2007 VCAA Mathematical Methods Exam 2 Solutions - Free download and print

Q: What does Section 1 provide for the exam?

Section 1 lists multiple-choice questions and gives the selected answer options (A–E) for each question from Q1 to Q22.

Q: How are solutions presented in Section 1 and early parts of Section 2?

Solutions show the key substitution or transformation steps and then compute the required result, such as solving for x, evaluating integrals, or determining domain and range.

Q: What kinds of problems are included in Section 2?

Section 2 includes multi-part tasks covering geometry and calculus applications, continuity checks, finding intersections using a horizontal line, and using graphing calculator steps to sketch and verify results.

Sep 20, 2026

Uploaded by Bulrr

Answer key and worked solutions for VCAA Mathematical Methods Exam 2, Section 1 multiple-choice questions and Section 2 multi-part problems. Includes final options (A–E) for each question, step-by-step algebraic, calculus, and inverse-function reasoning, definite integral results, and domain/range determinations. Probability and trigonometry questions use transformations to standard forms (e.g., Z-scores, angle solutions). Later parts include application problems involving geometry, optimization, graphing calculator guidance, and checking continuity and intersections.

LanguageEnglish

CategoryExam

FormatPDF

Pages3

Reading Time8 min

AccessFree to download

SECTION 1: Multiple-choice questions

Q1-Q22

SECTION 2

Q1 (a-d)

Q2 (a-g)

Q3 (a-c)

Q4 (a-d)

Q5 (a-b)

