LCHons2007

Scéimeanna Marcála Scrúduithe Ardteistiméireachta, 2007 Matamaitic Ardleibhéal Marking Scheme Leaving Certificate Examination, 2007 Maths Higher Level Coimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE MATHS HIGHER LEVEL MARKING SCHEME Page 1 Contents Page GENERAL GUIDELINES FOR EXAMINERS – PAPER 1 .............................................. 2 QUESTION 1......................................................................................................................... 3 QUESTION 2......................................................................................................................... 7 QUESTION 3....................................................................................................................... 11 QUESTION 4....................................................................................................................... 15 QUESTION 5....................................................................................................................... 18 QUESTION 6....................................................................................................................... 21 QUESTION 7....................................................................................................................... 25 QUESTION 8....................................................................................................................... 29 GENERAL GUIDELINES FOR EXAMINERS – PAPER 2 ............................................ 33 QUESTION 1....................................................................................................................... 34 QUESTION 2....................................................................................................................... 37 QUESTION 3....................................................................................................................... 40 QUESTION 4....................................................................................................................... 43 QUESTION 5....................................................................................................................... 45 QUESTION 6....................................................................................................................... 47 QUESTION 7....................................................................................................................... 50 QUESTION 8....................................................................................................................... 52 QUESTION 9....................................................................................................................... 56 QUESTION 10..................................................................................................................... 59 QUESTION 11..................................................................................................................... 61 Marcanna Breise as ucht freagairt trí Ghaeilge........................................................................ 63 Page 2 MARKING SCHEME LEAVING CERTIFICATE EXAMINATION 2007 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 of merit 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 (5, 5) marks Att (2, 2) Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (5, 5, 10) marks Att (2, 2, 3) Part (a) 10 (5, 5)marks Att (2, 2) 1. (a) Simplify 2 2 2 x yx xy −− . Factors: 5 marks Att 2 Cancellation: 5 marks Att 2 1 (a) ( ) ( )( ) x y x x y x y x x y x yx xy + = + − − = −− 2 2 2 Blunders (-3) B1 Factors once only B2 Indices B3 Incorrect cancellation Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 1 (b) Let f (x) = x2 + (k +1)x − k − 2, where k is a constant. (i) Find the value of k for which f (x) = 0 has equal roots. (ii) Find, in terms of k, the roots of f (x) = 0. (iii) Find the range of values of k for which both roots are positive. (i) b2 − 4ac = 0 applied: 5 marks Att 2 Finish: 5 marks Att 2 (ii) 5 marks Att 2 (iii) 5 marks Att 2 1 (b) (i) f (x) = 0⇒ x2 + (k +1)x + (− k − 2) = 0 Equal roots: b2 − 4ac = 0 ( ) ( )( )0k +1 2 − 4 1 − k − 2 = k 2 + 2k +1+ 4k + 8 = 0 k 2 + 6k + 9 = 0 (k + 3)2 = 0 ⇒ k = −3 Page 4 1 b(ii) Roots of f (x) = 0 ( ) ( ) ( )( ) 2 − +1 ± +1 2 − 4 1 − − 2 = k k k x ( ) 2 − +1 ± 2 + 6 + 9 = k k k ( ) ( ) 2 − +1 ± + 3 = k k 2 − −1+ + 3 x = k k or 2 − −1− − 3 x = k k 2x = 2 or 2 − 2 − 4 x = k x = 1 or x = −k − 2 or 1 b(ii) f (x) = (x −1)(x + k + 2) = 0 x −1 = 0 or x + k + 2 = 0 x = 1 or x = −k − 2 1b(iii) Positive roots : (− k − 2) > 0 − 2 > k Blunders (-3) B1 Equal roots condition B2 Expansion of (k +1)2 once only B3 Indices B4 Factors B5 Roots formula once only B6 Inequality sign B7 Deduction of root from factor or no root B8 Range Slips (-1) S1 Numerical Attempts A1 Equation not quadratic in b(i) gives A ≠ 2 at most in finish A2 Using remainder theorem Page 5 Part (c) 20 (5, 5, 10) marks Att (2, 2, 3) 1 (c) x + p is a factor of both ax2 + b and ax2 + bx − ac. (i) Show that ap2 = − b and that . b p b ac − − = (ii) Hence show that p2 + p3 = c. (i) Show p2 5 marks Att 2 Show p 5 marks Att 2 (ii) 10 marks Att 3 1 (c) (i) (x + p) factor of ax2 + b⇒ f (− p) = 0 a(− p)2 + b = 0 ap2 = −b ap b − 2 = (x + p) factor of ax2 + bx − ac⇒ f (− p) = 0 a(− p)2 + b(− p)− ac = 0 ap2 − bp − ac = 0 But ap2 = −b from above ⇒ −b − ac = bp p b b ac = − − or 1 (c) (i) ax apx p ax b −+ 2 + ax (b ap) x p ax bx ac + −+ 2 + − ax2 + apx ax2 + apx − apx + b (b − ap)x − ac − apx − ap2 (b − ap)x + p(b − ap) ap2 + b − p(b − ap)− ac Since (x − p) factor ⇒ ap2 + b = 0 Since (x + p) factor ap2 = −b − pb + ap2 − ac = 0 ap b − 2 = − pb − b − ac = 0 − b − ac = pb p b b ac = − − Page 6 1(c)(ii) p2 + p3 = p2 (1+ p) ⎟⎠⎞ ⎜⎝⎛ − − + − = b b ac ab 1 ⎟⎠⎞ ⎜⎝− ⎛ − − = b b b ac ab ab = bac = c Blunders (-3) B1 Deduction root from factor B2 Indices B3 Not in required form Slips (-1) S1 Not changing sign when subtracting in 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) 15 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 marks Att 3 2 (a) Without using a calculator, solve the simultaneous equations 2 2 15. 2 3 2− + = + + = + + =x y z x y z x y z (a) 10 marks Att 3 2 (a) (i) x + y + z = 2 (ii) 2x + y + z = 3 (iii) x − 2y + 2z = 15 (ii) 2x + y + z = 3 (i) x + y + z = 2 x = 1 (ii) 2x + y + z = 3 (iii) 1− 2y + 2z = 15 y + z = 1 − 2y + 2z = 14 − y + z = 7 (ii) y + z = 1 (iii) − y + z = 7 2z = 8 (ii) y + z = 1 z = 4 y + 4 = 1 y = −3 x = 1 y = −3 z = 4 Blunders (-3) B1 Multiplying one side of equation only B2 Not finding 2nd unknown (having found 1st ) B3 Not finding 3rd unknown (having found 1st and 2nd) Slips (-1) S1 Numerical S2 Not changing sign when subtracting Worthless W1 Trial and error only Page 8 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 2 (b) α and β are the roots of the equation x2 − 4x + 6 = 0. (i) Find the value of 1 1 . α β + (ii) Find the quadratic equation whose roots are α1 and β1 . Values of (α + β ) & αβ or solve quadr. 