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1 Q Integration. HSC 9 Mathematics Etension Solutions - Jan Hansen a Let u = ln. Then du = d/ and so ln d = udu = u /+c = (ln) /+c b I = e d. Use integration by parts; u =,dv = e d du = d,v = e / So, I = udv = uv vdu (Formula) I = e / e /d = e / e /4+c c As there is an term on top and bottom we manipulate to form a as shown, I = +4 d = d 6 = 4 6..tan +c = 4 8 tan +c 5 6 d I = +3 4 d 5 a = +4 + b d 5 a( )+b(+4) = d (+4)( ) 4 d + So a( )+b(+4) 6 and hence a+b = and4b a = 6. Solvingthesetwo simultaneously we obtain, a =,b = and hence, e I = 5 I = +4 d = ln(+4) ln( ) 5 = ln9 ln4 (ln6 ln) = 4ln3 ln ln ln3 = ln3 4ln ( ) 9 or ln d Use the substitution u = / =. Then du = 3 d and when = 3 = u = /3, when = = u = / = so we have I = = /3 = (u+) / Q 3 du +/u = du = u+ /3 /3 = / 3 Comple Numbers. du (+/u) (u+) / du a i 9 = (i ) 4 i = ( ) 4 i = i = +i (in the form a+ib). b +3i i (to realise the denominator) +i i = 4+6i+i 3i 4 i = 8i = i c.c.i.c.ii.c.iii c 9 by Jan Hansen. All rights reserved. janh@

2 d HSC 9 Mathematics Etension Solutions - Jan Hansen e.e.i We re solving z 5 =. Assume the form, z = r cisθ = rcosθ+isinθ. r 5 ( cisθ) 5 = = r 5 cis5θ = (By De Moivre s theorem) r = and cis5θ = = cis(π + πn) where n Z. 5θ = π +πn θ = (n+)π where n Z. (this is all of 5 them) The five with principal arguments are for n =,±,± we have the five roots, z = cis π 5, cis 3π π 3π 5, cis 5, cisπ, cis 5 and these are equally spaced around the circle differing in arguments by π/5,.e.ii f.f.i Solve (+iy) = 3+4i y +yi = 3+4i y = 3 and y = 4 Combining these by eliminating y we have, 4/ = = (quadratic in = (3± 9+6)/ = (3±5)/ = 4, Asisrealthen = ±, soy = ±/ = ± and therefore the two square roots are z = ± ± i. = ±( + i). (Check this by squaring it, you should get 3+4i.).f.ii We use the otherwise method; notice that z = is a zero of z + iz i, dividing out we get, z +i+ z )z +iz i z z (i+)z i (i+)z (+i) Thus the equation becomes, (z )(z+i+ ) = and so the solutions are z =, i c 9 by Jan Hansen. All rights reserved. janh@

3 Q3 3a Graphs. 3.a.i 3.a.ii 3.a.iii HSC 9 Mathematics Etension Solutions - Jan Hansen 3 Q3a y 3b Graphs. 3 4 y = f() y 4 Q3a (ii) y = f() +y+3y = 8. Differentiating implicitly regarding y as a function of, +y +y +6y.y = y (+3y) = (+y) y = y +3y So y = (for stat pts) = y = = y =, and substitution back into the equation gives, y Q3a (i) y = f() y 4 Q3a (iii) y = f( ) +3 = 8 = 8 = ±3 y = 3 and hence the stat points are ( 3,3),(3, 3) (and we have checked that at these points, +3y ] 3c Polynomials. We have, P() = 3 +a +b+5, a,b IR. If ( ) is a factor then by theory, P() = P () = and P () = 3 +a+b, so we c 9 by Jan Hansen. All rights reserved. 3 janh@

