MATHEMATICS
2012 HSC Course Assessment Task 3 (Trial Examination)
June 21, 2012
General instructions

SECTION I

• Working time – 3 hours.
(plus 5 minutes reading time)
• Write using blue or black pen. Where diagrams are to be sketched, these may be done in pencil.
• Board approved calculators may be used.
• Attempt all questions.

• Mark your answers on the answer sheet provided (numbered as page 9)
SECTION II
• Commence each new question on a new page.
Write on both sides of the paper.

• At the conclusion of the examination, bundle the booklets + answer sheet used in the • All necessary working should be shown in correct order within this paper and hand to every question. Marks may be deducted for examination supervisors. illegible or incomplete working.

STUDENT NUMBER:

..........................

# BOOKLETS USED: . . . . .

Class (please ✔)
12M2A – Mr Berry

12M3C – Ms Ziaziaris
12M3D – Mr Lowe
12M3E – Mr Lam

Marker’s use only.

QUESTION

MARKS

1-10

11

12

13

14

15

16

Total

10

15

15

15

15

15

15

100

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

2

Section I: Objective response
Mark your answers on the multiple choice sheet provided.
1.

What is the value of |−8| − |10|?
(A) 2

2.

(B) 1

(B) a < 0, ∆ > 0

(C) a > 0, ∆ < 0

(B) AAA

(C) SAS

(B) 1.9

3x5 − 2x2 + 7x − 3
.
x→∞
6x5 − 3x + 7

(D) AAS

(D) 2.01

(C) a circle

1
(C)

1
2

(D) 2

dP d2 P
Which of the following conditions for describe the slowing growth of and dt dt2 a variable P ?
(A)

dP d2 P
> 0 and
> 0. dt dt

(C)

dP d2 P
< 0 and
< 0. dt dt

(D)

1

dP d2 P
> 0 and
< 0. dt dt

(B)
9.

1

(D) an ellipse

Evaluate lim

(B) 0

1

1
(C) 2.09

(B) a hyperbola

1

(D) a > 0, ∆ > 0

Which of the following is the locus of a point that is equidistant from a fixed point and a fixed line?

(A) ∞
8.

(D) none of these

What is 1.9926 to two significant figures?

(A) a parabola
7.

(C) 360◦

1

Which of the following is not a condition for congruent triangles?

(A) 2.0
6.

(D) −2

Which conditions make the quadratic y = ax2 + bx + c positive definite?

(A) SSS
5.

(C) −1

(B) 180◦

(A) a < 0, ∆ < 0
4.

1

What is the sum of the exterior angles of a polygon?
(A) 90◦

3.

Marks

dP d2 P
< 0 and
> 0. dt dt

If a > b, which of the following is always true?
(A) a2 > b2

(B)

1 a >

1 b 1

(C) −a > −b

10. What is the exact value of b if the area beneath the curve y = and x = b (b > 1) is equal to 3
3

(A) e 2

JUNE 21, 2012

(B) e2

units2 ?
5

(C) e 2

(D) 2a > 2b
2
between x = 1 x 1

(D) e3

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

3

Section II: Short answer
Question 11 (15 Marks)

Commence a NEW page.

Marks

9
= 10. x2 (a)

Solve the equation x2 +

(b)

D(0, −2), E(4, 0) & F (2, 4) are three points on the number plane. y 3

F (2, 4)

x
E(4, 0)
D(0, −2)
i.

Calculate the length of the interval DF .

1

ii.

Calculate the gradient of DF .

1

iii.

Write the equation of the line DF in general form.

1

iv.

Calculate the perpendicular distance from E to the line DF .

2

v. Calculate the area of △DEF .
(c)

2

ABCDE is a regular pentagon. The diagonals AC and BD intersect at F .
A

B

E
F

C

D

Copy or trace this diagram into your writing booklet. By giving full reasons for your answer,
i. Prove that ∠ABC = 108◦ . ii. (d)

Find the size of ∠BAC.

Find the exact value of 3 tan 210◦ + 2 sin 300◦ .

NORTH SYDNEY BOYS’ HIGH SCHOOL

2
1
2

JUNE 21, 2012

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

4

Question 12 (15 Marks)
(a)

Commence a NEW page.

