Coimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 2010 MARKING SCHEME MATHEMATICS HIGHER LEVEL

Transcription

1 Coimisiún na Scrúduithe Stáit State Eaminations Commission LEAVING CERTIFICATE 00 MARKING SCHEME MATHEMATICS HIGHER LEVEL

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3 Contents Page GENERAL GUIDELINES FOR EXAMINERS PAPER... QUESTION...5 QUESTION...0 QUESTION... QUESTION...8 QUESTION 5... QUESTION QUESTION 7... QUESTION GENERAL GUIDELINES FOR EXAMINERS PAPER... QUESTION... QUESTION...6 QUESTION...9 QUESTION...5 QUESTION QUESTION QUESTION QUESTION QUESTION QUESTION QUESTION...76 MARCANNA BREISE AS UCHT FREAGAIRT TRÍ GHAEILGE...8 Page of 8

4 GENERAL GUIDELINES FOR EXAMINERS PAPER. Penalties of three types are applied to candidates work as follows: Blunders - mathematical errors/omissions (-) Slips - numerical errors (-) Misreadings (provided task is not oversimplified) (-). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B, B, B,, S, S,, M, M, etc. These lists are not ehaustive.. When awarding attempt marks, e.g. Att(), 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.. Worthless work is awarded zero marks. Some eamples of such work are listed in the scheme and they are labelled as W, W, etc.. 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 bo containing the relevant solution. 7. The sample solutions for each question are not intended to be ehaustive lists there may be other correct solutions. Any eaminer unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising eaminer. 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. 0. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most.. A serious blunder, omission or misreading results in the attempt mark at most.. Do not penalise the use of a comma for a decimal point, e.g may be written as 5,50. Page of 8

5 QUESTION Part (a) 0 (5, 5) marks Att (, ) Part (b) 0 (5, 0, 5) marks Att (,, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) (5, 5) marks Att (, ). (a) ( ) 6 t k, where t and k are constants. Find the value of k and the value of t. (a) Equating coefficients 5 marks Att Values 5 marks Att (a) 6 t ( k) 6 t k k. k 6 and t k k and t 9. Or (a) Perfect square 5 marks Att Values 5 marks Att (a) ( k) 6 t ( 6 t) is a perfect square ( ) 6 9 k and t 9 Blunders (-) B Epansion ( a) once only B Not like-to-like in equating coefficients B Indices Page 5 of 8

6 Part (b) 0 (5, 0, 5) marks Att (,, ) (b) Given that p is a real number, prove that the equation real roots. p p 0 has (b) Equation arranged 5 marks Att Correct substitution in b ac 0 marks Att Finish 5 marks Att (b) ( ) p p 0 p p 0. ( ) ( ) b ac p p p p p p p for all. Roots are real. Blunders (-) B Epansion of ( a b) once only B Incorrect value a B Incorrect value b B Incorrect value c B5 Inequality sign B6 Indices B7 Incorrect deduction or no deduction Page 6 of 8

7 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) (c) ( ) and ( ) are factors of (i) Epress c in terms of b. (ii) Epress d in terms of b. b c d. (iii) Given that b, c and d are three consecutive terms in an arithmetic sequence, find their values. f() and f(-) 5 marks Att c in terms of b 5 marks Att d in terms of b 5 marks Att Values 5 marks Att (c) (i) f 0. 8 b c d 0 b c d 8. ( ) is a factor ( ) ( ) is a factor ( ) f 0. b c d 0 b c d. b c 9 b c c b. (c) (ii) By part (i) b c d 8 b c d 6b d 6 b d d b. (c) (iii) An arithmetic sequence b, c, d c b d c c b d. b 6 b b b. c and d 6. Blunders (-) B Indices B Deduction root from factor B Statement of AP Slips (-) S Numerical Worthless W Geometric Sequence Or Page 7 of 8

8 Division & remainder 0 5 marks Att c in terms of b 5 marks Att d in terms of b 5 marks Att Values 5 marks Att (c) (i) factor ( )( ) ( ) (c) (ii) b ( b ) c d ( b ) ( c ) d ( b ) ( b ) ( b ) ( c ) ( b ) d ( b ) 0 since ( ) is a factor [( c ) ( b ) ] d ( b ) 0 Equating Coefficients [ ] ( ) ( 0) (i) b c 0 c b (ii) d b 0 d b (c) (iii) As in previous solution Blunders (-) B ( )( ) once only B Indices B Not like-to-like when equating coefficients Slips (-) S Not changing sign when subtracting Attempts A Any effort at division Worthless W Geometric sequence Page 8 of 8

9 Other linear factor & multiplication 5 marks Att c in terms of b 5 marks Att d in terms of b 5 marks Att Values 5 marks Att (c) (i) (ii) ( )( ) ( ) d ( ). b c d factor d d d b c d d d d b c Equating Coefficients ( ) ( ) ( d ) (i) : d c d c (ii) : d b d b b d Put this value of d into (i) (i) ( b ) c b c 6 b c c b (c) (iii) As in previous solution Blunders (-) B Indices B ( )( ) once only B Not like to like when equating coefficients Attempts A Other factors not linear in () only Worthless W Geometric sequence Page 9 of 8

10 QUESTION Part (a) 0 (5, 5) marks Att (, ) Part (b) 0 (0, 0) marks Att (, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) 0 (5, 5) marks Att(, ) (a) Solve the simultaneous equations y 0 y z 0 y z 9. (a) One unknown 5 marks Att Other values 5 marks Att (a) y z 0 y z 9 7 6y 9 6y 0 9. y and z. Blunders (-) B Multiplying one side of equation only B Not finding nd value, having found st value B Not finding rd value, having found other two Slips (-) S Numerical S Not changing sign when subtracting Worthless W Trial and error only Page 0 of 8

11 Part (b) 0(0, 0) marks Att (, ) (b) The equation 6 0 has roots α and β, where α > 0 and β > 0. (i) Find the value ofαβ. (ii) Hence, find the value of α β. (b) (i) Value of αβ 0 marks Att α β 0 marks Att (b) (ii) Value of ( ) (b) (i) α β 6 αβ. (b) (ii) α β and αβ. ( α β ) α α β Blunders (-) B Indices B Incorrect sum B Incorrect product B Incorrect statements B5 Ecess value each time Slips (-) S Numerical β 0 αβ Page of 8