5 marks Att 2 Finish 5 marks Att 2 Correct Statement 5 marks Att 2 Finish 5 marks Att 2 2 (b) (i) x2 − (4)x + (6) = 0 x2 − (α + β )x + (αβ ) = 0 ∴ α + β = 4 αβ = 6 32 61 1 = 4 = + + = αββ α α β 2 (b) (ii) x2 − (sum of roots) x + (product roots) = 0 0 1 1 1 1 2 = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − + α β α β x x 0 61 32 2 = ⎟⎠⎞ ⎜⎝+ ⎛ ⎟⎠⎞ ⎜⎝x − ⎛ x 6x2 − 4x +1 = 0 Blunders (-3) B1 Indices B2 Incorrect sum B3 Incorrect product B4 Statement incorrect Slips (-1) S1 Numerical S2 Not as equation Attempts A1 Not quadratic equation Page 9 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 2 (c) (i) Prove that 4, 2 9 ≥ + + x x where x + 2 > 0. (ii) Prove that 9 6 a, x a x ≥ − + + where x + a > 0 . (i) Quadratic Inequality 5 marks Att 2 Finish 5 marks Att 2 (ii) Quadratic Inequality 5 marks Att 2 Finish 5 marks Att 2 2 (c) (i) 4 2 9 ≥ + + x x or 4 2 9 − + + x x ⇔ x(x + 2)+ 9 ≥ 4(x + 2), [given x + 2 > 0 ] ( ) ( ) ( 2) 2 9 4 2 + + + − + = x x x x ⇔ x2 + 2x + 9 ≥ 4x + 8 2 2 2 9 4 8 + + + − − = x x x x ⇔ x2 − 2x +1 ≥ 0 2 2 2 1 + − + = x x x ⇔ (x −1)2 ≥ 0 True ( ) ( ) 0 2 1 2 ≥ + − = x x , which is true, [given x + 2 > 0 ] 2 (c) (ii) a x a x ≥ − + + 9 6 or ( a) x a x − − ⎟⎠⎞ ⎜⎝⎛ − + 9 6 ⇔ x(x + a)+ 9 ≥ (6 − a)(x + a) ( ) ( )( ) x a x x a a x a + + + − − + = 9 6 [given (x + a) > 0 ] ( ) x a x ax x ax a a + + + − − + − = 2 9 6 6 2 ⇔ x2 + ax + 9 ≥ 6x − ax + 6a − a2 ( ) x a x ax x a a + + − + − + = 2 6 2 6 9 2 ⇔ x2 + 2ax − 6x + a2 − 6a + 9 ≥ 0 ( ) ( ) (x a) x a x a + + − + − = 2 2 3 3 2 ⇔ x2 + 2(a − 3)x + (a − 3)2 ≥ 0 Let y = a − 3 Let y = a − 3 = x a x yx y + 2 + 2 + 2 ⇔ x2 + 2yx + y2 ≥ 0 ( ) x a x y + + = 2 ⇔ (x + y)2 ≥ 0 ( ) (x a) x a++ − = 3 2 ⇔ (x + a − 3)2 ≥ 0 ≥ 0 , given (x + a) > 0 . which is true. Page 10 Blunders (-3) B1 Inequality sign B2 Factors B3 Incorrect deduction or no deduction Attempts A1 Multiplication by (x + 2)2 A2 Multiplication by (x + a)2 Worthless W1 Squares both sides Page 11 QUESTION 3 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 3 (a) Let . 3 2 341 21 ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎝⎛ A = Find A2 − 2A.. A2 5 marks Att 2 Finish 5 marks Att 2 3 (a) ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 2 3 41 213 A ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = = 6 3 1 3 3 . 21 2 341 21 2 341 21 A2 A A ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = 6 3 1 3 2 2 21 2 341 21 A ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − = 0 00 0 6 3 1 6 3 1 2 21 21 A2 A Slips S1 Incorrect element S2 Numerical Page 12 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 3 (b) Let z = −1+ i, where i2 = −1. (i) Use De Moivre’s theorem to evaluate z5 and z9. (ii) Show that z5 + z9 = 12z. (i) z in Polar Form 5 marks Att 2 z5 5 marks Att 2 z9 5 marks Att 2 (ii) Show 5 marks Att 2 3 (b) (i) z = −1+ i = r[cosθ + i sinθ ] [ ] 4 3 4 z = 2 cos 3π + i sin π (i) [ ( )]5 4 3 4 z5 = 2 cos 3π + i sin π ( ) 4 15 4 22 cos 15 sin 5 = π + i π ( ) 4 7 4 22 cos 7 sin 5 = π + i π [ ] 2 2 2 1 5 = 2 − i = 22 (1− i) = 4 − 4i [ ( )]9 4 3 4 9 22 cos 3 sin 1 z = π + i π [ ] 4 27 4 22 cos 27 sin 9 = π + i π [ ] 4 3 4 22 cos 3 sin 9 = π + i π [ ] 2 2 2 1 9 = 2 − + i = 24 (−1+ i) = −16 +16i . 3b(ii) z5 + z9 = (4 − 4i)+ (−16 +16i) = 4 − 4i −16 +16i = −12 +12i = 12(−1+ i) = 12z Blunders (-3) B1 Argument B2 Modulus B3 Trig definition B4 Indices B5 i B6 Statement De Moivre once only B7 Application De Moivre Slips (-1) S1 Trig value z = −1+ i r α 1 θ r = 12 +12 = 2 tan 1 1 1 α = = 4 3 4 ∴α = π θ = π Page 13 Part (c) 20 (5, 5, 5, 5) Att (2, 2, 2, 2) 3(c) (i) Find the two complex numbers a + bi for which (a + bi)2 = 15 + 8i. (ii) Solve the equation iz 2 + (2 − 3i)z + (− 5 + 5i) = 0 . (i) Quadratic Equation 5 marks Att 2 Complex numbers 5 marks Att 2 3(c)(i) (a + bi)2 = 15i + 8i ( ) ( )ia2 + 2abi + b2i2 = 15 + 8 ( ) ( ) ( ) ( )ia2 − b2 + 2ab i = 15 + 8 (i): a2 − b2 = 15 (ii): 2ab = 8 (ii): 2ab = 8 ⇒ ab = 4 ⇒ a b = 4 Substitute into (i): ⇒ ( ) 15 2 − 4 2 = a a 15 2 2 − 16 = a a Let y = a2 , (so y > 0 ) ∴ − 16 = 15 y y y2 −16 = 15y y2 −15y −16 = 0 (y −16)(y +1) = 0 y −16 = 0 or y +1 = 0 y = 16 or y = −1 y = a2 ≠ −1 ∴ a2 = 16 a = ±4 a = 4 : 1 4 4 b = = 1⇒ 4 + i = z a = −4 : 4 2 b = 4 = −1⇒ −4 − i = z − Blunders (-3) B1 Expansion (a + ib)2 B2 Indices B3 i B4 Not like to like B5 Factors B6 Quadratic formula B7 Excess values (not real) B8 Only one complex number found B9 Incorrect deduction of root from factor Page 14 (ii) Using square root value 5 marks Att 2 Completion 5 marks Att 2 3 (c) (ii) ( ) ( )0iz 2 + 2 − 3i z + − 5 + 5i = a = i ; b = (2 − 3i) ; c = (− 5 + 5i) a z b b ac 2 − ± 2 − 4 = ( ) ( ) ()( ) i i i i i 2 − 2 − 3 ± 2 − 3 2 − 4 − 5 + 5 = i i i i i i 2 − 2 + 3 ± 4 −12 + 9 2 + 20 − 20 2 = i i i 2 − 2 + 3 ± 15 + 8 = ( ) i i i 2 − 2 + 3 ± 4 + = ( ) i z i i 2 2 3 4 1 − + + + = ( ) i z i i 2 2 3 4 2 − + − + = i 2 + 4i = i i 2 − 6 + 2 = ii i i ⋅ + = 1 2 ii i i ⋅ − + = 3 12 −− = i 1 3 1 − − − = i z = 2 − i 1 z 1 3i 2 = + Blunders (-3) B1 Indices B2 i B3 Expansion of (2 − 3i)2 once only B4 Root formula once only B5 i in denominator Slips (-1) S1 Numerical Page 15 QUESTION 4 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) 4 (a) Show that ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ 2 1 2 1 n n n for all natural numbers n ≥ 2. (a) L.H.S. 5 marks Att 2 R.H.S. 5 marks Att 2 4 (a) L.H.S.= ( ) 2 1 2 1 n n nn n + = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ 2 2n + n2 − n = 2 n2 + n = R.H.S.= ( )( ) 2 2 1 2 n 1 n n n2 + n = + = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ∴ 2 1 1 2 n n n Blunders (-3) B1 Indices B2 Value ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ rn Attempts A1 Correct by using values for n . Page 16 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 4 (b) ≥ ∈ + = = + u n x n u n n n for all 1, 1 5 and u 1 1 N. (i) Write down the values of u2 , u3 , and u4. (ii) Hence, by inspection, write an expression for un in terms of n. (iii) Use induction to justify your answer for part (ii). (i) values 5 marks Att 2 (ii) un 5 marks Att 2 (iii) P(1) and P(k) 5 marks Att 2 P(k+1) 5 marks Att 2 4 (b) (i) 15 1 u = 5 = ( ) 25 1 1 1 2 = 5 = + u ( ) 35 25 3 2 3 u = = ( ) 45 35 4 3 4 u = = (ii) n n u = 5 (iii) To prove : n n u = 5 P(1): n = 1: = ⇒ 15 1 u true for n = 1 Assume P(k): i.e., assume true for n = k k k ⇒u = 5 . Deduce P(k + 1): i.e., prove true for n = k +1: ( ) ( ) 1 5 5 +1 +1 +1 + = = = k k k k k k k k u u ∴ truth of P(k) implies truth of P(k+1), and P(1) is true. ∴ true for all n ≥ 1. * Note: Accept P(1) as given Blunders (-3) B1 Incorrect term once only B2 Incorrect deduction B3 Incorrect P(k) B4 Incorrect P(k + 1) Page 17 Part (c) 20 (5, 5, 5, 5) marks Att(2, 2, 2, 2) 4 (c) The sum of the first n terms of a series is given by 2log 3 Sn = n e . (i) Find the nth term and prove that the series is arithmetic. (ii) How many of the terms of the series are less than12 loge 27 ? (i) un 5 marks Att 2 Prove AP 5 marks Att 2 (ii) Inequality 5 marks Att 2 No of terms 5 marks Att 2 4 (c) (i) S n2ln3 n = ( 1)2 ln 3 1 = − − S n n 2 ln 3 ( 1)2 ln 3 1 = − = − − − u S S n n n n n = (ln3)[n2 − (n2 − 2n +1)] = (ln3)[2n −1] u = (2n −1)ln 3 n [2( 1) 1]ln 3 (2 1)ln 3 1 = + − = + + u n n n (2 1)ln 3 (2 1)ln 3 1 = − = + − − + d u u n n n n = ln 3[2n +1− 2n +1] = 2ln 3 constant ∴ arithmetic, with d = 2ln 3 4(c)(ii) 12ln 27 = 12ln(33 )= 12[3ln 3] = 36ln 3 Let (2n −1)ln 3 ≤ 36ln 3 2n −1 ≤ 36 2n ≤ 37 2n ≤ 18 1 So the first 18 terms are less than 12ln 27 . Blunders (-3) B1 AP formula once only B2 Incorrect terms (must be consecutive) B3 Log laws B4 Indices B5 Incorrect ln 27 or no ln 27 B6 Inequality sign B7 Incorrect value or no value B8 n n n U = S − S +1 Slips (-1) S1 Numerical Page 18 QUESTION 5 