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PENRITH HIGH SCHOOL 2015 HSC TRIAL EXAMINATION M athematics E xtension 1 General Instructions: โ€ข Reading time โ€“ 5 minutes โ€ข Working time โ€“ 2 hours โ€ข Write using black or blue pen Black pen is preferred โ€ข Board-approved calculators may be used โ€ข A table of standard integrals is provided at the back of this paper โ€ข In questions 11 โ€“ 14, show relevant mathematical reasoning and/or calculations โ€ข Answer each question on a new sheet of paper Total marksโ€“70 Pages 3โ€“5 10 marks โ€ข Attempt Questions 1โ€“10 โ€ข Allow about 15 minutes for this section Pages 6โ€“9 60 marks โ€ข Attempt Questions 11โ€“14 โ€ข Allow about 1 hours 45 minutes for this section Student Number: Teacher Name: This paper MUST NOT be removed from the examination room Assessor: T Bales SECTION II SECTION I 1

PENRITH HIGH SCHOOL - FC2mathext1.web.fc2.com/test/2015PenrithTrialSolutions.pdfQ A projectile is fired from ๐‘‚๐‘‚, with speed ๐‘‰๐‘‰๐‘ ๐‘  โˆ’๐‘š๐‘š1, at an angle of elevation

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PENRITH HIGH SCHOOL 2015 HSC TRIAL EXAMINATION

Mathematics Extension 1

General Instructions: โ€ข Reading time โ€“ 5 minutes โ€ข Working time โ€“ 2 hours โ€ข Write using black or blue pen

Black pen is preferred โ€ข Board-approved calculators may

be used โ€ข A table of standard integrals is

provided at the back of this paper โ€ข In questions 11 โ€“ 14, show

relevant mathematical reasoning and/or calculations

โ€ข Answer each question on a new sheet of paper

Total marksโ€“70

Pages 3โ€“5

10 marks โ€ข Attempt Questions 1โ€“10 โ€ข Allow about 15 minutes for this section Pages 6โ€“9 60 marks โ€ข Attempt Questions 11โ€“14 โ€ข Allow about 1 hours 45 minutes for this section

Student Number: Teacher Name:

This paper MUST NOT be removed from the examination room

Assessor: T Bales

SECTION II

SECTION I

1

Section I 10 marks Attempt Questions 1โ€“10 Allow about 15 minutes for this section Use the provided multipleโ€“choice answer sheet for Questions 1โ€“10 1 For ๐‘ฅ๐‘ฅ > 1, ๐‘’๐‘’๐‘ฅ๐‘ฅ โˆ’ ln๐‘ฅ๐‘ฅ is:

(A) = 0

(B) > 0

(C) < 0

(D) = ๐‘’๐‘’

2 Consider the function ๐‘“๐‘“(๐‘ฅ๐‘ฅ) given in the graph below: f(x) โˆ’๐‘Ž๐‘Ž 0 ๐‘Ž๐‘Ž ๐‘ฅ๐‘ฅ Which domain of the function f(x) above is valid for the inverse function to exist?: (A) ๐‘ฅ๐‘ฅ > 0

(B) โˆ’๐‘Ž๐‘Ž < ๐‘ฅ๐‘ฅ < ๐‘Ž๐‘Ž

(C) 0 < ๐‘ฅ๐‘ฅ < ๐‘Ž๐‘Ž

(D) ๐‘ฅ๐‘ฅ < 0

3 What is the acute angle between the lines 2๐‘ฅ๐‘ฅ โˆ’ ๐‘ฆ๐‘ฆ โˆ’ 7 = 0 and 3๐‘ฅ๐‘ฅ โˆ’ 5๐‘ฆ๐‘ฆ โˆ’ 2 = 0 ?

(A) 4โˆ˜24โ€ฒ

(B) 32โˆ˜28โ€ฒ

(C) 57โˆ˜32โ€ฒ

(D) 85โˆ˜36โ€ฒ

3

4 A particle is moving in a straight line with velocity ๐‘ฃ๐‘ฃ ๐‘š๐‘š/๐‘ ๐‘  and acceleration ๐‘Ž๐‘Ž ๐‘š๐‘š/๐‘ ๐‘ 2. Initially the particle started moving to the left of a fixed point ๐‘‚๐‘‚. The particle is noticed to be slowing down during the course of the motion from 0 to ๐ด๐ด. It turns around at ๐ด๐ด, keeps speeding up for the rest of the course of motion, passing ๐‘‚๐‘‚and ๐ต๐ต and continues. The particle never comes back. Take left to be the negative direction.

