SUCCESS ONEHSC · 2020-04-17 · 463 2018 HIGHER SCHOOL CERTIFICATE—EXAMINATION PAPER P P ISN 8 1...

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past HSC questions

Past HSC papers and worked answers

1994–2018

SUCCESS ONE ®

HSC*

MATHEMATICS EXTENSION 1

Free-to-download Sample Tests with answers

462

2018 HIGHER SCHOOL CERTIFICATE—EXAMINATION PAPER

NSW Education Standards Authority

2018 HIGHER SCHOOL CERTIFICATE EXAMINATION

Mathematics Extension 1 General Instructions

• Reading time – 5 minutes • Working time – 2 hours • Write using black pen • Calculators approved by NESA may be used • A reference sheet is provided at the back of this paper • In Questions 11–14, show relevant mathematical reasoning

and/ or calculations

Total marks: 70

Section I – 10 marks (pages 2–5) • Attempt Questions 1–10 • Allow about 15 minutes for this section

Section II – 60 marks (pages 6–13) • Attempt Questions 11–14 • Allow about 1 hour and 45 minutes for this section

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2018 HSC Mathematics Extension 1 Examination Paper

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iv Exce l S U C C E S S O N E H S C • M A T H E M A T I C S E X T E N S I O N 1

PAST HSC EXAMINATION PAPERSThe past HSC Examination papers contained in this publication have been reproduced under licence from the NSW Education Standards Authority (NESA) in whom copyright is vested.

NESA takes no responsibility for errors in the reproduction of past HSC Examination papers contained in this publication.

NESA was the first publisher of each examination paper in the year indicated on the first page of each examination paper.

HSC Examination papers in 3 Unit Mathematics 1994–2000 © NSW Education Standards Authority for and on behalf of the Crown in right of the State of New South Wales 1994–2000.

HSC Examination papers in Mathematics Extension 1 2001–2018 © NSW Education Standards Authority for and on behalf of the Crown in right of the State of New South Wales 2001–2018.

WORKED ANSWERSThe worked answers contained in this publication are examples of answers which the authors believe would score full marks. They are not necessarily ideal or model answers, nor are they the only answers which would score full marks. They are not endorsed by NESA.

Worked answers for the 1994–2003 HSC Examination papers were written by José Carreno.

Worked answers for the 2004–2008 HSC Examination papers were written by Barbara D’Angelo and Megan McKeown.

Worked answers for the 2009 HSC Examination paper were written by Barbara D’Angelo.

Worked answers for the 2010–2018 HSC Examination papers were written by Lyn Baker.

THE HSC MATHEMATICS EXTENSION 1 EXAMINATIONFrom 2001, HSC Mathematics Extension 1 has been the new name for the former HSC 3 Unit Mathematics course.

Since 2012 the HSC Mathematics Extension 1 Examination paper has consisted of two sections. Section I has 10 objective-response questions worth 10 marks in total. Section II has four questions worth 15 marks each. Each of these four questions consists of a number of parts where working should be shown.

There is reading time of 5 minutes and working time of 2 hours.

Students should note that marks are shown for each part of each question and should allocate their time accordingly.

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464


4 The diagram shows the graph of y = a(x + b)(x + c)(x + d)2.

O

−6

x1−1−2

y

What are possible values of a, b, c and d ?

A. a = −6, b = −2, c = −1, d = 1

B. a = −6, b = 2, c = 1, d = −1

C. a = −3, b = −2, c = −1, d = 1

D. a = −3, b = 2, c = 1, d = −1

5 The diagram shows the number of penguins, P(t), on an island at time t.

t

P

3000

1500

– 3 –

Which equation best represents this graph?

A. P(t) = 1500 + 1500e−kt

B. P(t) = 3000 − 1500e−kt

C. P(t) = 3000 + 1500e−kt

D. P(t) = 4500 − 1500e−kt

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465


– 4 –

6 The diagram shows the graph of y = â ( x ). The equation â ( x ) = 0 has a solution at x = w .

O x

y

a b c d w

y = â (x )

Newton’s method can be used to give an approximation close to the solution x = w.