Marks

Find the values of p, p > 0 for which the roots of the equation x2 − px + p = 0 are i. opposite in sign.

1

ii. real

2

i. Sketch the parabola with equation

2

(b)

(y − 2)2 = 2(x + 2)
Show the vertex of the parabola on your sketch. ii. Find the coordinates of the focus and the equation of the directrix of the parabola. (c)

The sum to n terms of a sequence of numbers is given by Sn = 102n − 2n2 .

2

i. Find an expression for Tn , the n-th term of the sequence.

ii. What type of a sequence is this?
(d)

2
1

Differentiate the following expressions:
2
i. x3 ii. 3 cos 4x

1

iii.

loge (2x)

2

Question 13 (15 Marks)
(a)

2

Commence a NEW page.

Marks

For the curve y = x3 (4 − x)

i. Find the stationary point(s) and determine their nature.

3

ii. Find the point(s) of inflexion.

2

iii.

3

Draw a neat sketch of the curve showing the intercepts with the coordinate axes, any stationary points and any point(s) of inflexion.

(b)

Find f (x) if f ′ (x) = 2x +

(c)

Find the primitive of
√
i. 3 x x ii. 3 sec2
3

JUNE 21, 2012

1 and the curve passes through the point (1, 2). x2 3

2
2

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

Question 14 (15 Marks)

5

Commence a NEW page.

Marks

3

(a)

Use Simpson’s Rule with five function values to evaluate

f (x) dx given the

2

1

following table: x 1 1.5 2 2.5 3 f (x) 0 3 5 2 1
(b)

The diagram below shows the graphs of y = −x2 + 2x + 8 and y = x + 6. y B y =x+6

A

y = −x2 + 2x + 8

x

i.

2

ii.

Hence or otherwise, find the shaded area bounded by the curves and the straight line.

3

i.

State the period and amplitude of y = 3 sin 2x.

2

ii.

Draw a neat sketch of y = 3 sin 2x, where 0 ≤ x ≤ 2π.

3

iii.

(c)

Show that the x coordinate of A and B are x = −1 and x = 2 respectively.

Hence or otherwise, state the number of solutions to the equation

1

3 sin 2x =

2
3

within the domain 0 ≤ x ≤ 2π.
(d)

Find the angle subtended at the centre of the circle of sector with radius 4 cm and area 20 cm2 . Give your answer correct to the nearest degree.

NORTH SYDNEY BOYS’ HIGH SCHOOL

2

JUNE 21, 2012

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

6

Question 15 (15 Marks)

Commence a NEW page.

Marks

8x = 16x+1 × 4−x .

(a)

Solve

(b)

In this diagram, ∠BCD + ∠BED = 180◦ .
A

2

3m
E
B

5m

2m
C

D

i. Prove that △ABE is similar to △ADC.

3

ii. Given that AE = 3 m, ED = 5 m and BC = 2 m, calculate the length of
AB.

3

(c)

i. Evaluate

d
(loge (sin x)). dx ii. Hence or otherwise, find
(d)

1 cot x dx.

2

A liquor bottle is obtained by rotating about the y axis the part of the curve
1
y = 2 − 2 between y = −1 and y = 2. x y

4

2 y= 1
−2
x2 x −1
Find the exact volume of the bottle.

Examination continues overleaf. . .

JUNE 21, 2012

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

Question 16 (15 Marks)
(a)

7

Commence a NEW page.

Marks

ABCD is a quadrilateral inscribed in a quarter of a circle centred at A with radius 100 m. The points B and D lie on the x and y axes and the point C moves on the circle such that ∠CAB = α as shown in the diagram below. y D
C

α
A

100 m

x
B

sin (x + 15◦ ) = cos 24◦ .

i.

Solve the equation

ii.

Show that the area of the quadrilateral ABCD can be expressed as

1
3

A = 5 000(sin α + cos α)
√
Show that the maximum area of this quadrilateral is 5 000 2 m2 .

4

i.

Sketch the curve y = 4e−2x .

2

ii.

Consider the series 2ex + 8e−x + 32e−3x + · · · .
α) Show that this series is geometric.

