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Section B continues with stepwise solution parts covering tangents, perpendicular lines, areas, maximum/minimum analysis, periodicity in population models, and probability computations, with frequent use of CAS to streamline calculations and verify 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Multiple-choice questions  \n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| B | A | E | D | C | C | E | E | D | D |\n\n\n| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| C | A | C | B | E | B | A | B | D | A |\n\nQ1 Period = ~~ ~~2π2 = π B  \nQ2 As x → ∞ , y → 4 A  \nQ3 When x = 0 , dydx = 3e3x = 3 E  \nQ4 f (x) = tan ~~ ~~x3 is undefined at x3 = ~~ ~~π2 , i.e. x = ~~ ~~3π2 \u003C 5 D Q5 f ′(x) = 2x + 3 = 0 when x = − ~~ ~~32 , i.e. turning point at x = − ~~ ~~32 .: a = −1 .5 C  \nQ6 Inverse of y = x 2 for x \u003C 0 is x = y 2 for y \u003C 0  \ni.e. y = − ~~ ~~x  C  \nQ7 Locate the points of zero gradient, after the second point the gradient is positive.  \nb a b  \nQ8 ∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx ,  \n0 0 a  \nb  \n10 = −4 + ∫ f (x)dx a  \nb.: ∫ f (x)dx = 14  \na  \nQ9 Shortest distance is straight line distance. d = x 2 + y 2  = x 2 + 2x + 1 = (x + 1)2  = x + 1  \nE  \nE  \nD  \nQ10 n = 1000 , ˆp = 0.55 , approx 95% confidence interval for ˆP 􀀇􀀅     0.55 × 0.45       0.55 × 0.45 􀀄􀀂  \nis 􀀆􀀅 0.55 −1.96 1000 , 0.55 + 1.96 1000 􀀃􀀂  \n≈ (0.519, 0.581) or (51.9%, 58.1%) D Q11 xcos x = ~~ ~~ddx (x sin x) − sin x , 1k~~ ~~ ∫ x cos xdx = ~~ ~~1k~~ ~~ 􀀇􀀆􀀅 ∫ ~~ ~~ddx (xsin x)dx − ∫ sin xdx􀀄􀀃􀀂  \n= ~~ ~~1k~~ ~~ (xsin x − ∫ sin xdx)+ c CQ12 Pr (RB ) + Pr(BR ) = ~~ ~~3~~ ~~x~~ ~~ × ~~ ~~2~~ ~~~~ ~~x + ~~ ~~3~~ ~~~~ ~~x~~ ~~ × ~~ ~~2~~ ~~~~ ~~x = ~~ (~~3+~~ ~~x6~~)~~x~~(~~2~~ ~~+~~ ~~x~~) ~~A  \nQ13 xx~~ ~~~~ ~~aa > 0 ,  \neither both x + a and x − a are both positive, i.e. x > a or both are negative, i.e. x \u003C −a .: x ∈ R \\ [− a, a] C  \nQ14  \nQ15  \nQ16  \n100  \n∫ ~~ ~~29 x2e − 19x2 dx ≈ 2.659 0  \nx2 − 2x − 3 ≥ 0 , (x + 1)(x − 3) ≥ 0 ,  \nSolve f ′(− 3) = 9 − 6m + n = 0 , f ′(1) = 9 + 2m + n = 0  \n.: m = 1 and n = −3 , .: f (x) = ~~ ~~13 x3 + x2 − 3x + p , (3, 4) .: p = −5  \nB E  \nB  \nQ17 Property 1 gives (a, L) is below (b, L) . Property 2 shows is a cubic shape function with turning points.  \nQ18 Pr (X ≥ 16 | X ≥ a) = 0.9175 and a ≤ 15  \nPr (X ≥ 16 ∩ X ≥ a)  Pr (X ≥ 16)  0.91728   \nPr (X ≥ a) = Pr (X ≥ a) = Pr (X ≥ a) = 0.9175 .: Pr (X ≥ a) ≈ 0.99976 , a = 12  \nQ19 V = x(a − 2x)(b − 2x) = abx − 2(a + b)x2 + 4x3 ddVx~~ ~~ = ab − 4(a + b)x + 12x2 = 0 , x = ~~ ~~a~~ ~~+~~ ~~b~~ ~~−~~ ~~a26~~ ~~−~~ ~~ab~~ ~~+~~ ~~b2~~  ~~  \nQ20 50 sin 2θ > 40 , 26.565 \u003C θ \u003C 63.435 , Pr (26 .565 \u003C θ \u003C 63 .435) ≈ 0.96947  \nSECTION B  \nQ1a x = 0  \nQ1b f ′(x) = ~~ ~~16 x  \nQ1c f ′(x) = ~~ ~~16 x = −2 , x = −12 , y = f (−12) = 12 Tangent at M : y −12 = −2(x − (−12)) .: y = −2x −12  \ng (x)  \nA  \nB  \nD  \nA  \nQ1di Perpendicular at M : y −12 = ~~ ~~12 (x − (−12)) , y = ~~ ~~12 x + 18  \nQ1dii Point N : y = ~~ ~~12 x + 18 and y = ~~ ~~x212 .: x = 18 812􀀇􀀆􀀅 12 x + 18 − ~~ ~~x212~~ ~~􀀄􀀃􀀂 dx = 375  \nQ1e All parabolas are similar. To simplify calculations, consider  \ny = x2 . M (− p, p 2 ) and the perpendicular at M is y = ~~ ~~21p~~ ~~ x + p 2 + ~~ ~~12 which intersects y = x 2 at x = p + ~~ ~~21p .  \np+2~~1~~p  \nEnclosed area A = ∫ 21p~~ ~~ x + p 2 + ~~ ~~12 − x 2 dx , Amin when p = ~~ ~~12 and − p  \n.: M 􀀅􀀇􀀆 − ~~ ~~12 , ~~ ~~14~~ ~~ 􀀄􀀃􀀂 . For y = ~~ ~~x4a22~~ ~~ , M 􀀅􀀇􀀆− b, ~~ ~~b4a22~~ ~~􀀄􀀃􀀂 . Q parabolas are similar,∴ same dilation in both directions .: − ~~ ~~12 : −b = ~~ ~~14 : ~~ ~~b4a22 .: b = 2a2 1  \n2022 VCAA Mathematical Methods Exam 2 Solutions © itute 2022  \nQ2ai r (0) = 2500  \nQ2aii Max = 1700 + 2500 = 4200 , min = −1700 + 2500 = 800 Q2aiii 200 − 40 = 160  \nQ2b (20, 700) and (100, 2500) , a = ~~ ~~2500~~ ~~700 = 900  \n π  \nSame period as rabbit population .: b =  \n80  \nQ2c By CAS , max value of r (t)+ f (t)≈ 5339  \nQ2d By CAS, 160  \nQ2e  \n􀀍t′ 􀀊 􀀍 90  t + 60 􀀊 􀀍t 􀀊 􀀍 π  (t′ − 60) 􀀊 For fox population, 􀀌􀀋 y′􀀉􀀈 = 􀀌􀀋 90π0y + 1600 􀀉􀀈 , 􀀌􀀋 y 􀀉􀀈 = 􀀌􀀋 9~~1~~0900(y′ − 1600)􀀉􀀈 9100 (y′ −1600) = sin ~~ ~~π90 (t′ − 60)  \n.: f (t) = 900sin ~~ ~~π90 (t − 60) + 1600  \n300  \n∫ ( f (t) + r(t ","cbCaicN3bMHygIw5","https://ap.wps.com/l/cbCaicN3bMHygIw5","pdf",68996,"English","# Section A – Multiple-choice questions\n## Q1–Q20\n# Section B\n## Q1a–Q1e\n## Q2ai–Q2h\n## Q3ai–Q3ciii\n## Q4a–Q5b","[{\"question\":\"What kinds of topics are covered in Section A?\",\"answer\":\"Section A contains multiple-choice questions spanning calculus (derivatives, gradients, turning points), function properties (including inverses), geometry distance, and statistics (e.g., confidence intervals).\"},{\"question\":\"How are results obtained quickly in these solutions?\",\"answer\":\"The solutions explicitly note using CAS to save time and also reference CAS for values such as maxima and combined rates, indicating CAS-assisted computation throughout.\"},{\"question\":\"What does Section B include beyond the multiple-choice section?\",\"answer\":\"Section B provides multi-part written solutions, including steps for tangents and perpendiculars, enclosed area calculations, optimisation ranges, population model analysis with periodicity, and detailed probability/distribution work.\"}]","2022 VCAA Mathematical Methods Exam 2 Solutions - Use CAS to save time - Section A and B | PDF"]