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) 5 (a) Plot, on the number line, the values of x that satisfy the inequality x +1 ≤ 2, where x ∈ Z. Inequality 5 marks Att 2 Solution set plotted 5 marks Att 2 5 (a) x +1 ≤ 2 ⇒ − 2 ≤ x +1 ≤ 2. ∴ − 3 ≤ x ≤ 1. or 5(a) x +1 ≤ 2 (x +1)2 ≤ (2)2 x2 + 2x +1 ≤ 4 x2 + 2x − 3 ≤ 0 Graph: y = x2 + 2x − 3 roots: (x + 3)(x −1) = 0 x = −3, x = 1 ∴ − 3 ≤ x ≤ 1. Blunders (-3) B1 Upper limit B2 Lower limit B3 Expansion (x +1)2 once only B4 Inequality sign B5 Indices B6 Factors once only B7 Root formula once only B8 Deduction root from factor B9 Incorrect range B10 Set not plotted Slips (-1) S1 Numerical Attempts A1 One inequality sign A2 Inequality signs ignored A3 Scaled and numbered line −4 −3 −2 −1 0 1 2 -4 -3 -2 -1 0 1 2Page 19 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 5 (b) In the expansion of 2 1 , 9 2 ⎟⎠ ⎞ ⎜⎝ ⎛ − x x (i) find the general term. (ii) find the value of the term independent of x. General Term 5 marks Att 2 ur+1 5 marks Att 2 Power of x 5 marks Att 2 Value 5 marks Att 2 5 (b) [ ( 1 )]9 2 2 x x + − (i) General Term: ( ) ( )r x r n x r u 2 9 1 1 2 9 − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = − + ( ) r x r ( x )r r 2 9 . 9 . 2 9 − − − − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = = k x9−r .x−2r = k.x9−3r (ii) Term independent or x is the term with xo : 9 − 3r = 0⇒ r = 3 ( ) ( ) ( )( ) 2 6 6 1 1.2.36 1 3 9.8.7 4 2 64 39 x x x x u − = − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = = -5376 or (ii) [ ( 1 )]9 2 2 x x + − ( ) ( ) ( ) ( ) ( ) (2 ) ( ) ...... 39 2 29 2 19 2 9 8 1 1 7 1 2 6 1 3 2 2 2 + − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = + x x x x x x x 4 u has ( )6 ( 1 )3 4 2 2 39 x x u x − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ o ⇒ = ( ) ( )( ) 6 6 6 3 1 1.2.39.8.7 2 1 x = x − = -5376 Blunders (-3) B1 General term B2 Error Binomial expansion once only B3 Indices B4 Value ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ rn or no value ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ rn B5 x° ≠ 1 B6 Correct term in expansion not identified. Slips (-1) S1 Numerical Page 20 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 5 (c) The nth term of a series is given by nxn , where x < 1. (i) Find an expression for Sn , the sum to n terms of the series. (ii) Hence, find the sum to infinity of the series. (i) xSn 5 marks Att 2 Correct GP evaluated 5 marks Att 2 Finish 5 marks Att 2 (ii) Sum to infinity 5 marks Att 2 5 (c) (i) ( ) n n n s = x + 2x2 + 3x3 + .................. + n −1 x −1 + nx = n xs x2 + 2x3 + ...................................... + (n −1)xn + nxn+1 − = + 2 + 3 + .................................... + n − n+1 n n s xs x x x x nx = [x + x2 + x3 + .................................. + xn ]− nxn+1 = [G.P to n terms with a = x and r = x] − nxn+1 ( ) ( ) 1 11 1 + −− = − − n xx x n s x nx n ( ) ( ) ( x ) nx xx x n n n s − − − + = − 1 1 1 1 2 5(c)(ii) x < 1. Hence, as n →∞, xn →0 . ( ) ( x) x s x − − −− = ∞ 1 0 1 1 02 (1 x)2 s x − = ∞ Blunders (-3) B1 Indices B2 GP formula B3 Incorrect ‘a’ B4 Incorrect ‘r’ B5 n S not isolated B6 xn →0 in (ii) Slips (-1) S1 Not changing sign when subtracting Page 21 QUESTION 6 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 6 (a) Differentiate 11 22 +− xx with respect to x. (a) 10 marks Att 3 6 (a) 11 22 +− = x y x ( )( ) ( )( ) ( ) ( 2 )23 3 2 22 2 1 2 2 2 2 1 1 2 1 2 + + − + = + + − − = x x x x x x x x x x dx dy ( 2 1)2 4+ = x x Blunders (-3) B1 Indices B2 Differentiation Attempts A1 Error in differentiation formula Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 6 (b) (i) Differentiate x1 with respect to x from first principles. (ii) Find the equation of the tangent to x y = 1 at the point . 21 , 2 ⎟⎠⎞ ⎜⎝⎛ (i) f (x + h)− f (x) simplified 5 marks Att 2 Finish 5 marks Att 2 6 (b) (i) ( ) x f x = 1 ( ) x h f x h + + = 1 ( ) ( ) x h x f x h f x 1 − 1 + + − = ( ) x(x h)x x h +− + = ( ) ( ) ( ) x x h f x h f x h + − + − = ( ) ( ) h x(x h) f x h f x + − = + − 1 ( ) ( ) 0 2lim 1 h x f x h f x h − = + − → Page 22 or 6(b)(i) x y = 1 x x y y + Δ + Δ = 1 x x x y 1 − 1 + Δ Δ = ( ) x x x x+ Δ − Δ = ( ) x x x x y + Δ − = ΔΔ 1 0 2lim 1 x x y x − = ΔΔ Δ → Blunders (-3) B1 f (x + h) B2 Indices B3 No limits shown or implied or no indication →0 B4 h →∞ Worthless W1 Not 1st principles (ii) Slope 5 marks Att 2 Equation 5 marks Att 2 6 (b) (ii) 21 dx x dy − = At ( ) 41 2, slope 1 22, 1 2 − = − = = ⎟⎠⎞ ⎜⎝⎛ dx dy ∴ Tangent is line through ( ) 22, 1 with slope 4m = −1 ( ) 1 1 y − y = m x − x ( 2) 41 2y − 1 = − x − 4y − 2 = −x + 2 x + 4y = 4 Blunders (-3) B1 Differentiation B2 Indices B3 Equation formula line B4 Substituting values into formula once only Slips (-1) S1 Numerical Page 23 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 6 (c) Let ( ) 2 f x = tan−1 x and ( ) = tan−1 2 , for x > 0. x g x (i) Find f ′(x) and g′(x). (ii) Hence, show that f (x)+ g(x) is constant. (iii) Find the value of f (x)+ g(x). (i) f ′(x) 5 marks Att 2 g′(x) 5 marks Att 2 (ii) 5 marks Att 2 (iii) 5 marks Att 2 6 (c)(i) ( ) ⎟⎠⎞ ⎜⎝= − ⎛ 2 f x tan 1 x [tables: a = 2] ( ) 4 2 2x f x + ′ = ( ) ⎟⎠⎞ ⎜⎝= − ⎛ x g x tan 1 2 ( ) 2 2 4 2 2 2 1 2 1 x x x g x +− = ⎟⎠⎞ ⎜⎝⋅ ⎛− ⎟⎠⎞ ⎜⎝+ ⎛ ′ = 6(c)(ii) ( ) ( ) 0 4 2 4 2 2 2 = +− + + ′ + ′ = x x f x g x . Derivative of ( f + g) is 0, so ( f + g) is constant. or 6(c)(ii) f ′(x)+ g′(x) = 0⇒ ∫[ f ′(x)+ g′(x)]dx = k ⇒ f (x)+ g(x) = k (constant) 6(c)(iii) Let k x x = ⎟⎠⎞ ⎜⎝+ ⎛ ⎟⎠⎞ ⎜⎝− ⎛ tan− 2 2 tan 1 1 [x > 0] Let x = 2: tan−1 (1)+ tan−1 (1) = k + = k 4 4 π π 2 π k = * Note: Any value of x in the domain can be used in place of x = 2 above. or Page 24 6(c)(iii) 2 π α + β = 2 tan x = α ⇒ ⎟⎠⎞ ⎜⎝= − ⎛ 2 α tan 1 x x2 tan = β ⇒ ⎟⎠⎞ ⎜⎝= − ⎛ xβ tan 1 2 β α + = ⎟⎠⎞ ⎜⎝+ ⎛ ⎟⎠⎞ ⎜⎝− ⎛ − x x tan 2 2 tan 1 1 2 π = Blunders (-3) B1 Differentiation B2 Indices B3 Integration Slips (-1) S1 Trig Value S2 Numerical β α x 2 Page 25 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) Taking 1 as a first approximation of a root of x3 + 2x − 4 = 0, use the Newton Raphson method to calculate a second approximation of this root. (a) 10 marks Att 3 7 (a) f (x) = x3 + 2x − 4 f ′(x) = 3x2 + 2 f (1) = (1)3 + 2(1)− 4 = −1 f ′(1) = 3(1)2 + 2 = 5 ( ) ( ) n n n n f x f x x x ′ = − +1 ( ) ( ) 1 1 2 1 f x x x f x ′ = − 1 1 x = ( ) 51 1 − = − 5= 1+ 1 5= 6 Blunder (-3) B1 Newton-Raphson formula once only B2 Differentiation B3 Indices B4 1 1 x ≠ once only Slips (-1) S1 Numerical S2 Answer not tidied up Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 7 (b) (i) Find the equation of the tangent to the curve 3x2 + y2 = 28 at the point (2, – 4). (ii) x = etcost and y = etsint. Show that . x yx y dx dy −+ = (i) Differentiation 5 marks Att 2 Equation 5 marks Att 2 7 (b)(i) 3x2 + y2 = 28 6 + 2 = 0 dx x y dy x dx 2y dy = −6 ⎟⎠⎞ ⎜⎝⎛ y x yx dx dy 3 26 − = − =Page 26 At (2, -4), ( ) 23 4 slope 3 2 = − − = = dx dy Tangent is line through (2, -4) with slope 2m = 3 ( ) ( ) 1 1 y − y = m x − x ( ) ( 2) 2y − − 4 = 3 x − 2(y + 4) = 3(x − 2) 2y + 8 = 3x − 63x − 2y −14 = 0 Blunders (-3) B1 Differentiation B2 Incorrect values or no values B3 Indices B4 Equation of tangent B5 Substituting values into formula once only Slips (-1) S1 Numerical Worthless W1 Integration W2 No differentiation in 1st 5 marks (ii) dt dx and dt dy 5 marks Att 2 dx dy 5 marks Att 2 7 (b)(ii) x = et (cost) y = et (sin t) et ( t) t(et ) dt dx = − sin + cos et ( t) t(et ) dt dy = cos + sin e t e t dt dx = t cos − t sin e t e t dt dy = t cos + t sin ( ) ( ) x yx y e t e te t e t dx dy t tt t dt dx dt dy −+ = −+ = = cos sin cos sin * Note: oversimplified differentiation in first 5 marks leads to Att 2 at most in second marks Blunders (-3) B1 Differentiation B2 Indices B3 Incorrect dx dy B4 Answer not in required form Attempts A1 Blunder in differentiation formula Worthless W1 Integration Page 27 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 7 (c) f(x) = loge3x – 3x, where x > 0. (i) Show that ( 31 , – 1) is a local maximum point of f(x). (ii) Deduce that the graph of f(x) does not intersect the x-axis. (i) Differentiation 5 marks Att 2 Max Value 5 marks Att 2 (ii) Only one root for f ′(x) = 0 5 marks Att 2 Absolute max pt. 5 marks Att 2 7 (c)(i) f (x) = ln(3x)− 3x x > 0 ( ) (3) 3 3′ = 1 − x f x = 1 − 3 x . ( ) 21 x f x − ′′ = Local max/min: ′( ) = 0⇒ 1 − 3 = 0 x f x ⇒ 1 = 3 x ⇒ 3x = 1 . ( ) ( ) 1 1 0 31 2 32 1 < − = − = ⇒ ′′ = x x f x ⇒ local max at 3x = 1 31 = x ⇒ ( ) ( ) ( ) x x x f 3 3 ln − = ( ) ( ) 1 1 ln − = 1 0 − = 1 − = ⇒ Local max at ⎟⎠⎞ ⎜⎝⎛ −1 , 31 or 7(c)(i) f (x) = ln 3x − 3x ′( ) = 1 − 3 x f x 3x = 1 ⇒ ( ) (1) 3 3 3 0 31 f ′ x = − = − = ⇒ turning pt at 3x = 1 . ( ) 1 0 2 < − ′′ = x f x for all x ⇒ local max pt at 3x = 1 31 = x ⇒ ( ) x x y 3 3 ln − = ( ) ⎟⎠⎞ ⎜⎝= − ⎛ 31 3 1 ln 1 − = ⇒ local max is at ⎟⎠⎞ ⎜⎝⎛ ,−1 31 Page 28 (c)(ii) f ′(x)has only one root. This implies that the local max. above is the only turning point. And f (x) is continuous, so the local max pt above is an absolute max. point. Since max pt ⎟⎠⎞ ⎜⎝⎛ ,−1 31 is below x-axis, the whole graph must lie below x-axis Thus, f (x) = 0 has no roots, since graph does not cut the x-axis. * Accept work showing max point to be the only turning point and below x-axis, with or without a diagram. * No need to mention “absolute” in answer. * No need to mention continuity Blunders (-3) B1 Differentiation B2 Not testing in f ′′(x) for max B3 Incorrect deduction or no deduction from test B4 Incorrect y value or no y value B5 Factors once only. Slips (-1) S1 ln1 ≠ 0 Worthless W1 No differentiation Page 29 QUESTION 8 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) Part (a) 10 (5, 5) marks Att (2, 2) 8. (a) Find (i) ∫x3dx (ii) 1 . 3 dx x ∫ (i) 5 marks Att2 (ii) 5 marks Att 2 Q8 (a) (i) ∫ x dx = x + c 44 3 (ii) c x dx x dx x c x + − + = − ∫ = ∫ = − − 2 2 3 3 2 1 2 1 * If c shown once, then no penalty Blunders (-3) B1 Integration B2 Indices B3 No ‘c’ (penalize 1st integration) Attempts A1 Only ‘c’ correct Worthless W1 Differentiation instead of integration Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 8 (b) (i) Evaluate x x 9 dx 4 0 2 ∫ + . (ii) f is a function such that f ′(x) = 6 − sinx and 2 . 3 π π = ⎟⎠⎞ ⎜⎝f ⎛ Find f (x). (i) Integration 5 marks Att 2 Value 5 marks Att 2 8 (b) (i) x x 9 dx 4 0 2 ∫ + = ∫ ( x2 + 9)xdx w2 .dw 1 2= 1 ∫ ⎥⎦⎤ ⎢⎣= ⎡ 2 32 3 21 w [ ] 2 3 3= 1 w ( ) 40 2 31 2 3 9 ⎥⎦ ⎤ ⎢⎣ ⎡ = x + [( ) ( ) ] 2 3 2 3 25 9 3= 1 − [ ] 3 125 27 98 3= 1 − = * Incorrect substitution and unable to finish yields attempt at most. Let w = x2 + 9 x dx dw = 2 dw = xdx 2 Page 30 Blunders (-3) B1 Integration B2 Indices B3 Differentiation B4 Limits B5 Incorrect order in applying limits B6 Not calculating substituted limits B7 Not changing limits Slips (-1) S1 Answer not “tidied up”. Worthless W1 Differentiation instead of integration except where other work merits attempt (ii) f (x) 5 marks Att 2 Value of c 5 marks Att 2 8 (b) (ii) f ′(x) = 6 − sinx f (x) = 6x + cos x + c (π ) 6(π ) cos(π ) 2π 3 3 3 f = + + c = 2π 2π 2+ 1 + c = 2c = −1 ( ) 2⇒ f x = 6x + cos x − 1 Blunders (-3) B1 Integration B2 No ‘c’ Slips (-1) S1 Trig Value Worthless W1 Differentiation instead of integration, except where other work merits attempt. Page 31 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 8 (c) The line 2x – y – 10 = 0 is a tangent to the curve y = x2 − 9 , as shown. The shaded region is bounded by the line, the curve and the x-axis. Calculate the area of this region. Point (1, -8) 5 marks Att 2 Points (3, 0) and (5, 0) 5 marks Att 2 Area under curve between 1 and 3 5 marks Att 2 Finish 5 marks Att 2 8 (c) Co-ords of b: y = x2 − 9 y = 0 : x2 − 9 = 0 x2 = 9 x = ±3 b(3, 0) Shaded area = area Δact − area under curve ac = 4 and at = 8 ⇒ area . (4)(8) 16 21 2Δact = 1 ac at = = or Area (2 10) [ 10 ] (25 50) (1 10) 51 2 51 51 Δact = ∫ ydx = ∫ x − dx = x − x = − − − = (− 25)− (− 9) =16. Area under curve ∫ =∫( − ) 31 2 31 y.dx x 9 dx = 31 3 9 3 ⎥⎦ ⎤ ⎢⎣ ⎡ x − x ( ) 3 9 28 39 27 1 = ⎟⎠⎞ ⎜⎝= − − ⎛ − . Shaded area 3 20 3 48 28 3 16 28 = − = − = y = x2 − 9 2x − y −10 = 0 x y {y = 2x −10}∩{y = x2 − 9} 2x −10 = x2 − 9 0 = x2 − 2x +1 0 = (x −1)2 ⇒ x = 1 y = 2(1)−10 y = −8 ⇒t(1,−8) and a(1, 0) b (3, 0) c(5, 0) t(1, -8) a(1, 0) Co-ords of c: y = 2x −10 y = 0 : 0 = 2x −10 x = 5 c(5,0) Page 32 Blunders (-3) B1 Integration B2 Indices B3 Factors once only B4 Calculation point of tangency B5 Calculation of point where curve cuts x-axis B6 Calculation of point where line cuts x-axis B7 Error in area triangle B8 Error in area formula B9 Incorrect order in applying limits B10 Not calculating substituted limits B11 Error with line B12 Error with curve B13 Uses π ∫ ydx for area formula Attempts A1 Uses volume formula A2 Uses y2 in formula Worthless W1 Differentiation instead of integration except where other work merits attempt W2 Wrong area formula and no work Page 33 MARKING SCHEME LEAVING CERTIFICATE EXAMINATION 2007 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 of merit 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 34 QUESTION 1 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) 1. (a) The following parametric equations define a circle: x = 5 + 7cosθ , y = 7sinθ , where θ ∈ R. What is the Cartesian equation of the circle? Isolates 7cosθ 5 marks Att 2 Finish 5 marks Att 2 1 (a) (x − 5)2 = 49cos2θ and y 2 = 49sin2θ . ( ) ( ) ( 5) 49 or 10 24 0. 5 49 cos sin .2 2 2 2 2 2 2 2 ∴ − + = + − − = − + = + x y x y x x y θ θ Blunders B1 Error in transposition. B2 Fails to square. B3 Sin2θ + cos2θ ≠ 1. Slips S1 Arithmetic errors. Attempts A1 Writes down equation of a circle without any further work. Page 35 Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 1 (b) x2 + y2 − 4x − 6y + 5 = 0 and x2 + y 2 − 6x − 8y + 23 = 0 are two circles. (i) Prove that the circles touch internally. (ii) Find the coordinates of the point of contact of the two circles. (i) One centre and radius (same circle) 5 marks Att 2 Distance between centres 5 marks Att 2 Conclusion 5 marks Att 2 1 (b) (i) x2 + y 2 − 4x − 6y + 5 = 0 has centre c1(2, 3) and radius = r1 = 4 + 9 − 5 = 8 = 2 2. x2 + y2 − 6x − 8y + 23 = 0 has centre c2 (3, 4) and radius = r2 = 9 +16 − 23 = 2. (2 3)2 (3 4)2 2. c1c2 = − + − = r1 − r2 = 2 2 − 2 = 2 = c1c2 . ∴ Circles touch internally. Blunders B1 Error in finding centre or radius. B2 Error in distance formula. B3 Fails to show internal touching Slips S1 Arithmetic (ii) 5 marks Att 2 1 (b) (ii) x2 + y 2 − 4x − 6y + 5 = 0 x2 + y 2 − 6x − 8y + 23 = 0 2x + 2y −18 = 0 ⇒ x + y − 9 = 0 ⇒ x = 9 − y. ( ) ( ) ( ) ( ).2 20 50 0 10 25 0 5 0. 