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Many answers include intermediate steps, CAS-assisted solving for simultaneous equations, and area computations via definite integrals, ending with selected options or final numeric results for each 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used whenever possible to speed up the solution process.  \nSECTION 1  \n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| A | A | C | B | E | C | C | D | B | D | A |\n\n\n| 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| E | E | D | B | D | D | A | C | B | D | B |\n\nQ1 Period = ~~ ~~2π3 , the range is [− 2 − 3, 2 − 3], i.e. [− 5, −1] A  \nQ2 A  \nx = ~~ ~~y1+~~ ~~2~~  ~~ , y = ~~ ~~x12~~ ~~ − 2 , .: f −1(x) = ~~ ~~x12~~ ~~ − 2  \nQ3 C  \nQ4 y = x2 , mt = ~~ ~~dydx = 2x . At (2, 4) , mt = 2(2) = 4 .  \nConsider (3, 8) and (2, 4) , m = ~~ ~~83−~~ ~~42 = 4 = mt B  \nQ5 Reflection in the line y = x E  \nQ6 P(3) = 33 − a.32 − 4(3)+ 4 = 10 , .: a = 1 C  \nQ7 C  \nQ8 Area under the straight line = p = , .: p = D  \n 1 2  25   5  \n2 8 2  \nQ9 16 (a − 2) = 1 , .: a = 8 , E (X ) = ∫28~~ ~~x6 dx = 􀀆􀀅􀀄 x212~~ ~~ 􀀃􀀂􀀁 82 = 5 B  \nQ10 np = 2 , npq = ~~ ~~43 , .: n = 6 , p = ~~ ~~13 and q = ~~ ~~23 Pr (X = 1)=6C1 􀀋􀀍 1  􀀊􀀈 1 􀀍􀀋 2  􀀊􀀈 5  \n􀀌3 􀀉 􀀌3 􀀉  \n􀀍 x 􀀊3  \nQ11 x3 + 1 =   8􀀋 ~~ ~~ 􀀈 + 1  \n 􀀌2 􀀉  \nD  \nA  \nQ12 Pr (at least one) = 1 − Pr(none) = 1 − ~~ ~~81C3~~ ~~ = ~~ ~~5556 Q13  \n∫01 aex dx + ae.1 = 1 , [aex ]10 + ae = 1 , 2ae − a = 1 , a = ~~ ~~2e1−~~ ~~1 Q14 p + 2 p + 3 p + 4 p + 5 p = 1 , .: p = ~~ ~~115  \nMean of X = 1× ~~ ~~115 + 2 × ~~ ~~125 + 3× ~~ ~~135 + 4 × ~~ ~~415 + 5 × ~~ ~~155 = ~~ ~~113 Q15 ∫05(2g(x) + ax)dx = ∫520g (x)dx + ∫5ax0dx  \n= 2∫5g0(x)dx + 􀀆􀀅􀀄 ax22~~ ~~􀀃􀀂􀀁 50 = 2 × 20 + ~~ ~~252a = 90 , .: a = 4  \nE  \nE  \nD  \nB  \nQ16 f ′(x) = ∫ g (x) dx , maxm−1 = ~~ ~~bnx+11  \n.: ba = m(n + 1)xm−n−2 .: m − n − 2 = 0 .: ba = m(n + 1) is an integer  \nsince m and n are positive integers. D  \nQ17 By CAS y = x3 − 3x2 has a local max at (0, 0) and a local min at (2,− 4) . To have three distinct x-intercepts, y = x3 − 3x2 must translate upwards by less than 4 units. Hence c ∈ (0, 4) . D  \nQ18 f (x) = x2 satisfies f (x + y) − f (x − y) = 4f (x)f (y) .  \nL.H.S. = (x + y)2 − (x − y)2  = 4xy = 4xy  \nR.H.S. = 4x 2 y 2  = 4(xy)2  = 4xy = L.H.S. A  \nQ19 Let F(t ) = ∫ 􀀋􀀍􀀌t 2 + 4􀀊􀀉􀀈dt  \n.: f (x) = ∫0x 􀀋􀀍􀀌t 2 + 4􀀊􀀉􀀈dt = [F (t)]x0 = F (x) − F(0)  \n.: f ′(x) = F ′(x) = x 2 + 4 , f ′(− 2) = (− 2)2 + 4 = 22  C  \nQ20 f (x −1) = x2 − 2x + 3  \nReplace x with x + 1 : f (x + 1 −1) = (x + 1)2 − 2(x + 1)+ 3  \n.: f (x) = x 2 + 2 B  \nQ21 Let ax2 = mx + c , ax2 − mx − c = 0  \nNo intersections: ∆ \u003C 0 , (− m)2 − 4a(− c)\u003C 0 , m2 + 4ac \u003C 0 If a > 0 , c \u003C − ~~ ~~m24a~~ ~~ ; if a \u003C 0 , c > − ~~ ~~m24a~~ ~~ D  \nQ22 Let f (x) = −ax where a ∈R + .  \n.: g (− f (x)) = g (ax) = g (a− x) ≥ 0 and g (− f (x)) is even. B  \n1  \n© 2015 [itute.com](itute.com) 2015 VCAA Mathematical Methods Exam 2 Solutions  \nSECTION 2  \nQ1a f (x) = ~~ ~~15 (x − 2)2 (5 − x)  \nf ′(x) = ~~ ~~25 (x − 2)(5 − x) − ~~ ~~15 (x − 2)2 = ~~ ~~35 (x − 2)(4 − x) Q1bi At P􀀋􀀍􀀌1, ~~ ~~45􀀊􀀉􀀈 , m = f ′(1) = − ~~ ~~95  \ny − ~~ ~~45 = − ~~ ~~95 (x −1) , 9x + 5 y = 13 or y = − ~~ ~~95 x + ~~ ~~135  \n13 􀀍 13 􀀊  \nQ1bii 9x + 5y = 13 When x = 0 , y =  5  , .: S 􀀌􀀋 0,  5 􀀉􀀈 When y = 0 , x =  13 , .: Q 􀀍􀀋 13 , 0􀀊􀀈  \n9 􀀌 9 􀀉  \nQ1c Distance PS = (0 −1)2 + 􀀍􀀋 13 −  4  􀀊􀀈 2  =  106  \n 􀀌 5 5 􀀉 5  \nQ1d Use CAS to solve simultaneous equations  \ny = ~~ ~~15 (x − 2)2 (5 − x) and 9x + 5 y = 13 to find the second point of intersection at x = 7 .  \nArea of the shaded region = ∫17 􀀋􀀍􀀌~~ ~~15 (x − 2)2 (5 − x) − 􀀍􀀌􀀋 − ~~ ~~95 x + ~~ ~~135􀀊􀀉􀀈􀀊􀀉􀀈 dx = 21.6 square units by CAS  \n3 2 dy 3 Q2a y = 60 −  80 x , dx = −  40 x  \nAt A(− 40, 0) , m = − ~~ ~~340 (− 40) = 3 , tanθ = 3 , θ ≈ 72° Q2b y = ~~ ~~25x3600 − ~~ ~~3x16 + 35 , dydx = ~~ ~~23x25600 − ~~ ~~136  \ndy 􀀍 3 􀀊 The turning point of ~~ ~~ is 􀀋 0, − ~~ ~~ 􀀈 .  \ndx 􀀌 16 􀀉  \n 3   \n.: the max downward slope is − .  \n16  \nQ2c Vertical distance D = 60 − ~~ ~~380 x 2 − 􀀍􀀋􀀌~~ ~~25x3600 − ~~ ~~3x16~~ ~~ + 35􀀊􀀉􀀈 Let ddDx = 0 . − ~~ ~~340 x − ~~ ~~23x25~~ ~~600 + ~~ ~~136 = 0  \n.:","cbCain2lyN8rv52I","https://ap.wps.com/l/cbCain2lyN8rv52I","pdf",108857,"English","# Section 1\n## Q1-Q22\n# Section 2\n## Q1-Q5\n## Q3-Q4","[{\"question\":\"What problem-solving tools are emphasized in the solutions?\",\"answer\":\"CAS is used whenever possible to speed up the solution process, especially for tasks like solving simultaneous equations, evaluating integrals, and finding numerical roots.\"},{\"question\":\"What topics appear in Section 1?\",\"answer\":\"Section 1 includes differentiation and gradient calculations, reflections across lines, polynomial and function transformations, inverse-function style reasoning, and several probability-related computations.\"},{\"question\":\"What types of answers are provided for Section 2 questions?\",\"answer\":\"Section 2 provides full worked steps for each question, including analytic derivations (e.g., derivatives, turning points), geometric distances, shaded-area integrals, and probability/normal or binomial distribution calculations where applicable.\"}]","2015 VCAA Mathematical Methods Exam 2 Solutions - Section 1 | PDF"]