Mathematics Extension 2

Transcription

1 NORTH SYDNEY GIRLS HIGH SCHOOL Mathematics Etension Trial HSC Eamination General Instructions Reading time 5 minutes Working time hours Write using black or blue pen. Black pen is preferred Board approved calculators may be used A table of standard integrals is provided at the back of this paper In Questions -6, show relevant mathematical reasoning and/or calculations Student Number Total marks Section Pages 5 marks Attempt Questions - Allow about 5 minutes for this section Section II Pages 6-9 marks Attempt Questions - 6 Allow about hours and 5 minutes for this section Class Student Name QUESTION MARK / /5 /5 /5 /5 5 /5 6 /5 TOTAL / []

2 Section I marks Attempt Questions - Allow about 5 minutes for this section Use the multiple-choice answer sheet for Questions -. Let z aibwhere a and b are real and non-zero. Which of the following is not true? (A) z z (B) z z is real is non-real (C) z z is real (D) zz is real and positive Which of the following corresponds to the set of points in the comple plane defined by zi z? (A) the point given by z i (B) the line Im z (C) the circle with centre i and radius (D) the line Re z The equation has roots, and. What is the value of (A) (B) (C) (D) ? The polynomial equation P has real coefficients, and has roots which include i and. What is the minimum possible degree of P? (A) (B) (C) (D) []

3 5 Using a suitable substitution, what is the correct epression for in terms of u? sin cos d 6 (A) u u du 6 (C) u u du 6 (B) u u du 6 (D) u u du 6 There are 5 pairs of socks in a drawer. Four socks are randomly chosen from the drawer. Which epression represents the probability that all four of the socks come from different pairs? (A) (B) (C) (D) The equation y yis differentiated implicitly with respect to. Which of the following epressions is dy d? (A) (B) (C) (D) y y y y y y []

4 8 The graph y f is shown. y Which of the following graphs best represents y f? (A) y (B) y (C) y (D) y []

5 9 A body is moving in a straight line and, after t seconds, it is metres from the origin and travelling at v ms -. Given that v and that t where, what is the equation for in terms of t? (A) (B) t e t e (C) t 5 (D) t 5 y y P (, y ) O The region bounded by the ais, the curve y around the y ais. and the line is rotated The slice at P, y on the curve is perpendicular to the ais of rotation. What is the volume V of the annular slice formed? (A) y y (B) y (C) (D) y [5]

6 Section II 9 marks Attempt Questions -6 Allow about hours and 5 minutes for this section In Questions -6, your responses should include relevant mathematical reasoning and/or calculations. Question (5 marks) Use a SEPARATE writing booklet. (a) Let z 5iand i. Find in the form iy (i) i (ii) z i (b) By first writing w i in modulus argument form, show that w 8i. (c) By completing the square, find d. (d) Use the substitution u to evaluate d. (e) (i) Without using calculus, sketch the curve important features. y showing all (ii) Find the area bounded by the curve and the -ais between and 5. [6]

7 Question (5 marks) Use a SEPARATE writing booklet. a) Consider the comple number z iywhere z a ib. (i) Sketch on the same set of aes, the graphs of y aand y b where both a and b are positive. The foci and directrices of the curves need NOT be found. (ii) Use the graphs to eplain why there are two distinct square roots of the comple number a ibif a and b. (iii) Consider how the sketch changes when b is negative. What is the relationship between the new square roots and those found when b was positive? (b) The region enclosed by the curves y y and the ordinates and is rotated about the y ais. Using the method of cylindrical shells, find the volume of the solid formed. (c) A particle s acceleration is given by where is the displacement in metres. Initially the particle is at the origin with velocity metres per second. (i) Show that v. (ii) Find the velocity and acceleration at. (iii) Describe the motion of the particle. (iv) Find the maimum speed and where it occurs. [7]

8 Question (5 marks) Use a SEPARATE writing booklet. (a) (i) Show that (ii) If y zshow that y y where and y are real numbers. y z. (b) The comple numbers z and w each have a modulus of. The arguments of z and w are 7 and respectively. 9 9 (i) Sketch vectors representing z, w and z w on the Argand diagram, showing any geometrical relationships between the three vectors. (ii) Find argz w. (iii) Evaluate z w. (c) (i) Use the substitution tan t to show that d. sin a a f f f a d. (ii) Show that d (iii) Hence, or otherwise, evaluate sin d. [8]