12 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) (c) (i) Prove that for all real numbers a and b, a ab b ab. (ii) Let a and b be non-zero real numbers such that a b 0. a b Show that. b a a b (c) (i) 5 marks Att (ii) Factors 5 marks Att Use of part (i) 5 marks Att Finish 5 marks Att (c) (i) ( ) a b a ab b 0 0. a ab b ab. (c) (ii) (c) (ii) ( a b)( a ab b ) a b a b. b a ab ab ( a b )( a ab b ) ab But ( a b ), by part (i) ab ab ab( a b) a b a b. a b ab ab ab b a a b. b a a b OR a b b a a b Multiply across by ab, which is positive: a b ab ba a b a ab b ab a b ( )( ) ( ) a ab b ab, since a b 0 true, by part (i). Blunders (-) B Epansion ( a b) once only B B B B5 Factors a b Indices Inequality sign Incorrect deduction or no deduction Page of 8

13 Slips (-) S Numerical Attepmts a b a b a b A ( )( ) Worthless W Particular values (c) (i) 5 marks Att (ii) Common denominator 5 marks Att Factorised 5 marks Att Finish 5 marks Att (c) (i) ( a ab b ) ab, ( ab b ) ab 0 ( a ab b ) ab a ab b a. ( a b) 0 (c) (ii) a b a b, 0. b a a b b a a b a b a b ab a b b a a b a b ( a a b) ( ab b ) a b a ( a b) b ( a b) a b ( a b)[ a b ] a b ( a b) [( a b)( a b) ] ( ab) ( a b) ( a b) ( ab) 0, since a b 0 Blunders (-) B Indices B Inequality Sign B Factors ( a b ) once only B Incorrect deduction or no deduction Worthless W Particular values Page of 8

14 QUESTION Part (a) 0 (5, 5) marks Att (, ) Part (b) 0 (5, 5, 0) marks Att (,, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) 0 (5, 5) marks Att(, ) (a) Find and y such that y Inverse of A evaluated 5 marks Att Finish 5 marks Att (a) y 6 y y Or. One unknown 5 marks Att Other unknown 5 marks Att (a) (i) y y 0 (ii) 5 6y. 0 y 8 Blunders (-) B Formula for inverse B Matri multiplication (i) y 0 y 0 y 8 y 8 Slips (-) S Each incorrect element in matri multiplication S Numerical S Not changing sign when subtracting Page of 8

15 Part (b) 0 (5, 5, 0) marks Att (,, ) (b) Let z s 8i and z t 8, where s, t and i. i (i) Given that z 0, find the values of s. (ii) π Given that arg( z ), find the value of t. (b) (i) Values for modulus 5 marks Att Values of s 5 marks Att (ii) Value of t 0 marks Att (b) (i) s 8i 0 s 6 0 s 6. s ± 6. π 8 (b) (ii) tan t 8 t 8. t (b) (i) z s i z 0 8 s s s 6 s ±6 Or (b) (ii) π tan α tan 8 t t 8 t 8 z 8 t α 8i θ π π θ α Blunders (-) B Formula for modulus B Indices B Only one value for s B Diagram for z once only B5 Incorrect argument B6 Trig Definition B7 Mod Values B8 π tan Slips (-) S Trig value S Numerical Page 5 of 8

16 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) (c) (i) Use De Moivre s theorem to find, in polar form, the five roots of the equation z 5. (ii) Choose one of the roots w, where w. Prove that w w is real. nπ (c) (i) z cis 5 5 marks Att Five roots 5 marks Att (c) (ii) w w as sum of cos and sin 5 marks Att Show real 5 marks Att (c) (i) 0 nπ 0 nπ z ( cos0 isin0) 5 cos isin, for n 0,,,,. 5 5 n 0 z0. n π π z cos isin. 5 5 n π π z cos isin. 5 5 n 6π 6π z cos isin. 5 5 n 8π 8π z cos isin. 5 5 (c) (ii) π π Let w z cos isin. 5 5 w w π π π π cos i sin cos i sin π π 6π 6π cos i sin cos i sin π π 6π π cos cos i sin sin π π cosπcos i sinπ cos 5 5 π cos i(0), 5 π cos, which is real 5 Blunders (-) B Formula De Moivre once only B Application De Moivre B Indices Page 6 of 8

17 B B5 B6 Trig Formula Polar formula once only i Slips (-) S Trig value S Root omitted Note: Must show (0)i Attempt A Use of decimals in c(ii) Worthless W w used in c(ii) Page 7 of 8

18 QUESTION Part (a) 0 (5, 5) marks Att (, ) Part (b) 5 (5, 5, 5) marks Att (,, ) Part (c) 5 (5, 5, 5, 5, 5) marks Att (,,,, ) Part (a) 0 (5, 5)marks Att (, ) (a) Write the recurring decimal as an infinite geometric series and hence as a fraction. (a) Series 5 marks Att Fraction 5 marks Att (a) a r Blunders (-) B Infinity formula once only B Incorrect a B Incorrect r Slips (-) S Numerical Page 8 of 8

19 Part (b) 5 (5, 5, 5) marks Att (,, ) (b) In an arithmetic sequence, the fifth term is 8 and the tenth term is. (i) Find the first term and the common difference. (ii) Find the sum of the first fifteen terms of the sequence. (b) (i) Terms in a and d 5 marks Att Values of a and d 5 marks Att (b) (ii) Sum 5 marks Att (b) (i) T5 8 a d 8 T0 a 9d 5d 0 d 6 and a (b) (ii) S n n S 5 5 { a ( n ) d}. { 8 ( 6) } ( 0) 0. 5 Blunders (-) B Term of A.P. B Formula A.P. once only (term) B Incorrect a B Incorrect d B5 Formula for sum arithmetic series once only Slips (-) S Numerical Worthless W Treats as G.P. Page 9 of 8

20 Part (c) 5 (5, 5, 5, 5, 5) marks Att (,,,, ) (c) (i) Show that ( r ) ( r ) 6r. (ii) Hence, or otherwise, prove that 0 (iii) Find ( r. ) r n r r n ( n )( n ). (c) (i) 5 marks Att (c) (i) ( r ) ( r ) r r r ( r r r ) 6r. 6 (c) (i) OR [ ] [ r r ][ r r r r ] ( r ) ( r ) [( r ) ( r ) ]( r ) ( r )( r ) ( r ) r r ( )( ) 6r Blunders (-) B Epansion of ( r ) once only B Epansion of ( r ) once only B B Formula Indices a B5 Epansion of ( r ) once only B6 Epansion of ( ) B7 b r once only Binomial epansion once only Page 0 of 8