Which initial approximation, x1, will give the second approximation that is closest to the solution x = w?

A. x1 = a

B. x = b1

C. x1 = c

D. x1 = d

7 The velocity of a particle, in metres per second, is given by v = x2 + 2, where x is its displacement in metres from the origin.

What is the acceleration of the particle at x = 1?

A. 2 m s−2

B. 3 m s−2

C. 6 m s−2

D. 12 m s−2

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466


– 5 –

8 Six men and six women are to be seated at a round table.

In how many different ways can they be seated if men and women alternate?

A. 5! 5!

B. 5! 6!

C. 2! 5! 5!

D. 2! 5! 6!

9 1

Which of the following is a general solution of the equation sin 2x = − ?2

pA. x = np + (−1)n

12

np pB. x = + (−1)n

2 12

n+1 pC. x = np + (−1)

12

np pD. x = + (−1)n+1

2 12

10 A particle is moving in simple harmonic motion. The displacement of the particle is x and

its velocity, v, is given by the equation v2 = n2 ( 2kx − x2 ), where n and k are constants.

The particle is initially at x = k.

Which function, in terms of time t, could represent the motion of the particle?

A. x = k cos (nt)

B. x = k sin (nt) + k

C. x = 2k cos (nt) − k

D. x = 2k sin (nt) + k

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Question 13 (continued)

(c) An object is projected from the origin with an initial velocity of V at an angle q to the horizontal. The equations of motion of the object are

x(t) = Vt cosq gt2

y(t) = Vt sinq − . (Do NOT prove these.) 2

(i) Show that when the object is projected at an angle q, the horizontal range V 2

is sin 2q . g

2

p (ii) Show that when the object is projected at an angle − q, the horizontal

2 2 V

range is also sin 2q . g

1

(iii) The object is projected with initial velocity V to reach a horizontal distance d, which is less than the maximum possible horizontal range. There are two angles at which the object can be projected in order to travel that horizontal distance before landing.

p Let these angles be a and b, where b = − a.

2

Let ha be the maximum height reached by the object when projected at the angle a to the horizontal.

Let hb be the maximum height reached by the object when projected at the angle b to the horizontal.

3

ab

V V

hb

ha

d

h + hShow that the average of the two heights, a b, depends only

2on V and g.

End of Question 13

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© 2018 NSW Education Standards Authority

– 13 –

Question 14 (continued)

(c) In triangle ABC, BC is perpendicular to AC. Side BC has length a, side AC has length b and side AB has length c. A quadrant of a circle of radius x, centred at C, is constructed. The arc meets side BC at E. It touches the side AB at D, and meets side AC at F. The interval CD is perpendicular to AB.

C

BD

F

c

x

xx1

a

b

E

A

C

B

F

G

D

H

A. . .

(i) Show that rABC and rACD are similar.

1

ab(ii) Show that x = .

c1

From F, a line perpendicular to AC is drawn to meet AB at G, forming the right‑angled triangle GFA. A new quadrant is constructed in triangle GFA touching side AB at H. The process is then repeated indefinitely.

C

BD

F

c

x

xx1

a

b

E

A

C

B

F

G

D

H

A. . .

pab2

(iii) Show that the limiting sum of the areas of all the quadrants is .4(2c − a)

4

p 2c − a(iv) Hence, or otherwise, show that < .

2 b 1

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475

2018 Higher School CertificateWorked Answers

(Total 10 marks)

1. B 2. A 3. A 4. D 5. A 6. C 7. C 8. B 9. D 10. B

1. P(x) = 2x3 + 6x2 – 7x – 10

αβγ = da

= 102

= 5

α + β + γ = ba

= 62

= –3

So αβγ (α + β + γ) = 5 × –3

= –15

Answer B

2. y = 3x has gradient m1 = 3

y = 5x has gradient m2 = 5

Now tan θ = m1 m2

1+m1m2

= 3 5

1+ 3 5

= 18

Answer A

3. limx 0

sin3xcos3x12x

= limx 0

sin6x24x

= 14limx 0

sin6x6x

= 14

Answer A

4. There are single roots at –2 and –1 and a double root at 1.

So possible values of b and c are 2 and 1 and a possible value of d is –1.

