LCHons2005_p2

Coimisiún na Scrúduithe Stáit State Examinations Commission Scéim Mharcála Scrúduithe Ardteistiméireachta, 2005 Matamaitic Ardleibhéal Marking Scheme Leaving Certificate Examination, 2005 Mathematics Higher Level Page 1 Contents Page GENERAL GUIDELINES FOR EXAMINERS – PAPER 1............................................... 2 QUESTION 1......................................................................................................................... 3 QUESTION 2......................................................................................................................... 7 QUESTION 3....................................................................................................................... 11 QUESTION 4....................................................................................................................... 14 QUESTION 5....................................................................................................................... 19 QUESTION 6....................................................................................................................... 23 QUESTION 7....................................................................................................................... 28 QUESTION 8....................................................................................................................... 32 GENERAL GUIDELINES FOR EXAMINERS – PAPER 2............................................. 36 QUESTION 1....................................................................................................................... 37 QUESTION 2....................................................................................................................... 41 QUESTION 3....................................................................................................................... 44 QUESTION 4....................................................................................................................... 49 QUESTION 5....................................................................................................................... 53 QUESTION 6....................................................................................................................... 56 QUESTION 7....................................................................................................................... 61 QUESTION 8....................................................................................................................... 65 QUESTION 9....................................................................................................................... 69 QUESTION 10..................................................................................................................... 72 QUESTION 11..................................................................................................................... 76 BONUS MARKS FOR ANSWERING THROUGH IRISH.............................................. 80 Page 2 MARKING SCHEME LEAVING CERTIFICATE EXAMINATION 2005 MATHEMATICS – HIGHER LEVEL – PAPER 1 GENERAL GUIDELINES FOR EXAMINERS – PAPER 1 1. Penalties of three types are applied to candidates’ work as follows: · Blunders -mathematical errors/omissions (-3) · Slips -numerical errors (-1) · Misreadings (provided task is not oversimplified) (-1). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not exhaustive. 2. When awarding attempt marks, e.g. Att(3), note that · any correct, relevant step in a part of a question merits at least the attempt mark for that part · if deductions result in a mark which is lower than the attempt mark, then the attempt mark must be awarded · a mark between zero and the attempt mark is never awarded. 3. Worthless work is awarded zero marks. Some examples of such work are listed in the scheme and they are labelled as W1, W2,…etc. 4. The phrase “hit or miss” means that partial marks are not awarded – the candidate receives all of the relevant marks or none. 5. The phrase “and stops” means that no more work is shown by the candidate. 6. Special notes relating to the marking of a particular part of a question are indicated by an asterisk. These notes immediately follow the box containing the relevant solution. 7. The sample solutions for each question are not intended to be exhaustive lists – there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising examiner. 8. Unless otherwise indicated in the scheme, accept the best of two or more attempts – even when attempts have been cancelled. 9. The same error in the same section of a question is penalised once only. 10. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most. 11. A serious blunder, omission or misreading results in the attempt mark at most. 12. Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50. Page 3 QUESTION 1 Part (a) 10 marks Att 3 Part (b) 20 (5, 15) marks Att (2, 5) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 marks Att 3 1(a) Solve the simultaneous equations: 0 5 4 -= x y 17 2 3 + = y x Part (a) 10 marks Att 3 1(a) (i) 0 5 4 -= x y ´ 204x -5y = 0 (ii) 17 2 3 + = y x ´1030x + 5y = 170 34x = 170 x = 5 (i) 4x – 5y = 0 4(5) = 5y x = 5 y = 4 y = 4 Blunders (-3) B1 second variable not found. Slips (-1) S1 numerical. S2 not changing sign in subtraction. Attempts A1 no solution. A2 correct solution by trial and error. Worthless W1 values for x and y. Part (b)(i) 5 marks Att 2 1(b)(i) Express 41 41 41 41 2 + 2 + 2 + 2 in the form qp 2 , where p, q ÎZ. Part (b)(i) 5 marks Att 2 1(b)(i) + + + 4 = 1 41 41 41 2 2 2 2 ( ) 41 4 2 = ( ) 4 1 22 2 = 49 2 Blunders (-3) B1 indices. Slips (-1) S1 not elements of Z. Page 4 Part (b)(ii) 15 marks Att 5 1(b)(ii) Let f (x) = ax3 + bx2 + cx + d . Show that (x -t )is a factor of f (x)-f (t ) Part (b)(ii) 15 marks Att 5 1(b)(ii) f (x) = ax3 + bx2 + cx + d f (t ) = at 3 + bt 2 + ct + d f (x)-f (t ) = a(x3 -t 3 )+ b(x2 -t 2 )+ c(x -t ) = a(x -t)(x2 + tx + t 2 )+ b(x -t)(x + t )+ c(x -t ) = (x -t)[a(x2 + tx + t 2 )+ b(x + t )+ c] or 1(b)(ii) f (x) = ax3 + bx2 + cx + d f (t ) = at 3 + bt 2 + ct + d f (x)-f (t ) = ax3 + bx2 + cx -at 3 -bt 2 -ct ( ) ax (at b)x (at bt c) x t ax bx cx at bt ct + + + + +-+ + ---2 2 3 2 3 2 ax3 -atx2 (at + b)x2 (at + b)x2 -(at + b)tx (at 2 + bt + c)x -at 3 -bt 2 -ct (at 2 + bt + c)x -at 3 -bt 2 -ct 0 * Accept solution by division by (x -t ) for full marks. Blunders (-3) B1 indices. B2 factors Slips (-1) S1 numerical. S2 not changing sign when subtracting in division. Page 5 Part (c) 20(5, 5, 5, 5) marks Att (2, 2, 2, 2) 1(c) (x -p)2 is a factor of x3 + qx + r Show that 27r 2 + 4q3 = 0 Express the roots of 3x2 + q = 0 in terms of p. Factor 5 marks Att 2 Values 5 marks Att 2 Show 5 marks Att 2 Express 5 marks Att 2 1(c) (Show) x px px p x qx r 2 2 2 2 3 + -+ + + x3 -2 px2 + p2 x 2 px2 -p2 x + qx + r 2 px2 -4 p2 x + 2 p3 3p2 x + qx + r -2 p3 = 0 Remainder must = 0 since ( )2 x -p is a factor (3p2 + q)x + (r -2 p3 )= (0)x + (0) (i) : 3p2 + q = 0 q = -3p2 (ii) : r -2 p3 = 0 r = 2 p3 \ 27r 2 + 4q3 = 27(2 p3 )2 + 4(-3p2 )3 = 108 p6 -108 p6 = 0 or If ( )2 x -p is a factor of f (x), then let (x + a)be other factor. \ (x2 -2 px + p2 )(x + a) = x3 + qx + r x3 + (-2 p + a)x2 + (p2 -2 pa)x + p2 (a) = x3 + (0)x2 + (q)x + r Equating like to like (i) -2 p + a = 0 (ii) p2 -2 pa = q (iii) p2a = r (i) a = 2 p q = p2 -2 p(2 p) = -3p2 r = p2 (2 p) = 2 p3 27r 2 + 4q3 = 27(2 p3 )2 + 4(-3p2 )3 = 108 p6 -108 p6 = 0 Page 6 (Express) 3x2 + q = 0 3x2 = -q 3x2 = -(-3p2 ) 3x2 = 3p2 x2 = p2 x = ± p Blunders (-3) B1 indices. B2 not like to like. B3 root from equation. B4 r not found, having found q. B5 roots from equation (in “ express” part). Slips (-1) S1 numerical. S2 not changing sign in subtraction (division). Attempts A1 remainder ¹ 0 in division. A2 any attempt at division. Page 7 QUESTION 2 Part (a) 10 marks Att 3 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 marks Att 3 2(a) Solve for x: x -1 < 7 , where xÎR Part (a) 10 marks Att 3 2(a) x -1 < 7 -7 < x -1 < 7 \ -7 < x -1 and x -1 < 7 -6 < x x < 8 \ -6 < x < 8 or 2(a) x -1 < 7 ( )2 ( )2 x -1 < 7 x2 -2x +1 < 49 x2 -2x -48 < 0 Solve: x2 -2x -48 = 0 (x + 6)(x -8) = 0 x + 6 = 0 or x -8 = 0 x = -6 x = 8 + x2 U-shaped: f (x) < 0 when -6 < x < 8 or 2(a) x -1 < 7 ( 1) 49 x -2 < x2 -2x -48 < 0 (x -8)(x + 6) < 0 case I: (x -8) > 0 and (x + 6) < 0 x > 8 and x < -6 not possible case II: (x -8) < 0 and (x + 6) > 0 x < 8 and x > -6 \ -6 < x < 8 Blunders (-3) B1 upper limit. B2 lower limit. B3 expansion of ( )2 x -1 , once only. -6 8 Page 8 B4 inequality sign. B5 indices. B6 factors once only. B7 root formula, once only. B8 deduction root from factor. B9 incorrect range. B10 answer not stated. Slips (-1) S1 numerical. Attempts A1 one inequality only. A2 inequality signs ignored. Part (b) 20(5, 5, 5, 5) marks Att (2, 2, 2, 2) 2(b) The cubic equation 4x3 +10x2 -7x -3 = 0 has one integer root and two irrational roots. Express the irrational roots in simplest surd form. Test 5 marks Att 2 Linear Factor 5 marks Att 2 Other Factor 5 marks Att 2 Roots 5 marks Att 2 2(b) f (x) = 4x3 +10x2 -7x -3 Integral root must be ±1, ± 3 f (1): 4 +10 -7 -3 ¹ 0 f (-1): ¹ 0 f (3):108 + 90 -21-3 ¹ 0 f (-3): -108 + 90 + 21-3 = 0 x = -3 is a root (x + 3) is a factor 4 2 13 4 10 7 3 2 3 2 --+ + --x x x x x x 4x3 +12x2 -2x2 -7x -2x2 -6x -x -3 -x -3 So, need to solve: 4x2 -2x -1 = 0 2(4) = 2 ± 4 +16 x 8 = 2 ± 20 8 = 2 ± 2 5 4 = 1± 5 Irrational roots: 4 1+ 5 , 4 1-5 or Page 9 2(b) Finds root x = -3 as above, and continues as follows: x = -3 is a root (x + 3) is a factor \ other factor = (4x2 + ax -1) \ (x + 3)(4x2 + ax -1) = 4x3 +10x2 -7x -3 4x3 +12x2 + ax2 + 3ax -x -3 = 4x3 +10x2 -7x -3 4x3 + (a +12)x2 + (3a -1)x -3 = 4x3 +10x2 -7x -3 Equating coefficients: (i) a +12 = 10 and/or (ii) (3a -1) = -7 a = -2 3a = -6 a = -2 f (x) = (x + 3)(4x2 -2x -1)= 0 Irrational roots: 4 1± 5 , as above. Blunders (-3) B1 indices. B2 root formula, once only. B3 not like to like.. B4 deduction factor from root or no factor. Slips (-1) S1 numerical. S2 not changing sign in subtraction (Division). S3 roots not in simplest form, once only. Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 2(c)(i) Let ( ) mxx k f x 2 + 2 = , where k and m are constants and m ¹ 0 (i) Show that ( ) = mk f km f (ii) a and b are real numbers such that a ¹ 0, b ¹ 0 and a ¹ b. Show that if f (a) = f (b), then ab = k 2 . Page 10 (i) f (km) 5 marks Att 2 mk f 5 marks Att 2 (ii) f (a) = f (b) 5 marks Att 2 ab 5 marks Att 2 2(c) ( ) mxx k f x 2 + 2 = , [ k , m constants] (i) show that ( ) = mk f km f ( ) ( )( ) ( ) ( ) ( 1) 1 2 2 2 2 2 2 2 = + = + = m + mk k m k m m kmkm k f km ( ) ( ) m k k m k k k m k mk f mk m km k 2 2 2 2 2 2 2 2 2 = + + = + = ( ) ( 2 )2 1 2 k m= k + m ( 2 1) 2 = m + mk ( ) = mk f km f (ii) ( ) maa k f a 2 + 2 = ( ) mbb k f b 2 + 2 = f (a) = f (b) mbb k maa2 + k 2 = 2 + 2 multiply across by mab: b(a2 + k 2 )= a(b2 + k 2 ) a2b + bk 2 = ab2 + ak 2 a2b -ab2 = ak 2 -bh2 ab(a -b) = k 2 (a -b) (a -b) ¹ 0 ab = k 2 Blunders (-3) B1 indices Page 11 QUESTION 3 Part (a) 10 (5, 5)marks Att (2, 2) Part (b) 20 (5, 10, 5) marks Att (2, 3, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) A3 5 marks Att 2 A-1 5 marks Att 2 3(a) Given that -= 0 1 1 0 A , show that A3 = A-1 . Part (a) A3 5 marks Att 2 A-1 5 marks Att 2 3(a) -= 0 1 1 0 A ( ) A A = -= --= ----= 0 1 1 0 0 11 0 1 0 11 0 1 0 1 1 A3 = A.A.A = A2 .A I A = = --= 0 11 0 0 1 1 0 0 1 1 0 2 A3 = A2 .A = I.A = A = A-1 or A-1 = A as above, and: A3 = A.A.A = A-1.A.A = IA = A-1 or A-1 = A as above, and: 3 3 0 1 1 0 -A = ( ) ( ) ( ) ( ) -= 3 3 3 3 0 1 1 0 -= 0 1 1 0 = A Blunders (-3) B1 formula inverse. Slips (-1) S1 each incorrect element. S2 numerical. Page 12 Part (b) 20(5, 10, 5) marks Att (2, 3, 2) 3(b) Solve the quadratic equation: 2iz 2 + (6 + 2i)z + (3 -6i) = 0, where i 2 = -1 Part (b) Values in formula 5 marks Att 2 Evaluate b2 -4ac 10 marks Att 3 Roots 5 marks Att 2 3(b) Solve: 2iz 2 + (6 + 2i)z + (3 -6i) = 0 ( ) ( ) ( )( ) ( i) i i i i z 2 2 6 2 6 2 4 2 3 6 2 -+ ± + --= ( ) i i i i i i 4 = -6 + 2 ± 36 + 24 + 4 2 -24 + 48 2 ( ) i i 4 = -6 + 2 ± 36 -52 ( )i i4 = -6 + 2 ± -16 ( ) i i i 4 = -6 + 2 ± 4 i i i 4 = -6 -2 + 4 or i i i 4 -6 -2 -4 i i 4 = -6 + 2 or i i 4 -6 -6 i i 2= -3 + or i i 2 -3 -3 21 3 2 3 1 2 3 23 2 2 1 i i i i i ii i i z = + -= -+ × = -+ = --2 3 3 23 3 2 3 3 2 3 3 2 2 2 i i i i i ii i i z = -+ -= --× = --= -Blunders (-3) B1 indices. B2 i. B3 expansion ( )2 6 + 2i once only. B4 root formula, once only Slips (-1) S1 numerical. S2 i in denominator Attempts A1 3 marks for a + bi and stops A2 2 marks for z = a + bi and stops. Page 13 Part (c) 20(5, 5, 5, 5) marks Att (2, 2, 2, 2) 3(c) (i) z = cosq + i sinq . Use De Moivre’s theorem to show that nq z z n n 2cos + 1 = , for n Î N. (ii) Expand 4 1 + z z and hence express cos4q in terms of cos4q and cos2q . Part (c) (i) z n 1 5 marks Att 2 Value 5 marks Att 2 (ii) Expansion 5 marks Att 2 Express 5 marks Att 2 3(c)(i) z = cosq + i sinq z n = cos nq + i sin nq ( ) ( ) n n n i z i = + -+ = q q q q cos sin cos sin 1 1 = cos(-nq )+ i sin(-nq ) = cos nq -i sin nq ( nq i nq ) ( nq i nq ) z z n n cos sin cos sin + 1 = + + -= 2cos nq 3(c)(ii) ( ) ( ) 1 cos sin cos sin + 1 = q + q + q + i q -i z z = cosq + i sinq + cosq -i sinq = 2cosq 2 3 4 4 3 2 4 1 1 31 4 21 4 11 4 + + + + = + z z z z z z z z z z ( ) 2 4 4 4 2 1 1 2cos 4 6 4 z z z z + q = + + + 6 1 4 1 16cos 2 2 4 4 4 + + + = + z z z q z 16cos4q = (2cos 4q )+ 4[2cos 2q ]+ 6 16cos4q = 2cos 4q + 8cos 2q + 6 [2cos 4 8cos 2 6] 16 1 cos4q = q + q + [cos 4 4cos 2 3] 8= 1 q + q + Blunders (-3) B1 statement De Moivre, once only. B2 application De Moivre. B3 binomial expansion. B4 i B5 answer not in required form. B6 indices. Slips (-1) S1 numerical Worthless W1 not using De Moivre. W2 not using “ hence” in part (ii). Page 14 QUESTION 4 Part (a) 10 marks Att 3 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (15, 5) marks Att (5, 2) Part (a) 10 marks Att 3 4(a) Write the recurring decimal 0.636363… … as an infinite geometric series and hence as a fraction. Part (a) 10 marks Att 3 4(a) 0.63= 0.636363… … … … … = 0.63 + 0.0063 + 0.000063 + .......... ....... 1000000 63 10000 63 100 = 63 + + + \ 100 = 63 a 100 = 1 r r a S -¥ = 1 ( ) ( ) 11 7 99 63 1 100 99 100 63 100 1 100 63 = = = -= Blunders (-3) B1 indices. B2 formula for infinite series. B3 incorrect a . B4 incorrect r. B5 not as infinite series. Slips (-1) S1 numerical. Attempts A1 correct answer with no work or by other method (i.e. not using geometric series). Page 15 Part (b)(i) 10 (5, 5) Att (2, 2) 4(b)(i) The first three terms in the binomial expansion of (1+ kx)n are 1-21x +189x2 Find the value of n and the value of k. Part(b)(i) equations 5 marks Att 2 values 5 marks Att 2 4 (b)(i) ( ) ( ) ( ) ........ 1 2 1 1 2 + + + = + kx n kx n kx n ( ) ( ) ......... 2! 1 1 × 2 2 + = + + -x k n n nk x ( ) ( ) ............ 2 1 . 1 2 2 + -= + + x n n k nk x = 1-21x +189x2 ................................. [i]: nk = -21 [ii]: ( ) 189 2 1 . 2 n n -k = [i] n k = -21 [ii]n(n -1)k 2 = 378 sub. in: ( ) 378 21 1 2 = --n n n (n2 -n)(441) = 378n2 441n2 -441n -378n2 = 0 63n2 -441n = 0 63n(n -7) = 0 n = 0 or n = 7 \ 3 721 7 = -= k = -n n = 7; k = -3 * Since must be integers, accept correct values by observation from nk = -21, with verification. Blunders (-3) B1 errors in binomial expansion, once only. B2 rn B3 indices. B4 not like to like B5 factors B6 value from factor. B7 second value not found, having found first. Slips (-1) S1 numerical. Page 16 Part (b) (ii) (5, 5) marks Att (2, 2) 4 (b) (ii) A sequence is defined by = (2 -)2n-1. un n Show that un+2 -4un+1 + 4un = 0, for all n Î N. Part (b) (ii) Terms simplified 5 marks Att 2 Show 5 marks Att 2 4 (b)(ii) = (2 -)2n-1 un n [ ( )] (n ) ( ) n n u 2 n 1 2 1 1 1 n 2 1 = -+ + -= -+ [ ( )] ( 2) 1 ( ) 1 2 2 2 2 + -2 + + = -+ n = -n n u n n 4 4 n 2 n 1 n u -u + u + + = (-n.2n+1 )-4[(1-n)2n ]+ 4[(2 -n)2n-1 ] = -n.2n+1 -(22 )(2n )(1-n)+ 22 (2n-1 )(2 -n) = -n.2n+1 -2n+2 + 2n.2n+1 + 2.2n+1 -n.2n+1 = 2.2n+1 -2n+2 = 2.2n+1 -2.2n+1 = 0 or 4 (b)(ii) n n n n ( n ) n n u n n 2 2 = (2 -)2 -1 = 2 -.2 -1 = 2 -[ ( )] (n ) ( ) n n u 2 n 1 2 1 1 1 n 2 1 = -+ + -= -+ [ ( )] (n ) ( ) n n n u 2 n 2 2 2 1 n 2 1 2n.2 2 = -+ + -= -+ = -+ Let a = 2n \ ( ) [ ] 2 4 1 4 2 4 1 4 n2a un+ -un+ + un = -na --n a + a -= -2na -4a + 4na + 4a -2na = 0 Blunders (-3) B1 indices. B2 factors. Slips (-1) S1 numerical. Attempts A1 must do some correct relevant work with indices in “ show” . Page 17 Part (c)(i) 20(15, 5) marks Att (5, 2) 4 (c) (i) Show that 2 2 a b a2 + b2 + £ , where a and b are real numbers. (ii) The lengths of the sides of a right-angled triangle are a, b and c, where c is the length of the hypotenuse. Using the result from part (i), or otherwise, show that a + b £ c 2 . Part (c)(i) 15 marks Att 5 Part (c) (ii) 5 marks Att 2 4(c)(i) 2 2 a b a2 + b2 + £ case: (a + b) positive: 2 2 2 2 2 a b £ a + b Û+ 4 2 a2 + 2ab + b2 £ a2 + b2 a2 + 2ab + b2 £ 2a2 + 2b2 0 £ a2 -2ab + b2 ( )2 0 £ a -b True when (a + b) positive. case: (a + b) negative: (a + b) < 0 ( ) 0 2 + < a b 2 2 a b a2 + b2 + £ , since x > 0 always. True when (a + b) negative. or 4(c)(i) ( ) 0 a -b 2 ³ for all a,bÎR . a2 -2ab + b2 ³ 0 (a2 -2ab + b2 )+ (a2 + b2 )³ (a2 + b2 ) 2a2 + 2b2 ³ a2 + 2ab + b2 2(a2 + b2 )³ (a + b)2 divide across by 4: ( ) 2 4 a2 b2 a + b 2 + ³ 2 2 2 2 2 a + b ³ a + b 2 2 2 a2 b2 a b 2 a + b ³ + ³ + Page 18 4(c)(ii) From (i) above, 2 2 a b a2 + b2 + £ 2c2 = 2 c = . 2 2 a b c + £ 2 a + b £ 2c a + b £ c 2 Blunders (-3) B1 indices B2 inequality sign. B3 deduction. B4 a and b both positive. B5 expansion ( )2 a -b . B6 right angled triangle. Slips (-1) S1 numerical. Worthless W1 particular values for a and b. a2 + b2 = c2 a b c Page 19 QUESTION 5 Part (a) 15 (5, 5, 5) marks Att (2, 2, 2) Part (b) 15 (5, 5, 5) marks Att (2, 2, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 15(5, 5, 5) marks Att (2, 2, 2) 5(a) Solve for x: 10 -x = 4 -x Part 5(a) Quadratic 5 marks Att 2 Factors 5 marks Att 2 Solution 5 marks Att 2 5(a) 10 -x = 4 -x ( )2 10 -x = 4 -x 10 -x = 16 -8x + x2 0 = x2 -7x + 6 0 = (x -1)(x -6) x -1 = 0 or x -6 = 0 x = 1 x = 6 Test : x = 1 L.H.S.: 10 -x = 10 -1 = 9 = 3 R.H.S.: 4 -x = 4 -1 = 3 \ L.H.S.=R.H.S. x = 6 L.H.S.: 10 -x = 10 -6 = 4 = 2 R.H.S.: 4 -x = 4 -6 = -2 \ L.H.S.¹ R.H.S. \ 6 is not a solution. Ans: x = 1. Blunders (-3) B1 indices. B2 expansion ( )2 4 -x once only. B3 factors. B4 root formula once only. B5 deduction values from factors. Slips (-1) S1 numerical. S2 excess value. Attempts A1 x = 1 and no other works merits 2 marks. Page 20 Part (b) 15(5, 5, 5) marks Att (2, 2, 2) 5(b) Prove by induction that ( ) ( ) =-= -n r n n r 1 3 1 2 3 2 Part (b) P(1) 5 marks Att 2 Assume 5 marks Att 2 P(k + 1) 5 marks Att 2 5(b) ( ) ( ) =-= -n r n n r 1 3 1 2 3 2 Test 1: 3(1) 2 1 1 n = u = -= ( ) ( ) (2) 1 21 3 1 21 3 1 2 n -= -= = n \ True for n = 1 P(1) Assume true for n = k (3 1) 2 = k -k Sk P(k ) To prove: ( )[3( 1) 1] 2 1 1 = + + -+ k k Sk [3 2] 2 = +1 k + k ( 1)(3 2) 2= 1 + k + k Proof: +1 +1 = + k k k S S U (3 1) [3( 1) 2] 2 = k -+ k + -k (3 1) (3 1) 2 = k -+ k + k 2 = 3 2 -+ 6 + 2 k k k [3 5 2] 2= 1 2 + k + k [( 1)(3 2)] 2= 1 + k + k P(k +1) So, P(k +1) true whenever P(k ) true. Since P(1) true, then by induction P(n) true for all positive integers n (nÎN, n ³ 1). Blunders (-3) B1 indices. B2 n ¹ 1(must prove n = 1 not enough to say true for n = 1) B3 factors. Slips (-1) S1 numerical. Page 21 Part (c) 20(5, 5, 5, 5)marks Att (2, 2, 2, 2) 5(c) (i) Show that log , log1 a b b a = where a, b > 0 and a,b ¹ 1 (ii) Show that + + + 2c 3c log4c 1 log1 log1 , log1 log1 ........ rc r!c + = where c > 0, c ¹ 1. Part (c) (i) 5 marks Att 2 (ii) logxc to a new base 5 marks Att 2 log(2.3.4…r) 5 marks Att 2 completion 5 marks Att 2 5(c)(i) b b a a a a a b log1 log log log = = 5(c)(ii) From (i): c c log2 1 log 2 = Similarly c r c r c c log1 , , log log1 log 3 3 = = \ + + + 2c 3c log4c 1 log1 log1 log1 ........ c r + r c c c c = log 2 + log 3 + log 4 + ....... + log ( r ) c = log 2.3.4.............. log (r!) c = c r! log= 1 . or 5(c)(ii) = log 2 log log !! 2 rr c c c c r r log log 2 log1 ! 2 = Similarly, c r c r log log 3 log1 ! 3 = , etc. \ + + + 2c 3c log4c 1 log1 log1 log1 ........ c r + = + c rr!! log log 2 + c rr!! log log 3 ........... log log 4 !! + c rr cr rr!! log log + c r r r r r r ! ! ! ! ! log log 2 + log 3 + log 4 + ...........log = ( ) c r r r ! ! log log 2.3.4......... = ( ) c r r r ! ! log log ! = c r! log= 1 or Page 22 5(c)(ii) log 2 log log 10 10 2 c c = , log 3 log log 10 10 3 c c = , etc. \ + + + 2c 3c log4c 1 log1 log1 log1 ........ c r + = + c 10 10 log log 2 + c 10 10 log log 3 ........... log log 4 10 10 + c cr 10 10 log log + c r 10 10 10 10 10 log log 2 + log 3 + log 4 + ...........log = ( ) c r 10 10 log log 2.3.4......... = ( ) (c) r 10 10 log log ! = log r! c = c r! log= 1 Blunders (-3) B1 log laws. B2 factorial. B3 change of base. Worthless W1 no change of base. Page 23 QUESTION 6 Part (a) 10 (5, 5)marks -Part (b) 20 marks -Part (c) 20 (10, 5, 5) marks -Note: The marking of Question 6 is not based on slips, blunders and attempts. In the case of each part, descriptions or typical examples of work meriting particular numbers of marks are given. The mark awarded must be one of the marks indicated. For example, in part (a)(i), descriptions are given for work meriting 0, 2 or 5 marks. It is therefore not permissible to award, 1, 3 or 4 marks for this part. Part (a) 10 (5, 5) marks -6 (a) Differentiate with respect to x (i) (1+ 7x)3 (ii) . 5 sin 1 -x Part (a) (i) 5 marks -6(a)(i) 3(1 7 ) .(7) 21(1 7 ) . 2 2 x x dx dy = + = + 5 marks: correct derivative in any form. (e.g. middle step above is acceptable, as is expansion followed by correct differentiation, unsimplified). 2 marks: differentiates with one or more errors, provided at least some aspect correct. 0 marks: no correct differentiation done. (e.g. integrates or expands the given expression). Part (a) (ii) 5 marks -6(a)(ii) sin ( ) sin 1 ( ) 5 5 y = -1 = -a = a xx 2 2 25 2 1 1 dx a x x dy -= -= or 6(a)(ii) y (x ) 1[f (x)] 5 = sin-1 = sin-( ) ( ) ( ) -¢ = -= 51 . 1 1 . 1 1 2 5 2 x f x dx f x dy 25 25 2 5 1 -x = 25 2 1 -x = or 6(a)(ii) ( ) 5 y = sin-1 x 5 sin y = x \ 5 1 cos = dx y dy \ 51 cos 1 . dx y dy = 5 1 5 25 2 × = -x 25 2 1 -x = 5 25 cos 5 sin x2 y x y = = -5 x 25 -x2 y Page 24 5 marks: correct derivative in terms of x, simplified or otherwise. 2 marks: differentiates with at least some aspect correct; fails to give answer in terms of x. 0 marks: no correct differentiation done. (e.g. integrates or rearranges the given expression, or gives only the first step in the second method above) Part (b) 20 marks -6 (b) . 1 cos 1 cos Let xx y += -Show that t t 3 dx dy = + , where . 2 tan x t = Part (b) 20 marks -6(b)(ii) vu xx y = += -1 cos 1 cos ( )( ) ( )( ) ( )2 1 cos 1 cos sin 1 cos sin x x x x x dx dy + = + ---( )2 1 cos sin sin cos sin sin cos x x x x x x x + = + + -( )2 1 cos 2sin x x + = ( ) ( )2 2 2 2 2 2cos 2 2sin cos x x x = 2 4 2 2 4cos4sin cosx x x = 2 2 22 cos1 cos sin x x x = × ( ) 2 2 2 = tan x sec x ( ) 2 2 2 = tan x 1+ tan x = t(1+ t 2 ) = t + t 3 or 6(b)(ii) ( ) 2 1 tan 1 tan 1 tan 2 tan 2 2 2 2 22 22 12 1 cos 2sin + = + = +-+ xx x x x x dx dy [( ) ( )]2 1 1 1 142 2 2 2 t t t tt + + + -= + ( )[ ]2 1 2 2 2 1 4 t t t + + = = t(1+ t 2 ) = t + t 3 or Page 25 6(b)(ii) ( ) [ ]2 2 2 1 tan 2 tan 2 2cos 2 1 cos 2sin 2 22xx x x x dx dy = + = + ( ) 2 2 2 2 2 2 2 1 tan .4cos .cos 4 tan x x x x + = 2 2 2 2 sec1 sec1 2 sec2 .4 . 4 x x x t = ( ) 2 = t. sec2 x ( ) 2 = t 1+ tan2 x = t(1+ t 2 ) = t + t 3 or 6(b)(ii) xx y 1 cos 1 cos += -2222 1111 11 tttt +-+-+-= ( ) ( ) ( 2 ) ( 2 ) 2 2 1 11 1 t tt t + + -= + --2 2 2 2 t t y = = y = t 2 ( )2 2 y = tan x or dx dt t dx dy = 2 ( ) ( ) 21 2 1 2 2 2 tan x . sec x . dx dy = = 2 sec 21 2 2 x t ( )( ) 2 2 2 tan 1 tan x x + = = + 2 1 tan 21 2 2 x t = t(1+ t 2 ) = t(1+ t 2 ) = t + t 3 = t + t 3 20 marks: fully correct solution. 17 marks: correct expression for dx dy in terms of t alone, but not simplified to required form or solution with one or two non-critical errors, simplified fully. [critical error = one that significantly alters the nature or complexity of the task]. 14 marks: correct expression for dx dy in terms of x , simplified or correctly establishes that y = t 2 or that (1 2 ) 21 t dx dt = + 7 marks: correct or almost-correct expression for dx dy in terms of x or correct expression for dx dt in terms of x or correct but unsimplified expression for y in terms of t or tan 2x 0 marks: no relevant work. Page 26 Part (c) 20 (10, 5, 5) marks -Part (c) (i) 10 marks -6 (c) The equation of a curve is , -1 = x x y where x ¹ 1. (i) Show that the curve has no local maximum or local minimum point. 6 (c) (i) -1 = x x y ( )( ) ( )( ) ( )2 1 1 1 1 -= --x x x dx dy ( )2 1 1 -= --x x x ( ) 0 11 2 ¹ -= -x \No local max/local min 10 marks: Correct solution, including assertion that derivative ¹ 0 or <0 or similar conclusion. 7 marks: Correct derivative. 3 marks: Substantial error(s) in differentiation. 0 marks: No relevant work Part (c) (ii) 5 marks -6 (c) (ii) Write down the equations of the asymptotes and hence sketch the curve. 6 (c) (ii) Vertical asymptote: x–1 = 0 x = 1 Horizontal asymptote: ® ® ±¥ -= -= x x x y x 1 as 1 1 1 1 y = 1 5 marks: Correct solution, (equations of both asymptotes, and sketch). 2 marks: One or two equations correct, or sketch of correct form. 0 marks: No significant work of merit. x-axis y-axis x = 1 b(2 -x,2 -y) p(1, 1) a(x, y) y = 1 Page 27 Part (c) (iii) 5 marks -6 (c) (iii) Show that the curve is its own image under the symmetry in the point of intersection of the asymptotes. 6 (c) (iii) S (a) b p = p: point of intersection of asymptotes = (1, 1). a = (x, y) b = (2 -x,2 -y) Test to see if b(2 -x,2 -y) is on curve -1 = x x y : ( ) ( ) (2 ) 1 2 2 ---= -x x y xx y --= -12 2 y xx = ---12 2 ( ) ( ) x x x x x y -= --Û = ---1 1 2 1 2 -1 Û = x x y (i.e. b is on the curve if and only if a is.) or 6 (c) (iii) p(1, 1): point of intersection of asymptotes -1 , x x a x is on curve -1 = x x y ( ) [ ] 1 2 ,2 -= --xx p S a b b x ( ( ) ) 1 2 , 2 1 ----x b x x x ( ) 12 , 2 ---xb x x Symmetry if ( ) 1 1 2 , 2 ----Î = xx xb x x y : ( ) ( ) 12 12 2 1 2 -= --= ----xx xx x x . 5 marks: Fully correct solution. 2 marks: Correctly finds image of general point on the curve, or Identifies general point on the curve in terms of one variable, or Fully or partially works a particular case, or Identifies (1, 1) as the point of intersection of the asymptotes. 0 marks: no relevant work. Page 28 QUESTION 7 Part (a) 10 marks Att 3 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 marks Att 3 7 (a) Find from first principles the derivative of x2 with respect to x. Part (a) 10 marks Att 3 7(a) f (x) = x2 ( ) ( )2 f x + h = x + h f (x + h)-f (x) = (x2 + 2hx + h2 )-x2 f (x + h)-f (x) = 2hx + h2 ( ) ( ) x h h f x h f x + -= + 2 ( ) ( ) x h f x h f x h lim 2 0 + -= ® or 7(a) y = x2 ( )2 y + Dy = x + Dx ( )2 2 Dy = x + Dx -x = x2 + 2xDx + Dx2 -x2 = 2x.Dx + Dx2 x x xy = + D DD 2 0 limDx® x xy = 2 DD Blunders (-3) B1 expansion of ( )2 a + b once only. B2 indices. B3 no limit shown or implied, or no indication ®0 . B4 x.Dx = Dx2 Worthless W1 not from 1st principles. Page 29 Part (b) (i) 10 (5, 5) marks Att (2, 2) 7 (b) (i) The parametric equations of a curve are: x = 8 + lnt 2 y = ln(2 + t 2 ), where t > 0. Find dx dy in terms of t and calculate its value at t = 2 . Part (b)(i) dt dy dt dx , 5 marks Att 2 value 5 marks Att 2 7 (b) (i) x = 8 + lnt 2 y = ln(2 + t 2 ), x = 8 + 2ln t t dt t dy .2 2 1 + 2 = dt t t dx 1 2 2 = = 2 2 2 t t + = ( ) ( ) ( ) ( ) ttt dt dx dt dy dx dy 2 22 = = + 2 2 2 2 t t+ = At t = 2 : t 2 = 2 21 2 2 2 2 2 2 = + = + = t t dx dy * f ¢(x) must be expressed as a function of t for second 5 marks. Blunders (-3) B1 differentiation. B2 logs. B3 indices B4 definition of dx dy B5 incorrect value or no value. Attempts A1 error in differentiation formula. Worthless W1 integration. W2 no differentiation. Page 30 Part (b) (ii) 10 (5, 5) marks Att (2, 2) 7 (b) (ii) Find the slope of the tangent to the curve xy2 + y = 6 at the point (1, 2). Part (b)(ii) Differentiation 5 marks Att 2 Slope 5 marks Att 2 7 (b) (ii) xy2 + y = 6 0 2 . 2 = + + dx dy y dx dy x y (2xy 1) y2 dx dy + = -2 1 2 + = -xyy dx dy At p(1, 2) x = 1 and y = 2 ( ) ( )( ) 54 2 1 2 1 2 2 -= + = = -dx dy m Blunders (-3) B1 differentiation. B2 indices. B3 incorrect value of x or no value of x. B4 incorrect value of y or no value of y. Slips (-1) S1 numerical. Attempts A1 error in differentiation formula. A2 dx dy y dx dy xy dx dy + + = 2 2 and uses all three dx dy terms. Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 7 (c) (i) Write down a quadratic equation whose roots are ± k . (ii) Hence use the Newton-Raphson method to show that the rule ( ) n n n u u k u 22 1 + = + can be used to find increasingly accurate approximations for k . (iii) Using the above rule and taking 23 as the first approximation for 3, find the third approximation, as a fraction. Page 31 Part (c) (i) 5 marks Att 2 (ii) Newton-Raphson 5 marks Att 2 Finish 5 marks Att 2 (iii) 5 marks Att 2 7(c) (i) Roots ± k Equation: x2 -k = 0 . 7(c)(ii) Equation: x2 = k or x2 -k = 0 , so let f (x) = x2 -k . \ f (u ) u k n n = 2 -( ) n n f ¢ u = 2u Newton-Raphson: ( ) ( ) n n n n f u f u u u ¢ = -+1 n n n u u k u 22 -= -( ) nn n u u u k 2 2 2 -2 -= n n n u u k u 22 1 + = + Hence the given rule is the Newton-Raphson method applied to f (x) = x2 -k . Thus it can be used with a suitable initial value to find increasingly accurate approximations for k . 