5, 4. Point is 4, 5 9 4 9 6 5 0 81 18 36 4 6 5 0 2 2 2 2 2 2 2 − + = ⇒ − + = ⇒ − = ∴ = = − + − − − + = ⇒ − + + − + − + = y y y y y y x y y y y y y y y y * Accept correct answer without work. Blunders B1 Error in finding equation of the radical axis (common tangent) B2 Error in substitution B3 Error in solving quadratic Slips S1 Arithmetic error. Page 36 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 1 (c) (c) A circle has its centre in the first quadrant. The x-axis is a tangent to the circle at the point (3, 0). The circle cuts the y-axis at points that are 8 units apart. Find the equation of the circle. x-value of centre 5 marks Att 2 ab is perp bisector 5 marks Att 2 Radius 5 marks Att 2 Equation 5 marks Att 2 1 (c) x value of centre is 3 as x-axis tangent at (3, 0). ab is perpendicular bisector of chord, ∴⏐bc⏐=4. Triangle abc is right-angled ⇒ ac = r = 32 + 44 = 5. ⎟ac⎥ = 5 = ⎥ad⎟ ⇒ y value of centre is 5. Circle has centre (3, 5) and radius length 5. ∴ Circle: (x − 3)2 + (y − 5)2 = 25 or x2 + y2 − 6x −10y + 9 = 0. Blunders B1 Incorrect x-coordinate of centre B2 bc ≠ 4 B3 Error in Pythagoras B4 Error in forming equation of the circle. Slips S1 Arithmetic 8 (3, 0) d(3, 0) 4 5 b 3 a c Page 37 QUESTION 2 Part (a) 10 marks Att 3 Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (a) 10 marks Att 3 2. (a) → → → x = −2 i + 5 j and 6 8 . −→ → → xy = − i − j Express →y in terms of →iand . →j (a) 10 marks Att 3 2 (a) 8 3 . → −→ → → → x + xy = y = − i − j Or 6 8 6 8 . 6 8 2 5 8 3 . −→ → → → → → → → → → → → → → xy = − i − j ⇒ y − x = − i − j ∴ y = − i − j − i + j = − i − j Blunders B1 xy ≠ y − x uur r r B2 Error in transposing Slips S1 Arithmetic Part (b) 20 (10, 10) marks Att (3, 3) (b) → → a = 5 i and 3 3 . → → → b = i + j (i) Show that −→ ab is not perpendicular to . →b (ii) Find the value of the real number k, given that → → c = k b and . −→ → ac ⊥ b (b) (i) 10 marks Att 3 2 (b) (i) ( ) . 3( 3 5) 9 0. Not perpendicular. 3 3 5 3 5 3 . = − + ≠ = − = + − = − + −→ → −→ → → → → → → → ab b ab b a i j i i j Blunders B1 ab ≠ b − a uur r r B2 Error in transposing B3 No conclusion Slips S1 Arithmetic Page 38 (b) (ii) 10 marks Att 3 2 (b) (ii) ( ) ( ) . 12 . 0. 3 5 3 9 0 12 5 3 5 3 3 3 . 3 5 3 . ⊥ ⇒ = ∴ − + = ⇒ = ⇒ = = = + = − = − + → → −→ → → → → → −→ → → → → ac b ac b k k k k c k b k i k j ac c a k i k j Blunders B1 ac ≠ c − a uur r r B2 Transposition errors B3 No use of ac.b = 0 uur r Slips S1 Arithmetic Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) (c) → → → p = 3 i + 4 j and 5 12 . → → → q = i + j ⎟ ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎜ ⎝⎛ = + →→ →→ → 16 | | | | 65 qq pr t p , where t > 0 . (i) Express →r in terms of →i and . →j (ii) Find → →p. r and . . → →q r (iii) Hence, show that r is on the bisector of ∠poq, where o is the origin. Part (c) (i) 10 marks Att 3 2 (c) (i) 64 112 . 4 7 . 65 16 39 52 25 60 16 65 135 12 53 4 16 65 16 65 ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + = ∴ ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ = + ⎟ ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎜ ⎝⎛ + + + = ⎟ ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎜ ⎝⎛ + + + = ⎟ ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎜ ⎝⎛ = + → → → → → → → → → → → → → → →→ →→ → r t i j i j t i j r t i j t i j i j | q | q | p | r t p Blunders B1 Error in p or q r r B2 Ignores t or t = some value. S1 Arithmetic Page 39 Part (c) (ii) 5 marks Att 2 2 (c) (ii) . 5 12 4 7 20 84 104 . . 3 4 4 7 12 28 40 . q r i j t i t j t t t p r i j t i t j t t t = + = ⎟⎠⎞ ⎜⎝⎛ + ⎟⎠⎞ ⎜⎝= ⎛ + = + = ⎟⎠⎞ ⎜⎝⎛ + ⎟⎠⎞ ⎜⎝= ⎛ + → → → → → → → → → → → → Blunders B1 Error in calculating scalar product Slips S1 Arithmetic Part (c) (iii) 5 marks Att 2 2 (c) (iii) is on bisector of . . 65 8 13 65 . cos 13 65 104 cos 104 . 65 8 5 65 . cos 5 65cos 40 cos 40 r poq q r q r t t p r p r t t ∴ ∠ = ⇒ = ⇒ = = = ⇒ = ⇒ = = → → → →→ → → → θ θ θ θ θ Blunders B1 Error in scalar product formula B2 Error in modulus. Slips S1 Arithmetic errors Page 40 QUESTION 3 Part (a) 10 marks Att 3 Part (b) 20 (5, 5, 10) marks Att (2, 2, 3) Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (a) 10 marks Att 3 3. (a) Find the area of the triangle with vertices (1, 1), (8, − 5) and (5, − 2). (a) 10 marks Att 3 3 (a) Map (1, 1) onto (0, 0), (8, − 5) onto (7, − 6) and (5, − 2) onto (4, − 3). Area of triangle . 221 24 3 21 21 = x1y2 − x2 y1 = − + = or Use 21 base × perp. height: Taking base: [(1,1), (8,–5)] [(1,1), (5,–2)] [(8,–5), (5,–2]] Length of base: 85 5 18 Equation base-line: 6x+7y = 13 3x+4y = 7 x+y = 3 Distance from other corner: 85 3 53 2 1 ∴Area = 3 3 3 Blunders B1 Error in translation B2 Error in formula for area of a triangle. B3 Incorrect subst Slips S1 Arithmetic errors Page 41 Part (b) 20 (5, 5, 10) marks Att (2, 2, 3) 3(b) f is the transformation (x, y)→(x′, y′), where x′ = 4x + 2y and y′ = −3x − y. K is the line x + y = 0 . (i) Show that K is its own image under f. (ii) p (1, −1) and q (3, −3) are two points. Find the ratio pq : f ( p) f (q) , giving your answer in its simplest form. (i) Evaluate x and y 5 marks Att 2 Find image 5 marks Att 2 3 (b) (i) y x y x x y2 6 2 4 2 ′ = − − ′ = + ( ) ( ) ( ) ( ) ( ) ( ) ( ): 0 ( ) . : 2 3 4 0 2 2 0 0. 3 4 0 22 1 2: 0 : 1 3 4 . 22 1 22 1 2But 1 2 . 22 2 1 f K x y f K K f K x y x y x y x y K x y f K x y x y y x x y x x y y x y x y x x x y + = ⇒ = − ′ − ′ + ′ + ′ = ⇒ ′ + ′ = ⇒ ′ + ′ = + = ⇒ − ′ − ′ + ′ + ′ = = ′ − ⇒ = ′ + ′ + ′⇒ = ′ + ′ ′ + ′ = − ⇒ = − ′ − ′ Blunders B1 Error in setting up or solving simultaneous equations. B2 Error in finding image. Slips S1 Arithmetic errors (b) (ii) 10 marks Att 3 3 (b) (ii) x′ = 4x + 2y and y′ = −3x − y. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) : ( ) ( ) 2 2 : 4 2 1: 2. 2 6 2 6 32 4 2. 1, 1 2, 2 and 3, 3 6, 6 1 3 1 3 8 2 2. 2 2 2 2 ∴ = = ⇒ = − + − + = = = − = − = − = − = − + − + = = pq f p f q f p f q f p f f q f pq Blunders B1 Error in distance formula B2 Error in finding images B3 No ratio shown or incorrect order. Slips S1 Arithmetic errors Page 42 Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) 3 (c) Consider the equation k(3x − 5y + 6)+ l(5x − 7y + 4) = 0. (i) Show that, for any k, l ∈ R, the given equation represents a line passing through the point of intersection of 3x − 5y + 6 = 0 and 5x − 7 y + 4 = 0. (ii) Find the relationship between k and l for which the given equation represents a line of slope 2. (iii) If k = 1, what line through the point of intersection cannot be represented by the given equation? Justify your answer. Part (c) (i) 10 marks Att 3 3(c)(i) The given equation is of first degree in x and y and is therefore a line. It remains to show that it passes through the point of intersection. Let( ) x1, y1 be the point of intersection of 3x − 5y + 6 = 0 and 5x − 7y + 4 = 0. ( ) ( , ) is on the line 5 7 4 0 5 7 4 0. , is on the line 3 5 6 0 3 5 6 0. 1 1 1 1 1 1 1 1 − + = ⇒ − + = − + = ⇒ − + = x y x y x y x y x y x y Blunders B1 Fails to show expression represents a line. B2 Fails to show passes through point of intersection. Slips S1 Arithmetic Part (c) (ii) 5 marks Att 2 3 (c) (ii) k(3x − 5y + 6)+ l(5x − 7y + 4) = 0 ⇒ x(3k + 5l)+ y(− 5k − 7l)+ (6k + 4l) = 0 2 10 14 3 5 7 9 0. 