1

β) Find the values of x for which this series has a limiting sum.

2

γ) Find the limiting sum of this series in terms of x.

2

iii.
(b)

End of paper.

NORTH SYDNEY BOYS’ HIGH SCHOOL

JUNE 21, 2012

STANDARD INTEGRALS

1 xn+1 + C, n+1 xn dx

=

1 dx x

= ln x + C,

eax dx

=

1 ax e + C, a a=0

cos ax dx

=

1 sin ax + C, a a=0

sin ax dx

1
= − cos ax + C, a a=0

sec2 ax dx

=

1 tan ax + C, a a=0

sec ax tan ax dx =

1 sec ax + C, a a=0

1 dx a2 + x2

1 x tan−1 + C, a a

a=0

=

√

1 dx a2 − x2

= sin−1

√

1 dx 2 − a2 x = ln x +

√

√

1 dx 2 + a2 x = ln x +

√

n = −1;

x = 0 if n < 0

x>0

x
+ C, a a > 0, −a < x < a

x2 − a2 + C, x > a > 0 x2 + a2 + C

NOTE: ln x = loge x, x > 0

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination

9

Answer sheet for Section I
Mark answers to Section I by fully blackening the correct circle, e.g “●”
STUDENT NUMBER: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Class (please ✔)
12M2A – Mr Berry

12M3C – Ms Ziaziaris
12M3D – Mr Lowe
12M3E – Mr Lam

1–

B

C

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 –

NORTH SYDNEY BOYS’ HIGH SCHOOL

A

A

B

C

D

JUNE 21, 2012

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

10

Suggested Solutions

v. (2 marks)
[1] for using parts (iii) & (iv)

Section I

[1] for final answer.

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

1
× DF × d⊥
2
√
√
1
= × 2 10 × 10
2
= 10 units2

A=

Question 11 (Lowe)
(a)

(3 marks)
[1] for quartic.
[1] for final solutions.

(c) x2 +

×x2

i.

9
= 10 x2 ×x2

(2 marks)
• Divide pentagon into equilateral triangles.

×x2

five

A

4

x + 9 = 10x2 x4 − 10x2 + 9 = 0

x2 − 9

x2 − 1 = 0

B

∴ x = ±1, ±3

(b)

i.

E
F

(1 mark)
(2 − 0)2 + (4 − (−2))2
√
= 22 + 62 = 40
√
= 2 10

DF =

C
D
• Apex angle of one of the triangles
360◦
◦
5 = 72 .
• Angle sum of the two base angles is thus

ii. (1 mark) mDF =

6
=3
2

180◦ − 72◦ = 108◦ ii. iii. (1 mark) y−4 =3 x−2 y − 4 = 3x − 6

3x − y − 2 = 0 iv. (2 marks)
[1]
for correctly recalling perpendicular dist formula.
[1] for final answer.

(1 mark)
• △BAC is isosceles.
◦
◦
• ∴ ∠BAC = 180 −108 = 36◦ .
2

(d) (2 marks)
3 tan 210◦ + 2 sin 300◦
1
=3× √
+2×
3
√
3
= √ − 3=0
3

√
3
−
2

|ax1 + by1 + c|
√
a2 + b2
|3(4) + (−1)(0) − 2|
√
=
32 + 12
√
10
= √ = 10
10

d⊥ =

LAST UPDATED JUNE 26, 2012

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

Question 12 (Lowe)
(a)

i.

(c)

i. (2 marks)
Sn = 102n − 2n2

(1 mark)

Tn = Sn − Sn−1

= 102n − 2n2

x2 − px + p = 0 α = −β

∴α+β =0=−

− 102(n − 1) − 2 (n − 1)2

b
=p
a

✘
= ✘✘ − 2n2
102n

✘
− ✘✘ − 102 − 2(n2 − 2n + 1)
102n

∴p=0

✟

= ✟✟ 2 + 102 + ✟✟ − 4n + 2
−2n
2n2
= 104 − 4n

But as p > 0, therefore there are no real solutions. ii. 11

ii.