7(c)(iii) 1 2 1 2 2u u k u = + 3 k = ; 23 1 u = ( )( ) 47 12 21 3 3 2 3 4 9 2 3 2 2 3 2 = = + = + u = ( ) ( )( ) ( ) ( ) 27 16 97 27 16 49 47 2 47 2 2 2 3 3 2 3 2 = + = + = + = u u k u 56 = 97 Blunders (-3) B1 equation B2 Newton-Raphson formula; apply once only to second 5 marks in (ii) or to 5 marks in (iii). B3 differentiation. B4 indices. B5 k ¹ 3. B6 23 1 U ¹ , once only B7 3 U not found. Slips (-1) S1 numerical. S2 not as fraction. Misreadings (-1) M1 takes “ above rule” in c(iii) to mean “Newton-Raphson method” and uses this in (iii). Page 32 QUESTION 8 Part (a) 10 (5, 5) marks Att (2, 2) Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 20 (10, 10) marks Att (3, 3) Part (a) 10 (5, 5) marks Att (2, 2) 8 (a) Find (i) (2 + x3 )dx (ii) e3x dx. Part (i) 5 marks Att 2 (ii) 5 marks Att 2 8 (a) (i) c x + x dx = x + + 4 (2 ) 2 4 3 (ii) = + c e e dx x x 33 3 * If c shown once, then no penalty Blunders (-3) B1 integration. B2 no ‘c’ (Penalise 1st integration) B3 indices. Attempts A1 anything + c. Worthless W1 differentiation instead of integration. Part (b) 20 (10, 10) marks Att (3, 3) 8 (b) (i) Evaluate + + + 4 1 2 1 2 1 dx x x x . (ii) Evaluate sin 2 . 8 0 2 q q p d Part (b) (i) 10 marks Att 3 (ii) 10 marks Att 3 8(b)(i) + + + 4 1 2 1 2 1 dx x x x ( ) ( ) + + = + 1 2 1 x2 x x dx = u du = lnu ln( 1) ] ln(16 4 1) ln(1 1 1) 41 = x2 + x + = + + -+ + 3 21 = ln = ln 7 Let u = x2 + x +1 = 2x +1 dx du du = (2x +1)dx Page 33 8(b)(ii) [ ]8 4 0 sin 4 8 0 2 21 sin 2 . p q p q dq = q -[Tables page 42] ( )--= -0 0 4 sin 2 8 1 8 p 4p = -41 2 8 1 p 81 16 = p -or 8(b)(ii) q q p sin 2 d 8 0 2 = (1-cos 4q )dq 21 8 4 0 sin 4 21 p q q = -( )--= -0 0 4 sin 2 8 1 8 p 4p = -41 2 8 1 p 81 16 = p -Blunders (-3) B1 integration. B2 indices. B3 limits. B4 no limits. B5 incorrect order in applying limits. B6 not calculating substituted limits. B7 not changing limits. B8 differentiation. B9 trig formula. Slips (-1) S1 numerical. S2 trig value. Worthless W1 differentiation instead of integration except where other work merits attempt. Note: Incorrect substitution and unable to finish yields attempt at most. Note: (-3) is maximum deduction when evaluating limits Note: In 8(b)(ii), do not penalise = 11.25° 16 p , etc. q (1 cos2q ) 21 sin2 = -q (1 cos 4q ) 21 sin2 2 = -Page 34 Part (c)(i) 10 marks Att 3 8 (c) (i) Evaluate + -2 1 3 2 2 1 dx x x . Part (c)(i) 10 marks Att 3 8 (c) (i) + -2 1 3 2 2 1 dx x x ( ) --= 3 + 2 -2 22 x 1 2 dx x x dx 22 -u2 du [Let u = x -1 = 1 dx du du = dx ] = -2 sin 1 u ( ) ]22 1 = sin-1 x-1 sin (0) 21 sin 1 1 ---= 0 6 = p -6 = p or 8 (c) (i) + -2 1 3 2 2 1 dx x x ( ) --= 3 + 2 -2 22 1 x 2 dx x x dx --22 w2 dw [Let w = 1-x = -1 dx dw -dw = dx ] = --2 sin 1 w ( )]22 1 = -sin-1 1-x ( )-= ----0 sin 21 sin 1 1 = ---0 6p 6 = p Blunders (-3) B1 integration B2 completing square once only. B3 limits B4 no limits B5 incorrect order in applying limits B6 not calculating substituted limits B7 not changing limits. B8 differentiation. Slips (-1) S1 numerical S2 trig value. 3 + 2x -x2 = 4 -(x2 -2x +1) ( )2 ( )2 = 2 -x -1 3 + 2x -x2 = 4 -(1-2x + x2 ) = (2)2 -(1-x)2 Page 35 Worthless: W1 no effort at completing square W2 differentiation instead of integration except where other work merits attempt. W3 puts u = 3 + 2x -x2 Note: Incorrect substitution and unable to finish yields attempt at most. Note: (-3) is maximum deduction when evaluating limits Part (c) (ii) 10 marks Att 3 8 (c) (ii) Use integration methods to derive a formula for the volume of a cone. Part (c)(ii) 10 marks Att 3 8 (c) (ii) Vol of cone, with height = h , and base-radius = r Equation op: slope hr = ; through (0, 0) (x) hr y = = h V y dx 0 2 p = h dx h rx 0 2 p = h x dx hr 0 2 2p 2 [ ]h x hr 0 3 22 3= p [ ] 0 3 3 22 = h -hpr r 2h 3= 1p Blunders (-3) B1 integration B2 slope of line. B3 equation of line. B4 volume formula provided it is quadratic B5 limits B6 no limits. B7 incorrect order in applying limits. B8 indices. Slips (-1) S1 numerical Attempts A1 uses v =py Worthless W1 differentiation instead of integration. Note: (-3) is maximum deduction when evaluating limits. y dx h r p(h ,r) y = h rx Page 36 MARKING SCHEME LEAVING CERTIFICATE EXAMINATION 2005 MATHEMATICS – HIGHER LEVEL – PAPER 2 GENERAL GUIDELINES FOR EXAMINERS – PAPER 2 1. Penalties of three types are applied to candidates’ work as follows: · Blunders -mathematical errors/omissions (-3) · Slips -numerical errors (-1) · Misreadings (provided task is not oversimplified) (-1). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B1, B2, B3,… , S1, S2,… , M1, M2,… etc. These lists are not exhaustive. 2. When awarding attempt marks, e.g. Att(3), note that · any correct, relevant step in a part of a question merits at least the attempt mark for that part · if deductions result in a mark which is lower than the attempt mark, then the attempt mark must be awarded · a mark between zero and the attempt mark is never awarded. 3. Worthless work is awarded zero marks. Some examples of such work are listed in the scheme and they are labelled as W1, W2,… etc. 4. The phrase “ hit or miss” means that partial marks are not awarded – the candidate receives all of the relevant marks or none. 5. The phrase “ and stops” means that no more work is shown by the candidate. 6. Special notes relating to the marking of a particular part of a question are indicated by an asterisk. These notes immediately follow the box containing the relevant solution. 7. The sample solutions for each question are not intended to be exhaustive lists – there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising examiner. 8. Unless otherwise indicated in the scheme, accept the best of two or more attempts – even when attempts have been cancelled. 9. The same error in the same section of a question is penalised once only. 10. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most. 11. A serious blunder, omission or misreading results in the attempt mark at most. 12. Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50. Page 37 QUESTION 1 Part (a) 15 marks Att 5 Part (b) 20 (5, 5, 10) marks Att (2, 2, 3) Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (a) 15 marks Att 5 1 (a) Circles S and K touch externally. Circle S has centre (8, 5) and radius 6. Circle K has centre (2, -3). Calculate the radius of K. Radius of K 15 marks Att 5 1 (a) (8 2) (5 3) 36 64 10. (8, 5) and (2, 3). = -2 + + 2 = + = -ab a b But r + 6 = 10. \ r(radius K) = 4. Blunders (-3) B1 Error in distance formula. Slips (-1) S1 Arithmetic error. Attempts ( 5 marks) A1 Distance between centres. A2 Correct condition for circles touching externally. Part (b) 20 (5, 5, 10) marks Att (2, 2, 3) Part (b) (i) 10 marks (5, 5) Att (2, 2) 1(b) (i) Prove that the equation of the tangent to the circle x2 + y2 = r 2 at the point ( ) 1 1 x , y is 2 . 1 1 xx + yy = r Slope of tangent 5 marks Att 2 Finish 5 marks Att 2 1(b) (i) Equation of tangent T : ( ). 1 1 y -y = m x -x Slope of normal op . 00 11 11 xy xy = --= \Slope of T at point p ( ) 1 1 x , y . 11 yx = -Equation of T : ( ) 2 1 1 2 1 1 1 11 1 x x yy y xx x yx y y --= -+ --= . 2 1 2 1 1 1 xx + yy = x + y But ( , ) 2 2 . 1 2 1 2 2 2 1 1 x y Î x + y = r x + y = r \Equation of tangent T : 2 . 1 1 xx + yy = r or K S x2 + y2 = r 2 o (0, 0) p ( , ) 1 1 x y 900 T K S Page 38 1(b) (i) 2 2 2 2 2 0 . yx dx dy dx dy x y r x y = -+ = + = Slope of tangent T at point p ( ) 1 1 x , y . 11 y-x = Equation of T : ( ) 2 1 1 2 1 1 1 11 1 x x yy y xx x yx y y --= -+ --= . 2 1 2 1 1 1 xx + yy = x + y But ( , ) 2 2 . 1 2 1 2 2 2 1 1 x y Î x + y = r x + y = r \Equation of tangent T : 2 . 1 1 xx + yy = r Blunders (-3) B1 Incorrect sign in slope formula. B2 Slope formula inverted. B3 Incorrect perpendicular slope. B4 Error in differentiation. B5 Fails to show that 2 2. 1 2 x1 + y = r Slips (-1) S1 Arithmetic error. Attempts ( 2, 2 marks) A1 Correct slope of normal. A2 Correct differentiation. A3 Correct substitution into tangent formula and stops. A4 Stops at . 2 1 2 1 1 1 xx + yy = x + y Part (b) (ii) 10 marks Att 3 1 (b) (ii) Hence, or otherwise, find the two values of b such that the line5x + by = 169 is a tangent to the circle x2 + y2 = 169. Values of b 10 marks Att 3 1 (b) (ii) By part (i) the line 5x + by = 169 is a tangent to the circle x2 + y2 = 169 at the point (5, b). But (5, b) Î x2 + y2 = 169 25 + b2 = 169. b2 = 144 b = ±12. or 1 (b) (ii) Perpendicular distance from centre of circle to tangent 5x + by = 169 equals radius. 25 13 25 169 144. 12. 13 169 13 25 25 5(0) (0) 169 2 2 2 2 2 + = + = = \ = ± = -= + + + -b b b b b b b Page 39 Blunders (-3) B1 Error in solving for b other than slip. B2 Only one correct value of b given. B3 Incorrect radius. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 (5, b) point of tangency. A2 Perpendicular distance formula with substitution. Part (c) 15 marks (5, 5, 5) Att (2, 2, 2) 1 (c) A circle passes through the points (7, 2) and (7, 10). The line x = -1 is a tangent to the circle. Find the equation of the circle. Two equations in g, f and c 5 marks Att 2 Value of f 5 marks Att 2 Finish 5 marks Att 2 1 (c) Circle: x2 + y2 + 2gx + 2 fy + c = 0. 16 96 6. (7 10) 49 100 14 20 0 14 20 149 (7 2) 49 4 14 4 0 14 4 53 \ = -= -Î + + + + = + + = -Î + + + + = + + = -f f , C g f c g f c , C g f c g f c ( ) Circle : 8 12 27 0.16 64. 4 and 27. But 14 4 53 14 29. But 2 35. 36 2 1 36 2 35. 1 1 Perpendicular distance from , to 1 0 equals radius. 1 0 is a tangent. 