21 (c) (ii) Cancellation & addition shown 5 marks Att Prove 5 marks Att (c) (ii) ( ) ( n ) n 0 6 ( ) ( ) 6 6 ( n ) ( n ) 6 ( n ) ( n ) ( n ) ( n ) 6 ( ) 6n n n n 6 r n r n r n n n n n n n n 6 6 r ( ) ( ) ( ) ( )( ) n n n n n n. 6 6 OR (c) (ii) Prove by induction that n... n ( n )( n ) 6 P(): Test n : ( )( ) True for n. 6 P(k): Assume true for n k: k S k ( k )( k ) 6 k To prove : S k ( k )( k ) 6 Proof: ( ) k S ( )( ) ( ) k... k k k k k, using P(k) 6 ( k ) [ k( k ) 6( k ) ] 6 ( k ) [ k k 6k 6] 6 k [ k 7k 6] 6 k [( k )( k ) ] 6 Formula true for n ( k ) if true for n k It is true for n true for all n * Must show three terms at start and two at finish or vice versa in first method. Page of 8

22 Blunders (-) B Indices B Cancellation must be shown or implied B Term omitted B Epansion ( n ) once only (c) (iii) Substitution of r 0 and r 0 5 marks Att Sum 5 marks Att (c) (iii) ( r ) r r 0 0 r Blunders (-) B Formula B Not (Σ0 Σ0) B Value n Slips (-) S Numerical ( )( )( 6) ( 0)( )( ) Page of 8

23 QUESTION 5 Part (a) 0 (5, 5)marks Att(, ) Part (b) 0 (5, 5, 0) marks Att (,, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) 0 (5, 5)marks Att (, ) (a) Solve log ( 6) log ( ). (a) Log law applied 5 marks Att Value 5 marks Att log ( 6) log ( ). 6 6 log 6. Blunders (-) B Log laws B Indices Page of 8

24 Part (b) 0 (5, 5, 0) marks Att (,, ) (b) Use induction to prove that where n is a positive integer. ( ) ( ) ( )... ( ) n n Part (b) P() 5 marks Att P(k) 5 marks Att P(k ) 0 marks Att 5 (b) Test for n, P ( ). True for n. Assume P( k ). (That is, assume true for n k.). k i.e., assume S, where S k is the sum of the first k terms. k Deduce P( k ). (That is, deduce truth for n k.) k i.e. deduce that S k. k k k k Proof: Sk Sk Tk ( ). True for n k. So, P ( k ) is true whenever P ( k) is true. Since ( ) P ( n) is true, for all positive integers n. Blunders (-) B Indices B Not T k added to each side B Not n Worthless W P(0), P is true, then, by induction, Page of 8

25 Page 5 of 8 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) (c) (i) Epand and. (ii) Hence, or otherwise, find the value of, given that. (c) (i) 5 marks Att 5 marks Att (c) (ii) Terms collected 5 marks Att Value 5 marks Att 5 (c) (i).. 6 C C C OR ( ) 6 5 (c) (ii) But. 75

26 Blunders (-) B Binomial Epansion once only B Indices B n n Value or no r r B 0 B5 Epansion once only B6 Epansion once only B7 Value or no value OR (c) (ii) Roots 5 marks Att Value 5 marks Att 5 (c) (ii) ( ) 7 0 7± Similarly, when, 7. Note: must test two roots. Blunders (-) B Roots formula once only B Indices B Epansion once only Attempts A Decimals used Page 6 of 8

27 QUESTION 6 Part (a) 0 (5, 5) marks Att (, ) Part (b) 0 (5, 5, 0) marks Att (,, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) 0 (5, 5) marks Att (, ) (a) The equation 0 has only one real root. Taking as the first approimation to the root, use the Newton-Raphson method to find, the second approimation. (a) Differentiation 5 marks Att Value 5 marks Att 6 (a) f ( ) f. f ( ) 7 9 f ( ) f f ( ) f Blunders (-) B Newton-Raphson formula once only B Differentiation B Indices B Page 7 of 8

28 Part (b) 0 (5, 5, 0) marks Att (,, ) (b) Parametric equations of a curve are: t t t y, where t \ { }. t (i) dy Find. d (ii) What does your answer to part (i) tell you about the shape of the graph? (b)(i) d dt dy d 6 (b) (i) or dy dt t t t y t dy d dy. dt dt d 5 marks Att 5 marks Att ( t ) ( t ) 5 ( t ) ( t ). ( t ) t( ). ( t ) ( t ) ( t ).. ( t ) 5 5 d dt dy dt OR (b) (i) Elimination of t 5 marks Att dy d 5 marks Att 6 (b) (i) t t y t t ( ) y t t y ( ) t ( ) ( ) 5y dy d 5 ( y) ( y ) Page 8 of 8

29 Blunders (-) B Indices B Differentiation B dy Incorrect d Attempts A Error in differentiation formula (b) (ii) 0 marks Att 6 (b) (ii) Since the slope is constant, it is a (subset of a) straight line. If line is not mentioned in the answer, can only get Att at most. Page 9 of 8

30 Part (c) 0(5, 5, 5, 5) marks Att (,,, ) (c) A curve is defined by the equation y y. (i) dy Find in terms of and y. d (ii) Show that the tangent to the curve at the point ( 0, 6) is also the tangent to it at the point (, 0). (c) (i) Differentiation 5 marks Att Isolate dy d 5 marks Att (c) (ii) Equation st Tangent 5 marks Att Equation nd Tangent 5 marks Att 6 ( c) (i) dy dy y y.y y. 0. d d dy dy y ( y ) y. d d y 6 (c) (ii) dy y d y Slope of tangent at (0, 6) is Equation of tangent at (0, 6) is y 6 y 6. Slope of tangent at (, 0) is y y 6. Equation of tangent at (, 0) is ( ) same tangent. Blunders (-) B Differentiation B Indices B Incorrect value of or no value of in slope B Incorrect value of y or no value of y in slope B5 Equation of tangent B6 Incorrect conclusion or no conclusion Slips (-) S Numerical Attempts A Error in differentiation formula dy dy dy A y and uses the three d d d OR dy term d Page 0 of 8

31 (c) (ii) 0 marks Att 6 (c) (ii) dy y d y Slope of tangent at A(0, 6) is Slope of tangent at B(, 0) is m m 6 Slope of the line [AB] is m So, m m m the line through A and B is the tangent at both points. Blunders (-) B Slope omitted B Incorrect deduction or no deduction Page of 8