4 HSC 9 Mathematics Etension Solutions - Jan Hansen have +a+b+5 = and 3+a+b =. b = 6 a = 3+a+ 6 a = = a = 3 = b = 6 3 = 9 3d Volumes cylindrical shells. From theory we have the volume of the infinitesimal shell δv = π(r r)h = (r +r )(r r )(y y ) and r = +δ, r = δv = π(++δ)(+δ )(y y ) δv = π(+ ( ) )δ (Ignoring δ terms) V = π 3 (+) ( ) d (Limits were obtained as; + = + = 3 = = =,3) V = π V = π d 3 3 d = π [ 3 4 /4 ] 3 = π(7 8/4) = 7π units 3 Q4 4a Conics. 4.a.i Get y : a + yy = b = y = b a y = b. ( at P) a y So the normal gradient is the negative reciprocal of the tangent gradient, and we have y y = a y ( b ) NORMAL 4.a.ii When y = (on ais) then y = a y ( b ) Solving this for gives, = (a b ) a Recallthatforanellipseb = a ( e ) = b = a a e = a b = a e andtherefore we have = a e = e and so N(e,) as required.// 4.a.iii By the focus directri definition we have PS = epm and PS = epm. Therefore, PS = epm = PM PS epm PM and in terms of the coordinates we have PS = a/e PS +a/e = a e e +a = e(a e ) e(e +a) = NS NS (since NS = ae e = e(a e ) and NS = e +ae = e(e +a) ) 4.a.iv We have α = S PN and β = NPS and by the sine rule we have PS sin PNS = NS sinβ and PS = NS sin PNS sinα So by (iii) since PS/NS = PS /NS then we have c 9 by Jan Hansen. All rights reserved. 4 janh@ a

5 HSC 9 Mathematics Etension Solutions - Jan Hansen sin PNS = sin PNS sinβ sinα but PNS = π PNS (since SNS isa straight angle) and since sin(π PNS) = sin PNS we have sin PNS sin(π PNS) = = sinβ sinα sin PNS and therefore sinα sinβ = = sinα = sinβ and as sinα α,β < π/ (by construction) then we conclude that α = β as required. 4b Mechanics. 4.b.i Resolve vertical forces: T cosα+n sinα = mg...() Resolve horizontal forces: T sinα N cosα = mrω...() 4.b.ii () cosα+() sinα gives T = mgcosα+mrω sinα = m(gcosα+rω sinα) as required. () sinα () cosα gives N = mgsinα mrω cosα = m(gsinα rω cosα) (For both calculations we used cos α + sin α = ) 4.b.iii If T = N then m(gcosα+rω sinα) = m(gsinα rω cosα) gcosα+rω sinα = gsinα rω cosα ω = g r sinα cosα sinα+cosα = g ( ) tanα as required.// r tanα+ (here we dividing top/bottom by cosα and using tanα = sinα/cosα) 4.b.iv We know that ω > and so the right hand side, in (iii), must be positive also. Hence tanα tanα+ > tanα > (since α > and so tanα+ > ) α > tan and hence α > π/4 and hence as obviously α < π/ we arrive at Q5 π/4 < α < π/ 5a Circle Geometry. 5.a.i c 9 by Jan Hansen. All rights reserved. 5 janh@

6 HSC 9 Mathematics Etension Solutions - Jan Hansen ADB = π/ (angle subtended by a diameter AB) KXD = Y XB (vertically opposite angles are equal) KDX BY X ( AA - two corresponding angles are equal ) Hence the third corresponding angle is also equal and thus α = AKY = XKD = XBY = ABD = β, as required. 5.a.ii We know α = β (corresponding angles in the similar triangles of part (i) are equal) Also, DCA = DBA = α (angles on the circle subtended by same segment AD are equal) DKCX is a cyclic quadrilateral as the interval DY subtends the same angle α at K and at B. (Circle result: Angles on circumference of a circle subtended by same segment are equal) 5.a.iii Firstly from the right-angled triangle XYB we know that γ +β = π/. But α = β = γ +α = π/. Also, ACB = π/ (as AB is a diameter). Hence, KCB = KCA + ACB = α + γ +π/ = π/+π/ = π. Thus KCB = π is a straight angle and therefore K,C,B are on the same straight line and so are collinear, as required. 5b Integration. 5.b.i I n = n+ e d. First we will use a simple substitution and then we use integration by parts to finish. Let w =, then dw = d and the integral becomes I n = w n e w dw. Now for integration by partsusingtheformula udv = uv vdu we choose u = w n and dv = e w dw. Then du = nw n dw and v = e w, so that I n = wn e w n w n e w dw = (n.e ) ni n = e ni n as required. 5.b.ii I = e/ I I = e/.i I =.5 w e w dw =.5 e w = e Thus I = e/ (e )/ = / and I = e/./ = e/ c 9 by Jan Hansen. All rights reserved. 6 janh@

