Blacktown Boys 2019 3U Trials & Solutions PDF

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BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Office Use Only Question

Blacktown Boys’ High School 2019 HSC Trial Examination

Mathematics Extension 1 General Instructions

x x x x x x x

Reading time – 5 minutes Working time – 2 hours Write using black pen NESA approved calculators may be used A reference sheet is provided for this paper All diagrams are not drawn to scale In Questions 11 – 14, show relevant mathematical reasoning and/or calculations

Total marks: Section I – 10 marks (pages 3 – 7) x Attempt Questions 1 – 10 70 x

Allow about 15 minutes for this section

Section II – 60 marks (pages 8 – 12) x Attempt Questions 11 – 14 x Allow about 1 hour and 45 minutes for this section Assessor: X. Chirgwin

Student Name: _________________________________________ Teacher Name: _________________________________________

Mark

Q1

/1

Q2

/1

Q3

/1

Q4

/1

Q5

/1

Q6

/1

Q7

/1

Q8

/1

Q9

/1

Q10

/1

Q11 a)

/1

Q11 b)

/3

Q11 c)

/3

Q11 d)

/3

Q11 e)

/3

Q11 f)

/2

Q12 a)

/2

Q12 b)

/3

Q12 c)

/5

Q12 d)

/3

Q12 e)

/2

Q13 a)

/5

Q13 b)

/5

Q13 c)

/5

Q14 a)

/3

Q14 b)

/2

Q14 c)

/5

Q14 d)

/5 Total

/70

Students are advised that this is a trial examination only and cannot in any way guarantee the content or format of the 2019 Higher School Certificate Examination.

-2-

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Section I

3

10 marks Attempt Questions 1–10

The diagram shows the graph  = () .

Use the multiple choice answer sheet provided on page 13 for Questions 1–10.

1

The acute angle between the lines  = 1 + 5 and  = 9 is . What is the value of tan  ?

A.

B.

C.

D.

2

Which diagram shows the graph  =   () ?

5 9

A.

B.

C.

D.

7

22

2 23

1 11

The diagram below shows a circle with tangents at  and .  is a point on the circle such that  = 55° .

NOT TO SCALE 4 Which of the following is true?

A. B. C. D.

What are the asymptotes of

A.

 = 70°

B.

 = 55°

C.

 = 50°

D.

=

10 ? (  5)( + 2)

 = 0,  = 2,  = 5

 = 0,  = 2,  = 5

 = 10,  = 2,  = 5

 = 10,  = 2,  = 5

 = 35°

-3-

-4-

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

5

When the polynomial () is divided by (  2)( + 3) the remainder is (1  7). What is the remainder when () is divided by ( + 3) ? A. B. C. D.

6

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

8

20

13 15

22

A particle is moving in simple harmonic motion with period 6 and amplitude 5. Which of the following is a possible equation for the velocity of the particle?  6

A.

 = 5 cos

B.

5   = cos 6 3

C.

 = 5 cos

D.

=

5

3

 3

cos

 3

What is the general solution of the equation 2 cos   9 cos   5 = 0 ? A.

 = 2 ±

 6

9

The diagram shows the graph of a function.

5

B.

 = 2 ±

C.

2  = 2 ± 3

D.

 = 2 ±

6

NOT TO SCALE

 3

Which function does the graph represent? 7

Cameron, Vishaal and five other friends sit randomly around a table. How many arrangements are possible if Cameron and Vishaal sit away from each other? A. B. C. D.

120

240

480

720

-5-

  sin(2  1) 2

A.

=

B.

=

C.

 = cos (2 + 1)

D.

 = cos  (2  1)

 + sin(2 + 1) 2

-6-

BBHS 2019 HSC Mathematics Extension 1 Trial Examination 

10

Given that   

=

 8 + 38 +   

The value of    is: A. B. C. D.

5

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Section II 60 Marks Attempt Questions 11–14 Answer each question on a SEPARATE writing booklet. Extra writing booklets are available. In Questions 11–14, your responses should include relevant mathematical reasoning and/or calculations.

4

(15 marks)

Question 11

3

2

1   + 8

a)

Find 

b)

Evaluate  sin 3 

c)

Solve



End of Section I

d)

ii)

i) ii)

f)

1

 

3

3 1 2 + 5

3

The letters A, E, I, O, and U are vowels. i)

e)

Use a SEPARATE writing booklet.

How many arrangements of the letters in the word BINOMIAL are possible? How many arrangements of the letters in the word BINOMIAL are possible if the vowels must occupy the first, third, fifth, and eighth positions?