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10 Question (5 marks) Use a SEPARATE writing booklet. (a) The diagram shows the graph of y f( ) which is only defined over the domain. Draw separate one-third page sketches of the graphs of the following: (i) y f (ii) ln y f (b) y T P O The point P ct, c t lies on the hyperbola y c. The point T lies at the foot of the perpendicular drawn from the origin O to the tangent at P. (i) Show that the tangent at P has equation (ii) If the coordinates of T are, y show that (iii) Show that the locus of T is given by. t y ct y t.. y c y Question continues on page []

11 Question (continued) (c) The horizontal base of a solid is an isosceles triangle OEF where OE OF r and EF b. HAE is the parabolic arc with equation y r where E lies on the -ais. HBF is another parabolic arc, congruent to HAE, so that the plane OHBF is vertical. A rectangular slice ABCD of width is taken perpendicular to the base, such that CD lies in the base and CD EF. (i) Show that the volume of the slice ABCD is b r. r (ii) Hence show that the solid HOEF has volume br. (iii) Suppose now that EOF and that n identical solids HOEF are arranged n about O as centre with common vertical ais OH to form a solid S. Show that the volume V n of S is given by Vn sin n r n. (iv) When n is large, the solid S approimates the volume of the solid of revolution formed by rotating the region bound by the ais and the curve y r about the y ais. Using the fact that sin as find limv n. n End of Question []

12 Question 5 (5 marks) Use a SEPARATE writing booklet. (a) The equation has roots,,. Find the equation whose roots 5 are,, and. n n (b) (i) For z cos isin, show that z z cosn. (ii) If z u, find an epression for z z in terms of u. z (iii) It can be shown that Show that 5 5 z u 5u 5u. (Do not prove this). 5 z 5 cos 6cos cos 5cos (c) Q P R In the Argand diagram, the points P, Q and R represent the comple numbers p, q and r. (i) Given that the triangle PQR is equilateral, eplain why rq cis q p (ii) Hence, or otherwise show r p q i q p Question 5 continues on page []

13 Question 5 (continued) (d) P M A N B L C In the diagram, P is any point on the circle ABC. The point N lies on AB such that PN is perpendicular to AB. Similarly, points M and L lie at the foot of the perpendiculars drawn from P to CA (produced) and BC respectively. (i) State why BLNP is a cyclic quadrilateral. (ii) Prove that the points L, M and N are collinear. End of Question 5 []

14 Question 6 (5 marks) Use a SEPARATE writing booklet. (a) Let Pacos, bsin be a point on the ellipse a y b Is Sae,where e is the eccentricity and O is the origin.. The focus of the ellipse (i) Find the coordinates of the centre C and the radius of the circle of which SP is a diameter. (ii) Show that OC ecos a (b) (i) Show that the polynomial is divisible by P ( ) 8. (ii) Hence epress the polynomial in the form P A Q where is a real quadratic polynomial. Q (c) Using the fact that y tan tan tan y y prove by mathematical induction that for all positive integers n, for and y tan tan tan... tan tan n n Question 6 continues on page 5 []

15 Question 6 (continued) (d) Consider f log. (i) Show that f for all. (ii) Consider the set of n positive numbers p, p, p,... p n such that p p p... p n. By using the result in part (i), deduce that n r r log np np np np... np n n (iii) Show that n log npr. r n (iv) Hence deduce that n p p p... p n End of paper [5]

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18 Etension Trial HSC Solutions. (A) z+ z ( a+ ib) + ( a ib) a ( which is real) (But you should know that z+ z Rez ) (B) z a+ ib a+ ib z a ib a+ ib a b ab i ab ( which is not real since, ) + a + b a + b Alternatively z r cisθ z r cis θ cis θ ( ) (C) z ( z) ( a+ ib) ( a ib) which is real if θ kπ ( where k is an integer) k θ π But this is not the case, since neither a nor b is zero a + abi b a + abi+ b ( ie. z is either real or pure imaginary) ( ab ) abi which is never real since, (D) zz z which by definition of modulus is real and positive (C). Without doing any algebra, this is the set of point which are equidistant from (, ) and (, ). ie. y (B). α+β+γ + + βγ αγ αβ αβγ (A). Since ( ) P has real coefficients, the conjugate of the root + i must also be a root. (C) So the polynomial must have at least roots (and there is not enough information to conclude more.) π 5. ( ) sin cos d cos cos sind π ( ) ( ) u u du 6 ( ) u u du Let u cos du sind, u π, u (B)