A ๐‘‚๐‘‚ B ๐‘ฅ๐‘ฅ

During the course of the particleโ€™s motion from ๐‘‚๐‘‚ to ๐ด๐ด, which statement of the following is correct?

(A) ๐‘ฃ๐‘ฃ > 0 and ๐‘Ž๐‘Ž > 0

(B) ๐‘ฃ๐‘ฃ > 0 and ๐‘Ž๐‘Ž < 0

(C) ๐‘ฃ๐‘ฃ < 0 and ๐‘Ž๐‘Ž > 0

(D) ๐‘ฃ๐‘ฃ < 0 and ๐‘Ž๐‘Ž < 0

5 A particle is moving in a straight line with velocity, ๐‘ฃ๐‘ฃ = 11+๐‘ฅ๐‘ฅ

๐‘š๐‘š/๐‘ ๐‘  where ๐‘ฅ๐‘ฅ is the displacement of the particle from a fixed point ๐‘‚๐‘‚. If the particle was observed to have reached the position ๐‘ฅ๐‘ฅ = โˆ’2 ๐‘š๐‘š at a certain moment of time, then this particle:

(A) will definitely reach the position ๐‘ฅ๐‘ฅ = 1๐‘š๐‘š (B) may reach the position ๐‘ฅ๐‘ฅ = 1๐‘š๐‘š (C) will never reach the position ๐‘ฅ๐‘ฅ = 1๐‘š๐‘š (D) will come to rest before reaching ๐‘ฅ๐‘ฅ = 1๐‘š๐‘š

6 If the rate of change of a function ๐‘ฆ๐‘ฆ = ๐‘“๐‘“(๐‘ฅ๐‘ฅ) at any point is proportional to the value of the function at that point then the function ๐‘ฆ๐‘ฆ = ๐‘“๐‘“(๐‘ฅ๐‘ฅ) is a:

(A) Polynomial function (B) Trigonometric function (C) Exponential function (D) Quadratic function

4

7 Let ๐›ผ๐›ผ and ๐›ฝ๐›ฝ be any two acute angles such that ๐›ผ๐›ผ < ๐›ฝ๐›ฝ. Which of the following statements is correct ?

(A) ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐›ผ๐›ผ < ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐›ฝ๐›ฝ (B) ๐‘๐‘๐‘๐‘๐‘ ๐‘ ๐›ผ๐›ผ < ๐‘๐‘๐‘๐‘๐‘ ๐‘ ๐›ฝ๐›ฝ (C) ๐‘๐‘๐‘๐‘๐‘ ๐‘ ๐‘’๐‘’๐‘๐‘๐›ผ๐›ผ < ๐‘๐‘๐‘๐‘๐‘ ๐‘ ๐‘’๐‘’๐‘๐‘๐›ฝ๐›ฝ (D) ๐‘๐‘๐‘๐‘๐‘๐‘๐›ผ๐›ผ < ๐‘๐‘๐‘๐‘๐‘๐‘๐›ฝ๐›ฝ

8 Which of the following is a primitive function of ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 2๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘ฅ2? (A) ๐‘ฅ๐‘ฅ โˆ’ 1

2๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 2๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘ฅ3

3+ ๐‘๐‘

(B) 1

2๐‘ฅ๐‘ฅ โˆ’ 1

4๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 2๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘ฅ3

3+ ๐‘๐‘

(C) ๐‘ฅ๐‘ฅ โˆ’ 1

2๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 2๐‘ฅ๐‘ฅ + 2๐‘ฅ๐‘ฅ + ๐‘๐‘

(D) 1

2๐‘ฅ๐‘ฅ โˆ’ 1

4๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 2๐‘ฅ๐‘ฅ + 2๐‘ฅ๐‘ฅ + ๐‘๐‘

9 Consider the binomial expansion (1 + ๐‘ฅ๐‘ฅ)๐‘›๐‘› = 1 + ๐‘ ๐‘ ๐ถ๐ถ1๐‘ฅ๐‘ฅ+๐‘ ๐‘ ๐ถ๐ถ2๐‘ฅ๐‘ฅ2 + โ‹ฏ+ ๐‘ ๐‘ ๐ถ๐ถ๐‘›๐‘›๐‘ฅ๐‘ฅ