2 2 2 2 2 \ + --+ = = -\ = -= + + = -+ = --= -\ -+ = + --+ = + --= -\ --+ = + = x y x y g g c g f c g c g c g c g g g c g c g g f x x or y value of centre 5 marks Att 2 ‘Quadratic’ in x 5 marks Att 2 Finish 5 marks Att 2 1 (c) a(7, 2) and b(7,10). \Mid-point of [ab] is (7, 6). Equation of mediator of chord [ab] is y = 6. Centre point of circle is c(x, 6). As x = -1 is a tangent then point of tangency is d(-1, 6). ( ) ( ) ( ) Equation of circle is ( 4) ( 6) 25. Centre is 4 6 and radius 5 2 1 14 49 16 16 64 4. 1 7 16.2 2 2 2 2 2 2 2 \ -+ -= \ = \ + + = -+ + = = = + = -+ x y , x x x x x x cd ca x x Page 40 Blunders (-3) B1 Error in mid-point formula. B2 Error in perpendicular distance formula. B3 Error in radius formula. B4 Circle equation formula error. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2, 2 marks) A1 One equation in f, g and c. A2 Mid-point of [ab]. A3 Attempt at solving simultaneous equations. A4 ca, cbor cdfound. A5 Distance from centre to tangent with substitution. A6 Attempt at solving quadratic for x. A7 Value of third unknown. A8 Length of radius. Page 41 QUESTION 2 Part (a) 5 marks -Part (b) 25 (10, 5, 10) marks -Part (c) 20 (15, 5) marks -Note: The marking of Question 2 is not based on slips, blunders and attempts. In the case of each part, descriptions or typical examples of work meriting particular numbers of marks are given. The mark awarded must be one of the marks indicated. For example, in part (a) (i), descriptions are given for work meriting 2, 4 or 5 marks. It is therefore not permissible to award, 1 or 3 marks for this part. Part (a) 5 marks -2 (a) Copy the parallelogram oabc into your answerbook. Showing your work, construct the point d such that , 21 2 ® 1 ® ® ® d = a + b -c where o is the origin. Point d 5 marks -2 (a) * Accept any labelled parallelogram with vertices o, a, b, c. 5 marks: point d shown in correct position in diagram. Point d need not be joined to origin. 4 marks: Correct work with one error or omission e.g. + ® ®a b 21 or ® ® a -c 21 or ® ® b -c 21 correctly on diagram. 2 marks: One correct significant step e.g. ® ® ® a b or -c 21 or 21 or + ® ®a b correctly shown on diagram. 0 marks: No significant work of merit. Part (b) 25 (10, 5, 10) marks -Part (b) (i) 15 (10, 5) marks -2 (b) (i) 3 4 . ® ® ® p = i + j ®qis the unit vector in the direction of ®p. (i) Express ® ®^ q and q in terms of and . ® ®i j ®qExpress 10 marks -2 (b)(i) . 54 53 9 16 3 4 ® ® ® ® ®® ® = + + = = + j i i j pp q c b o a c b o a ® ® a + b 21 21 d Page 42 10 marks: Correct solution for®q, simplified or otherwise. 7 marks: Calculates ®p correctly but does not give unit vector or writes ®®pp and stops or divides ® ® 3 i + 4 j by any number. 3 marks: Unit vector expressed as . a2 b2a i b j + + ® ® 0 marks: No significant work of merit. ^ ®qExpress 5 marks -2 (b) (i) ®^ ® ® q = -i + j 53 54 , or equivalent from candidates . ®q 5 marks: Fully correct answer. 2 marks: Gives as solution, 53 5®^ 4 ® ® q = i -j or equivalent from candidates . ®q 0 marks: Any other answer. Part (b) (ii) 10 marks -2 (b) (ii) Express ® ® 11 i -2 j in the form ® ®^ k q + l q , where k, l Î R. Express 10 marks -2 (b) (ii) ® ®^ k q + l q 11 2 . ® ® = i -j 11 2 . 53 54 54 53 11 2 . 53 54 54 53 ® ® ® ® ® ® ® ® ® ® -= + + --= + -+ + k l i k l j i j k i j l i j i j \ 3k -4l = 55 and 4k + 3l = -10. 16 12 40 9 12 165 + = --= k l k l 25k = 125 \k = 5. But 3(5)-4l = 55 l = -10. 11 2 5 10 . ® ® ® ®^ \ i -j = q-q 10 marks: Correct k and l found. 7 marks: Solves for k and/or for l with minor error(s). 3 marks: One equation in k and l allowing for minor error(s). 0 marks: No significant work of merit. Page 43 Part (c) 20 marks (15, 5) -Part (c) (i) 15 marks -2 (c) 5 and 4 4 . ® ® ® ® ® ® u = i + j v = i + j (i) Find cosÐuov , where o is the origin. 2 (c) (i) . 13 3 8 13 24 26 32 4 20 5 . 4 4 5 4 4 cos = = = + + + + + Ð = ® ® ® ® ® ® ® ®i j i j i j i j uov 15 marks: cosÐuov expressed as fraction of real numbers, simplified or otherwise. 10 marks: Correctly evaluates and either or allowing for minor error(s) . ® ® ® ®u . v u v 5 marks: Correctly evaluates or or . . ® ® ® ®u v u v 0 marks: No significant work of merit. Part (c ) (ii) 5 marks -2 (c) (ii) (1 ) , ® ® ® r = -k u + k v where k Î R and k ¹ 0. Find the value of k for which Ðuov = Ðvor . 2 (c) (ii) (1 ) 5 4 4 (1 3 ) (5 ) . ® ® ® ® ® ® ® -+ + = + + r = -k i + j k i j k i k j ( ) ( ) ( ) ( ) ( ) ( ) 3 0 3 0 as 0. 3. 568 312 52 468 72 180 128 384 0 13 6 2 3 2 26 4 1013 24 8 12 2 26 4 10 13 3 4 2 26 4 10 4 12 20 4 . 13 3 32 1 3 5 4 4 1 3 5 cos 2 2 2 2 2 2 2 2 2 \ -= -= ¹ \ = + + = -+ -= + = -+ + = -+ = -+\ + + -= + + --+ + + Ð = ® ® ® ®k k k k k k k k k k k k k k k k k k k k k k i j k i k j vor 5 marks: Fully correct solution. 4 marks: Complete solution with minor error(s). 3 marks: Correct or substantially correct equation in k (without i jand ). 2 marks: ® ® ® r expressed in the form a i + b j , allowing for minor error(s). 0 marks: No significant work of merit. Page 44 QUESTION 3 Part (a) 15 marks Att 5 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (c) 15 (10, 5) marks Att (3, 2) Part (a) 15 marks Att 5 3 (a) The line : 3 2 7 0 1 L x -y + = and the line : 5 3 0 2 L x + y + = intersect at the point p. Find the equation of the line through p perpendicular to . 2 L Equation of line 15 marks Att 5 3 (a) 5 3 0 10 2 6 3 2 7 0 3 2 7+ + = + = --+ = -= -x y x y x y x y 13x = -13 x = -1. \y = 2. p(-1, 2). ( 1) 5 11 0. 51 Equation of line : 2 . 51 : 5 3 slope 5. perpendicular slope 2 2 -= + -+ = = --= -\ = = y x x y L y x L m or 3 (a) ( ) ( ) ( ) ( ) 0 Required line : 5 11 0. 2 11 25 21 . 21 15 25 2 26 13. 51 2 3 5 . 51 : 5 3 slope 5. Slope of required line . 2 3 5 Slope 3 5 2 7 3 0 Required line : 3 2 7 5 3 0. 2 2 \ -+ = -+ = = + = -= -\ = --+ = --= -\ = -= + \ + + -+ + = -+ + + + = x y x y L y x L x y x y x y l l l l ll ll l l l l Blunders (-3) B1 Error in slope of L2 other than slip. B2 Incorrect perpendicular slope. Slips (-1) S1 Arithmetic error. Attempts ( 5 marks) A1 x or y coordinate of point p. A2 Correct slope of L2. A3 Correct perpendicular slope. Page 45 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (b) (i) 10 marks Att 3 3 (b) (i) The line K passes through the point (-4, 6) and has slope m, where m > 0. Write down the equation of K in terms of m. Equation of K 10 marks Att 3 3 (b) (i) y -6 = m(x + 4). Blunders (-3) B1 Error in equation line formula. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Equation of line with some substitution. Part (b) (ii) 5 marks Att 2 3 (b) (ii) Find, in terms of m, the co-ordinates of the points where K intersects the axes. Co-ordinates 5 marks Att 2 3 (b) (ii) y -6 = m(x + 4) mx -y + 6 + 4m = 0. Cuts x-axis at p(x, 0). . 6 4 6 4 m m mx m x = --= --. , 0 6 4 --m m p Cuts y-axis at q(0, y). y = 6 + 4m. q(0, 6 + 4m). Blunders (-3) B1 Equation of axes incorrect. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 One correct coordinate. Page 46 Part (b) (iii) 5 marks Att 2 3 (b) (iii) The area of the triangle formed by K, the x-axis and the y-axis is 54 square units. Find the possible values of m. Values of m 5 marks Att 2 (b) (iii) Area triangle opq = 54 square units. ( )( ) ( ) ( )( ) ( )( ) or 3. 43 4 15 9 0 4 3 3 0. 16 48 36 108 16 60 36 0 6 4 6 4 108 . 6 4 54. 6 4 0 0 21 . 21 Area triangle 2 2 2 1 2 2 1 \ = = -+ = --= \ + + = -+ = + + = = + \ ---= -m m m m m m m m m m m m m m m m m opq x y x y Blunders (-3) B1 Error in triangle area formula. B2 Error in factors or quadratic formula. B3 Misuse of modulus in formula. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Triangle area formula with some substitution. A2 Quadratic in m. o --, 0 6 4 m m p q(0, 6+4m) (-4, 6) Page 47 Part (c) 15 (10, 5) marks Att (3, 2) Part (c) (i) 10 marks Att 3 3 (c) (i) f is the transformation (x, y)®(x¢, y¢), where x¢ = 3x -y and y¢ = x + 2y. (i) Prove that f maps every pair of parallel lines to a pair of parallel lines. You may assume that f maps every line to a line. Prove 10marks Att 3 3(c)(i) ¢¢ -= -= ¢¢ yx yx yx yx 1 32 1 71 1 23 1 Let L have equation: ax + by + c = 0 , and M: ax + by + d = 0 . ( ) ( ) (2 ) 0 ( ) : ( 2 ) (3 ) 7 0 7 3 7 \ : -¢ + ¢ + x¢ + y¢ + c = f L -a + b x¢ + a + b y¢ + c = b x y a f L and ( ) ( ) (2 ) 0 ( ) : ( 2 ) (3 ) 7 0 7 3 7 : -¢ + ¢ + x¢ + y¢ + d = f L -a + b x¢ + a + b y¢ + d = b x y a f M So, f (L) || f (M) , since (-a + 2b)(3a + b) = (3a + b)(-a + 2b) , [i.e. a1b2 = a2b1 ] or 3 (c) (i) L : y = mx + c and M : y = mx + k are two parallel lines. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) , parallel. 3 1 2 Both lines have same slope, . 37 3 1 2 Similarly : . 37 3 1 2 : 3 1 2 7 : 2 : 3 2 7 . 7 3 71 : 3 . 71 2 2 71 But 2 2 . 71 2 2 . 2 7 3 2 6 2 \ -+ -+ ¢ -¢ = + -+ ¢ -\ -¢ = + ¢ + ¢ = + \ -¢ + ¢ = ¢ + ¢ + -¢ + ¢ = ¢ + ¢ + ¢ = + ¢ = ¢ + ¢ + = -¢ + ¢ ¢ = + ¢ = + \ ¢ + ¢ = = ¢ + ¢ ¢ = -¢ = -mm m k x mm f M y m c x mm f L m y m x c f L y x y c f L x y mx my c m f L x y y x y y x y y y x y y x y y x y x y x x x y x x y x x y or Let L and M pass through p and q respectively and both be in the direction m . L p tm M q tm\ = + and = + , where t ÎR f (L) f ( p tm) f ( p) tf (m) \ = + = + and f (M) f (q tm) f (q) tf (m) = + = + \ f (L) and f (M) are both lines in the direction of f (m) , and hence are parallel. * Note: second method above fails to deal with the case where L and M are vertical, or where they have slope 3. Do not penalise this. Page 48 Blunders (-3) B1 Error in determining slope other than slip. B2 Incorrect matrix or matrix multiplication. B3 Failure to establish image lines parallel. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Expressing x or y in term of primes. A2 correct matrix for f. A3 Finds image of one line and stops. Part (c) (ii) 5 marks Att 2 3 (c) (ii) oabc is a parallelogram, where [ob] is a diagonal and o is the origin. Given that f (c) = (-1, 9), find the slope of ab. Slope ab 5 marks Att 2 3 (c) (ii) ( ) ( ) ( ) ( 3 ). 71 2 and 71 f c = -1, 9 . x = x¢ + y¢ y = -x¢ + y¢ ( ) Slope 4 slope 4 as is parallel to . 