So y = a(x + 2)(x + 1)(x – 1)2

When x = 0, y = –6

–6 = a(0 + 2)(0 + 1)(0 – 1)2

–6 = 2a

a = –3

Answer D

5. P(0) = 3000

Now 1500e0 = 1500

So possible equations are A and D.

As t → ∞ , 1500e–kt → 0

So the equation that best represents the graph is P(t) = 1500 + 1500e–kt.

Answer A

6. The tangent at c cuts the x-axis at the closest point to w, so the initial approximation x1 = c, will give the second approximation that is closest to x = w.

O x

y

a b c d w

y = ƒ (x )

Answer C

7. v = x2 + 2

Now a = vdvdx

= (x2 + 2) × 2x

= 2x3 + 4x

When x = 1,

dvdt

= 2 × 13 + 4 × 1

= 6

The acceleration of the particle is 6 m s–2.

Answer C

8. The men can sit in (6 – 1)! or 5! ways.

For each of those the women can sit in 6! ways.

Number of ways = 5! 6!

Answer B

Section ISection I

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476

2018 HIGHER SCHOOL CERTIFICATE—WORKED ANSWERS

9. sin 2x = 12

2x = nπ + (–1)n sin 1 12

= nπ + (–1)n 6

= nπ + (–1)n+1 6

x = n2

+ (–1)n+1 12

Answer D

10. v2 = n2(2kx – x2)

= n2(k2 – k2 + 2kx – x2)

= n2(k2 – (x2 – 2kx + k2))

= n2(k2 – (x – k)2)

So x – k = k cos (nt + α)

x = k cos (nt + α) + k

But x = k when t = 0.

So x = k sin (nt) + k could represent the motion.

Answer B

QUESTION 11

(a) P(x) = x3 – 2x2 – 5x + 6

(i) P(1) = 13 – 2 × 12 – 5 × 1 + 6 = 0 ∴ x = 1 is a zero of P(x).

(1 mark)

(ii)

x 1 x3 2x2 5x+ 6x2 x 6

x3 x2

x2 5xx2 + x

6x+ 66x+ 6

0

So P(x) = (x – 1)(x2 – x – 6) = (x – 1)(x + 2)(x – 3) ∴ the other zeros are x = –2 and x = 3.

(2 marks)

(b) log25 + log2(x – 2) = 3

log2(5x – 10) = 3

∴ 5x – 10 = 23

5x – 10 = 8

5x = 18

x = 3.6

(2 marks)

Section II

(c) 3 sin x + cos x ≡ R sin (x + α) Now R sin (x + α) = R sin x cos α + R cos x sin α ∴ R cos α = 3 and R sin α = 1

RsinRcos

= 13

tan α = 13

α = 6

02

R cos 6

= 3

R × 3

2 = 3

R = 2

∴ 3 sin x + cos x ≡ 2 sin x+6

(2 marks)

(d) The products of the intercepts of secants from an external point are equal.

∴ x(x + 2) = 3 × (3 + 5)

x2 + 2x = 24

x2 + 2x – 24 = 0

(x – 4)(x + 6) = 0

x = 4 or x = –6

But x > 0

∴ x = 4

(2 marks)

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477


(e) f(x) = 1

4x 1

(i) Now 4x – 1 ≠ 0 4x ≠ 1

x ≠ 14

Domain is all real x except x = 14

.

(1 mark)

(ii) f(x) < 1

If x < 14

, the denominator is negative and so f(x) is also negative and hence less than 1.

If x > 14

,

1

4x 1 < 1

1 < 4x – 1

2 < 4x

12

< x

So f(x) < 1 when x < 14

and when x > 12

.