32 QUESTION 7 Part (a) 0 (5, 5) marks Att (, ) Part (b) 0 (0, 0) marks Att (, ) Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) Part (a) 0 (5, 5) marks Att (, ) (a) Differentiate with respect to from first principles. f ( h) f ( ) simplified 5 marks Att Finish 5 marks Att 7 (a) ( ) f ( h) ( h). f ( h) f ( ) ( h) f dy d limit Limit h 0 h 0 h ( h). limit h 0 h Blunders (-) B f ( h) B Indices B Epansion of ( h) once only B h B5 No limits shown or implied or no indication of h 0 h h limit h 0 h Page of 8

33 Part (b) 0(0, 0) marks Att (, ) cos sin (b) Let y. cos sin dy (i) Find. d (ii) Show that dy d y. (b) (i) Differentiation 0 marks Att (ii) Show 0 marks Att 7 (b) (i) cos sin dy ( cos sin)( sin cos) ( cos sin)( sin cos) y. cos sin d cos sin dy d 7 (b) (ii) dy d ( ) ( cos sin) ( cos sin) ( cos sin) ( cos sin). 7 (b) (i) & 7 (b) (ii) ( sin) ( cos sin) ( cos sin) Page of 8 ( cos sin) ( cos sin) cos y. OR cos sin cos sin. cos sin cos sin cos sin. cos sin sin co cos sin sin cos y ( ) ( ) dy d ( ) ( ) ( ) ( ) ( ) ( cos sin ) cos sin ( cos sin) cos sin cos sin cos sin y Blunders (-) B Differentiation B Indices B Trig formula Attempts A Error in differentiation Formula Worthless W Integration

34 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) (c) The function f ( ) ( ) loge ( ) (i) Show that the curve f ( ) (ii) is defined for >. e y has a turning point at,. e e Determine whether the turning point is a local maimum or a local minimum. (c) (i) f'() 5 marks Att Value of 5 marks Att Value of y 5 marks Att (c) (ii) Turning points 5 marks Att 7 (c) (i) f ( ) ( ) loge( ) f ( ) ( ). loge( ) loge( ). f ( ) 0 e loge ( ) e.. e e e y loge y ( log ee). So turning point is, e e e e e e OR log 7 (c) (i) ( ) [ ( )] f e e At, ( ) e e e f log e log e log e e e e e log ( ) e log e( e ) 0 0. e So f ( ) 0 at. e e Also, at, y loge y ( log ee). e e e e e e So turning point is, e e. 7 (c) (ii) e e ( ) e f f e > 0., is a local minimum. e e e e e Blunders (-) B Differentiation B f ( ) 0 B Indices B Incorrect deduction or no deduction Slips (-) S log e e Attempts A Error in differentiation formula Worthless W Integration Page of 8

35 QUESTION 8 Part (a) 0 marks Att Part (b) 0 (5, 5, 5, 5) marks Att (,,, ) Part (c) 0 (5, 5, 0) marks Att (,, ) Part (a) 0 marks Att (a) Find ( sin e ) d. (a) 0 marks Att 8 (a) ( sin ) cos e d e c Blunders (-) B Integration B No c Attempts A Only c correct Att Worthless W Differentiation instead of integration Page 5 of 8

36 Part (b) 0 (0, 5, 5) marks Att (,,) (b) The curve y 8 6 crosses the -ais at 0, and, as shown. y Calculate the total area of the shaded regions enclosed by the curve and the -ais. (b) First area 5 marks Att Second area 5 marks Att Total Area 5 marks Att 8 (b) Required area ( 8 6 ) ( 8 6 ) ( 8 6 ) d 0 ( ) d d 0 ( ) ( ) d the required area is 5 7. Blunders (-) B Integration B Indices B Error in area formula B Incorrect order in applying limits B5 Not calculating substituted limits B6 Uses yd for area formula Attempts A Uses volume formula A Uses y in formula Worthless W Wrong area formula and no work Page 6 of 8

37 Part (c) 0 (5, 5 0) marks Att (,, ) (c) (i) Find, in terms of a and b (ii) I b cos d. sin Find in terms of a and b J a b sin d. cos π (iii) Show that if a b, then I J. a (c) (i) 5 marks Att (ii) 5 marks Att (iii) 0 marks Att 8 (c) (i) 8 (c) (ii) I b cos d. Let u sin du cosd. sin J J J a I I b sinb du u sina sinb loge. sina sinb [ log u] log ( sinb) log ( sina). e sina sin d. Let u cos du sind. cos a cosb cosa du u cosa loge. cosb e cosb [ log u] log ( cosb) log ( cosa). e cosa e e e 8 (c) (iii) π When a b, then π sin a sinb cosa I log log e loge sina π cosb sin b e J. Page 7 of 8

38 Blunders (-) B Integration B Differentiation B Trig Formula B Logs B5 Limits B6 Incorrect order in applying limits B7 Not calculating substituted limits B8 Not changing limits B9 Incorrect deduction or no deduction Slips (-) S S Numerical Trig value Page 8 of 8

39 Coimisiún na Scrúduithe Stáit State Eaminations Commission LEAVING CERTIFICATE 00 MARKING SCHEME MATHEMATICS PAPER HIGHER LEVEL Page 9 of 8

40 Page 0 of 8

41 GENERAL GUIDELINES FOR EXAMINERS PAPER. Penalties of three types are applied to candidates work as follows: Blunders - mathematical errors/omissions (-) Slips - numerical errors (-) Misreadings (provided task is not oversimplified) (-). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B, B, B,, S, S,, M, M, etc. These lists are not ehaustive.. When awarding attempt marks, e.g. Att(), 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.. Worthless work is awarded zero marks. Some eamples of such work are listed in the scheme and they are labelled as W, W, etc.. 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 bo containing the relevant solution. 7. The sample solutions for each question are not intended to be ehaustive lists there may be other correct solutions. Any eaminer unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising eaminer. 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. 0. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most.. A serious blunder, omission or misreading results in the attempt mark at most.. Do not penalise the use of a comma for a decimal point, e.g may be written as 5,50. Page of 8

42 QUESTION Part (a) 0 marks Att Part (b) 5 (5, 0) marks Att (, ) Part (c) 5 (0, 5) marks Att (, 5) Part (a) 0 marks Att (a) A circle with centre (, ) passes through the point (, ) Find the equation of the circle. 7. (a) 0 marks Att (a) Centre is Circle : (, ) and r ( 7) ( ) ( ) ( y ) 7. or 6 7. (a) y -6 8y c 0 But (7,-) Circle c 0 c 8 Equation of circle y -6 8y 8 0 Blunders (-) B Error in substituting into distance formula B Incorrect sign assigned to centre in equation of circle Slips (-) S Arithmetic error Attempts (marks) A Radius length A Equation of circle without radius evaluated A Equation of circle without substitution for c A Substitution of (7,-) and stops A5 y 7 Misreading(-) M (7,-) as centre of circle Page of 8