19 6. st sock can be any of them. nd sock cannot be the only matching sock 8 possibilities of 9 socks remaining rd sock cannot be either of the two matching socks 6 possibilities of the 8 socks remaining th sock cannot be either of the three matching socks possibilities of the 7 socks remaining (D) 7. + y y dy dy + y y d d dy ( y ) y d dy y d y + ( product rule) Alternative setting out: + y y d+ y dy ( dy+ y d) ( ) ( ) y dy y d dy y d y (A) 8. Negative root (single root) must become a vertical point. f is ve, so can t be square rooted. To the left of the negative root, ( ) Positive root is a multiple root, so we can t determine the nature of the corresponding point on the new graph. (But we are only asked for the best answer.) ( ) becomes y f ( ) y f ±, hence symmetry in the -ais. (D) 9. Easiest method check the options by differentiating to get v. The only options whose derivatives are the function itself (ie. v ) are (A) and (B). But (B) is the only option that also allows to equal. Alternative method: d dt d dt d dt ln t+ c ( t ), + c ln t e t ± e t c But for to equal, we need the ve case: t e (B)

20 . δ π( ) ( ) V R r h π δy But y y y + ( ) π( y ) δy δ V π y + δy (A) Question (a) (i) ( + i) ω ( + i)( + i) + i+ i + i (ii) z 5 i + i i i + i 5i i i i (b) 5π w cis 6 5π w 8i cis 8i 6 5π 8cis 8i 5π 5π 5π 8i 8i hopefully you don't need to write cis as cos + i sin to see this (c) d d + + ( ) d ( ) + + sin + c + +

21 (d) (e) (i) d d + + u du u u u du u u 8 + Let u + du d, u, u (ii) A Let 5 ( ) ( + ) d ( )( + ) + A B ( )( ) ( ) ( ) + A + + B ( ) A A B B 5 A + d + ln ln + + ln + ln8 ln8 8 ln 5 5 ln 5 ( ) 5

22 Question (a) (i) When y, b ( b> ) ± y b y y b b (, ) a (, y ) a (, y ) y b, b> y b, b< ( part iii) (, ) b b y a (ii) Let z + iy ( ) ( ) z a+ ib y + yi a+ ib + iy a+ ib Equating real and imaginary parts: y a y b Solving simultaneously for and y, we get the graphs of part (i). The graphs show that there are two distinct points of intersection (, y ) and (, y ) corresponding to two distinct comple roots z + iy and z + iy of the comple number a+ ib. (iii) When b is negative, the graph of y b lies in the nd and th quadrants. So the points of intersection with are (, y ) and (, y ) y a. ie. the new square roots are the conjugates of the roots found in part (ii).

23 (b) (, ) P y (, ) P y y + y + h π δ h y y + + Volume of shell δv πh δ π δ + + Volume V lim π δ δ π d + + ( ) ( ) π ln + ln + ( ln 8 ln 5 ln ) π + 8 πln 5 6 πln units 5

24 (c) (i) ( )( + ) d v d, v c ( ) v + c v + v 6 + Check RHS: ( )( + ) ( )( + + ) v + ( )( ) (ii) : v + ( )( ) 9 ms LHS (iii) v ẋ Firstly, the particle cannot ever be to the right of, as v would be ve. Secondly, the particle can possibly change direction only when v, ie. at and. Initially, the velocity is +, so the particle moves to the right, speeding up until it reaches, then slowing to a stop at. Since the acceleration at is ve, it then changes direction, speeds up until it again reaches, then slowing to a stop at. At the velocity and acceleration are both zero (and dependent only on position, not time), so the particle remains at. (iv) From the graphs, the ma speed (over the restricted domain ) occurs at v v ma ma ( )( ) + 8 m/s. ( )