Show that a root of the continuous function () = cos 4  ln  lies between  = 1.2 and  = 1.3 . Hence use one application of Newton’s method with an initial estimate of  = 1.2 to find a second approximation to the zero. Write your answer correct to three significant figures.

Find the exact value of

tan 󰇧2 tan

3 󰇨 5

End of Questions 11 -7-

-8-

1 2

1 2

2

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Question 12 a)

b)

c)

Use the substitution  =  + 5 to evaluate 





3  8 



 + 5

.

 .

Question 13 2

iii)

Show that  = 23 +   satisfies this equation, where  is a constant.

If the temperature of the cup of hot chocolate is 50° after 10 minutes. Find the temperature of the cup of hot chocolate after 20 minutes. Round your answer to the nearest degree. How long would it take for the cup of hot chocolate to cool to one third of its initial temperature? Round your answer to the nearest minute.

a)

3

b)

(15 marks)

Use a SEPARATE writing booklet.

A particle moves in a straight line and is  metres from a fixed point  after  seconds, where  = 10 + 3 cos 2  sin 2. i)

A cup of hot chocolate, which is initially at a temperature of 75°, is placed on a table in the dining room to cool. The dining room has a constant temperature of 23°. The cooling rate of the cup of hot chocolate is proportional to the difference between the dining room temperature and the temperature, , of the cup of hot chocolate. That is,  satisfies the equation  = (  23),  where  is the number of minutes after the cup of hot chocolate is placed on the dining table.

ii)

e)

Use a SEPARATE writing booklet.

Find the constant term in the expansion of 2 

i)

d)

(15 marks)

 Show that 3 cos 2  sin 2 = 2 cos 󰇡2 + 󰇢. 6

1

ii)

Prove that the acceleration of the particle is 4(  10).

2

iii)

Between which two points does the particle oscillate?

1

iv)

At what time does the particle first pass through the point  = 10?

1

The point (2,  ) lies on the parabola   = 4. The focus  is the point (0, ). The tangent at  meets the -axis at . Derive the equation of the tangent at  and show that it is  =   

1

i)

3

ii)

Find the co-ordinates of .

iii)

Show that the distance of  is ( + 1).

1

iv)

Hence prove that  is an isosceles triangle.

1

1

The acceleration of a particle moving in a straight line is given by 󰇘 = 3  , where  is the displacement from the origin. Initially the object is at the origin with velocity  = 3  .

c)

2

1 

Consider the function () = 2 cos    cos (2  1) +  for 0    1.

i)

Prove that  = 3  .

2

i)

Show that  󰆒 () = 0 for 0 <  < 1.

3

ii)

What happens to  as  increases without bound?

1

ii)

Sketch the graph of  = ().

2

Show that lim



1  cos 4 1 = . 16  2

2

End of Questions 12

-9-

End of Questions 13

-10-

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Question 14 a)

b)

c)

(15 marks)

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Use a SEPARATE writing booklet.

Use mathematical induction to prove that 9  4 is divisible by 5, for all positive integers .

The circles below touch at .  is a straight line.  is a tangent to circle  and  is a tangent to circle . Prove that  +   =  .

Question 14 (continued) 3

Use  = 10 /  , show that the equations of the motion of this object is  =  cos  and  =  sin   5  , where   50.

A light hangs at a vertical distance  metres above the centre of a circular table of radius 2 metres. At any point on the table where the angle of incidence is  and the distance from the light is , as shown in the diagram, assume that the illumination  is

2

given by  =

i)

An object is projected from ground level at an angle  to the horizontal, with a velocity of  /. The object returns to the ground after 50 seconds and 2  from its point of projection. i)

d)

ii)

Hence find the exact value of , and find angle  to the nearest minute.

2

iii)

What is the maximum height reached by the object?

1

Show that, at the edge of the table,  =

iii)

iv)

The vertical height of the light above table is varied. Given that 1 , 3 find the value of  that gives the maximum illumination at the edge of the table. If the light is raised vertically at 0.16  , find an expression for

 . 

Hence, or otherwise, find

 at the edge of the table when the light 

is 2 metres above the table.

Question 14 continues on page 12 End of Paper

-11-

 cos  sin  . 4

1 1

cos  sin  has a maximum value when  = cos

2

ii)

 cos  , where  is a positive constant. 