7 5c Graphs. HSC 9 Mathematics Etension Solutions - Jan Hansen 5.c.i f() = e e f () = e ( )e = e +e f () = e e f () = e +e 5.c.ii f () > since e >,e > for all Also, f () = and so for > the function f () is increasing for all > (since the sign of f tells us how f is changing), whilst being equal to zero at =, hence f () > for >. [It is also obvious graphicallyifyougraphandcombiney = e and y = e.] 5.c.iii By a similar argument as in (ii) f () = and f () > for > (established in part (i)) Hence f () is an inceasing function so that f () > for > ( as it started at zero when = ) Hence f() is an increasing function for >, and since it starts at = (ie f() = ), then we must have f() > for > as required. Q6 6a The cross sectinal area is A() = y(4 ) = (4 ) 4 = (4 ) 3/ V = 4 A()d = 4 (4 ).5 d = [ 5 (4 ).5] 4 = ( 4/5)[ 4 5/ ] = 4 3 = 8/5 units 5 6b 6.b.i Assume that α is a zero we have P(α) = α 3 +qα +qα+ =. Now, P( α ) = α 3 + q α + q α + = α 3(+qα+qα +α 3 ) = α 3P(α) = α 3 = and hence α is a zero.// 6.b.ii Since α is not real and the coefficients of P arerealthenbytheorytherootsmustoccur in comple conjugate pairs, say, α, ᾱ. Q6.b.ii. Product of roots is α ᾱ ( ) = d/a = / =, and hence α = = α = (since for any comple number z, z and z z = z ). Q6.b.ii. Sumrootsisα+ᾱ+( ) = b/a = q/ = q = α+ᾱ = q. So Re(α) = q (since for a comple number z, z + z = Re(z) ) 6c 6.c.i PQ = OP r (Pythagoras) PQ = +y r (using distance formula on OP) Finaly, PQ = +y r. 6.c.ii PQ = PR = +y r = c (PR is just the difference in coordinates) +y r = c + c y = r +c c as required. c 9 by Jan Hansen. All rights reserved. 7 janh@

8 HSC 9 Mathematics Etension Solutions - Jan Hansen 6.c.iii We put the parabola into standard form to read off the verte and focal length; y = c+r ( +c) = c (r +c ) c This the focal( length ) is a = c/4 = c/ and r the verte is +c. c, Therefore, ( the) focus( is ) r S +c c c, r = S c, 6.c.iv The directri of the locus is = r +c c +c/ = r +c c. But we know that PM = PS (definition of parabola, with M on directri) and PQ = PR (from the locus definition) Hence, PS PQ = PM PR (difference in the -values of points M and R) = r +c c c which is independant of as required. Q7 7a 7.a.i Q7.a.i. ẍ = g rv v dv d = g rv (since by theory ẍ = d ( v / ) = v dv d d ) v dv d = g rv v d dv = v g rv v = g rv dv = rv r g rv dv = g rv r g rv g g rv dv = r [v + g ln g rv ]+c r Using intital conditions, = (g/v )lng +c = c = g r lng Updating, = r [v + g r ln g rv ]+ g r lng = g ( ) g r ln v g rv r Q7.a.i. Hence. L = 9.8 ( ). ln = 8m ( sf) 7.a.ii = e t/ (9s t c t ) + 9 (where s t = sint,c t = cost) ẋ = e t/ (9s t c t )/+e t/ (9c t + s t ) = e t/ (9c t +s t 9s t /+c t ) = e t/ (3c t +7s t /) Solve for t where ẋ = 3c t +7s t / = 3c t = 7s t / tant = 3/7 = t = nπ tan (3/7) and we need t > so require that nπ > tan (3/7) = n > tan (3/7)/π = = n. Choosing n = we have t = π tan (3/7)/ =.8 (dp) Hence = e.8/ (9sin.8 cos.8)+ 9 = = and as he is m tall his head reaches down a distance of = 9.487m and c 9 by Jan Hansen. All rights reserved. 8 janh@