1 and 4 1, 4 . oc ab ab oc x y c = = \ = = or 3 (c)(ii) Slope 4 slope 4. . 41 28 7 71 91 1 32 1 71 91 1 23 1 1 23 1 Matrix 1 \ = = = = = --= ---= -oc ab f c or 3 (c) (ii) f (c) = (-1, 9). x¢ = 3x -y and y¢ = x + 2y. ( ) But is parallel to slope 4. 1, 4 and slope 4. 7 7 1 and hence 4. 2 9 2 9 3 1 6 2 2 = \ = = = = + = + = -= --= -ab oc ab c oc x x y x y x yx y x y Blunders (-3) B1 Slope oc and stops. B2 Incorrect matrix. B3 Incorrect matrix multiplication other than slip. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Two simultaneous equations. A2 Correct point c and stops. Page 49 QUESTION 4 Part (a) 10 marks Att 3 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (a) 10 marks Att 3 4 (a) Evaluate 3 sin4 lim® 0 . * Accept correct answer without work. If candidate’s answer is correct, ignore the work. Evaluate 10 marks Att 3 4 (a) . 34 34 4sin4 lim 3 4 4sin4 lim 3sin4 lim 0 0 0 = ´ = ´ = ® ® ® q q or . 34 3 4cos(0) (0)(0) ( )( ) ( ) sin4 and ( ) 3 . lim0 = = ¢¢ = = \ = ® gf g f f g q q q q q Blunders (-3) B1 sin4q = 4sinq. B2 Error in differentiation. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Has in solution. 4sin4 qq A2 Correct differentiation. Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (b) (i) 10 marks Att 3 4 (b) (i) Using cos2A = cos2 A -sin2 A, or otherwise, prove (1 cos2 ). 21 cos2 A = + A Prove 10 marks Att 3 4 (b) (i) ( ) (1 cos2 ). 21 2cos 1 cos2 cos cos2 cos sin cos 1 cos 2 2 2 2 2 2 A A A A A A A A A \ = + = + = -= --Page 50 Blunders (-3) B1 Error in cos2A formula. B2 Error in sin2A formula. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Correct substitution for cos2A. A2 Sin2A =1-cos2A. Part (b) (ii) 10 (5, 5) marks Att (2, 2) 4 (b) (ii) Hence, or otherwise, solve the equation 1+ cos2x = cosx, where 0o £ x £ 360o. Quadratic in Cosx 5 marks Att 2 Solution for x 5 marks Att 2 4 (b) (ii) ( ) 90 , 270 or 60 , 300 . solution {60 , 90 , 270 , 300 }. . 21 cos 2cos 1 0 cos 0 or cos 1 cos2 cos 2cos cos . o o o o o o o o 2 \ = = \ = -= = = + = = x x x x x x x x x x Blunders (-3) B1 Incorrect substitution for 1+cos2x or cos2x. B2 Error in factors. B3 Each incorrect solution or missing solution. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2 marks) A1 cos 2x = cos2 x -sin2 x. A2 Correct factors. A3 One correct solution. Page 51 Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (c) (i) 15 (10, 5) marks Att (3, 2) 4 (c) (i) 1 S is a circle of radius 9 cm and 2 S is a circle of radius 3 cm. 1 2 S and S touch externally at f. A common tangent touches 1 S at point a and 2 S at b. (i) Find the area of the quadrilateral abcd. Give your answer in surd form. Find ec10 marks Att 3 Area quadrilateral abcd 5 marks Att 2 4 (c) (i) 144 36 108. 108 6 3. ec 2 = dc 2 -de 2 ec 2 = -= \ec = = But ec = ab. Area of the quadrilateral abcd [ ] (6 3)[9 3] 36 3 21 2= 1 + bc = + = ab ad cm2. or 4 (c) (i) Area of quadrilateral abcd = triangle dce + rectangle ecba = (6)(6 3) 3(6 3) 36 3. 21 + = Blunders (-3) B1 Incorrect application of Pythagoras. B2 Error in area formula. Slips (-1) S1 Arithmetic error. Attempts ( 3, 2 marks) A1 Correct length of dcor de. A2 Area of triangle dce or rectangle ecba correct. A3 Area formula for trapezium abcd with some substitution. a b d f c e S1 S2 a b d f c e S1 S2 6 3 9 3 Page 52 Part (c) (ii) 5 marks Att 2 4 (c) (ii) Find the area of the shaded region, which is bounded by [ab] and the minor arcs af and bf. Area of shaded region 5 marks Att 2 4 (c) (ii) 60 . 21 12 6 cosÐedc = = Ðedc = o \ Ðbcf = 30o + 90o = 1200. Area of sector ( ) . 2 27 3 81 21 21 2q p = p adf = r = Area of sector ( ) 3 . 3 2 9 21 21 2q p = p bcf = r = \ Area of shaded region . 2 33 3 36 3 2 27 36 3 = -p -p = -p Blunders (-3) B1 Error in sector area formula. B2 Finds area of both sectors but fails to finish. B3 Incorrect conversion from degree to radians. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Ðedc = 600 or Ðecd = 300 or Ðbcf = 1200. A2 12 6 cosÐedc = or . 12 6 sinÐecd = Page 53 QUESTION 5 Part (a) 15 marks Att 5 Part (b) 20 (15, 5) marks Att (5, 2) Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (a) 15 marks Att 5 5(a) The area of an equilateral triangle is 4 3 cm2 . Find the length of a side of the triangle. Length of side 15 marks Att 5 5 (a) Area of triangle . 3 Sin , where and 2= 1 = ÐC = p ab C a b 16 4. Length of side 4 cm. 4 3. 23 21 4 3 3 sin 21 2 2 2 \ = = = \ = = a a a a p Blunders (-3) B1 Error in triangle area formula. B2 Incorrect evaluation of sin60o. B3 sin60o in decimal form. Slips (-1) S1 Arithmetic error. Attempts ( 5 marks) A1 Triangle area formula with substitution. Part (b) 20 (15, 5) marks Att (5, 2) Part (b) (i) 15 marks Att 5 5 (b) (i) In the triangle xyz, Ðxyz = 2b and Ðxzy = b . xy = 3 and xz = 5. (i) Use this information to express sin2b in the form a bÎ ba sinb , where , N. Express 15 marks Att 5 5 (i) sin . 35 sin2 3 sin 5sin2b b b = b = y z 23 5 x Page 54 Blunders (-3) B1 Error in substitution into Sine rule. Slips (-1) S1 Arithmetic error. Attempts ( 5 marks) A1 . sin 2 5 or sin3 b b Part (b) (ii) 5 marks Att 2 5 (b) (ii) Hence express tan in the form , dc b where c, d Î N. Express tanb 5 marks Att 2 5 (b) (ii) . 511 tan 65 cos sin . 35 sin 2sin cos 35 sin2 \ = = = = b b b b b b b Blunders (-3) B1 Error in sin2b formula. B2 Incorrect ratio of sides for cosb or tanb. B3 Incorrect application of Pythagoras. B4 and stops. 65 cosb = Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Equation in b. 90o 6 5 11Page 55 Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) 5 (c) pqrs is a vertical wall of height h on level ground. p is a point on the ground in front of the wall. The angle of elevation of r from p is q and the angle of elevation of s from p is 2q. pq = 3 pt . Find q. Tanq or Tan2q in terms of h and x 5 marks Att 2 Equation in tan3q or tan2q 5 marks Att 2 Find q 5 marks Att 2 5 (c) ( ) ( ) . 6 3 1 tan , 0. 31 3 3 2 3 0. 3 1 0 3 1 2 , where tan . 1 tan 2tan 3 tan tan2 3tan 3 tan . Also tan2 tan2 . 3 tan 3 3 2 2 2 2 q q p q q q q q q q q q q \ = = = \ -= -= -= = ¹ -= = -\ = = = = = = t t t t t t t t t t x x t t t t h x xh h x x h Blunders (-3) B1 Incorrect ratio of sides for tan. B2 Error in tan2q formula. B3 Incorrect factors. B4 Incorrect value forq. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2, 2 marks) A1 Tanq or tan2q expressed as ratio of sides. A2 Tan2q expressed in terms of tanq. A3 Correct value for tan2q. h s t r p 3x q x q 2q Page 56 QUESTION 6 Part (a) 10 (5, 5) marks Att (-, 2) Part (b) 25 (5, 5, 5, 5, 5) marks Att (2, 2, 2, 2, 2) Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (a) 10 (5, 5) marks Att (-, 2) Part (a) (i) 5 marks Hit/Miss 6 (a) (i) How many three-digit numbers can be formed from the digits 1, 2, 3, 4, 5, if (i) the three digits are all different 6 (a) (i) Answer 3 5 4 3 60. =5P = ´ ´ = Part (a) (ii) 5 marks Att 2 6 (a) (ii) How many three-digit numbers can be formed from the digits 1, 2, 3, 4, 5, if (ii) the three digits are all the same? 6 (a) (ii) Answer = 5´1´1 = 5. Blunders (-3) B1 5´5´1. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 5´5´5. Page 57 Part (b) 25 (5, 5, 5, 5, 5) marks Att (2, 2, 2, 2, 2) Part (b) (i) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 6 (b) (i) Solve the difference equation 4 8 0, 2 1 --= n+ n+ n u u u where n ³ 0, given that u0 = 0 and u1 = 2. Characteristic equation 5 marks Att 2 Characteristic roots 5 marks Att 2 Simultaneous equations 5 marks Att 2 Solution 5 marks Att 2 6 (b) (i) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (2 2 3) . 63 2 2 3 63 . 63 . 63 2 3 1 4 3 2 2 2 3 2 2 3 2 2 2 3 2 2 3 2 2 2 2 3 2 2 3 2 0 0. . 2 2 3 2 2 3 . 2 2 3. 2 4 4 3 2 4 48 2 4 16 32 4 8 0 4 8 0. 1o 2 2 1 n n n n n n n n n u k k l k k k k k k u k l u k l l k u k l x u u u x x \ = + --\ = = = \ = -\ + --= + -+ = = + + -= = + = = -= + + -\ = ± + = ± = ± = ± + -+ -= --= Blunders (-3) B1 Error in characteristic equation. B2 Error in quadratic formula. B3 Incorrect use of initial conditions. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2, 2, 2 marks) A1 An equation in k and l. A2 Correct value for k or l. Page 58 Part (b) (ii) 5 marks Att 2 6 (b) (ii) Verify that your solution gives the correct value for . 2 u Verify 5 marks Att 2 6 (b) (ii) ( ) ( ) ( ) (16 3) 8. Verified. 63 4 8 3 12 4 8 3 12 63 2 2 3 63 2 2 3 63 But 8 0 8. 4 8 0. But 2 and 0. 2 2 2 2 2 2 1 0 1 0 = = \ = + --= + + -+ -\ = + = --= = = u u u u u u u u or 6 (b) (ii) ( ) ( ) ( ) ( ) ( ) ( ) gives 8 4(2) 0 0. Verified. Substituting 0, 2 and 8 into 4 8 , 16 3 8. 63 4 8 3 12 4 8 3 12 63 2 2 3 . 63 2 2 3 63 2 2 3 . 63 2 2 3 63 o 1 2 2 1 2 2 2 2 2 2 --= \ = = = --= + + -+ -= = \ = + --= + --n+ n+ n n n n u u u u u uu u u u u Blunders (-3) B1 Error in calculating u2 other than slip. B2 Finds u2 but fails to verify. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct value for u2. Page 59 Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (c) (i) 5 marks Att 2 6 (c) (i) Nine cards are numbered from 1 to 9. Three cards are drawn at random from the nine cards. (i) Find the probability that the card numbered 8 is not drawn. Probability 5 marks Att 2 6 (c) (i) Total outcomes (choose three cards from nine): 3 9C = 84. Outcomes of interest (choose three from the eight allowed): 3 8C = 56. \ Probability . 32 84 = 56 = or 6 (c) (i) (first card not 8) and (second card not 8) and (third card not 8) Probability . 32 76 87 9= 8 ´ ´ = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. Part (c) (ii) 5 marks Att 2 6 (c) (ii) Nine cards are numbered from 1 to 9. Three cards are drawn at random from the nine cards. (ii) Find the probability that all three cards drawn have odd numbers. Probability 5 marks Att 2 6 (c) (ii) Outcomes of interest (choose three from the five odd-numbered): 3 5C = 10. \ Probability = . 