = 20x3 − 20

P′�′�(1) = 20 − 20 = 0

∴ x = 1 is a root of multiplicity three.

ii) The coefficients of the polynomial are real, so the

complex roots of the polynomial occur in

conjugate pairs. Let the roots be a + ib, a − ib.

By sum of roots:

a + ib + a − ib + 1 + 1 + 1 = 0

2a + 3 = 0

a = − 32

By product of roots:

(a + ib)(a − ib)(1)(1)(1) = 6

a2 − b2 = 6

94 − b2 = 6

b = ± 152

Therefore the complex roots are

− 32

± 152 i

(b)x2

a2 + y2

b2 = 1

i) Differentiating with respect to x:

2xa2 + 2y

b2d yd x

= 0

d yd x

= − b2xa2y

d yd x

= − b cos θa sin θ

at P .

∴ mnorm = a sin θb cos θ

mOP = b sin θa cos θ

Using the angle between two lines formula:

tan ϕ = mnorm − mOP

1 + mnormmOP

=a sin θb cos θ − b sin θ

a cos θ

1 + ab sin2 θab cos2 θ

= a2 sin θ cos θ − b2 sin θ cos θab cos2 θ + ab sin2 θ

= (a2 − b2)(sin θ cos θ )ab(cos2 + sin2 θ )

= ( a2 − b2

ab ) sin θ cos θ , as a > b

ii) tan ϕ = 12 ( a2 − b2

ab ) sin 2θ

sin 2θ is maximum when sin 2θ = 1, i.e. 2θ = π2

∴ θ = π4

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(c)

i) There are 2 forces on the train - the constant driving

force F and a resistive force Kv2.

Hence the net force on the train is:

Net Force = F − Kv2

m ··x = F − Kv2

The terminal velocity is 300 and occurs when

··x = 0.

Subbing this information into the equation yields

0 = F − K(300)2

K = F3002

We sub K back into the original equation to obtain:

m ··x = F − ( F3002 ) × v2

= F [1 − ( v300 )

2

], as required.

ii)m ··x = F 1 − ( v2

3002 )

··x = Fm

1 − ( v2

3002 )

dvdt

= Fm

1 − ( v2

3002 )

dvdt

= Fm ( 300 + v

300 ) ( 300 − v300 )

dtdv

= mF ( 300

300 + v ) ( 300300 − v )

= 3002mF [ 1

(300 + v)(300 − v) ]

Let 1

(300 + v)(300 − v) = A300 + v

+ B300 − v

1 = A(300 − v) + B(300 + v)

When v = 300, 1 = 600B

B = 1600

When v = − 300, 1 = 600A

A = 1600

∴ ∫T

0dt = 3002m

600F ∫200

0

1300 + v

+ 1300 − v

dv

T = 150 mF [ln ( 300 + v

300 − v )]200

0

T = 150 mF

ln ( 500100 )

∴ time taken for train to reach a velocity of

200k m /h is T = 150 mF

ln 5 hours.

Kv2 F

- 11 -



(a) (a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc) = 1,

since a + b + c = 1.

Since a ≤ b ⟹ ab ≥ a2

Since a ≤ c ⟹ ac ≥ a2

Since b ≤ c ⟹ bc ≥ b2

Thus, we have:

a2 + b2 + c2 + 2(ab + ac + bc) = 1a2 + b2 + c2 + 2(a2 + a2 + b2) ≤ 1

∴ 5a2 + 3b2 + c2 ≤ 1

(b)

i) Consider (1 + i )n + (1 − i )n

1 + i = 2cisπ4

1 − i = 2cis(− π4 )

Applying De Moivre’s Theorem, and the conjugate

property, we obtain:

L HS = (1 + i )n + (1 − i )n

= 2Re [(1 + i )n]

= 2( 2)ncos nπ4

= R HS

ii) Consider (1 + i )n + (1 − i )n = 2( 2)ncos nπ4

(1 + i )n + (1 − i )n =n

∑r=0 [(n

r) ir + (nr) (−1)r ir]

Now this = {0 for odd r2(n

r)ir for even r

i.e.

2( 2)ncos nπ4 = 2 [(n

0) + i2 (n2) + i4 (n

4)⋯ + in (nn)]

(n0) + i2(n

2) + . . . + i2k( n2k) + . . . + in(n

n) = ( 2)ncos nπ4

If n is divisible by 4, i.e. n = 4p, then:

cos nπ4 = cos pπ = (−1) n

4

∴ (n0) − (n

2) + (n4) − (n

6) + . . . + (nn) = (−1) n

4 ( 2)n

(c)

i) ∑ F y = 0, as there is no vertical motion

∴ k v2 = mg + T sin θ − (1)

∑ Fx = − mv2

r, as horizontal forces sum to

centripetal forces

−T cos ϕ = − mv2

r

∴ T cos ϕ = mv2

r− (2)

But since r = l cos ϕ, cos ϕ = rl

∴ T cos2 ϕ = mv2

l− (3)

Performing the operation (1)(3) :

sin ϕcos2 ϕ

= k v2 − mgmv2

l

= lkm

− lgv2 , as required

- 12 -


k v2

mg

ϕ

T

ii)From assumption, sin ϕ

1 − sin2 ϕ< lk

m

m sin ϕ < lk (1 − sin2 ϕ), noting m > 0 and

1 − sin2 ϕ > 0

∴ lk sin2 ϕ + m sin ϕ − lk < 0

Using quadratic formula,

−m − m2 + 4l2k2

2lk< sin ϕ < −m + m2 + 4l2k2

2lk

iii)Let y = sin ϕcos2 ϕ

= sec ϕ . tan ϕ

∴ d ydϕ

= sec3 θ + sec ϕ . tan2 ϕ

= 1cos3 ϕ

(1 + sin2 θ )

But for − π2 < ϕ < π

2 :

cos ϕ > 0, ∴ cos3 ϕ > 0, 1 + sin2 ϕ > 0

∴ d ydϕ

> 0 for − π2 < ϕ < π

2

∴ sin ϕcos2 θ

is an increasing function for

− π2 < ϕ < π

2

iv) As v increases, sin ϕ

cos2 ϕ= ( lk

m− lg

v2 ) → lkm

from

below.

