Scholarship Calculus (93202) 2013 page 1 of 8. ( 6) ± 20 = 3± 5, so x = ln( 3± 5) 2. 1(a) Expression for dy = 0 [1st mark], [2nd mark], width is

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1 Sholarship Calulus 93) 3 page of 8 Assessent Shedule 3 Sholarship Calulus 93) Evidene Stateent Question One a) e x e x Solving dy dx ln x x x ln ϕ e x e x e x e x ϕ, we find e x x e y The drop is widest at x ln.693, and then y ϕ e ln e ln ϕ 4 ϕ.636. It is widest ln fro the rounded end C, and is exatly ϕ wide there. b) Now we need d y dx, so ex 6e x Solving as a quadrati in e x we find e x Sine < 3 5 <, x ln 3 5 6) ± 3± 5, so x ln 3± 5). < and so this point is not valid, and the only turning point is at x ln 3+ 5) ) ln p ln p π y dx!ϕ e x e x dx!ϕ e x e x ln p )!ϕ + e ln p e ln p + p p )!ϕ +!ϕ p p+ p!ϕ p ) p p!ϕ p Sine p < p and ln p for the odel, we an see that p p < and so V <!ϕ. As p gets larger, the volue approahes!ϕ In the diagra, the drop has 99.99% of the axial volue in the odel.) a) Expression for dy [st ark], dx [nd ark], widest at x ln width is ϕ [3rd ark]. b) Reognise quadrati a 6a + 4 and solve [st ark], find x ln 3+ 5 [nd ark]. ) Definite integral πϕ e x e x ln p [st ark], show required for [nd ark], πϕ p p explain upper liit [3rd ark]. In all questions, inor error ignored one if one single harater is inorret, inserted or oitted. Note that the -ark question is not always part a). Marks in eah question part are independent, with follow through arks.

2 Sholarship Calulus 93) 3 page of 8 Question Two a) The two funtions are orthogonal if θ π, so f,g. f,g kx +)x + k)dx kx + k +)x + k))dx k k + kx3 3 + k +)x + kx k 3 + k +) + k a) Definite integral [st ark], values k 7 4± 3 [nd ark]. b) Find at least two of f, g, f,g find osθ 7/ 7 7 [nd ark], [st ark], and θ π [3rd ark]. 3 ) Either trig identity used, or first step of integration by parts [st ark], show that when n [nd ark], test what happens when n [3rd ark]. k 8 3 ± ± ± 3 7 k 4 3 ± ± 7 3 b) The angle requires finding three inner produts, sine f,g. f, f f,g 3x 4)9x 5)dx 7x 5x + )dx 9x 3 5 x + x 3x 4)3x 4)dx 9x 4x +6)dx 3x 3 x +6x g,g 9x 5)9x 5)dx 8x 9x + 5)dx osθ f, f f,g g,g 7 7 so θ os π x 3 45x + 5x π ) sinx)sinnx)dx sin n)x) n However, if n π sinx)sinnx)dx sin + n)x) + n π π os n)x) os + n)x) ) dx π if n osnx) ) dx x sinnx) n So sinnx) and sinx) are othogonal if n. see also integration by parts in appendix) π π sin4nπ ) 4n π.

3 Sholarship Calulus 93) 3 page 3 of 8 Question Three a) i) A polynoial px) a + a x + a x + + a n x n is: even if and only if the odd oeffiients are all zero, and odd if and only if the even oeffiients are all zero. neither if there are both even and odd oeffiients whih are non-zero. Candidates ight note that px) is the unique polynoial whih is both even and odd. [It is insuffiient to say that px) ax n is even, and px) ax n+ is odd, unless also noting that the su of odd polynoials is odd, and the su of even polynoials is even.] ii) We are given that g is an even funtion, so g x) gx) for all x. We use the fats that g x + h) gx h) and g x) gx). b) g x + h) g x) g x) li h h li h li k li k g x) gx h) gx) h gx + k) gx) k gx + k) gx) k where k h Sine g x) g x) we have shown that dg dx is an odd funtion. We find the third derivative, then look to the oeffiients of the ters. dy dx e x sinkx) + ke x oskx) d y dx e x sinkx) ke x oskx) ke x oskx) k e x sinkx) d 3 y k )e x sinkx) ke x oskx) dx 3 k )e x sinkx) + k k 3 )e x oskx) + ke x oskx) + k e x sinkx) 3k )e x sinkx) + 3k k 3 )e x oskx) 3k )y + k3 k )e x oskx) To have d3 y dx 3 Cy we ust have k3 k ) so k ± 3 and then C 3k 8. 3a)i) Reognise building blok funtions powers) as odd and even [st ark], full desription of odd, even and neither [nd ark]. 3a)ii) Write expression for g x) [st ark], use g x + h) gx h) [nd ark], full proof [3rd ark]. 3b) Find d y dx [st ark], get for d3 y dx 3 Ay + Be x oskx [nd ark], find C 8 [3rd ark].