5 7Slope 3 5 = ⇒ + = + ⇒ + = ++ ∴ = k l k l k l k lk l Blunders B1 Error in finding slope B2 Transposing error Slips S1 Arithmetic errors. Part (c) (iii) 5 marks Att 2 3 (c) (iii) ( ) ( ) 5 7 4 0. If 1, the equation 3 5 6 5 7 4 0 cannot represent the line − + = = − + + − + = x y k k x y l x y Justification: If k = 1, the slope of the line will be ll 5 73 5 ++ . There is no value of l that can make this expression equal to 75 , (because attempting to solve this yields 21+ 35l = 25 + 35l , which has no solution). Blunders B1 Fails to justify answer. Page 43 QUESTION 4 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) 4. (a) Show that (cos A + sin A)2 = 1+ sin 2A. Square 5 marks Att 2 Tidy up 5 marks Att 2 4 (a) (cosA + sinA)2 = cos2A + sin2A + 2cosAsinA = 1+ sin2A. Blunders B1 cos2 A+ sin2 A ≠1 B2 2cos Asin A ≠ sin 2A Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 4 (b) Find all the solutions of the equation 6cos2 x + sin x − 5 = 0 , where 0o ≤ x ≤ 360o. Give the solutions correct to the nearest degree. cos2 x = 1 − sin2 x 5 marks Att 2 Quadratic form 5 marks Att 2 Solve quadratic 5 marks Att 2 Values for x 5 marks Att 2 4 (b) ( ) ( )( ) 30 , 150 , 199 , 341 . . 3or sin 1 22sin 1 3sin 1 0 sin 1 6cos2 sin 5 6 1 sin2 sin 5 0 6sin2 sin 1 0. ∴ = o o o o ∴ − + = ⇒ = = − + − = − + − = ⇒ − − = x x x x x x x x x x x Blunders B1 Incorrect substitution for cos2 x B2 Error in factors or quadratic formula. B3 Each incorrect or missing solution. Slips S1 Arithmetic/not rounded. Attempts A1 cos2 x =1− sin2 x Page 44 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 4 (c) [ab] is the diameter of a semicircle of centre o and radius-length r. [ac] is a chord such that ∠cab =α , where α is in radian measure. (i) Find ac in terms of r andα. (ii) [ac] bisects the area of the semicircular region. Show that . 2 2 sin 2 π α + α = (c) (i) 5 marks Att 2 4 (c) (i) [ab] is a diameter, ∴ ∠acb = 90o. 2 cos . 2 cos cosα ac r α r ac ab ac ∠cab = = = ⇒ = Blunders B1 Error in trig ratio B2 ab ≠ 2r B3 Transposing error (ii) Area of triangle 5 marks Att 2 Area of sector 5 marks Att 2 Show 5 marks Att 2 4 (c) (ii) Area of semicircle . 2= 1π r 2 Area of region abc = Area of triangle aoc + sector obc. Area of triangle aoc = ( ) sin2 . 22 cos sin 1 2. sin 1 21 ao ac α = r r α α = r 2 α Area of sector obc (2 ) . 2= 1 r 2 α = r 2α as ∠cob = 2α , since ao = oc ⇒ ∠aco =α. ∴ Area of region sin2 . 2= r 2α + 1 r 2 α This is half the semicircle, so . 2 2 sin2 4sin2 1 22 1 2 2 π r α + r α = π r ⇒ α + α = Blunders B1 Error in area of triangle B2 Error in area of sector Slips S1 Arithmetic o α a c b 90o o α a c bPage 45 QUESTION 5 Part (a) 10 marks Att 3 Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 20 (5, 5, 5,5) marks Att (2, 2, 2, 2) Part (a) 10 marks Att 3 5. (a) Evaluate . sin3Limsin2 0 xx x→ (a) 10 marks Att 3 5 (a) . 32 32 11 32 3 lim sin 3 2 lim sin 2 32 3sin3 2sin2 Lim sin3Limsin2 00 0 0 × = × = ⎟⎠⎞ ⎜⎝⎛ ⎟⎠⎞ ⎜⎝⎛ × = = ⎟ ⎟ ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎜ ⎜ ⎝⎛ = →→ → → x x x x x x x x xx xx x x * Accept correct answer with or without work; if answer is correct, ignore the work. Blunders B1 sin2x = 2sinx. B2 Error in differentiation Slips S1 Arithmetic Part (b) 20 (10, 10) marks Att (3, 3) 5 (b) (b) Using the formula cos(A + B) = cosAcosB − sinAsinB , derive a formula for cos(A − B) and hence prove that sin(A + B) = sinAcosB + cosAsinB. Formula for cos(A–B) 10 marks Att 3 Hence prove 10 marks Att 3 5 (b) ( ) ( ) ( ) cos( ) cos cos sin sin as cos( ) cos and sin( ) sin . cos cos cos sin sin cos cos sin sin . A B A B A B B B B B ( A B ) A B A B A B cos A B A B ∴ − = + − = − = − + = − ⇒ − = − − − ( ) ( [ ]) ([ ] ) ( ) ( ) as cos(90 ) sin and sin(90 ) cos . cos 90 cos sin 90 sin sin cos cos sin sin cos 90 cos 90 A A A A A B A B A B A B A B A B A B − = − = = − + − = + + = − + = − − o o o o o o Blunders B1 Fails to replace B with -B B2 Fails to show cos-B = cosB B3 Fails to show sin -B = -sin B B4 Hence not used Slips S1 Arithmetic error Page 46 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 5 (c) (c) p, q and r are three points on horizontal ground. [sr] is a vertical pole of height h metres. The angle of elevation of s from p is 60° and the angle of elevation of s from q is 30°. pq = cmetres. Given that 3c2 = 13h2 , find ∠prq . Calculates |pr| 5 marks Att 2 Calculates |rq| 5 marks Att 2 Cosine rule 5 marks Att 2 Solves equation 5 marks Att 2 5 (c) . 120 . 21 6 10 13 6 10 3 23 3 2 . cos 3. 3 tan30 1 tan30 . tan60 3 tan60 2 2 2 2 2 2 2 2 2 2 2 2 2 o o o o o = − ∴ ∠ = − = − = + − = + − ∠ = = ⇒ = = = = ⇒ = = prq h h h h h c h h h c pr rq pr rq pq prq rq h h h rq h pr h h pr h Blunders B1 Error in trig ratio B2 Error in Cosine Rule B3 Error in solving equation Slips S1 Arithmetic errors. s p 60° 30° q r h c Page 47 QUESTION 6 Part (a) 10 (5, 5) marks Att ( -, 2) Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) Part (a) 10 (5, 5) marks Att ( -, 2) 6. (a) Six people, including Mary and John, sit in a row. (i) How many different arrangements of the six people are possible? (ii) In how many of these arrangements are Mary and John next to each other? (a) (i) 5 marks Hit/Miss 6 (a) (i) Number of arrangements 6 720. =6P = (a) (ii) 5 marks Att 2 6 (a) (ii) Number of arrangements 2 240. 2 5 =5P × P = Blunders B1 Fails to rearrange Mary and John B2 Uses 5! + 2! Slips S1 Arithmetic Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 6 (b) α and β are the roots of the quadratic equation px2 + qx + r = 0. n n , un = lα + mβ for all n ∈ N. Show that pun+2 + qun+1 + run = 0, for all n ∈ N. Uses root property correctly 5 marks Att 2 Deduces 1 2 , n n u u + + 5 marks Att 2 Substitutes and tidies up 5 marks Att 2 Conclusion 5 marks Att 2 6 (b) 2 2 2 1 1 2 2 1 2 2 1 2 2 1 2 2 2 is a root of 0 0 : 0 : n n n n , n n n n n n n n n n n n n n n n px qx r p q r Similarly p q r Given u l m u l m u l m pu qu ru p l m q l m r l m l p q r m p q r α α α β β α β α β α β α β α β α β α α α β β β + + + + + + + + + + + + + + = ⇒ + + = + + = = + ⇒ = + = + ⇒ + + = ⎡⎣ + ⎤⎦ + ⎡⎣ + ⎤⎦ + ⎡⎣ + ⎤⎦ = ⎡⎣ + + ⎤⎦ + ⎡⎣ + + ⎤ [0] [0] 0lα n mβ n ⎦ = + = Page 48 Blunders B1 Fails to use root property B2 Error in expressing value of term B3 Error in substituting or tidying B4 No conclusion Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) 6 (c) w white discs and r red discs are placed in a box. Two of the discs are drawn at random from the box. The probability that both discs are red is p.. (i) Find p in terms of w and r. (ii) When w = 1, find the value of r for which 2 p = 1 (iii) There are other values of w and r that also give 2p = 1 . The next smallest such value of w is even. By investigating the even numbers in turn, find this value of w and the corresponding value of r. (c) (i) 10 marks Att 3 6 (c) (i) There are ( r + w ) discs in the box. ∴ Number of ways of picking two discs ( )( ). 2 2 1 + + − = r+wC = r w r w Number of ways of picking two red discs ( ). 2 2 1 − = rC = r r ∴ Probability ( ) ( )( ) . 1 1 + + − − = r w r w r r Blunders B1 Incorrect possible B2 Incorrect favourable B3 No division Slips S1 Arithmetic (c) (ii) 5 marks Att 2 6 (c) (ii) ( ) ( ) 2( 1) 1 3 . 21 11 = ⇒ − = + ⇒ = +− r r r r r r r Blunders B1 Error in solving the equation Slips S1 Arithmetic Page 49 (c) (iii) 5 marks Att 2 6 (c) (iii) Probability ( ) ( )( ) , for 2, 4, 6, etc. 21 1 1 = = + + − − = w r w r w r r ( ) ( )( ) No solution possible for a natural number. 2 2 3 2 5 2 0. 21 2 1 2 1 2 2 2 r r r r r r r r r w r r = ⇒ − = + + ⇒ − − = + + − = ⇒ ( ) ( )( ) No solution possible for a natural number. 2 2 7 12 9 12 0. 21 4 3 4 1 2 2 2 r r r r r r r r r w r r = ⇒ − = + + ⇒ − − = + + − = ⇒ ( ) ( )( ) ( 15)( 2) 0 15 as 2. 