(2 marks)

(1 mark)
T1 = 104 − 4(1) = 100

[1] for p ≤ 0 or p ≥ 4.

T2 = 104 − 4(2) = 96

[1] justify why p ≥ 4 only.

T3 = 104 − 4(3) = 92
T3 − T2 = T2 − T1

Arithmetic sequence.

∆≥0

∴ b2 − 4ac = p2 − 4p ≥ 0

(d)

p(p − 4) ≥ 0

i. (1 mark)

∴ p ≤ 0 or p ≥ 4

d
2x−3 = −6x−4 dx But as p > 0, hence p ≥ 4 only.
(b)

i.

ii.

(2 marks) d (3 cos 4x) = −12 sin 4x dx (2 marks) y iii.

(2 marks) d d
1
(loge 2x) =
(loge 2 + loge x) = dx dx x 4
2
x
−2
(y − 2)2 = 2(x + 2) ii. (2 marks)
4a = 2
1
∴a=
2
3
S − ,2
2
Directrix is x = − 5 .
2

NORTH SYDNEY BOYS’ HIGH SCHOOL

LAST UPDATED JUNE 26, 2012

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

12

y

Question 13 (Berry)
i.

(3 marks)

27
16

y = 4x3 − x4 = x3 (4 − x) dy = 12x2 − 4x3 dx Stationary pts occur when

dy dx |

(a)

y = x3 (4 − x)

x
−1

1

2

3

4

= 0:

4x2 (3 − x) = 0
∴ x = 0, 3

(b) (3 marks) x 0

dy dx +

0

3
+

0

[1] for correct integral.
[1] for correct value of C.

−

[1] for final answer.

27 y 0 f (x) =

Hence (0, 0) is a horizontal point of inflexion and (3, 27) is a local maximum. ii. (2 marks)
Points of inflexion occur when d2 y
= 0:
(c)
dx2

f (1) = 1 − 1−1 + C = 2
∴C=2

∴ f (x) = x2 −
i.

d2 y dx2 0
−
⌢

0

+

0

⌣

(2 marks)

1

x 3 dx = ii. 2

1
+2
x

[−1] if missing arbitrary constant.

d2 y
= 24x − 12x2 = 12x(2 − x) dx2 ∴ x = 0, 2

x

2x + x−2 dx = x2 − x−1 + C

3 4 x3 + C
4

(2 marks)
[−1] if missing arbitrary constant.

−
⌢

3 sec2

x x dx = 9 tan + C
3
3

When x = 2, y = x3 (4 − x)

x=2

= 23 (4 − 2) = 16
Hence points of inflexion occur at (0, 0) and (2, 16) as concavity changes at these two pts. iii. (3 marks)
[−1]
for each omission from requirements of the question, provided graph is correct.

LAST UPDATED JUNE 26, 2012

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

Question 14 (Berry)
(a)

y
3

(2 marks)
3

f (x) dx =
1

=

h
(y0 + 4
3

yodd + 2

i.

x

yeven + yℓ ) π 2

1
2

(0 + 1 + 4(3 + 2) + 2(5))
3
31
=
6
(b)

13

π

3π
2

2π

−3 iii. (1 mark)
4 solutions.

(2 marks)
(d) (2 marks)

y = −x2 + 2x + 8 y =x+6

[1] for answer in radians.
[1] for answer in degrees.

Solve by equating,
A=

2

−x + 2x + 8 = x + 6

1
× 42 × θ
2
5
5 180◦ θ= = ×
≈ 71◦
4
4 π x2 − x + 2 = 0

20 =

(x − 2)(x + 1) = 0
∴ x = −1, 2

ii.

1 2 r θ
2

(3 marks)
2

A=
−1

=

x2 − x + 2 dx

1
1
− x3 + x2 + 2x
3
2

2
−1

1
= − 23 − (−1)3
3
1 2
2 − (−1)2
+
2
+ 2(2 − (−1))
= −3 +
(c)

i.

(2 marks)
T =

ii.

3
9
+6 =
2
2

2π
=π
2

a=3

(3 marks)
[1] for shape.
[1] for correct period.
[1] for amplitude.