(2 marks)

(f) Let u = 1 – x x = 1 – u dx = –du

When x = 0, u = 1When x = –3, u = 4

∴ x

1 x dx

3

0

= 1 uu �

du4

1

= u12 u

12 du

1

4

= u12

12

u32

32 1

4

= 2 u2u u

31

4

= 2 42 4 4

32 1

2 1 13

= –223

(3 marks)

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478


QUESTION 12

(a) cos 2θ = cos2θ – sin2θ = 2 cos2θ – 1

So cos2θ = 12

(cos 2θ + 1)

cos2 3x( ) dx = 12

cos 6x+ 1( )dx

= 1

12 sin 6x +

x2

+ C

(2 marks)

(b) (i) sin θ = h20

h = 20 sin θ

dhd

= 20 cos θ

(1 mark)

(ii) ddt

= 1.5

dhdt

= dhd �

ddt

= 20 cos θ × 1.5 = 30 cos θ When h = 15,

sin θ = 1520

= 34

cos θ = 134

2

(θ is acute, so cos θ > 0)

So dhdt

= 30 134

2

= 19.843 1348… = 19.8 (1 d.p.)The top of the carriage is rising at 19.8 m per minute, correct to one decimal place.

(2 marks)

(c) (i) f(x) = sin–1x + cos–1x

f'(x) = 1

1 x2 + 1

1 x2

= 0

(1 mark)

20 m h

θ

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(ii) f'(x) = 0

∴ f(x) is constant.

When x = 0,

f(x) = sin–10 + cos–10 = 0 +

2 =

2 ∴ f(x) =

2 for all x in the domain.

∴ sin–1x + cos–1x = 2

(1 mark)

(iii) sin–1x and cos–1x have domain –1 ≤ x ≤ 1.

x1

y = sin–1x + cos–1x

y

π2

–1 0

(1 mark)

(d) P(finishes in 8 h) = 0.75

∴ P(does not finish in 8 h) = 0.25

P(at least 10 finish in 8 h)

= P(10 finish) + P(11 finish) + P(12 finish)

= 1210

(0.75)10(0.25)2 + 1211

(0.75)11(0.25) + 1212

(0.75)12

= 66 34

10 14

2

+ 12 34

11 14

+ 34

12

= 66 310 +12 311 + 312

412

= 311 22 +12 + 3( )

412

= 37 311

412

(2 marks)

(e) (i) The tangent at P has equation y = px – ap2.

The gradient of the tangent is m1 = p.

At A, y = 0

∴ px – ap2 = 0

px = ap2

x = ap (p ≠ 0)

A(ap, 0), S(0, a)

m = y2 y1

x2 x1

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480


m2 = a 0

0 ap

= 1p

Now m1m2 = p × 1p

= –1 ∴ AS is perpendicular to AP. ∴ ∠PAS = 90°

(2 marks)

(ii) Similarly, BS is perpendicular to BQ. So ∠TBS = 90°

O B

QST

A x

y

P

x2 = 4ay

So ∠TAS = ∠TBS Angles in same segment standing on

the same arc are equal. ∴ S, B, A and T are concyclic points.

(1 mark)

(iii) An angle in a semicircle = 90°. So ST is a diameter of the circle

through S, B, A and T. S(0, a), T(a(p + q), apq)

d = x2 x1( )2 + y2 y1( )2

= a p+q( ) 0( )2+ apq a( )2

= a2 p+q( )2 +a2 pq 1( )2

= a p2 + 2pq+q2 + p2q2 2pq+1

= a p2q2 + p2 +q2 +1

= a p2 q2 +1( )+1 q2 +1( )

= a p2 +1( ) q2 +1( )(2 marks)

QUESTION 13

(a) To prove 2 – 6 + 18 – 54 + … + 2(–3)n–1 = 1 3( )n

2.