๐‘›๐‘›. Which of the following expressions is correct? (A) ๐‘๐‘1๐‘›๐‘› + 2 ๐‘๐‘2๐‘›๐‘› + โ‹ฏ+ ๐‘ ๐‘  ๐‘๐‘๐‘›๐‘›๐‘›๐‘› = ๐‘ ๐‘ 2๐‘›๐‘›โˆ’1 (B) ๐‘๐‘1๐‘›๐‘› + 2 ๐‘๐‘2๐‘›๐‘› + โ‹ฏ+ ๐‘ ๐‘  ๐‘๐‘๐‘›๐‘›๐‘›๐‘› = ๐‘ ๐‘ 2๐‘›๐‘›+1 (C) ๐‘๐‘1๐‘›๐‘› + ๐‘๐‘2๐‘›๐‘› + โ‹ฏ+ ๐‘๐‘๐‘›๐‘›๐‘›๐‘› = 2๐‘›๐‘›โˆ’1 (D) ๐‘๐‘1๐‘›๐‘› + ๐‘๐‘2๐‘›๐‘› + โ‹ฏ+ ๐‘๐‘๐‘›๐‘›๐‘›๐‘› = 2๐‘›๐‘›+1

10 Using the substitution, ๐‘ข๐‘ข = 1 + โˆš๐‘ฅ๐‘ฅ, find the value of โˆซ 1

๏ฟฝ1+โˆš๐‘ฅ๐‘ฅ๏ฟฝ21โˆš๐‘ฅ๐‘ฅ๐‘‘๐‘‘๐‘ฅ๐‘ฅ4

1 is:

(A) 65

(B) 1

3

(C) 2

3

(D) 3

2

5

Section II 60 marks Attempt Questions 11โ€“14 Allow about 1 hour and 45 minutes for this section Begin each question on a new sheet of paper. Extra sheets of paper are available. In questions 11โ€“14, your responses should include relevant mathematical reasoning and/or calculations Question 11 (15 marks) Start a new sheet of paper. (a) (๐‘ฅ๐‘ฅ + 1) and (๐‘ฅ๐‘ฅ โˆ’ 2) are factors of ๐ด๐ด(๐‘ฅ๐‘ฅ) = ๐‘ฅ๐‘ฅ3 โˆ’ 4๐‘ฅ๐‘ฅ2 + ๐‘ฅ๐‘ฅ + 6. Find the third factor. 2 (b) Find the coordinates of the point ๐‘ƒ๐‘ƒ which divides the interval ๐ด๐ด๐ต๐ต internally in the ratio 2: 3 3 with ๐ด๐ด(โˆ’3, 7) and ๐ต๐ต(15,โˆ’6)

(c) Solve the inequality 1

4๐‘ฅ๐‘ฅโˆ’1< 2, graphing your solution on a number line. 3

(d) Use the method of mathematical induction to prove that, for all positive integers ๐‘ ๐‘ : 3 12 + 32 + 52 + 72 + โ‹ฏ+ (2๐‘ ๐‘  โˆ’ 1)2 = ๐‘›๐‘›

3(2๐‘ ๐‘  โˆ’ 1)(2๐‘ ๐‘  + 1)

(e) T P โ€ขB โ€ขQ A PT is the common tangent to the two circles which touch at T. PA is the tangent to the smaller circle at Q, intersecting the larger circle at points ๐ต๐ต and ๐ด๐ด as shown. i) State the property which would be used to explain why ๐‘ƒ๐‘ƒ๐‘ƒ๐‘ƒ2 = ๐‘ƒ๐‘ƒ๐ด๐ด ร— ๐‘ƒ๐‘ƒ๐ต๐ต 1

ii) If ๐‘ƒ๐‘ƒ๐‘ƒ๐‘ƒ = ๐‘š๐‘š,๐‘„๐‘„๐ด๐ด = ๐‘ ๐‘  ๐‘Ž๐‘Ž๐‘ ๐‘ ๐‘‘๐‘‘ ๐‘„๐‘„๐ต๐ต = ๐‘Ÿ๐‘Ÿ, prove that nrm

n r=

โˆ’ 3

Proceed to next page for question (12)