43 Part (b) 5 (5, 0) marks Att (, ) (b) (i) Find the centre and radius of the circle y 8 0y 0. (ii) The line y k 0 is a tangent to the circle y 8 0y 0. Find the two possible values of k. (b)(i) 5 marks Att (b) (i) Centre is(, 5). r or (b) (i) ( 8 6) ( y - 0y 5) ( ) ( y 5) 9 Centre (, 5) Radius 9 or Both correct One correct None correct 5 marks marks 0 marks (b) (ii) 0 marks Att (b) (ii) ( ) ( 5) k 9 6 k 5 k ± 5. k 5 or k 5 k 7 or k 7. * Accept candidates centre and radius from (b)(i) or k (b) (ii) y k k ( ) 8 0( ) 0 5 (6k-8) k 0k 5 0 Equal roots (6k-8) 00 (k 0k 5) 6k 096 k 56 0 k 6k (k 7)(k 7) 0 k -7 and k - 7 Blunders (-) B Error in substitution into perpendicular distance formula B One value of k only B Incorrect squaring B Error in factors Slips (-) S Arithmetic error Page of 8

44 Attempts (marks) A Some correct substitution into perpendicular formula A Some correct substitution of either or y from linear equation into circle Part (c) 5 (0, 5) marks Att (, 5) (c) A circle has the line y as a tangent at the point (, ). The circle also,. Find the equation of the circle. contains the point ( ) (c) First equation in two variables 0 marks Att Finish 5 marks Att 5 (c) Slope of tangent slope of normal at (, ). Equation of normal: ( y ) ( ) y 8 y 0. Mid-point of chord joining (, ) and (, ) is (,). Slope of chord. Equation of mediatior is y ( ) y y 0. y 0 y 0 5y 0 y and 6 Centre is 6,. r ( 6) ( ) Equation of circle is Or 6 y g fy c 0. (,) Circle 0 g 8f c 0 (,-) Circle 0 8g -f c 0 Slope of tangent slope of normal at (, ) 0. ( 6) ( y ) 0. ( ) g f centre (-f, -f). y y 8 y 0. -f (-f) 0 f - Centre is (6, ) Equation of normal: ( ) ( ). r ( 6) ( ) 6 0. equation of circle is ( 6) ( y ) 0. Page of 8

45 First Equation: Blunders (-) B Error in substituting into slope formula B Error in substituting into midpoint formula B Error in substituting into equation of line formula B Incorrect signs for centre of circle Slips (-) S Arithmetic error Attempts (marks) A Slope of Tangent A Midpoint of chord Finish: Slips and blunders do not apply. Award 0, 5 or 5 marks, as follows: Fully correct: 5 marks Attempt (5 marks) A Second equation in two variables Page 5 of 8

46 QUESTION Part (a) 0 marks Att Part (b) 0 (5, 5) marks Att (, 5) Part (c) 0 (5, 5, 0) marks Att (,, - ) Part (a) 0 marks Att (a) A, B and C are points and O is the origin. a i j, b i 6 jand AC OB. Epress c in terms of i and j. (a) 0 marks Att (a) AC OB c a b c b a i 6j i j. c i j. Blunders(-) B Error in AC c a or equivalent B Answer not epressed in correct form Slips (-) S Arithmetic error Attempts (marks) A AC c a and stops Part (b) 0 (5, 5) marks Att (, 5) (b) u i j and v i kj where k. (i) Epress v and uv. in terms of k. (ii) Given that cosθ, where θ is the angle between u and v, find the two possible values of k. (b) (i) 5 marks Att (b) (i) v i kj k. u. v i j i kj k. Both correct: 5 marks One correct: marks None correct: 0 marks ( )( ) Page 6 of 8

47 (b) (ii) 5 marks Att5 u. v k cosθ. u. v 5 k k ( k) 5 k ( k) 8k 0 Attempt (5 marks) A Substitutes correctly 5 5k ( k )( k ) 0 k, k. 5k 5 8 8k k Part (c) 0 (5, 5, 0) marks Att (,, - ) (c) OABC is a parallelogram where O is the origin. Q is the midpoint of [ BC ]. R AQ is etended to R such that AQ QR. [ ] (i) Epress q in terms of a and c. C Q B (ii) Epress AQ in terms of a and c. (iii) Show that the points O, C and R are collinear. O A (c) (i) 5 marks Att (c) (i) q c a. Blunders (-) B CQ OA B Answer not in required form Slips (-) S Arithmetic error Attempts ( marks) A A correct epression with q Page 7 of 8

48 (c) (ii) 5 marks Att (c) (ii) AQ q a a c a c. Blunders (-) B B a AQ q Answer not in required form Slips (-) S Arithmetic error Attempts (marks) A A A correct epression with AQ q a and stops AQ (c) (iii) 0 marks Hit /Miss (c) (iii) r a AR a AQ a c a c. As r c, then points O, C and R are collinear. Page 8 of 8

49 QUESTION Part (a) 0 marks Att Part (b) 0 (5, 5) marks Att (, ) Part (c) 0 (0, 0) marks Att (, 6) Part (a) 0 marks Att (a) The line y 7 0 is perpendicular to the line a 6 y 0. Find the value of a. (a) 0 marks Att (a) a Slope of y 7 0 is. Slope of a 6y 0 is. 6 a 6 a a 8. Blunders (-) B Error in slope B Product of slopes - B Product of slopes - but fails to finish Slips (-) S Arithmetic error Attempts (marks) A Slope of one line found Part (b) 0(5, 5) marks Att (, ) (b) (i) The line 5y k 0 cuts the -ais at P and the y-ais at Q. Write down the co-ordinates of P and Q in terms of k. (ii) The area of the triangle OPQ is 0 square units, where O is the origin. Find the two possible values of k. (b) (i) 5 marks Att (b) (i) k k P, 0, Q 0,. 5 Blunders (-) B P and Q not in coordinate form B P or Q only correct Slips (-) S Arithmetic error Page 9 of 8