15 and 6. 2 2 11 30 13 30 0. 21 6 5 6 1 2 2 2 ∴ − + = ⇒ = ≠ − ∴ = = = ⇒ − = + + ⇒ − − = + + − = ⇒ r r r r r w r r r r r r r r w r r Blunders B1 Error in setting up or solving the equation Slips S1 Arithmetic Page 50 QUESTION 7 Part (a) 10 (5, 5) marks Att ( -, 2) Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 20 (5,5,5,5) marks Att (2, 2, 2, 2) Part (a) 10 (5, 5) marks Att( -, 2) 7. (a) (i) How many different selections of four letters can be made from the letters of the word FLORIDA ? (ii) How many of these selections contain at least one vowel? (a) (i) 5 marks Hit/Miss 7 (a) (i) 4 35. 7C = (a) (ii) 5 marks Att 2 7 (a) (ii) Number of selections of four letters with no vowel 4 1. =4C = Number of selections with at least one vowel = 35 −1 = 34. Blunders B1 Error in deriving solution with no vowel B2 Does not subtract Part (b) 20 (10, 10) marks Att (3, 3) 7 (b) Two dice are thrown. (i) What is the probability of getting two identical numbers or a total of five? (ii) What is the probability that the product of the two numbers thrown is at least twice their sum? (b) (i) 10 marks Att 3 7 (b) (i) Number of possible outcomes = 6×6 = 36. Outcomes of interest: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4), (4, 1), (2, 3), (3, 2). So there are 10 outcomes of interest. Probability = . 18 5 36 10 = Blunders B1 Incorrect possible B2 Incorrect favourable B3 No division Slips S1 Arithmetic Page 51 (b) (ii) 10 marks Att 3 7 (b) (ii) Outcomes of interest Product Sum (6, 6) 36 12 (6, 5) or (5, 6) 30 11 (6, 4) or (4, 6) 24 10 (6, 3) or (3, 6) 18 9 (5, 5) 25 10 (5, 4) or (4, 5) 20 9 (4, 4) 16 8 11 outcomes of interest: ∴ Probability . 36 = 11 Blunders B1 Incorrect possible B2 Incorrect favourable B3 No division Slips S1 Arithmetic Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 7 (c) (i) Find, in terms of a and d, the mean of the first seven terms of an arithmetic sequence with first term a and common ratio d. (ii) Show that the standard deviation of these seven numbers is 2d. (c)(i) Correct Total 5 marks Att 2 Mean 5 marks Att 2 7 (c) (i) ( ) 77 2 6 7 21 mean 7 21 3 . 2 7S a d a d x a d a d + = + = + ⇒ = = = + Blunders B1 Error in finding total B2 Error in finding mean Slips S1 Arithmetic (c)(ii) Correct total devs 5 marks Att 2 Correct std dev 5 marks Att 2 7 (c) (ii) a, a+d, a+2d, a+3d, a+4d, a+5d, a+6d x = a + 3d ⇒ –3d, –2d, –d, 0, d, 2d, 3d Σd 2 = 28d 2 . ∴Standard deviation 2 . 7 28 2 d = d Blunders B1 Error in finding total B2 Error in finding Std Dev Slips S1 Arithmetic Page 52 QUESTION 8 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) 8. (a) p and q are real numbers such that p + q = 1. Find the value of p that maximizes the product pq. Expression in one variable 5 marks Att 2 Finishes 5 marks Att 2 8 (a) pq = p(1− p) = p − p2 ( ) ( ) . 22 0 maximum value at 1 . 21 2 0 for maximum 1 2 2 2 2 − = − < ⇒ = − = − = ⇒ =p p p dp d p p p p dp d Blunders B1 Error in finding expression B2 Error in finding derivative Slips S1 Arithmetic Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 8 (b) (i) Derive the Maclaurin series for f (x) = (1+ x)m up to and including the term containing x3. (ii) Given that the general term of the series f (x) is , ! ( 1)( 2).........( 1) xr r m m − m − m − r + show that the series converges for −1 < x < 1. (i) Differentiation 5 marks Att 2 Evaluates at 0 5 marks Att 2 Correct series 5 marks Att 2 8 (b) (i) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 1 2 3 2 3 2 3 ( ) (1 ) 0 1. 1 0 . 1 1 0 1 . 1 2 0 1 2 . 0 0 0 0 ............ 1! 2! 3! 1 1 2 1 1 ... 2! 3! m m m m m f x x f f x m x f m f x m m x f m m f x m m m m f m m m f x f x f x f x f m m x m m m x f x x mx − − − = + ⇒ = ′ = + ⇒ ′ = ′′ = − + ⇒ ′′ = − ′′′ = − − ⇒ ′′′ = − − ′ ′′ ′′′ = + + + + − − − ∴ = + = + + + + .. Page 53 Blunders B1 Error in finding expression B2 Error in finding Derivative B3 Error in establishing series Slips S1 Arithmetic Part (b) (ii) 5 marks Att 2 8 (b) (ii) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) 1 1 1 1 ( 1)( 2).........( 1) 1 2 ......... ! 1! 1 2 ........ 1 1 ! 1 Lim Lim . ! 1 2 ....... 1 Lim Lim 1 for convergency. r r r rr r r r r r r r m m m m r m m m m r u x u x r r u m m m m r x r u r m m m m r x m r x mx x x x x r r r x − + + →∞ →∞ − →∞ →∞ − − − + − − − = ⇒ = − − − − + − = × − − − − + = = − + = − ⇒ < <1 ⇒ −1< x <1. Blunders B1 Error in finding expression B2 Error in finding Derivative B3 Error in establishing range. Slips S1 Arithmetic Page 54 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 8 (c) Evaluate tan . 1 0 ∫ 1xdx − Set up integration by parts 5 marks Att 2 Parts stage done 5 marks Att 2 ∫ + dx x x 1 2 5 marks Att 2 Evaluation 5 marks Att 2 8 (c) ∫udv = uv − ∫vdu. Let and . 1 tan and . 1 2 1 dx v x x u x dv dx du = + = − = ∴ = [ ] [ ] [ ] log 2. 21 4 log 2 2log 2 tan 1 1 2tan tan 1 log 2. 2log 2 log 1 1 2log 1 21 21 1 . Let 1 2 . 1 . 1 tan tan 1 10 1 1 0 1 21 2 1 1 0 2 2 1 0 2 1 0 2 1 1 0 1 e e e e e e e xdx x x x w dw x xdx dx w x dw xdx x x dx x xdx uv vdu x x x ∴ = − = − = − = = = − = + ∴ = + ⇒ = + + ∴ = − = − − − − − − ∫ ∫ ∫ ∫ ∫ ∫ ∫ π Blunders B1 Error in setting up expression B2 Error in integrating B3 Error in establishing value Slips S1 Arithmetic Page 55 Note: Candidates may attempt to use a Maclaurin expansion to answer this question. They are unlikely to make substantial progress. The solution below is presented so as to facilitate the award of relevant partial credit. Expand tan-1x 5 marks Att 2 Integrate & evaluate at limits 5 marks Att 2 Partial fractions & separate 2 series 5 marks Att 2 Finish 5 marks Att 2 L +L + = − + − + + − + − 2 1 ( 1) 3 5 7 tan 3 5 7 2 1 1 n x x x x x x n n ∴ L +L + + = − + − + + − + − ∫ (2 1)(2 2) ( 1) 1.2 3.4 5.6 7.8 tan 1 2 4 6 8 2 2 0 1 n n xdx x x x x x n n L +L + + = − + − + + − (2 1)(2 2) ( 1) 1 7.8 1 5.6 1 3.4 1 1.2 1 n n n L L + ⎟⎠⎞ ⎜⎝⎛ + − + − + + ⎟⎠⎞ ⎜⎝− ⎛ − ⎟⎠⎞ ⎜⎝+ ⎛ − ⎟⎠⎞ ⎜⎝− ⎛ − ⎟⎠⎞ ⎜⎝= ⎛ − 2 2 1 2 1 ( 1) 1 81 71 61 51 41 31 21 11 n n n = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + + − + + − + − − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + + − − + − +L+ L L L 2 2 ( 1) 81 61 41 21 2 1 ( 1) 71 51 31 1 n n n n = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + + − − − − + − + + L L 1 ( 1) 41 31 21 1 2tan 11 1 n n ln(1 1) 2= tan−11− 1 + ln 2 21 4 = − π . Page 56 QUESTION 9 Part (a) 10 (5, 5) marks Att (2, 2) Part (b) 20 (10, 10) marks Att (3, 3) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) Part (a) 10 (5, 5) marks Att (2, 2) 9. (a) Two events E1 and E2 are independent. . 7 and ( ) 1 5(E ) 1 2 P 1 = P E = Find (i) ( ) 1 2 P E ∩ E (ii) ( ) 1 2 P E ∪ E . (a) (i) 5 marks Att 2 9 (a) (i) ( ) ( ) ( ) . 35 1 71 5. 2 1 P E1 ∩ E2 = P E1 P E = × = Blunders B1 Addition for multiplication Slips S1 Arithmetic (a) (ii) 5 marks Att 2 9 (a) (ii) ( ) ( ) ( ) ( ) . 35 11 35 1 71 51 P E1 ∪ E2 = P E1 + P E2 − P E1 ∩ E2 = + − = Blunders B1 Multiplication for addition B2 Double counts Slips S1 Arithmetic Page 57 Part (b) 20 (10, 10) marks Att (3, 3) 9 (b) Five unbiased coins are tossed. (i) Find the probability of getting three heads and two tails. (ii) The five coins are tossed eight times. Find the probability of getting three heads and two tails exactly four times. Give your answer correct to three places of decimals. Part (b) (i) 10 marks Att 3 9 (b) (i) . 