4 Sholarship Calulus 93) 3 page 4 of 8 Question Four a) For eah n 9 we list the solutions of z n z ; these are the solutions of z n ; roots of unity. a We need to be areful not to ount any roots twie; this is easiest when writing the solutions in the for z is n π. n n 3 n 4 n 5 n 6 n 7 n 8 n 9 z is! z is! z is! 3 z is 4! 3 z is z is 3! z is 5 z is 4! 5 z is 6! 5 z is 8! 5 z is! 3 z is 5! 3 z is 7 z is 4! 7 z is 6! 7 z is 8! 7 z is! 4 z is 3! 4 z is 5! 4 z is 7! 4 z is! 7 z is! a There are solutions in the for z is n!, and also the solution z. There are 3 solutions in total. b)i) We rearrange to ollet ) u ) u u Δv + Δv Δv + Δv u ) u Δv Δ v ters together. + Δv Δv Δv ) u ) u ) u + + u u ) exp ln exp ln tanh u ln + 4a) Understanding of roots of unity ould be shown in a diagra) [st ark], find all non-zero roots repetition allowed) [nd ark], 3 solutions in total allow MEI for ) [3rd ark]. 4b)i) Multiply out Δv ) [st ark], get Δv as subjet, any for [nd ark], required for [3rd ark]. OR an equivalent for in the opposite diretion, with substituting tan h orretly as [st ark]. 4b)ii) Differentiate both sides, or integrate with partial frations [st ark], into required for [nd ark]. Also possible to work this bakwards.

5 Sholarship Calulus 93) 3 page 5 of 8 b)ii) We differentiate the given equation and show it satisfies the differential equation. d dv ln M ) u d + v / ln dv v / dm M dv u v / + v / dm dv M u + v / M u v / + v / v / ) v / v / )

Sholarship Calulus 93) 3 page 6 of 8 Question Five a) We find the ritial points where da dθ : da dθ R + 4!R sinθ) osθ) + R os θ R sin θ +!sinθ osθ) + os θ sin θ +!sinθ osθ) + sin θ sin θ!

6 Sholarship Calulus 93) 3 page 6 of 8 Question Five a) We find the ritial points where da dθ : da dθ R + 4!R sinθ) osθ) + R os θ R sin θ +!sinθ osθ) + os θ sin θ +!sinθ osθ) + sin θ sin θ!sinθ osθ) sin θ sinθ! osθ) sinθ 4sin θ os θ ) sinθ! sin θ sinθ sin θ!sin θ os θ So either sinθ giving θ n!, whih are the axia at and! ), or:!sin θ os θ sin θ os θ! tan θ! θ tan! θ tan! 5a) Find any for of da [st ark] dθ take fator of sinθ out [nd ark], half-angle forula step [3rd ark], find solution [4th ark]. b) EITHER The feasible region has verties at x, y) a,a) and x, y) a,3a). The required non-linear objetive funtion Px, y) ust take its largest value at both points, like the iddle line in the diagra. Larger values the outer urve) are not et, and saller values the inner urve) are not axial. a,3a) a,a) 5b) linear prograing Find verties a,3a) and a,a) [st ark], diagra of feasible region [nd ark], urved objetive funtion through verties [3rd ark], justified answer [4th ark].