2007 VCAA Mathematical Methods Exam 2 Solutions Free download & print [from www.itute.com](from www.itute.com) Do not photocopy ©[Copyright 2007 itute.com](Copyright 2007 itute.com) SECTION 1: Multiple-choice questions | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | A | D | A | C | D | B | A | B | E | E | E | | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | C | D | A | B | E | C | D | A | A | A | E | Q1 f (x ) = 6 − 2x . 12 = 6 − 2x −4 = 6 − 2x , , x = −3 . x = 5 . ∴ D is [− 3,5] Q2 f (g (x )) = e 2g (x )+3 = e 2x2 +4x−3 Q3 ∫ 􀂧􀂩􀂨 x12~~ − ~~co1s2~~ ~~x~~ ~~ 􀂷􀂹􀂸dx = ∫ 􀂨􀂧􀂩 ~~ ~~x12~~ ~~ − sec 2 x􀂷􀂹􀂸dx 1 = − − tan x x Q4 f (x ) = x 3 − x +1 , f (0) = 03 − 0 +1 = −1 , f (3) = 33 − 3 +1 = 25 . Average rate = ~~ ~~253−−(−0~~ ~~1) = ~~ ~~263 . Q5 ∫ (sin 2x + 24x3 )dx = − ~~ ~~12~~ ~~ cos 2x + 6x4 + c Q6 By Graphics calculator. 2 0 ~~ ~~π3~~ ~~ x The range off is [2,5] . Q7 Pr (X < 10 .5) = Pr (X < 11− 2× 0.25) = Pr (X < µ − 2σ) = Pr (Z < −2) = Pr (Z > 2) Q8 f ′(x ) is undefined at 2x + 4 = 0 , i.e. x = −2 . ∴ f ′(x ) is discontinuous at x = −2 . Q9 k = 1xdx = − ~~ ~~1xdx = − [log e x]21 = −loge 2 = loge 􀂨􀂧􀂩~~ ~~12􀂷􀂹􀂸 . k 1 ∴e = . 2 Q10 f (x ) is an increasing function. f (0) = −2 , f (1) = e 2 − 3 . The range of f (x ) is [0, e 2 − 3) . ∴ the domain of f −1 (x ) is [0, e 2 − 3) . Let y = e 2 x − 3 be the equation of f (x ) . y + 3 = e 2x , ∴ x = ~~ ~~12~~ ~~ loge (y + 3) ∴ the equation of f −1 (x ) is y = ~~ ~~12~~ ~~ loge (x + 3) . Q11 (e 2x )2 − 5(e 2x )+ 4 = 0 ,(e 2x − 4)(e 2x −1) = 0 . e 2x = 1 , x = 0 , or e 2x = 4 , 2x = loge 4 , x = loge 2 . Solution set is {0, loge 2} . Q12 f ′(x ) 0 3 x f (x ) 0 3 x Q13 By graphics calculator. Local maximum at x = −5 , local minimum at x = ~~ ~~12 . Gradient is negative for x ∈ 􀂧􀂨 − 5, 1 􀂷􀂸 . 􀂩 2 􀂹 Q14 f (x ) = log e x − 3 + 6 is defined for x ≠ 3 . Maximal domain is R \ {3} . Q15 y = 3x52 → y = −3x52 → y = −3(x − 3)52 → y = −3(x − 3)52 − 4 . Q16 f (x ) = (x − a)2 g (x ) , f ′(x ) = 2(x − a)g (x ) + (x − a)2 g ′(x ) f ′(x ) = (x − a)(2g (x) + (x − a)g ′(x )) . © Copyright [itute.com 2007](itute.com 2007) Not to be photocopied or reproduced by other means Free download and print [from www.itute.com](from www.itute.com) 1 Q17 E(X ) = x􀂨􀂧􀂩 ~~ ~~x2~~ ~~ 􀂷􀂹􀂸dx = ~~ ~~x22~~ ~~dx = 􀂫􀂪􀂬 ~~ ~~x63~~ ~~􀂺􀂼􀂻 20 = ~~ ~~43 . Q18 Pr (X < x) = 0.35 , x = invNorm(0.35,130,2.7) = 129 . Q19 a + b = 1− (0.2 + 0 .2 + 0 .3) = 0.3 . E(X ) = 0a + 2× 0.2 + 4× 0.2 + 6× 0.3 + 8b = 5 , ∴ 8b = 2 , b = 0.25 and a = 0.05 . Q20 tan 2 ~~ ~~θ3 = 1 and θ ∈ [0,2π] . ∴ tan ~~ ~~θ3 = 1 , θ3 = ~~ ~~π4~~ ~~ , θ = ~~ ~~3π4~~ ~~ . Note: When ∴ tan ~~ ~~θ3 = −1 , θ3 = ~~ ~~3π4~~ ~~ , θ = ~~ ~~9π4~~ ~~ ∉ [0,2π] . Q21 cos 2 x + 2cos x = 0 , cos x(cos x + 2) = 0 . Since cos x + 2 ≠ 0 , ∴ cos x = 0 . Q22 Let f (x ) = x(x −1) and g (x ) = − x . y = f (g (x )) = g (x )(g (x ) −1) = − x(− x −1) = x 2 + x Graphics calculator: 0 x SECTION 2: 2 2 1000 Q1a V = πr h , 1000 = πr h , h = πr2 . Q1b Area of top plus bottom = 2 × πr2 . Area of curved surface = 2πrh = 2πr􀂧􀂨 1000 􀂷􀂸 = 2000 . 􀂩 πr2 􀂹 r ∴ A = 2000 + 2πr 2 cm2. r Q1c dAdr = − ~~ ~~20r020 + 4πr . Let dAdr = 0 . 1 − ~~ ~~20020 + 4πr = 0 , r 3 = 500 , r = 􀂧􀂨 500 􀂷􀂸 3 . r π 􀂩 π 􀂹 2 Q1d Amin = 2000 + 2π􀂧􀂨 500 􀂷􀂸 3 = 553.58 cm2. 􀂧 500 􀂷 13 􀂩 π 􀂹 􀂨 ~~ ~~ 􀂸 􀂩 π 􀂹 Q2a When t = 8 , C = 1000(cos0 + 2)2 −1000 = 8000 . When t = 16 , C = 1000(cos 4π + 2)2 −1000 = 8000 . ∴ m = 8000 for C(t ) to be continuous. Q2b Use graphics calculator to sketch the middle section. Q2c Cmin = 0 when t = 10 or 14. Q2d Use graphics calculator to find the first intersection of the middle section and the horizontal line C = 1250 . t = ~~ ~~283 hours after midnight, i.e. 9.20 am. Q2e t 283~~ ~~ 10

FAQs

What does Section 1 provide for the exam?

Section 1 lists multiple-choice questions and gives the selected answer options (A–E) for each question from Q1 to Q22.

How are solutions presented in Section 1 and early parts of Section 2?

Solutions show the key substitution or transformation steps and then compute the required result, such as solving for x, evaluating integrals, or determining domain and range.

What kinds of problems are included in Section 2?

Section 2 includes multi-part tasks covering geometry and calculus applications, continuity checks, finding intersections using a horizontal line, and using graphing calculator steps to sketch and verify results.

Sep 20, 2026

Uploaded by Bulrr