When n = 1, LHS = 2(–3)1–1

= 2

RHS = 1 3( )1

2 = 2 \ it is true for n = 1. Assume true for n = k

i.e. assume 2 – 6 + 18 – 54 + … + 2(–3)k–1 = 1 3( )k

2 We want to prove that it is true for n = k + 1.

i.e. 2 – 6 + 18 – 54 + … + 2(–3)k = 1 3( )k+1

2

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481


LHS = 2 – 6 + 18 – 54 + … + 2(–3)k – 1 + 2(–3)k

= 1 3( )k

2 + 2(–3)k

= 1 3( )k + 4 3( )k

2

= 1+ 3 3( )k

2

= 1 3( )k+1

2 = RHS So, if true for n = k, it is also true for n = k + 1. It is true for n = 1, so it is true for n = 1 + 1 = 2. It is true for n = 2, so it is true for n = 2 + 1 = 3. And so on. By the process of induction it is true for all integers n ≥ 1.

(3 marks)

(b) (i) f(x) = x

x2 +1, x ≥ 1

Range of f(x) is 0 < y ≤ 12

.

So the domain of f –1(x) is 0 < x ≤ 12

.

Range of f –1(x) is y ≥ 1.

(2 marks)

(ii)

x

y = ƒ –1x

y

2

0

( , 1)1

(1 mark)

(iii) y = x

x2 +1

Inverse function:

x = y

y2 +1

xy2 + x = y

xy2 – y + x = 0

y = b± b2 4ac

2a

= 1( )± 1( )2 4 x x

2x

= 1± 1 4x2

2x

But, y ≥ 1

∴ y = 1+ 1 4x2

2x(3 marks)

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482


(c) (i) y = Vt sin θ – gt 2

2

When y = 0,

Vt sin θ – gt 2

2 = 0

t 2V sin θ –

gt22

= 0

t = 0 or V sin θ – gt2

= 0

gt = 2V sin θ

t = 2V sin

g

Now x = Vt cos θ

When t = 2V sin

g,

x = 2V 2 sin cos

g

= V 2 sin2

g

So the horizontal range of the projectile

is V 2

g sin 2θ.

(2 marks)

(ii) When the object is projected at 2

:

x = V 2

g sin 2

2

= V 2

g sin (π – 2θ)

= V 2

g sin 2θ

So the horizontal range of the projectile is

also V 2

g sin 2θ.

(1 mark)

(iii) When projected at angle α:

y = Vt sin α – gt 2

2

y = V sin α – gt

Maximum height occurs when y = 0

i.e 0 = V sin α – gt

gt = V sin α

t = V sing

When t = V sing

,

hα = V 2 sin2

g gV sing

2

2

= V 2 sin2

g V 2 sin2

2g

= V 2 sin2

2g

Similarly, when projected at angle β:

hβ = V 2 sin2

2g

Now sin2β = sin2 2

= cos2α

So hβ = V 2 cos2

2g

h + h

2 =

V 2 sin2

2g+V 2 cos2

2g2

= V 2 sin2 + cos2( )

4g

= V 2

4g

So the average of the two heights depends only on V and g.

(3 marks)

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483


QUESTION 14

(a) ∠DAB + ∠ABC + ∠BCD + ∠CDA = 360° (angle sum of quadrilateral) So 2α + 2β + 2γ + 2δ = 360° α + β + γ + δ = 180° ∠BPQ = ∠PAB + ∠ABP (ext. ∠ of ∆) = α + β ∴ ∠SPQ = 180° – (α + β) (angles in st. line) ∠DRS = ∠RCD + ∠CDR (ext. ∠ of ∆) = γ + δ ∴ ∠QRS = 180° – (γ + δ) (angles in st. line) ∠SPQ + ∠QRS = 180° – (α + β) + 180° – (γ + δ) = 360° – (α + β + γ + δ) = 360° – 180° = 180°∴ PQRS is a cyclic quadrilateral because the opposite angles are supplementary.