25 Question (a) (i) ( y) OR real, y y+ y + y y If, y have the same sign, y>, so y> y, so If, y have opposite sign, y<, so y + y y + y y as + y y + y sum of perfect squares ( ) + y y. + y. (ii) + y + y z + y + y z + y + y+ y + y y y + y y ( + y y) ( + y y ) since from i

26 (b) (i) z+ w z w 8 (ii) Since this shape is a rhombus, the verte angles are bisected by the diagonals. arg( z+ w) is the average of π 7π and 9 9 π 7π arg( z+ w) + 7 OR arg( z w) π π π π 8 (iii) Since diagonals bisect each other at right angles, z+ w cos z+ w (c) (i) π d + sin dt + t + t t + + t + t dt ( ) + t + t Let t tan tan t tan t dt d t + t+ +t dt ( t ) + + ( t+ ) tan tan tan π π 6 π dt, t π, t

27 a a a f d f d f d Let nd u a ( in integral ) a (ii) ( ) ( ) + ( ) a f ( ) d f ( a u) ( du), a + ( ) ( ) ( so u a) du d u a a, u ( ) f ( a ) d ( since choice of var does not affect def int) a f d + f a u du a f d + a ( ) ( ) f + f a d a a (iii) π π π d + + sin + sin + sin π π ( ) d π + d + sin + sin + sin π d π + sin π π ( by part i) π ( by part ii)

28 Question (a) (i) y y f ( ) (ii) ln (, ln ) (b) (i) y c dy + y d dy y d at P ct, c t, c t mt ct t c y ct t t t y ct + ct Tangent: ( ) + t y ct

29 y (ii) OT has gradient, and OT PT. So m m y t OT PT y t (iii) Since T satisfies equation of tangent: + t y ct ( ) ( ) ( y ) y y + y c from part ii + y c c y + c y squaring ie. locus of T is ( y ) + c y y ( ) (c) (i) By similar triangles OCD and OFE in the base: O r D C DC OD EF OE DC b r b DC r E ( base of rectangular slice) b F Height of slice h y r Thickness of Slice δ b δ r δ r Volume of slice V ( )

30 r δ b r δ r (ii) Volume V lim ( ) r b ( r ) d r b r r b ( r r ) r br r (iii) O π n r Vn b r π n r sin r n π r nsin n E b π b sin n r π b r sin n F (iv) As n, π π π, so sinn n n π So limvn r n n n πr

31 Question 5 (a) Let ( ) P + P has roots α, β, γ n n n n (b) (i) z + z ( cosθ+ i sinθ ) + ( cosθ+ isinθ) cosnθ+ i sinnθ+ cos( nθ ) + isin( nθ) cosnθ+ i sinnθ+ cosnθ i sinnθ cosnθ (ii) z + z + z+ + z z z z z z z z z u u (iii) IF you HAD to show this result: z z + 5z + z z z + z+ 5 z + z+ z z z z ( ) 5 u 5 u u u + 5 u 5u 5u ( ) + cosθ + cos 5θ cos 5θ z z z ( cos5 θ) 5 5 ( z + z ) ( from part i) 5 ( u 5u + 5u) ( given) 5 ( cos θ) 5 ( cos θ ) + 5 ( cos θ) 5 ( cos θ cos θ+ cos θ) 5 ( ) 6cos θ cos θ+ 5cosθ

32 (c) (i) π π QP QR cis since angle in equilateral triangle is π p q ( r q) cis π r q p q π r q q p cis π to multiply by, add π to the argument π r q cis ( q p) OR ( ) cis ( to divide by a comple number, multiply by its conjugate) ( ) ( ) S π Q P π SQR r q QR π π cis QS anticlockwise rotation by π cis PQ π cis π π ( eterior angle of triangle opposite interior angle) ( q p) (ii) r q cos + isin ( q p) + i ( q p) ( i )( q p) r q + r q q+ p+ iq ip ( ) ( ) r p+ q + i q p R