9 HSC 9 Mathematics Etension Solutions - Jan Hansen so he stays dry. (i.e. out of the water). 7b Let z = cosθ+isinθ. 7.b.i Using De Moivre s Theorem we have z n = cosnθ+isinnθ and z n = cos( nθ)+isin( nθ) = cosnθ isinnθ (sincesineisanoddfunction and cosine is even function) So, z n + z n = cosnθ + isinnθ + cosnθ isinnθ = cosnθ as required. 7.b.ii By part (i), with n = we have, z +z = cosθ, and hence (z +z ) m = (cosθ) m and epansion of the left side using the binomial theorem gives m ( ) m (z +z ) m = z m r.z r r m ( ) m = z m r ( ) ( ) ( ) m m m = z m + z m + z m 4 + ( ) ( ) ( ) m m m z m z m+ + z m 3 m m But ( ) ( m = ( m), m ) ( = ( m ), m ) ( = m ),... and so ( ) ( ) m m (z +z ) m = (z m +z m )+ ( ) m (z m + z m+ ) + (z m 4 + z m+4 ) + + ( ) ( ) m m + (z +z )++ z m m [( ) ( ) m m = cosmθ + cos(m )θ+ ( ) ( ) ] ( ) m m m cos(m )θ+ + cosθ + as m m required. 7.b.iii Integrating the epression in (ii) and manipulating the contants we have π/ cos m θθ [ ( ) sinmθ m sin(m )θ = m+ m + m ( ) ( ) m sin(m 4)θ m sinθ m 4 m ( ) m m π/ θ m ( ) m π = m+ []+ m m = π ( ) m m Q8 m+ 8a 8.a.i Start ] π/ (cotθ/ tanθ/) = ( cosθ/ sinθ/ sinθ/ ) cos θ/ = ( cos θ/ sin ) θ/ sinθ/cosθ/ = cosθ (using double angle formulas) sinθ = cotθ and hence we are done as (cotθ/ tanθ/) = cotθ is equivalent to the identity we are asked to prove. 8.a.ii We are required to prove: n r tan r = n cot n cot r= Firstly, for n =, LHS =.tan = tan RHS = cot/ cot = cot/ cot = cot/ (cot/ tan/) by part (i). = tan/ = LHS Hence the statement is true for n =. Secondly, we prove the statement true for n = k + whenever the statement is true for n = k. So we assume that k r tan r = k cot k cot...( ) r= Now assuming (*) we prove that k+ r= r tan r = k cot cot...( ) k+ k+ Start, LHS = r tan r c 9 by Jan Hansen. All rights reserved. 9 janh@ r= +

10 HSC 9 Mathematics Etension Solutions - Jan Hansen = k cot k cot+ k tan (by induction hypothesis) k+ = ( k cot k +tan ) k+ cot = k cot cot k+ = RHS and hence the statement is true for n = k + when it is true for n = k. Finally, we conclude that by induction the statement is true for all positive integers. Remark: This is enough of a conclusion to a proof by induction contrary to the advice of many tetbooks. A reference is the 7 eaminers report, hsc_eams/eam-papers-7/pdf_doc/ mathematics-et-notes-7.pdf. The /3 unit syllabus confirms this view also - see page 47. JSH 8.a.iii Firstly a preliminary, Recall that Hence We have, n r= tan r r = n cot n cot n = tan cot n lim tan = n lim tan = n = cot by the above result. 8.a.iv Substituting = π/ into the result of part (iii) we have tanπ/4+ tanπ/8+ 4 tanπ/6+ = π = 4 π = 4 π 8b cot π Area under curve is n = lnn lnn = ln n n n Define A L =lower rectangle area= /n = /n Define A U =upper rectangle area= /(n ) = /(n ) Ths since by construction, A L < A < AU then we have /n < ln n n < /(n ) and eponentiating (i.e. take the power of e) and use the fact that e ln = to get e n < n n < e (n ) ( ) n e n < n n < e (n ) (by taking power of n) (n ) < ( ) n n < e e n n (We twice used the fact that for positive numbers a,b,c,d, that a/b < c/d = d/c < b/a) Thus e n (n ) < ( n) n < e as required.// 8c 8.c.i The probability of A winning first up equals p. The probability of A winning on the second round [after n people (including A s first attempt), failed to win] equals q n p TheprobabilityofA winniungaftertworoundsisq n q n p Thus, W = p+pq n +pq n +pq 3n + W = p+pq n (+q n +q n + ) = p+pq n (using G.P. summation formula) qn = p+ pqn q n = p pqn +pq n q n W = p q n Then manipulate this to get the requested format; W( q n ) = p W Wq n = p W = p+q n W as required. 8.c.ii W m = p+pq n +pq n + +pq (m )n = p(qmn ) q n W m W = p(qmn ) q n ( qn ) p = q ( mn + = n) nm + By part (b), taking the mth power, we have e nm (n ) < ( n) nm < e m If n is large, that is, as n, then nm (n ) m. So, ( n) nm e m as n. Thus ( n) nm + e m + as n. So Wm W is aproimately equal to e m, as required. Contact: janh@, Regards, Comments/corrections are appreciated, Jan Hansen 3 November 9. c 9 by Jan Hansen. All rights reserved. janh@

2013 HSC Mathematics Extension 2 Marking Guidelines

2013 HSC Mathematics Extension 2 Marking Guidelines 3 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer B A 3 D 4 A 5 B 6 D 7 C 8 C 9 B A 3 HSC Mathematics Extension Marking Guidelines Section II Question