-12-

1

2

2019 Mathematics Extension 1 AT4 Trial Solutions

BBHS 2019 HSC Mathematics Extension 1 Trial Examination

Student Name:_________________________________________________

Section 1 1

C

1 Mark

Line 1:  = 1 + 5,  = 5 Line 2:  = 9,  = 9 95 tan  =   1+9×5 2 tan  = 23

Multiple Choice Answer Sheet Select the alternative A, B, C or D that best answers the question. Fill in the response oval completely. Sample: 2

A

1 Mark

If you think you have made a mistake, put a cross through the incorrect answer and fill in the new answer.

If you change your mind and have crossed out what you consider to be the correct answer, then indicate the correct answer by writing the word ‘correct’ and drawing an arrow as follows.

Join   =  = 55° (Alternate Segment Theorem)  =  = 55° (Alternate Segment Theorem)  +  +  = 180° (Angle sum of )  = 180°  55°  55°  = 70° 3

A

1 Mark

4

B Vertical asymptotes at  = 2,  = 5 Horizontal asymptote at  = 0

1 Mark

10 (  5)( + 2) 10 =    3  10 10  =   3 10         ,   0 =

5

-13-

D () = (  2)( + 3)() + (1  7 ) (3) = (3  2)(3 + 3)(3) + (1  7 × 3) (3) = 22  the remainder is 22

1 Mark

6

7

8

C 2 cos    9 cos   5 = 0 (2 cos  + 1)(cos   5) = 0 1 cos  =  2 2  = 2 ± 3

1 Mark

Q11 a)

1 Mark C The arrangement of 7 people sitting around a table is 6! The arrangement of Cameron and Vishaal sit together is 5! × 2! The arrangement of Cameron and Vishaal sit apart is the complement to Cameron and Vishaal sit together, which is 6!  5! × 2! = 480 D =

Section 2

Q11 b)

10

B

=

 + sin (2 + 1) 2

A



  = 

8 +  +    38

1 + 2 =

3 Marks Correct solution

 

1

2 Marks Correct integral

 

1 =   sin 6 2 6  6 1  1 =   sin    0 2 9 6 9 1  1 3 = 󰇩  × 󰇪 2 9 6 2  3  = 18 24

1 Mark Q11 c)

3 1 2 + 5

1 Mark Find the correct relationship between cos 6 and sin 3

5     2

3 Marks Correct solution

3(2 + 5)  (2 + 5) 3(2 + 5)  (2 + 5)  0 (2 + 5)[3  (2 + 5)]  0 (2 + 5)(  5)  0 5  <  ,  5 2

1 Mark

8 × 1 +  × 1 +  × 1   1 38  +  = 31 … … … (1) 1 =

 

1 =  (1  cos 6) 2 

 2 = 6 1 = 3   = 5 sin 3  5   = cos 3 3

9

1 Mark Correct solution

 sin 3  

1 Mark

2

1   + 8 1  = tan   +  8 8 1  + = tan  22 22 2 2 󰇨+ tan 󰇧 = 4 4 

2 Marks Identifies both important values 1 Mark Multiplies both sides by the square of the denominator, or equivalent

8 × 2  +  × 2 +  × 2  2

38 256 + 16 + 8  2 38 16 + 8 = 392 2 +  = 49 … … … (2) 17 =

Q11 d) i)

(2)  (1)  = 18  = 13

Q11 d) ii)

BINOMIAL 8 letters to be arranged with 2 ‘I’s are identical. ! The number of arrangements = = 20160

1 Mark Correct solution

!

4 consonants can be arranged in 4! Ways. ! 4 vowels with 2 ‘I’s can be arranged in ! ways !

Total number of arrangements = 4! × = 288 !

   = 5

Q11 e) i)

2 Marks Correct solution 1 Mark Identifies the number of arrangements for consonants or vowels

() = cos 4  ln  1 Mark (1.2) = cos(4 × 1.2)  ln 1.2 Correct solution (1.2) = 0.0948 … (1.3) = cos(4 × 1.3)  ln 1.3 (1.3) = 0.206 … Since (1.2) < 0 and (1.3) > 0, a root exists between 1.2 and 1.3 .

Q11 e) ii)

1  󰆒 () = 4 sin 4    = 1.2   = 1.2 

2 Marks Correct solution

(1.2)

Q12 b)

1 Mark Correct ()

(1.2)

cos(4 × 1.2)  ln 1.2 1 4 sin(4 × 1.2)  1.2

Let  = tan tan 󰇧2 tan = tan 2 = = =

Q12 a)

3 5

3 󰇨 5

  ...