42 5 84 10 = or 6 (c) (ii) (first card odd) and (second card odd) and (third card odd) \ Probability . 42 5 504 60 73 84 9= 5 ´ ´ = = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. Page 60 Part (c) (iii) 5 marks Att 2 6 (c) (iii) Nine cards are numbered from 1 to 9. Three cards are drawn at random from the nine cards. (iii) Find the probability that the sum of the numbers on the cards drawn is greater than the sum of the numbers on the cards not drawn. Probability 5 marks Att 2 6 (c) (iii) Outcomes of interest: Sum of all the cards numbered 1 to 9 is 45. \ Sum of three drawn cards must be ³ 23, (i.e. more than half of total). Sum of cards 7, 8, 9 = 24 Sum of cards 6, 8, 9 = 23 No other possibilities. \Only two possible favourable outcomes. \ Probability = . 42 1 84 2 = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of favourable outcomes. A2 Correct number of possible outcomes. A3 One correct element properly identified e.g. 9+8+7=24 > 21. Page 61 QUESTION 7 Part (a) 10 (5, 5) marks Att (-, 2) Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, -, 2) Part (a) 10 (5, 5) marks Att (-, 2) Part (a) (i) 5 marks Hit/Miss 7 (a) (i) How many different groups of four can be selected from five boys and six girls? 7 (a) (i) Choose four from eleven answer 4 330. = 11C = Part (a) (ii) 5 marks Att 2 7 (a) (ii) How many of these groups consist of two boys and two girls? 7 (a) (ii) Choose two from five and choose two from six answer 2 10 15 150. 6 2 = 5C ´ C = ´ = Blunders (-3) B1 2 . 6 2 5C + C Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 or 2 . 6 2 5C C Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (b) (i) 5 marks Att 2 7 (b) (i) There are sixteen discs in a board-game: five blue, three green, six red and two yellow. Four discs are chosen at random. What is the probability that (i) the four discs are blue Part (b) (i) 5 marks Att 2 7 (b) (i) Total outcomes (choose four discs from sixteen): 4 16C = 1820. Outcomes of interest (choose four of the five blue): 4 5C = 5. \ Probability . 364 1 1820 = 5 = or 7 (b) (i) (first blue) and (second blue) and (third blue) and (fourth blue) \ Probability . 364 1 43680 120 13 2 14 3 15 4 16 5 = ´ ´ ´ = = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. Page 62 Part (b) (ii) 5 marks Att 2 7 (b) (ii) There are sixteen discs in a board-game: five blue, three green, six red and two yellow. Four discs are chosen at random. What is the probability that (ii) the four discs are the same colour Probability 5 marks Att 2 Part (b) (ii) Outcomes of interest: (four blue or four red): 4 6 4 5C + C = 5 + 15 = 20. \ Probability . 91 1 1820 20 = or 7 (b) (ii) Probability = P(4 blue) + P(4 red) . 91 1 43680 480 43680 120 360 13 3 14 4 15 5 16 6 13 2 14 3 15 4 16 5 = = + = ´ ´ ´ + = ´ ´ ´ Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. A3 P (4 red) correct. Part (b) (iii) 5 marks Att 2 7 (b) (iii) There are sixteen discs in a board-game: five blue, three green, six red and two yellow. Four discs are chosen at random. What is the probability that (iii) all four discs are different in colour Probability 5 marks Att 2 7 (b) (iii) Outcomes of interest: one blue and one green and one red and one yellow: 1 2 1 6 1 3 1 5C ´ C ´ C ´ C = 180. \ Probability = . 91 9 1820 180 = or 7 (b) (iii) (first blue) and (second green) and (third red) and (fourth yellow) or any permutation; \ Probability . 91 9 43680 4320 4! 13 2 14 6 15 3 16 5 = ´ ´ ´ ´ = = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. Page 63 Part (b) (iv) 5 marks Att 2 7 (b) (iv) There are sixteen discs in a board-game: five blue, three green, six red and two yellow. Four discs are chosen at random. What is the probability that (iv) two of the discs are blue and two are not blue? Probability 5 marks Att 2 7 (b) (iv) Of interest: (choose two of five blue and two of remaining eleven) 2 11 2 5C ´ C = 550. \ Probability . 182 55 1820 550 = = or 7 (b) (iv) (first blue) and (second blue) and (third not blue) and (fourth not blue), or any permutation thereof; \ Probability . 182 55 174720 52800 2!.2! 4! 13 10 14 11 15 4 16 = 5 ´ ´ ´ ´ = = Blunders (-3) B1 Incorrect number of possible outcomes. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Correct number of possible outcomes. A2 Correct number of favourable outcomes. A3 P (two blue) correct. A4 P (two are not blue) correct. Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, -, 2) Part (c) (i) 10 (5, 5) marks Att (2, 2) 7 (c) (i) On 1st September 2003 the mean age of the first-year students in a school is 12.4 years and the standard deviation is 0.6 years. One year later all of these students have moved into second year and no other students have joined them. (i) State the mean and the standard deviation of the ages of these students on 1st September 2004. Give a reason for each answer. Mean 5 marks Att 2 Standard deviation 5 marks Att 2 (c) (i) Mean = 13.4 years. As all the students are one year older, the mean is increased by one. Standard deviation = 0.6 years. The spread of ages in the group is still the same. or As they are each one year older and the mean is increased by one, each deviation from the mean is unchanged, and hence so is the standard deviation. Page 64 Blunders (-3) B1 Reason for new mean not given or incorrect reason. B2 Reason for new standard deviation not given or incorrect reason. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2 marks) A1 Correct new mean. A2 Correct new standard deviation. Part (c) (ii) 5 marks Hit/Miss 7 (c) (ii) A new group of first-year students begin on 1st September 2004. This group has a similar age distribution and is of a similar size to the first-year group of September 2003. (ii) State the mean age of the combined group of the first-year and second-year students on 1st September 2004. Combined mean 5 marks Hit/Miss Part (c) (ii) 12.9 years. 2 12.4 13.4 Mean = » + Part (c) (iii) 5 marks Att 2 7 (c) (iii) State whether the standard deviation of the ages of this combined group is less than, equal to, or greater than 0.6 years. Give a reason for your answer. State & reason 5 marks Att 2 7 (c) (iii) Standard deviation > 0.6 years. There is a greater spread of ages in the combined group than in a single year group. [or: Data more spread out.] Blunders (-3) B1 Incorrect reason given. B2 No reason given. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 States greater than 0.6 years. [Aside: the actual value is approximately 0.8; this is not required.] Page 65 QUESTION 8 Part (a) 15 marks Att 5 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (a) 15 marks Att 5 8 (a) Use integration by parts to find x2lnxdx . Integration by parts 15 marks Att 5 8 (a) x2lnxdx = uv -vdu. = = = = = . 31 . 1 ln dx dv x2dx v x2dx x3 x u x du constant. 91 ln 31 31 ln 31 1 31 ln 31 ln 3 3 2 3 3 3 2 = -+ -= \= -x x x dx x x x dx x x xdx x x x Blunders (-3) B1 Incorrect differentiation or integration. B2 Constant of integration omitted. B3 Incorrect ‘parts’ formula. Slips (-1) S1 Arithmetic error. Attempts ( 5 marks) A1 Correct assigning to parts formula. A2 Correct differentiation or integration. Page 66 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (b) (i) 10 marks Att 3 8 (b) (i) Derive the Maclaurin series for f (x) = ln(1+ x) up to and including the term containing x3. Maclaurin series 10 marks Att 3 8 (b) (i) ... 3!(0) 2!(0) 1! (0) ( ) (0) 2 3 + ¢¢ + ¢ + ¢ = + f x f x f x f x f ( )( ) ( ) ( ) ( ) .... 31 21 .... 31 21 ( ) ln 1 0 ( ) 2 1 (0) 2. ( ) 11 (0) 1. 1 (0) 1. 1 1 ( ) ( ) ln 1 (0) ln1 0. 2 3 2 3 32 1 \ = + = + -+ -= -+ -¢¢ = + ¢¢ = ¢ = -+ ¢ = -= + ¢ = + ¢ = = + = = ---f x x x x x x x x f x x ff x x fx f x f x f x x f Blunders (-3) B1 Incorrect differentiation. B2 Incorrect evaluation of f (n) (0). B3 Each term not derived. B4 Error in Maclaurin series. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Correct expansion of ln(1+ x) given but not derived. A2 f (0) correct. A3 Any one correct term derived. Part (b) (ii) 5 marks Att 2 8 (b) (ii) Use those terms to find an approximation for . 10 11 ln Find approximation 5 marks Att 2 8 (b) (ii) . 1500 143 3000 286 3000 300 15 1 3000 1 200 1 10 1 10 1 ln 1 10 11 ln = = + -= + -= = + Blunders (-3) B1 Error in simplification other than slip. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 . 10 1 1 10 11 = + A2 Correct value for x. Page 67 Part (b) (iii) 5 marks Att 2 8 (b)(iii) Write down the general term of the series f (x) and hence show that the series converges for -1 < x < 1. General term/converges 5 marks Att 2 (b) (iii) General term = ( ) ( ) 1 1 . 1 2 1 1 1 + = -\ = -+ + + + n x u n x u n n n n n n ( ) ( ) Series converges when 1 1 1. . 1 1 Limit 1 ( 1) Limit 1 1 1 Limit Limit 1 2 1 1 < -< < = + = + = --´ + \ = -+ ®¥ ®¥ + + ®¥ + ®¥ x x x n x n xn x n n x u u n n n n n n n n n n Blunders (-3) B1 Incorrect power in general term. B2 (-1) omitted from general term. B3 Error in un+1. B4 Error in evaluating limit other than slip. B5 Evaluates limit as x and stops. Slips (-1) S1 Arithmetic error. Attempts ( 2marks) A1 Power of x correct. A2 Denominator correct. A3 un+1correct, given that un is not worthless. A4 Correct substitution into ratio test and fails to finish. Page 68 Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) 8 (c) A cone has radius r cm, vertical height h cm and slant height 10 3 cm. Find the value of h for which the volume is a maximum. Volume in terms of h or r 5 marks Att 2 Correct differentiation 5 marks Att 2 Value of h 5 marks Att 2 8 (c) ( ) ( ) ( ) 10 cm gives maximum volume. 2 0 for 10. 300 3 0 10, (since 0). 300 3 0 for maximum volume. 31 300 . 31 300 31 31 300 300 . 2 2 2 23 2 2 2 2 2 2 \ = = -< = \ -= = > = -= \ = -= = -+ = = -h h h dxd V h h h h dh dV V h h V r h h h h r r h p pp p p * 0, for 10 cm not required. 