50 Attempts ( marks) k k A or written 5 k k A 0,,0 5 (b) (ii) 5 marks Att (b) (ii) Area OPQ 0 k k 0. k 00 5 k ± 0. Blunders (-) B Error in substitution into formula for area of triangle B One value of k only found Slips (-) S Arithmetic error Attempts ( marks) A Some correct substitution into formula for area of triangle A k -00 or equivalent Part (c) 0(0,0) marks Att (, 6) (c) (i) f is the transformation (, y) (, y ), where y and y y. The line l has equation y m c. (i) Find the equation of f ( l), the image of l under f. (ii) Find the value(s) of m for which f ( l) makes an angle of Page 50 of 8 5 with l. (c) (i) 0 marks Att (c) (i) y y y y ( y ). y ( y ) y ( y ). m f ( l) : ( y ) ( y ) c y m my c. f l : m y m c 0 ( ) ( ) ( ). Blunders (-) B Image of line not in the form a by c 0 or y m c. B Incorrect matri B Incorrect matri multiplication

51 Slips (-) S Arithmetic error Attempts (marks) A Epressing or y in terms of primes A Correct matri for f when finding f(l) A Correct image point on f(l) (c) (ii) 0 marks Att 6 (c) (ii) ( ) ( m ) m Slope l m and slope f l. m m m m ( ) tan5 m m m m. m m ( m ) m m m m m m m ± m m m m ( ). m m m m m 0 m 0. OR m m m m m m m. ± (m - gives denominator of 0 for slope of f(l), but is still a solution, since in this case f(l) is vertical and l makes an angle of 5 with it.) solutions are m 0, m, m -. Attempt (6 marks) A Substitutes correctly into formula. Note: all three solutions not found attempt mark at most. Page 5 of 8

52 QUESTION Part (a) 0 marks Att Part (b) 5 (5, 5, 5) marks Att (,, ) Part (c) 5 (5, 5, 5) marks Att (,, 5) Part (a) 0 marks Att (a) The area of a triangle PQR is 0 cm. PQ 0 cm and PR 8 cm. Find the two possible values of QPR. (a) 0 marks Att (a) Area PQR 0 ( 0)( 8) sin QPR 0. sin PQR PQR 0 or 50. Blunders (-) B Error in substitution into area formula B One angle only B Angle outside the range Slips (-) S Arithmetic error Attempts ( marks) A Substitution into formula Page 5 of 8

53 Part (b) 5 (5, 5, 5) marks Att (,, ) (b) Find all the solutions of the equation cos cos in the domain (b) Equation 5 marks Att Roots 5 marks Att Finish 5 marks Att (b) cos cos cos cos 0. cos cos 0 cos 0, 60 0,0, 0, 60. or or cos ( )( ) 0, 0. (b) cos cos cos cos 0 sin sin 0 sin 0 0, 0, 0, 60 sin 0 0, 60 0, 0, 0, 60 Blunders (-) B Incorrect substitution for cos B Error in factors B Error in substitution in quadratic formula B One value omitted for either root B5 Angle outside the domain Slips (-) S Arithmetic error Attempts (,, marks) A Cos sin and stops A Correct factors and stops Page 5 of 8

54 Part (c) 5(5, 5,5) marks Att (,,5) (c) ABC is a triangle with sides of lengths a, b and c, as shown. Its incircle has centre O and radius r. c (i) Show that the area of ABC is r ( a b ). A c O r b B a C (ii) (iii) The lengths of the sides of a triangle are a p q, b p q and c pq, where pand q are natural numbers and p > q. Show that this triangle is rightangled. Show that the radius of the incircle of the triangle in part (ii) is a whole number. (c) (i) 5 marks Att (c) (i) Area ABC ( ar ) ( br) ( cr) r( a b c ). Blunders (-) B Error in substitution into triangle area formula B Answer not in correct format Slips (-) S Arithmetic error Attempts ( marks) A Area of one triangle found (c) (ii) 5 marks Att (c) (ii) p q p p q q. ( ) ( p q) ( pq) p pq q pq p pq q ( p q). triangle is right-angled. Blunders (-) B Error in squaring B Incorrect application of Pythagoras B Conclusion not stated or implied Slips (-) S Arithmetic error Page 5 of 8

55 Attempts ( marks) A Squares any one side in terms of p and q (c) (iii) 5 marks Att 5 (c) (iii) Area of ( )( ) ( pq p q pq p ). q But, by part (i), area of ( ) ( ) ( r p q p q pq r p pq r p pq ). pq ( ) ( ) ( p q)( p q) p pq pq p q r r q( p q). p( p q) As p and q are natural numbers, and p>q, then p q is a natural number and thus r q p q is a whole number. ( ) Attempt (5 marks) A Correct epression for r in terms of p and q Page 55 of 8

56 QUESTION 5 Part (a) 0 marks Att Part (b) 5 (0, 5) marks Att (, ) Part (c) 5 (0, 5, 0) marks Att (,, ) Part (a) 0 marks Att 5 (a) Given that tan θ, show that tan θ. (a) 0 marks Att 5 (a) 9 tanθ tanθ tan θ or 5 (a) tan θ θ in st or rd quadrant sinθ and cosθ both positive in st quadrant and both negative rd quadrant sin θ ± and cos θ ±, (both having same sign) 0 0 In the case of both positive: ( ).( ) sin θ sinθ cosθ tan θ cos θ cos θ sin θ 8 ( ) ( ) 0 0 and in the case of both negative: ( ).( ) sin θ sinθ cosθ tan 0 0 θ cos θ cos θ sin θ ( ) ( ) 0 Blunders (-) B Error substituting into tanθ formula B Incorrect application of Pythagoras B Error substituting into Sin θ and/ or Cosθ formula(e) B One quadrant only Slips (-) S Arithmetic error Attempts ( marks) A Some substitution into tanθ formula A Effort at application of Pythagoras Page 56 of 8

57 Part (b) 5 (0, 5) marks Att (, ) 5 (b) A triangle has sides of lengths, 5 and 6. The angles of the triangle are A, B and C, as in diagram. C 5 B A 6 7 (i) Using the cosine rule, show that cos A cosc. 8 9 (ii) Show that cos( A C). 6 (b) (i) 0 marks Att 5 (b) (i) 5 cosa ( 5)( 6) cosc ( )( 5) 0 7 cos A cosc. 8 8 Blunders (-) B Error substituting into cosine formula B Cos A Cos C not indicated Slips (-) S Arithmetic error Attempts ( marks) A Some values substituted into cosine formula for either Cos A or Cos C A Cos A or Cos C formula epressed in terms of sides of a triangle A Cos A or Cos C only and stops Worthless (0) W Cos A Cos C Cos (AC) Page 57 of 8