6

Question 12 (15 marks) Start a new sheet of paper. (a) The equation ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ฅ๐‘ฅ = 1 โˆ’ 2๐‘ฅ๐‘ฅ has a root near ๐‘ฅ๐‘ฅ = 0.3. Use one application of Newtonโ€™s 3 methods to find a better approximation, giving your answer correct to 2 decimal places. (b) Five couples sit at a round table. How many different seating arrangements are possible if: i) there are no restrictions? 1 ii) each person sits next to their partner? 2

(c) In the expansion of (4 + 2๐‘ฅ๐‘ฅ โˆ’ 3๐‘ฅ๐‘ฅ2) ๏ฟฝ2 โˆ’ ๐‘ฅ๐‘ฅ5๏ฟฝ6, find the coefficient of ๐‘ฅ๐‘ฅ5 3

(d) i) Write the binomial expansion for (1 + ๐‘ฅ๐‘ฅ)๐‘›๐‘› 2

ii) Using part (i), show that ( )3 10 0

11 31

nn n kk

kx dx C

k+

=+ =

+โˆ‘โˆซ 2

iii) Hence show that 1 1

0

1 13 4 11 1

nn k n

kk

Ck n

+ +

=

= โˆ’+ +โˆ‘ 2

Proceed to next page for question (13) 7

Question 13 (15 marks) Start a new sheet of paper. (a) Use the substitution ๐‘ข๐‘ข = ๐‘ฅ๐‘ฅ

๏ฟฝ1โˆ’๐‘ฅ๐‘ฅ2 to show that ๐‘‘๐‘‘

๐‘‘๐‘‘๐‘ฅ๐‘ฅ๏ฟฝ๐‘๐‘๐‘Ž๐‘Ž๐‘ ๐‘ โˆ’1 ๏ฟฝ ๐‘ฅ๐‘ฅ

โˆš1โˆ’๐‘ฅ๐‘ฅ2๏ฟฝ๏ฟฝ = 1

โˆš1โˆ’๐‘ฅ๐‘ฅ2 3

(You can use the result that โˆซ 1

๐‘Ž๐‘Ž2+๐‘ฅ๐‘ฅ2๐‘‘๐‘‘๐‘ฅ๐‘ฅ = 1

๐‘Ž๐‘Ž๐‘๐‘๐‘Ž๐‘Ž๐‘ ๐‘ โˆ’1 ๐‘ฅ๐‘ฅ

๐‘Ž๐‘Ž+ ๐‘๐‘. You do not need to prove this).

(b) Give the exact value for 3

23 3

dxxโˆ’ +โˆซ 3

(c) The distinct points P, Q have parameters ๐‘๐‘ = ๐‘๐‘1 and ๐‘๐‘ = ๐‘๐‘2 respectively on the parabola ๐‘ฅ๐‘ฅ = 2๐‘๐‘,๐‘ฆ๐‘ฆ = ๐‘๐‘2. The equations of the tangents to the parabola at P and Q respectively are given by: ๐‘ฆ๐‘ฆ โˆ’ ๐‘๐‘1๐‘ฅ๐‘ฅ + ๐‘๐‘12 = 0 and ๐‘ฆ๐‘ฆ โˆ’ ๐‘๐‘2๐‘ฅ๐‘ฅ + ๐‘๐‘22 = 0 (You do not need to prove these) i) Show that the equation of the chord ๐‘ƒ๐‘ƒ๐‘„๐‘„ ๐‘ ๐‘ ๐‘ ๐‘  2๐‘ฆ๐‘ฆ โˆ’ (๐‘๐‘1 + ๐‘๐‘2)๐‘ฅ๐‘ฅ + 2๐‘๐‘1๐‘๐‘2 = 0 2 ii) Show that M, the point of intersection of the tangents to the parabola at P and Q, 2 has coordinates ๏ฟฝ๐‘๐‘1 + ๐‘๐‘2 , ๐‘๐‘1๐‘๐‘2๏ฟฝ.