NORTH SYDNEY BOYS’ HIGH SCHOOL

LAST UPDATED JUNE 26, 2012

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

14

are equal,

Question 15 (Ziaziaris)
(a)

AB
AE
=
AD
AC x 3
=
8 x+2 x(x + 2) = 24

(2 marks)
[1] for resolving into powers of 2.
[1] for final answer.

x2 + 2x − 24 = 0

8x = 16x+1 × 4−x

3 x

2

4 x+1

= 2

(x + 6)(x − 4) = 0

2 −x

∴ x = 4, −6

× 2

23x = 24x+4 × 2−2x

As x > 0 (length), ∴ x = 4 only.

23x = 22x+4

3x = 2x + 4

(c)

i.

(1 mark)

x=4
(b)

i.

d cos x
(loge (sin x)) = dx sin x

(3 marks)

ii.

[1] for each correct reason.

(2 marks) cos x dx sin x
= loge (sin x) + C

In △ABE and △ADC
A
3m

x θ B
2m

cot x dx =

(d) (4 marks)

E

180◦ − θ

1
−2
x2
1
y+2= 2 x 1
2
x = y+2 y=

5m

θ
C
•
•

D

2

∠CAD (common)
Let ∠BCD = θ. information, From the

−1

−1

dy y+2 2

= π loge (y + 2)
◦

∠BED = 180 − θ

•

2

x2 dy = π

V =π

−1

= π (loge 4 − loge 1)
= π loge 4

Hence ∠AEB = θ (supplementary), and ∠ACD = ∠AEB.
∴ ∠ABE = ∠ADC
(remaining ∠)

Hence △ABE △ACD (equiangular) ii. (3 marks)
[1] for ratio of lengths.
[1] for setting up quadratic.
[1] for final answer.

Let AB = x. As the ratio of the side lengths of corresponding sides
LAST UPDATED JUNE 26, 2012

NORTH SYDNEY BOYS’ HIGH SCHOOL

2012 Mathematics HSC Course Assessment Task 3 – Trial Examination SOLUTIONS

Question 16 (Lam)
(a)

i.

(b)

i. (2 marks)

6

x + 15◦ = 66◦

4

∴ x = 51◦

2

(loge 2, 1) x (3 marks)

1

1 ii. (α)
× 1002 sin α = 5 000 sin α
2
1 π A△CAD = × 1002 sin
− α = 5 000 cos α
2
2
∴ AABCD = 5 000 (sin α + cos α)
A△CAB =

iii.

y

(1 mark) sin(x + 15◦ ) = cos 24◦ = sin(90◦ − 24◦ )

ii.

15

(4 marks)
AABCD = 5 000 (sin α + cos α) dA = 5 000(cos α − sin α)
∴
dα
Stationary pts occur when
i.e.

dA dα = 0,

(1 mark)
2ex + 8e−x + 32e−3x · · ·
8e−x
T2
=
= 4e−2x
T1
2ex
32e−3x
T3
=
= 4e−2x
T2
8e−x
T3
T2
=
T1
T2
∴ 2ex + 8e−x + 32e−3x · · · is a geometric series with a = 2ex and r = 4e−2x .

(β) (2 marks)
[1] for 4e−2x < 1.

5 000(cos α − sin α) = 0

[1] for justification.

cos α = sin α

÷ cos α

A geometric series has a limiting sum when −1 < r < 1; i.e.

÷ cos α

tan α = 1 π ∴α=
4

By inspecting the graph in the previous part, −1 < 4e−2x < 1 when x > loge 2.

π
4

α dA dα

+

−

0

∴ limiting sum exists when

−

A
• α < π,
4
• α > π,
4

−1 < 4e−2x < 1

x > loge 2 dA dα dA dα

(γ) (2 marks)

< 0.
> 0.

[1] for recalling formula
[1] for final answer

Maximum area occurs when
A = 5 000 (sin α + cos α)
1
1
= 5 000 √ + √
2
2
√
2
= 5 000 2 m

NORTH SYDNEY BOYS’ HIGH SCHOOL

S= α= π
4

= 5 000

a
2ex
=
1−r
1 − 4e−2x

2
√
2

LAST UPDATED JUNE 26, 2012