58 (b) (ii) 5 marks Att 5 (b) (ii) cos( A C) cosb. ( )( 6) or 5 (b) (ii) Cos (AC) Cos A Cos C SinA Sin C Blunders (-) B Cos (AC) - Cos B B Incorrect ratio for Sin A or Sin C B Error substituting into epansion of Cos (A C) B Conclusion not stated or implied Slips (-) S Arithmetic error Attempts ( marks) A 0 Cos (A C) Cos ( 80 B) and stops A Some substitution into Cos (AC) epansion A Use of Pythagoras Worthless (0) W Cos (A C) Cos ACosC W Cos (A C) Cos A. Cos C Part (c) 5(0, 5, 0) marks Att (,, ) 5 (c) (i) Show that ( A B) ( A B) ( A B) (ii) Hence solve the equation ( ) ( ) cos cos sin sin cos. the domain cos cos sin sin sin in (c) (i) 0 marks Att 5 (c) (i) ( cos A cos B) ( sin A sin B) cos A cos Acos B cos B sin A sin Asin B sin B ( cos A sin A) ( cos B sin B) ( cos Acos B sin Asin B) ( A B) cos. Blunders (-) B Error in squaring B cos A sin A B cosacosb sinasinb cos (A B) Page 58 of 8

59 Slips (-) S Arithmetic error Attempts ( marks) A (cos A cos B) or equivalent correct (c) (ii) tan 5 marks Att Solutions 0 marks Att 5 (c) (ii) ( cos cos) ( sin sin) cos by part (i). cos sin sin cos sin. cos tan 0, 0, 90, 570, 750, 90. 0, 70,0,90, 50, 0. Or (c) (ii) cos ( 60 0 ) 5 marks Att Solutions 0 marks Att ( ) ( ) cos cos sin sin cos, by part (i). cos sin sin cos sin cos 0 cos sin 0 cos sin 0 cos( 60 ) , 70, 50, 60, 80, 990 Equation: Attempts ( marks) A A correct manipulation Worthless (0) W Cos Cos or equivalent 0, 70, 0, 90, 50, 0. Solutions: Attempt (marks) A Solution set with one omitted or incorrect value Worthless (0) W Solution set with more than one omitted or incorrect value Page 59 of 8

60 QUESTION 6 Part (a) 0 marks Att Part (b) 0 (5, 5, 5, 5) marks Att (,,, ) Part (c) 0 marks Att 6 Part (a) 0 marks Att 6 (a) One bag contains four red discs and si blue discs. Another bag contains five red discs and seven yellow discs. One disc is drawn from each bag. What is the probability that both discs are red? (a) 0 marks Att 6 (a) 5 Numbers of favourable outcomes C C 0. Numbers of possible outcomes C C 0. 0 Probability both discs are red. 0 6 Or 5 P(Both red discs). 0 6 Blunders (-) B Incorrect number of possible outcomes a B Answer not in form of, a, b b Slips (-) S Arithmetic error Attempts ( marks) A Correct number of possible outcomes A Correct number of favourable outcomes A 5 C C or equivalent, with or without further work 0 Page 60 of 8

61 Part (b) 0 (5, 5, 5, 5) marks Att (,,, ) 6 (b) α and β are the roots of the quadratic equation n n u lα mβ, for all n. n Prove that pu qu ru 0, for all n n n n p q r 0. (b) Uses root property correctly 5 marks Att Deduces u, n un 5 marks Att Substitutes and tidies up 5 marks Att Conclusion 5 marks Att 6 (b) α is a root of p q r 0 pα qα r 0 Similarly: p β qβ r 0 n n n n n n Given: un lα mβ un lα mβ, un lα mβ n n n n n n pun qun run p( lα mβ ) q( lα mβ ) r( lα mβ ) n n lα ( pα qα r) mβ ( pβ qβ r) n n lα (0) mβ (0) 0 Blunders (-) B Fails to use root property correctly B Error in epressing value of term B Error in substituting or tidying B Incorrect conclusion or no conclusion implied Slips (-) S Arithmetic error Attempts (,,, marks) A Effort at substituting either root into quadratic A Some correct substitution for u n or equivalent Page 6 of 8

62 Part (c) 0 marks Att 6 6 (c) In a café there are seats in a row at the counter. Si people are seated at the counter. How much more likely is it that all si are seated together than that no two of them are seated together? (c) 0 marks Att 6 6 (c) Taking arrangements as unordered: Number of possible ways of seating si people in a row of seats C 6 6. To seat si people together, seat them in seats to 6, or to 7, or to 8, or to 9, or 5 to 0, or 6 to. Number of favourable outcomes 6. 6 Probability of si seated together. 6 In order to seat si people with no two together, seat them in seat, seat, seat 5, seat 7, seat 9 and seat. There is no other possible way to seat them. There is only one favourable outcome. Probability of no two of them seated together. 6 It is si times more likely that all si people are seated together. OR 6 (c) Taking arrangements as ordered: Number of possible ways of seating si people in a row of seats P To seat si people together, seat them in seats to 6, or to 7, or to 8, or to 9, or 5 to 0, or 6 to. Number of favourable outcomes 6 6! 0. 0 Probability of si seated together. 60 In order to seat si people with no two together, seat them in seat, seat, seat 5, seat 7, seat 9 and seat. There is no other possible way to seat them. So there are 6! 70 favourable outcomes. 70 Probability that no two of them seated together , so it is si times more likely that all si people are seated together. Attempt (6marks) A Correct epression for one or other case * Note special case: If the candidate has both probabilities correct but subtracts them instead of dividing: award7 marks * Apart from the special case mentioned, award 0, 6, or 0 marks. Page 6 of 8

63 QUESTION 7 Part (a) 0 marks Att Part (b) 0 (5, 5, 0) marks Att (,, ) Part (c) 0 (0, 0) marks Att (, ) Part (a) 0 marks Att 7 (a) A password for a website consists of capital letters A, B, C,. Z and/or digits 0,,,. 9. The password has four such characters and starts with a letter. For eample, BA7A, C999 and DGKK are allowed, but 7DCA is not. Show that there are more than a million possible passwords. (a) 0 marks Att 7 (a) Numbers of possible passwords ,,056 >,000,000. or 7 (a) Numbers of possible passwords: ( ) ( ) ,,056 >,000,000. Attempt (marks) A Solution with one error Part (b) 0(5, 5, 0) marks Att (,, ) 7 (b) Karen is about to sit an eamination at the end of an English course. The course has twenty prescribed tets. Si of these are novels, four are plays and ten are poems. The eamination consists of a question on one of the novels, a question on one of the plays and a question on one of the poems. Karen has studied four of the novels, three of the plays and seven of the poems. Find the probability that: (i) (ii) Karen has studied all three of the tets on the eamination Karen has studied none of the tets on the eamination (iii) Karen has studied at least two of the tets on the eamination. (b) (i) 5 marks Att 7 (b) (i) Probability (studies all three tets) Blunders (-) B Incorrect number of possible outcomes B a Answer not epressed in form of, a, b or equivalent b Slips (-) S Arithmetic error Page 6 of 8