From (iii), we know sin ϕ

cos2 ϕ is an increasing

function, and thus ϕ must increase as v increases.

- 13 -



(a) Let α = ∠APX, β = BPY, γ = XPY, as shown on diagram.

i) ∠DPQ = ∠A DP (alt. angles on parallel lines A D ∥PQ)

∠APX = ∠A DP (angle subtended by chord AP and tangent at

P is equal to the angle subtended by chord AP

at the circumference in the alternate segment)

∴ ∠DPQ = ∠APX (transitivity property)

ii) Similarly to (i), we have ∠BPY = ∠PCB = ∠CPR = β

Since YPQ & XPR are collinear, ∠XPY = ∠QPR = γ (vert. opposite angles)

Now, all angles at P must sum to 2π (angles in a revolution)

∴ 2π = 2 (α + β + γ + π2 )

∴ α + β + γ = π2

Now, ∠APX + ∠XPY + ∠YPB + ∠BPC = α + β + γ + π2

∴ ∠APC = π

∴ A, P and C must be collinear.

iii) We know ∠A DP = α, and ∠PCB = β = ∠BPY

∠DPQ = β (vert. opposite angles)

∠A DP = ∠DPQ (alt. angles on A D ∥PQ)

i.e. α = β and thus ∠A DP = ∠PCB

∴ A BCD must be cyclic (A B is a chord & C, D lie on same circle, as angle subtended by A B to C, & to D are equal)

- 14 -


(b) Consider the sum of the following GP:

1 + (−x2)1 + (−x2)2 + . . . + (−x2)n−1 =((−x2)2 − 1)

−x2 − 1

but

1 + (−x2)1 + (−x2)2 + . . . + (−x2)n−1 =n

∑r=0

(−1)r x2r

−(1)

n

∑r=0

(−1)r x2r =((−x2)2 − 1)

−x2 − 1

=−(−x2)2 + 1

x2 + 1

=−(−x2)n

1 + x2 + 11 + x2

By rearranging,

(−x2)n

1 + x2 = 11 + x2 −

n

∑r=0

(−1)r x2r

= 11 + x2 −

n−1

∑r=0

(−1)r x2r + (−1)nx2n

∴ (−x2)n

1 + x2 − (−1)nx2n = 11 + x2 −

n−1

∑r=0

(−1)r x2r

∴ (−1)nx2n − (−1)nx2n − (−1)nx2n+2

1 + x2 = 11 + x2 −

n−1

∑r=0

(−1)r x2r

∴ (−1)n+1x2n+2

1 + x2 = 11 + x2 −

n−1

∑r=0

(−1)r x2r −(2)

Now as 0 ≤ x2

1 + x2 ≤ 1,

It follows that −1 ≤ (−1)nx2

1 + x2 ≤ 1

And so −x2n ≤ (−1)nx2n+2

1 + x2 ≤ x2n

And finally by subbing in equation (2),

−x2n ≤ 11 + x2 −

n−1

∑r=0

(−1)r x2r ≤ x2n

And subbing in (1),

−x2n ≤ 11 + x2 − (1 + (−x2)1 + (−x2)2 + . . . + (−x2)n−1) ≤ x2n

ii) ∫1

0− x2n d x ≤ ∫

1

0

11 + x2 d x −

n

∑r=0

∫1

0(−1)r x2r d x ≤ ∫

1

0x2n d x

[− x2n+1

2n + 1 ]1

0

≤ [tan−1 x]10 −

n

∑r=0 [ (−1)r x2r+1

2r + 1 ]1

0

≤ [ x2n+1

2n + 1 ]1

0

− 12n + 1 ≤ π

4 − (1 − 13 + 1

5 − . . . + (−1)n−1 12n + 1 ) ≤ 1

2n + 1

- 15 -


iii) As n → ∞, 1

2n + 1 → 0

Therefore, 0 ≤ π4 − (1 − 1

3 + 15 − . . . ) ≤ 0

And thus π4 = 1 − 1

3 + 15 − . . .

c) ∫ ln x

(1 + ln x)2 d x = ∫ ln x + 1 − 1

(1 + ln x)2 d x

= ∫ 11 + ln x

d x + ∫ −1

(1 + ln x)2 d x

Now for ∫ 11 + ln x

d x, apply integration by parts

∫ 11 + ln x

d x = x1 + ln x

+ ∫ 1

(1 + ln x)2 d x

Now subbing back in,

∫ ln x

(1 + ln x)2 d x = x1 + ln x

+ ∫ 1

(1 + ln x)2 d x − ∫ 1

(1 + ln x)2 d x

= x1 + ln x

+ C

u1

1 + ln x

1− 1

x

(1 + ln x)2

x

u′�

v

v′�

- 16 -


2014 hsc mathematics ext2 solutions

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