33 (d) (i) PB subtends equal angles at N and L on the same side of PB. OR BLP BNP ( given) BLNP is cyclic (converse of angles in same segment [or angles standing on same arc]) (ii) NOTE: You may NOT say BNL MNA ( vertically opposite) OR MNP PBL ( et angle of cyclic quad opposite interior angle) as these assume that LNM is a straight line, WHICH IS WHAT YOU ARE TRYING TO PROVE. ( ) PMA PNA 9 given PNAM is cyclic (eterior angle of cyclic quad equals opposite interior angle) ( both angles stand on chord of cyclic quad ) ( eterior angle of cyclic quad APBC opposite interior angle) MNP MAP PM PNAM PBC MNP is the eterior angle of cyclic quad BLNP (it equals the opposite interior angle) ie. LNM is straight (ie. L, M and N are collinear)

34 Question 6 (a) (i) C ( cos θ, bsinθ ) P a a M, bsinθ e S O S( ae, ) Centre: Diameter: ( + cosθ) ae+ a cosθ bsinθ a e bsinθ C, C, PS epm a a e a cosθ e a e θ ( cos ) radius ( ecosθ ) a e (ii) [The sneaky way] Since OS SS, CS PS and OSC SSC, then OSC and SSP are similar OC PS But PS+ PS a (sum of focal lengths length of major ais) CS+ CO a a ( ecos ) θ + OC a a a OC a + e cosθ a OC ( + ecosθ) [The hard slog] a b OC ( e+ cosθ ) + sin θ ( ae + aecosθ+ a cos θ+ b sin θ) ae + aecosθ+ a cos θ+ a ( e ) sin θ a ( e sin θ + e cosθ+ cos θ+ sin θ ) ( )

35 (b) (i) a e ( cos e cos ) θ+ θ+ a e ( cos ) θ+ Since e< for ellipse and cosθ then ecosθ < So + ecosθ> a OC ( ecosθ+ ) P ( ) P is divisible by ( ) (ii) Let zeros be, α, β 5 Sum: +α+β α+β Product: αβ αβ A polynomial with zeros α and β is + +. P But to get equal leading coefficients: ( ) ( )( ) [Alternatively: divide] (c) In case you had to prove the given result: tan tan + tan tan tan( tan + tan y) tan tan tan tan + y y tan + tan tan y ( ) ( y) ( ) ( y) + y y

36 The Induction Proof: Test n : RTP π tan + tan + tan tan tan n n+ LHS tan π RHS tan π LHS RHS tan + tan + π tan π tan ( using given rule) π π LHS RHS true for n Assume true for n k : π ie. tan + tan + tan tan tan k k+ Prove true for n k+ : π ie. RTP tan + tan + tan tan + tan tan k k+ k+ ( k+ ) k+ ( k+ ) ( k+ ) + ( k+ ) ( k+ ) ( k+ ) ( k+ ) ( k ) ( k ) ( k+ ) ( k+ ) ( ) π π LHS RHS tan + tan tan by assumption k+ k+ tan + tan tan k+ k+ + ( ) k k tan tan k+ tan ( k+ )( k + 6 k+ 5 ) tan k+ k + 6k+ 5 tan tan k + k + 6k+ 5 k+ k + 6k+ 5 tan tan k+ tan tan k+ k+ LHS RHS ( )

37 OR Using the fact that tan ( ) tan : π LHS tan + tan k+ + ( k ) ( k+ ) ( k+ ) ( k+ )( k+ ) + ( k ) π tan tan k+ ( k+ ) + + ( k )( k ) ( k )( k ) π k+ ( k+ ) tan + k π tan π k + k+ tan k + k + 8k+ π tan k+ RHS ( by part b) true for n k+ when true for n k by Mathematical Induction, true for all positive integers n. (d) (i) f ( ) log + f ( ) stationary point at f ( ) < minimum turning point at (, ) Also domain > (and continuous for all in the domain). y log + f > ( ) n n r r r r n n npr r r n n log( npr) npr n r r (ii) log( np ) ( np ) ( from part i - log > )

38 (iii) Continuing from part ii: n n r r r n r ( n ) lognp n p n since is a constant lognp r n n (iv) Continuiing from part iii: lognp + log np lognp r ( np np np ) log... n ( n pp pn) log... n n pp... p n n n n Also, since p, p,..., p n and n are all positive, then n pp... p n > n < n pp... p n