2 2 < h = dhd V Blunders (-3) B1 Incorrect application of Pythagoras. B2 Error in differentiation. B3 Error in solving for h or r, other than slip. Slips (-1) S1 Arithmetic error. S2 Correct value for r, but value of h not given. Attempts (2, 2, 2 marks) A1 h2 + r 2 = 300. A2 Some part of differentiation correct. A3 = 0 dh dV , given that candidate’s work is not worthless. 10 3 h r 10 3 h rPage 69 QUESTION 9 Part (a) 10 marks Att 3 Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (a) 10 marks Att 3 9 (a) z is a random variable with standard normal distribution. Find P (1 < z < 2). 9 (a) P (1 < z < 2) = 0.9772 -0.8413 = 0.1359. Blunders (-3) B1 P(z £ 1) or P(z < 2) incorrect. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 P(z £ 1) or P(z < 2) correct. Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) Part (b) (i) 10 marks Att 3 9 (b) (i) During a match John takes a number of penalty shots. The shots are independent of each other and his probability of scoring with each shot is . 54 (i) Find the probability that John misses each of his first four penalty shots. Probability 10 marks Att 3 9 (b) (i) Probability C . 625 1 51 54 or 625 1 51 0 4 4 4 4 = = = Blunders (-3) B1 Error in binomial. B2 Incorrect q. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 Correct q. Page 70 Part (b) (ii) 5 marks Att 2 9 (b) (ii) Find the probability that John scores exactly three of his first four penalty shots. Probability 5 marks Att 2 9 (b) (ii) Probability . 625 256 51 54 3 3 4 = = C Blunders (-3) B1 Error in binomial. B2 Incorrect q. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 . 51 . 54 3 Part (b) (iii) 5 marks Att 2 9 (b) (iii) If John takes ten penalty shots during the match, find the probability that he scores at least eight of them. Probability 5 marks Att 2 9 (b) (iii) P (scores at least eight) = P(scores eight) + P(scores nine) + P(scores ten). ( 0.678). 9765625 6619136 9765625 2949120 2621440 1048576 51 54 51 54 51 54 10 0 10 10 9 1 9 10 8 2 8 10 = + + = »+ + = C C C Blunders (-3) B1 Error in binomial. B2 Omits one essential probability. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 Finds one correct probability. A2 Probability = P (scoring eight) + P (scoring nine) + P (scoring ten). Page 71 Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) 9 (c) A survey was carried out to find the weekly rental costs of holiday apartments in certain country. A random sample of 400 apartments was taken. The mean of the sample was €320 and the standard deviation was €50. Form a 95% confidence interval for the mean weekly rental costs of holiday apartments in that country. Correct standard error 10 marks Att 3 Correct confidence interval 5 marks Att 2 Final solution 5 marks Att 2 9 (c) x = 320. = 50. n = 400. 2.5. 20 50 = = = n x The 95% confidence interval is [ ( ) ( )] x -1.96 x , x +1.96 x = [320 -1.96(2.5), 320 +1.96(2.5)] = [€315.10, €324.90] Blunders (-3) B1 Error in standard error of mean. B2 Error in confidence interval. B3 Answer not simplified. Slips (-1) S1 Arithmetic error. Attempts ( 3, 2, 2 marks) A1 Standard error of mean with some substitution. A2 Correct confidence with substitution. Page 72 QUESTION 10 Part (a) 15 (10, 5) marks Att (3, 2) Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (a) 15 (10, 5) marks Att (3, 2) 10 (a) Show that {0, 2, 4} forms a group under addition modulo 6. You may assume associativity. Show closure 10 marks Att 3 Identity and inverses 5 marks Att 2 10 (a) (i) Closed: No new element. Identity = 0. Inverses: 0-1 = 0, 2-1 = 4, 4-1 = 2. \ Group. Blunders (-3) B1 Identity not given. B2 Inverses not stated. Slips (-1) S1 Arithmetic error. S2 each inverse not given. Attempts ( 3, 2 marks) A1 Incomplete Cayley table or error in Cayley table. A2 Identity given. A3 One inverse given. + mod 6 0 2 4 0 0 2 4 2 2 4 0 4 4 0 2 Page 73 Part (b) 20 (10, 10) marks Att (3, 3) Part (b) (i) 10 marks Att 3 10 (b) (i) R90o and SM are elements of D4 , the dihedral group of a square. (i) List the other elements of the group. List elements 10 marks Att 3 10 (b) (i) R0o , R180o , R270o , SN , SL , SK . Blunders (-3) B1 Each incorrect element. B2 Each missing element. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 One correct element. Part (b) (ii) 10 marks Att 3 10 (b) (ii) Find ( ) C SM , the centralizer of SM . Find centralizer 10 marks Att 3 10 (b) (ii) ( ) , , , . 00 1800 C S R S S R M M N = Blunders (-3) B1 Each incorrect element. B2 Each missing element. Slips (-1) S1 Arithmetic error. Attempts ( 3 marks) A1 One correct element. a d b c o L M K N a d b c o L M K N Page 74 Part (c) 15 (5, 5, 5) marks Att (2, 2, 2) Part (c) (i) 5 marks Att 2 10 (c) A regular tetrahedron has twelve rotational symmetries. These form a group under composition. The symmetries can be represented as permutations of the vertices a, b, c and d. = b a d ca b c d a b c da b c d X , , is a subgroup of this tetrahedral group. (i) Write down one other subgroup of order 2. Subgroup of order two 5 marks Att 2 10 (c) (i) d c b aa b c d a b c da b c d , or c d a ba b c d a b c da b c d , . * If subgroup is not of order 2 then 0 marks. Blunders (-3) B1 Incorrect element. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 One correct element. Part (c) (ii) 5 marks Att 2 10 (c) (ii) Write down a subgroup of order 3. Subgroup of order three 5 marks Att 2 10 (c) (ii) , , . a d b ca b c d a c d ba b c d a b c da b c d or 10 (c) (ii) , , . d b a ca b c d c b d aa b c d a b c da b c d or 10 (c) (ii) , , . d a c ba b c d b d c aa b c d a b c da b c d or 10 (c) (ii) , , . c a b da b c d b c a da b c d a b c da b c d * If subgroup is not of order 3 then 0 marks. b c a d Page 75 Blunders (-3) B1 Each incorrect element. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 One correct element. Part (c) (iii) 5 marks Att 2 10 (c) (iii) Write down the only subgroup of order four. Subgroup of order four 5 marks Att 2 10 (c) (iii) , , , . c d a ba b c d d c b aa b c d b a d ca b c d a b c da b c d * If subgroup is not of order 4 then 0 marks. Blunders (-3) B1 Each incorrect element. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 One correct element. Page 76 QUESTION 11 Part (a) 10 (5, 5) marks Att (2, 2) Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 (5, 5) marks Att (2, 2) 11(a) Find the equation of an ellipse with centre (0, 0), eccentricity 65 and one focus at (10, 0). Value of a 5 marks Att 2 Finish 5 marks Att 2 11 (a) Focus ( ) ( ) 10 12. 65 = 10, 0 = ae, 0 ae = 10. \ a = a = ( ) 1. 144 44 x 1 x 44. 36 11 144 36 25 1 144 1 2 2 22 22 2 2 2 2 2 \ + = + = = = = -= -y by a b a e b b Blunders (-3) B1 Values for a and b found but final equation not given. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2 marks) A1 ae=10. A2 b2 = a2 (1-e2 ). A3 Correct value for b2 and stops. Page 77 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 11 (b) f is a similarity transformation having magnification ratio k. A triangle abc is mapped onto a triangle a¢b¢c¢ under f. Prove that Ðabc = Ða¢b¢c¢ . CosÐabc 5 marks Att 2 CosÐa¢b¢c¢ 5 marks Att 2 p¢q¢ = k pq 5 marks Att 2 Finish 5 marks Att 2 11 (b) cos cos , as 0 180 . cos . 2 . 2 . cos But , and as is a similarity transformation. . 2 . cos . 2 . cos o o 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Ð = Ð ¢ ¢ ¢ Ð = Ð ¢ ¢ ¢ £ Ð £ = Ð + -= + -\ Ð ¢ ¢ ¢ = ¢ ¢ = ¢ ¢ = ¢ ¢ = ¢ ¢ ¢ ¢ ¢ ¢ + ¢ ¢ -¢ ¢ Ð ¢ ¢ ¢ = + -Ð = abc a b c abc a b c abc abc ab bc ab bc ac k ab bc k ab k bc k ac a b c a c k ac a b k ab b c k bc f a b b c a b b c a c a b c ab bc ab bc ac abc Blunders (-3) B1 Error in cosine formula. B2 Error in definition of similarity transformation. B3 Fails to square k. B4 Reason why Ðabc = Ða¢b¢c¢ not given. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2, 2, 2 marks) A1 Use of cosine rule. A2 CosÐa¢b¢c¢ expressed in terms of sides of triangle abc. a¢ b¢ c¢ a b c f Page 78 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) (i) 5 marks Att 2 11 (c) (i) g is the transformation (x , y)®(x¢, y¢) where x¢ = ax and y¢ = by and a > b > 0. (i) C is the circle x2 + y 2 = 1. Show that g(C) is an ellipse. Show that g(C) is an ellipse 5 marks Att 2 11 (c) (i) C: x2 + y 2 = 1. x¢ = ax and and . by y ax y by x ¢ = ¢ ¢ = = ( ) 1. ( ) is an ellipse. 22 22 g C by ax g C = \ ¢ + ¢ \ = Blunders (-3) B1 Error in substitution. Slips (-1) S1 Arithmetic error. Attempts ( 2 marks) A1 x in terms of x¢ or y in terms of y¢. Page 79 Part (c) (ii) 15 (5, 5, 5) marks Att (2, 2, 2) 11 (c) (ii) L and K are tangents at the end points of a diameter of the ellipse g(C). Prove that L and K are parallel. g-1 mapping of g(C), D, L and K 5 marks Att 2 Showing g-1 (L) or g-1(K) ^ g-1(D) 5 marks Att 2 Prove L and K are parallel 5 marks Att 2 11 (c) (ii) By g-1, L, K and D map onto g-1(L), g-1(K) and g-1(D) respectively. But g-1(L) is perpendicular to g-1(D) and g-1(K) is perpendicular to g-1(D), as tangent to circle is perpendicular to diameter at point of contact. \ g-1(L) is parallel to g-1(K). \ L is parallel to K, as parallelism is invariant. Blunders (-3) B1 Error in mapping or mapping circle to ellipse.. B2 Reason why g-1 (L) or g-1(K) ^ g-1(D) not given. B3 Reason why L is parallel to K not given. Slips (-1) S1 Arithmetic error. Attempts ( 2, 2, 2 marks) A1 One correct mapping. A2 States g-1 (L) or g-1(K) ^ g-1(D) without reason given. A3 g-1 (L) parallel to g-1(K). g(C) L K D g-1(L) g-1(K) C g-1(D) g-1 Page 80 BONUS MARKS FOR ANSWERING THROUGH IRISH Bonus marks are applied separately to each paper as follows: If the mark achieved is less than 226, the bonus is 5% of the mark obtained, rounding down. (e.g. 198 marks ´ 5% = 9.9 bonus = 9 marks.) If the mark awarded is 226 or above, the following table applies: Marks obtained Bonus 226 – 231 232 – 238 239 – 245 246 – 251 252 – 258 259 – 265 266 – 271 272 – 278 279 – 285 286 – 291 292 – 298 299 – 300 11 10 9 8 7 6 5 4 3 2 1 0