16 5 32 10 21 2 Probability 1 2, 1 21 3 2 3 5 = = ⎟⎠⎞ ⎜⎝⎛ ⎟⎠⎞ ⎜⎝p = q = ⇒ = C ⎛ Blunders B1 Error in finding p or q B2 Error in finding Binomial Coefficients B3 Error in evaluation Slips S1 Arithmetic Part (b) (ii) 10 marks Att 3 9 (b) (ii) 0.149. 16 11 16 Probability 5 16 , 11 16 5 4 4 4 8 = ⎟⎠⎞ ⎜⎝⎛ ⎟⎠⎞ ⎜⎝p = q = ⇒ = C ⎛ Blunders B1 Error in finding p or q B2 Error in binomial coefficients Slips S1 Arithmetic Page 58 Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 9 (c) The amounts due on monthly mobile phone bills in Ireland are normally distributed with mean €53 and standard deviation €15. (i) If a monthly phone bill is chosen at random, find the probability that the amount due is between €47 and €74. (ii) A random sample of 900 mobile phone bills is taken. Find the probability that the mean amount due on the bills in the sample is greater than €53·3. (i) z1 and z2 5 marks Att 2 Finish 5 marks Att 2 9 (c) (i) ( ) ( ) ( ) ( ) ( ) [ ( )] 0.9192 (1 0.6554) 0.9192 0.3446 0.5746. 0.4 1.4 1.4 0.4 0.9192 1 0.41.4. 15 21 150.4. 74 53 15 6 1547 53 53, 15. 47 74 . 1 2 1 1 = − − = − = − < < = ≤ − > = − − ≤ = = − = − = − = − = − = = = < < = < < P z P z P z P z z x x z x P x P z z z σσ Blunders B1 Error in finding z1 or z2 B2 Error in setting up probability B3 Error in evaluation Slips S1 Arithmetic (ii) Std error 5 marks Att 2 Finish 5 marks Att 2 9 (c) (ii) ( ) ( ) ( ) 1 0.7257 0.2743. 0.6 1 0.6 0.5 53.3 53.3 53 21 30 15 900 15 900, 53, 15. = − = ≤ − = > = ⎟⎠⎞ ⎜⎝⎛ − > = > = = = = = = = P x P z P z P z n n x x σ σ σ Blunders B1 Error in finding standard error B2 Error in evaluation Slips S1 Arithmetic Page 59 QUESTION 10 Part (a) 10 (5, 5) marks Att ( -, -) 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 ( -, -) 10. (a) For each of the following, give a reason why it is not a group. (i) The set of natural numbers under subtraction. (ii) The set of real numbers under multiplication. (a) (i) 5 marks Hit/Miss 10 (a) (i) Not closed: e.g. 6−14 = −8, −8 ∉ N. (a) (ii) 5 marks Hit/Miss 10 (a) (ii) Not all elements have inverses: 0∈R, but 0 has no multiplicative inverse in R. Part (b) 20 (10, 10) marks Att (3, 3) 10 (b) G = { Iπ , R180o , SX , SY } is the set of symmetries of the rectangle abcd. (i) Show that G is a group under composition. You may assume that composition is associative. (ii) Find Z(G), the centre of the group. Part (b) (i) 10 marks Att 3 10 (b) (i) ο Iπ R180o S X SY Closed: No new element. Iπ Iπ R180o S X SY Associative: yes, given. R180o R180o Iπ SY S X Identity : Iπ S X S X SY Iπ R180o Inverses : ( ) , ( ) 180 , 1 180 1 o o I − = I R − = R π π SY SY S X R180o Iπ ( ) , ( ) 1 . 1 S X = S X SY − = SY − Blunders B1 Identity not given B2 Inverses not stated B3 Closure not defined Slips S1 each inverse not given a cb X Y d Page 60 Part (b) (ii) 10 marks Att 3 10 (b) (ii) In table elements are symmetrical about main diagonal. ∴ G is a commutative group ⇒ G(Z ) = G = { Iπ , R180o , S X , SY }. or from the table, x o y = y o x for all x,y ∈G i.e each element commutes with each other element so Z(G) = G. Blunders B1 Each element missing from set. Part (c) 20 (10, 10) marks Att (3, 3) 10 (c) Use Lagrange’s theorem to prove that (i) any group of prime order is cyclic. (ii) the order of any element of a finite group G divides the order of G. Part (c) (i) 10 marks Att 3 10 (c) (i) Let (G,∗) be a group of order k, where k is prime. Let a∈G and a ≠e. ∴ a , (the group generated by a,) is a subgroup of G. Hence, the order of a is a factor of k (by Lagrange’s theorem). But k is prime ⇒ order of a = k, (since the order of a ≠ 1, since a ≠ e). ∴ a = G. ∴ G is cyclic. Blunders B1 Fails to establish that < a > is a subgroup of G. B2 Fails to use Lagrange B3 No conclusion Part (c) (ii) 10 marks Att 3 10 (c) (ii) Let (G,∗) be a group of order n. Let a ∈G and let the order of a be m. Then m is also the order of , the subgroup generated by a. But the order of a subgroup divides the order of the group (by Lagrange’s theorem). ∴ m is a factor of n. That is, the order of the element a divides the order of the group G. Blunders B1 Fails to use Lagrange B2 No conclusion. Page 61 QUESTION 11 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 11. (a) Find the eccentricity of an ellipse with equation 1. 64 482 2 x + y = (a) 10 marks Att 3 11 (a) a2 = 64, b2 = 48 and b2 = a2 (1− e2 ). ( ) . 2. 1 448 = 64 1− e2 ⇒ 64e2 = 16 ⇒ e2 = 1 ∴ e = Blunders B1 Incorrect a2 B2 b2 ≠ a2 (1− e2 ) Slips S1 Arithmetic Part (b) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 11 (b) Prove that a similarity transformation maps the orthocentre of a triangle onto the orthocentre of the image triangle. Orthocentre 5 marks Att 2 Mapping 5 marks Att 2 Perp invariant 5 marks Att 2 Conclusion 5 marks Att 2 11 (b) f is a similarity transformation. [ps]⊥ [qr] and [rt]⊥ [pq] ⇒ h is the orthocentre of triangle pqr. By f, triangle pqr is mapped to triangle p′q′r′. To Prove: f (h) is orthocentre of triangle p′q′r′. By f, [ps] maps to [p′s′] and [rt] maps to [r′t′]. But [ps]⊥ [qr] ⇒ [p′s′]⊥ [q′r′] as perpendicularity is invariant. Similarly [ ] [ ].r′t′ ⊥ p′q′ ∴ h′ is orthocentre of triangle p′q′r′. But h = [ps]∩[rt] maps to f (h) = [p′s′]∩[r′t′]. ∴ f (h) = h′ ⇒ f (h) is orthocentre of triangle p′q′r′. 90o s′ 90o p′ q′ r′ t′ h′ 90o p q r s t 90o h f Page 62 Blunders B1 Fails to define orthocentre B2 Fails to state perpendicularity is invariant B3 Fails to show h maps onto f(h) Part (c) 20 (5, 5, 5, 5) marks Att (2, 2, 2, 2) 11 (c) E is the ellipse 1 4 2 2 x + y = and L is the line y = x. Using a transformation that maps E to the unit circle, or otherwise, find the equation of the diameter that is conjugate to L in E. Transformation 5 marks Att 2 f(L) 5 marks Att 2 Conjugate diameter 5 marks Att 2 Inverse map 5 marks Att 2 11 (c) Let f be the transfomation (x, y)→(x′, y′) where x = 2x′ and y = y′. ( ) ( ) ( ) ( ) ( ) is 0 ( ): 2 0. 1 1. 4 : 2 2 2 2 2 − = ⇒ ′ − ′ = + ′ = ⇒ ′ + ′ = ′ L x y f L x y f E x y x y But f (E) is a circle, so the conjugate diameter is the perpendicular diameter. Slope of f (L) is 2, so slope of f(K) is 12 − . Hence, equation of f (K) is x′ + 2y′ = 0. By f −1 , f (K) maps to K, the conjugate diameter of L in E. Applying f −1 , we get 2 0 2 K : x + y = . So, the conjugate diameter of L is x + 4y = 0. Blunders B1 Incorrect transformation B2 Incorrect image for x -y =0 B3 Error in mapping back to the ellipse. Slips S1 Arithmetic E L Page 63 Marcanna Breise as ucht freagairt trí Ghaeilge (Bonus marks for answering through Irish) Ba chóir marcanna de réir an ghnáthráta a bhronnadh ar iarrthóirí nach ngnóthaíonn thar 75% d’iomlán na marcanna don pháipéar. Ba chóir freisin an marc bónais sin a shlánú síos. Déantar an cinneadh agus an ríomhaireacht faoin marc bónais i gcás gach páipéar ar leithligh. Is é 5% an gnáthráta agus is é 300 iomlán na marcanna don pháipéar. Mar sin, bain úsáid as an ngnáthráta 5% i gcás marcanna suas go 225. (e.g. 198 marks × 5% = 9·9 ⇒ bónas = 9 marc.) Thar 225, is féidir an bónas a ríomh de réir na foirmle seo: [300 – bunmharc] × 15%, (agus an marc sin a shlánú síos). In ionad an ríomhaireacht sin a dhéanamh, is féidir úsáid a bhaint as an tábla thíos. Bunmharc Marc Bónais 226 11 227 – 233 10 234 – 240 9 241 – 246 8 247 – 253 7 254 – 260 6 261 – 266 5 267 – 273 4 274 – 280 3 281 – 286 2 287 – 293 1 294 – 300 0