iii) โˆ) Prove that for any value of ๐‘๐‘1, except ๐‘๐‘1 = 0, there are exactly two values 3 of ๐‘๐‘2 for which M lies on the parabola ๐‘ฅ๐‘ฅ2 = โˆ’4๐‘ฆ๐‘ฆ. ๐›ฝ๐›ฝ) Find these two values of ๐‘๐‘2 in terms of ๐‘๐‘1. 2

Proceed to next page for question (14)

8

Question 14 (15 marks) Start a new sheet of paper. (a) A particle ๐‘ƒ๐‘ƒ is moving in simple harmonic motion on the ๐‘ฅ๐‘ฅ axis, according to the law ๐‘ฅ๐‘ฅ = 4๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 3๐‘๐‘ where ๐‘ฅ๐‘ฅ is the displacement of ๐‘ƒ๐‘ƒ in centimetres from ๐‘‚๐‘‚ at time ๐‘๐‘ seconds. i) State the period and amplitude of the motion. 2 ii) Find the first time when the particle is 2cm to the positive side of the origin and 2 itโ€™s velocity at this time. iii) Find the greatest speed and greatest acceleration of ๐‘ƒ๐‘ƒ 3 (b) ๐‘ฆ๐‘ฆ ๐‘‰๐‘‰ ๐œƒ๐œƒ ๐‘‚๐‘‚ ๐‘ƒ๐‘ƒ ๐‘ฅ๐‘ฅ Q A projectile is fired from ๐‘‚๐‘‚, with speed ๐‘‰๐‘‰๐‘š๐‘š๐‘ ๐‘ โˆ’1, at an angle of elevation of ๐œƒ๐œƒ to the horizontal. After ๐‘๐‘ seconds, its horizontal and vertical displacements from ๐‘‚๐‘‚ (as shown) are ๐‘ฅ๐‘ฅ metres and ๐‘ฆ๐‘ฆ metres,

repectively. i) Prove that ๐‘ฅ๐‘ฅ = ๐‘‰๐‘‰๐‘๐‘๐‘๐‘๐‘๐‘๐‘ ๐‘ ๐œƒ๐œƒ and ๐‘ฆ๐‘ฆ = โˆ’1

2๐‘”๐‘”๐‘๐‘2 + ๐‘‰๐‘‰๐‘๐‘๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐œƒ๐œƒ 3

ii) Show that the time taken to reach ๐‘ƒ๐‘ƒ is given by ๐‘๐‘ = 2๐‘‰๐‘‰๐‘‰๐‘‰๐‘‰๐‘‰๐‘›๐‘›๐‘‰๐‘‰๐‘”๐‘”

2

iii) The projectile falls to ๐‘„๐‘„, where its angle of depression from ๐‘‚๐‘‚ is ๐œƒ๐œƒ. 3 Prove that, in its flight from ๐‘‚๐‘‚ to ๐‘„๐‘„, ๐‘ƒ๐‘ƒ is the half-way point in terms of time.

End of paper

9

STANDARD INTEGRALS

0nif,0x;1n,x1n

1dxx 1nn <โ‰ โˆ’โ‰ +

=โˆซ +

โˆซ >= 0x,xlndxx1

โˆซ โ‰ = 0a,ea1dxe axax

โˆซ โ‰ = 0a,axsina1dxaxcos

โˆซ โ‰ โˆ’= 0a,axcosa1dxaxsin

โˆซ โ‰ = 0a,axtana1dxaxsec2

โˆซ โ‰ = 0a,axseca1dxaxtanaxsec

โˆซ โ‰ =+

โˆ’ 0a,axtan

a1dx

xa1 1

22

โˆซ <<โˆ’โ‰ =โˆ’

โˆ’ axa,0a,axsindx

xa

1 122

โˆซ >>โˆ’+=โˆ’

0ax),axxln(dxax

1 2222

โˆซ ++=+

)axxln(dxax

1 2222

NOTE: ln x = loge x, x > 0

10

Student Number:_________________ Teacher Name:_____________________

11