7 Sholarship Calulus 93) 3 page 7 of 8 OR The ritial path in the first flowhart is 4 inutes 7 hours). When splitting the tasks out, the new flowhart would look soething like this: start ) buy aterial 3) ove furniture ) prepare 5) paint 4) tidy 5) paint dries 4) return furniture 5) end ) ove furniture ) prepare 5) paint 4) tidy 5) paint dries 4) return furniture 5) The new ritial path is 8 inutes shorter 34 inutes). This assues the roos are the sae size, and the tasks are divided equally between the roos. Splitting out the oving furniture task akes the buy aterial task part of the ritial path. However, splitting the tidying task does not help, beause the paint drying takes uh longer. The painters ight need to spend ore on aterial if they are painting two roos at one. There ight not be enough painters to paint two roos at one; or to ove the furniture twie as fast. Other deeper-thinking stateents about the new flowhart possible. Would expet at least four good stateents about the new flowhart in a full answer. Surfae-level understanding not suffiient. 5b) ritial path analysis Find the ritial paths are 4 and 34 inutes [st ark] insightful atheatial stateent [nd ark] insightful real-world stateent [3rd ark] insightful stateent onneting real-world and atheatial ontexts [4th ark]

Sholarship Calulus 93) 3 page 8 of 8 Question Six a) Using de Moivre s Theore, and olleting real and iaginary ters: os5θ + isin5θ is5θ is 5 θ 5 osθ + isinθ os 5 θ + 5ios 4 θ sinθ os 3 θ sin θ ios θ

8 Sholarship Calulus 93) 3 page 8 of 8 Question Six a) Using de Moivre s Theore, and olleting real and iaginary ters: os5θ + isin5θ is5θ is 5 θ 5 osθ + isinθ os 5 θ + 5ios 4 θ sinθ os 3 θ sin θ ios θ sin 3 θ + 5osθ sin 4 θ + isin 5 θ os5θ os 5 θ os 3 θ sin θ + 5osθ sin 4 θ isin5θ 5ios 4 θ sinθ ios θ sin 3 θ + isin 5 θ sin5θ 5os 4 θ sinθ os θ sin 3 θ + sin 5 θ Alternatively, with rather ore work, the su of angle forulas an get the sae results. Differentiating either identity gives the other. 6a) Use de Moivre s Theore [st ark] b) expand binoial theore or otherwise [nd ark] siplify, separate real / iaginary parts [3rd ark] EITHER well reasoned proof [4th ark]. Using x n+ x n, y n+ y n and z n+ z n, we get the equations x n.8x n +.7 y n +.6z n y n.x n +.y n +.4z n z n.x n +.y n yielding:.x n +.7 y n +.6z n.x n.8y n +.4z n.x n +.y n z n We also need the additional inforation that x n + y n + z n 99, as the first three are insuffiient. Using any of several ways to solve this syste of four equations, we find that x n 76 6a) linear systes y n 4 Set x n+ xn et [st ark] ollet ters [nd ark] z n 9 introdue extra equation x + y + z 99 [3rd ark] OR Consider the equation z + i + z i A. This is an ellipse, with foi at i and of points on the iaginary axis between i and i. If k < then there are no suh points. At k the points are on the iaginary axis between i and i. If < k the points for a solid ellipse, with foi at i and i. solution x, y,z) 76,4,9) [4th ark] i, so long as A >. When A it is the set If k > the points lie between two ellipses with foi at i and i ; the outer ellipse has an inner ellipse of points whih do not satisfy the inequality. For large k, both ellipses are approxiately irles, with the inner irle half the radius of the outer.) k < k. k.5 k k.5 k 3 6a) oni setions For of ellipse with foi orret [st ark] ring shape [nd ark], solid when k [3rd ark], no solutions when < k < [4th ark].

(Newton s 2 nd Law for linear motion)

(Newton s 2 nd Law for linear motion) PHYSICS 3 Final Exaination ( Deeber Tie liit 3 hours Answer all 6 questions You and an assistant are holding the (opposite ends of a long plank when oops! the butterfingered assistant drops his end If