(3 marks)

(b) (i) (1 + (1 + x))n

= n0

+ n1

(1 + x) + n2

(1 + x)2 + n3

(1 + x)3 +…. + nn

(1 + x)n

= n0

+ n1

10

+11x +

n2

20

+21x+

22x2

+ n3

30

+31x+

32x2 +

33x3 +…. +

nn

n0

+n1x+…+

nnxn

So the coefficient of xr will be nr

rr

+n

r+1r+1r

+n

r+ 2r+ 2r

+…+nn

nr

(2 + x)n = n0

2n + n1

2n–1x + n2

2n–2x2 + n3

2n–3x3 +…. + nn

xn

The coefficient of xr will be nr

2n – r.

But 1 + (1 + x) = 2 + x.

So the coefficient of xr in both expansions must be equal.

∴ nr

rr

+n

r+1r+1r

+n

r+ 2r+ 2r

+…+nn

nr

= nr

2n – r

(3 marks)

(ii) Selector A chooses at least 4 from 23. So the number of ways selector A can choose a group is

234

+235

+236

+237

+…+2323

If selector A chooses a group of 4 then selector B must choose 4 from 4. If selector A chooses a group of 5 then selector B must choose 4 from 5. So for each of the

groups of 5 that selector A has chosen there are 54

possible selections of the group of 4.

If selector A chooses a group of 6 then selector B must choose 4 from 6. So, for each of the

groups of 6 that selector A has chosen there are 64

possible selections of the group of 4, and

so on.

A

B C

D

Q

P R

S

α

γγ

δ δ

ββ

α

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484


So the number of ways for the selection process is given by:

234

44

+235

54

+236

64

+237

74

+…+2323

234

Using the result from part (i) the selection process could be carried out in 234

223 – 4 ways.

So the selection process could be carried out in 234

219 or 4 642 570 240 ways.

(2 marks)

(c) (i) In ∆ABC, ∆ACD ∠BAC = ∠CAD (common angle) ∠ACB = ∠ADC (both 90°) ∴ ∆ABC ||| ∆ACD (equiangular)

(1 mark)

(ii) CDBC

= ACAB

(corresponding sides similar triangles)

xa

= bc

x = abc

(1 mark)

(iii) Total area of all quadrants

= 14πx2 +

14πx1

2 + 14πx2

2 + …

= 4

(x2 + x12 + x2

2 + …)

Now ∆AHF ||| ∆ADC (equiangular)

So HFDC

= AFAC

x1

x = b xb

= b

abc

b

= 1 – ac

xx1

C

B

F

G

D

H

A. . .

ac

x b – xb

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485


Now as the same process of forming quadrants continues each subsequent triangle will be similar to all the previous ones and the ratios of the radii will be the same.

So x2

x1

= 1 – ac

and x3

x2

= 1 – ac

and so on.

x1

x

2

= x2x1

2

= x3

x2

2

…

So x2 + x12 + x2

2 + … is a geometric series.

r = 1ac

2

= 1 – 2ac

+a2

c2

= 1 – 2ac a2

c2

= 1 – a 2c a( )

c2

As 0 < r < 1, x2 + x12 + x2

2 + … is an infinite geometric series.

Limiting sum = x2

1 1a 2c a( )

c2

=

abc

2

a 2c a( )c2

= a2b2

c2

c2

a 2c a( )

= ab2

2c a

Limiting sum of all areas = 4

× ab2

2c a

= ab2

4 2c a( )(4 marks)

(iv) Limiting sum of all areas must be less than the area of the triangle.

So ab2

4 2c a( ) <

12

ab

b

2 2c a( ) < 1 (a > 0, b > 0)

2

<2c ab

< 2

<2c ab

(c > a so 2c – a > 0)

(1 mark)

© P

asca

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BN

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4125

695

6
















SUCCESS ONEHSC · 2020-04-17 · 463 2018 HIGHER SCHOOL CERTIFICATE—EXAMINATION PAPER P P ISN 8 1...

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Transcript of SUCCESS ONEHSC · 2020-04-17 · 463 2018 HIGHER SCHOOL CERTIFICATE—EXAMINATION PAPER P P ISN 8 1...