2020HSCQ14 (b)A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2\mathrm{m/s^2}m/s2, is modelled by a=10(1−(kv)2),a=10\bigl(1-(kv)^2\bigr), where is the velocity of the particle in and . Find the velocity of the particle after seconds.
2020HSCQ14 (a) (i)Let z1z_1z1 be a complex number and let z2=eiπ/3z1.z_2=e^{i\pi/3}z_1.z2=eiπ/3z1. The diagram shows points AAA and BBB, which represent z1z_1z1 and z2z_2z2, respectively, in the Argand plane. (i) Explain why triangle OABOABOAB is an equilateral triangle.2 marksShort answer
2020HSCQ15 (b) (iii)Let OPQROPQROPQR be a parallelogram with OP→=p\overrightarrow{OP}=\mathbf pOP=p and OR→=r\overrightarrow{OR}=\mathbf rOR=r. The point SSS is the midpoint of QRQRQR and TTT is the intersection of PRPRPR and OSOSOS, as shown in the diagram. (iii) Show that OT→=23r+13p.\overrightarrow{OT}=\frac23\mathbf r+\frac13\mathbf p.OT=323 marksShort answer
2020HSCQ15 (b) (iv)Using parts (ii) and (iii), or otherwise, prove that TTT is the point that divides the interval PRPRPR in the ratio 2:12:12:1.1 markShort answer
2020HSCQ15 (a) (ii)Using the proposition PPP from part (i), write down the contrapositive of PPP.1 markShort answer
2020HSCQ15 (a) (i)In the set of integers, let PPP be the proposition: “If k+1k+1k+1 is divisible by 333, then k3+1k^3+1k3+1 is divisible by $3”. (i) Prove that the proposition PPP is true.2 marksShort answer
2020HSCQ15 (a) (iii)Using the proposition PPP from part (i), write down the converse of PPP and state, with reasons, whether this converse is true or false.3 marksShort answer
2020HSCQ15 (b) (i)The point CCC divides the interval ABABAB so that CBAC=mn.\frac{CB}{AC}=\frac mn.ACCB=nm. The position vectors of AAA and BBB are a\mathbf aa and b\mathbf bb, respectively, as shown in the diagram. (i) Show that AC→=nm+n(b−a).\overrightarrow{AC}=\frac n{m+n}(\mathbf b-\mathbf a).AC=m+n2 marksShort answer
2020HSCQ15 (b) (ii)Using the same definitions as in part (i), prove that OC→=mm+na+nm+nb.\overrightarrow{OC}=\frac m{m+n}\mathbf a+\frac n{m+n}\mathbf b.OC=m+nma+m+nnb.1 markShort answer
2020HSCQ16 (a) (ii)Given that v<gmkv<\dfrac{gm}{k}v<kgm, show that when t=3mkln2,t=\frac{3m}{k}\ln2,t=k3mln2, the velocity of the larger mass is gm2k.\frac{gm}{2k}.2kgm.3 marksShort answer
2020HSCQ16 (b) (i)Let In=∫0π/2sin2n+1(2θ) dθ,n=0,1,….I_n=\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta,\qquad n=0,1,\ldots.In=∫0π/2sin2n+1(2θ)dθ,n=0,1,…. (i) Prove that In=2n2n+1In−1,n≥1.I_n=\frac{2n}{2n+1}I_{n-1},\qquad n\geq1.In=2n+12n3 marksShort answer
2020HSCQ16 (b) (ii)Using the result of part (i), deduce that In=22n(n!)2(2n+1)!.I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}.In=(2n+1)!22n(n!)2.3 marksShort answer
2020HSCQ16 (b) (iii)Let Jn=∫01xn(1−x)n dx,n=0,1,2,….J_n=\int_0^1x^n(1-x)^n\,dx,\qquad n=0,1,2,\ldots.Jn=∫01xn(1−x)ndx,n=0,1,2,…. Using the result of part (ii), or otherwise, show that Jn=(n!)2(2n+1)!.J_n=\frac{(n!)^2}{(2n+1)!}.Jn=(2n+1)!(n!3 marksShort answer
2020HSCQ16 (b) (iv)Prove that (2nn!)2≤(2n+1)!.\bigl(2^n n!\bigr)^2\leq(2n+1)!.(2nn!)2≤(2n+1)!.2 marksShort answer
2020HSCQ16 (a) (i)Two masses, 2m2m2m kg and 4m4m4m kg, are attached by a light string. The string is placed over a smooth pulley as shown. The two masses are at rest before being released and vvv is the velocity of the larger mass at time ttt seconds after they are released. The force due to air resistance on each mass has magnitude kvkvkv, where kkk is a positive constant. (i) Show that dvdt=gm−kv3m.\frac{dv}{dt}=\frac{gm-kv}{3m}.dtdv=3m2 marksShort answer