64 Attempts ( marks) A Correct number of possible outcomes A Correct number of favourable outcomes A 7 with or without further work 6 0 (b) (ii) 5 marks Att 7 (b) (ii) 6 Probability (studies none of the tets) Blunders (-) B Incorrect number of possible outcomes B a Answer not epressed in form of, a, b or equivalent b Slips (-) S Arithmetic error Attempts ( marks) A Correct number of possible outcomes A Correct number of favourable outcomes A with or without further work 6 0 (b) (iii) 0 marks Att 7 (b) (iii) Probability (studies at least two of the tets) Probability (studies two) Probability (studies three) Attempt (marks) A Correct epression all three terms in P( tets) or all three terms in P( tet). 9. Page 6 of 8

65 Part (c) 0(0, 0) marks Att (, ) (c) The mean of the real numbers a, a, a, a and 5a is µ and the standard deviation isσ. (i) Epress µ and σ in terms of a. (ii) Hence write down in terms of a, the mean and the standard deviation of a 5, 6a 5, 9a 5, a 5, 5a 5. (c) (i) 0 marks Att 7 (c) (i) a a a a 5a 5a µ a. 5 5 ( a a) ( a a) ( a a) ( a a) ( 5a a) a a a a σ 5 5 0a a. 5 σ a. Attempt (marks) A Epression for mean or standard deviation correct. (c) (ii) 0 marks Att 7 (c) (ii) Mean µ 5 9a 5. Standard deviation σ a. * If not Hence (i.e. otherwise) marks for mean and/or standard deviation correct Attempt (marks) A Mean or standard deviation correct Page 65 of 8

66 QUESTION 8 Part (a) 0 marks Att Part (b) 0 (5, 5) marks Att ( -, 5) Part (c) 0 (0, 5, 5) marks Att (, -, - ) Part (a) 0 marks Att 8 (a) Use integration by parts to find loge d. (a) 0 marks Att 8 (a) udv uv vdu. Let u log e du d. dv d v. log e d loge d loge d log e Blunders (-) B Incorrect differentiation or integration B Incorrect parts formula Slips (-) S Arithmetic error S Omits constant of integration Attempts ( marks) A One correct assigning to parts formula A Correct differentiation or integration C. Part (b) 0(5, 5) marks Att ( -,5) 8 (b) A rectangle is inscribed between the curve y 9 and the -ais, as shown. y (i) Write an epression for the area of the rectangle in terms of p. p (ii) Hence, calculate the area of the largest possible rectangle. (b) (i) 5 marks Hit/Miss 8 (b) (i) Length of rectangle p and its width 9 p. Area of rectangle A p( 9 p ) 8 p p. Page 66 of 8

67 (b) (ii) 5 marks Att 5 8 (b) (ii) da da 8 6 p. For maimum, p 0 p dp dp. d A p < 0 for p dp A 8 6. is largest possible rectangle. * Note: If candidate gets no marks for (b)(i), then cannot get any marks for (b)(ii). Attempt (5 marks) A Correct differentiation Part (c) 0 (0, 5, 5) marks Att (, -, - ) 8 (c) (i) Derive the Maclaurin series for f ( ) cos, up to and including the term 6 containing. (ii) Hence, and using the identity sin ( cos), show that the first three non zero terms of the Maclaurin series for sin are. 5 Sin, as a fraction. (iii) Use these terms to find an approimation for ( ) (c) (i) 0 marks Att 8 (c) (i) () f ( 0) f ( 0) f ( 0) f ( 0) f ( ) f ( 0 )...!!!! f ( ) cos f ( 0) cos0. f ( ) sin f ( 0) sin0 0. f ( ) cos f ( 0) cos0. f ( ) sin f ( 0) sin0 0. () f ( ) cos () f ( 0) cos0. (5) f ( ) sin (5) f ( 0) sin0 0. (6) f cos (6) f 0 cos0. f ( ) ( )!! 6! ( ) cos... Blunders (-) B Incorrect differentiation B (n) Incorrect evaluation of f (0) B Each term not derived (to ma of ) B Error in Maclaurin series 6 6 Slips (-) S Arithmetic error Page 67 of 8

68 Attempts ( marks) A Correct epansion for Cos given but not derived A f(0) correct A A correct differentiation A Any one correct term (c) (ii) 5 marks Hit/Miss 8 (c) (ii) 6 6 By part (i), cos sin cos 5 5 ( ). 6 6 (c) (iii) 5 marks Hit/Miss 8 (c) (iii) sin ( ) ( ) ( ) ( ) Page 68 of 8

69 QUESTION 9 Part (a) 0 marks Att Part (b) 0 (5, 5, 0) marks Att (,, ) Part (c) 0 (5,5, 5, 5) marks Att (,,, ) Part (a) 0 marks Att 9 (a) Z is a random variable with standard normal distribution. P < Z. Find ( ) Part (a) 0 marks Att 9 (a) P( < Z ) P( Z ) P( Z ) ( 0 8) or 9 (a) P ( Z ) < (P(Z ) P(Z 0)) ( ) Blunders ( ) B P(z ) incorrect or P(Z 0) incorrect B Mishandles P(- < Z) Slips ( ) S Arithmetic error. Attempts ( marks) A P(Z ) correct. Part (b) 0 (5, 5, 0) marks Att (,, ) 9 (b) A test consists of twenty multiple-choice questions. Each question has four possible answers, only one of which is correct. Seán decides to guess all the answers at random. Find the probability that: (i) (ii) Seán gets none of the answers correct Seán gets eactly five of the answers correct (iii) Seán gets four, five or si of the answers correct. Give each of your answers correct to three decimals places. (b) (i) 5 marks Att 9 (b) (i) p, q. Probability (none correct) Page 69 of 8

70 Blunders (-) B Incorrect p or q B Binomial error B Answer not in required form Slips (-) S Arithmetic error S Answer not to two decimal places Attempts ( marks) A Correct p or q (b) (ii) 5 marks Att 9 (b) (ii) Probability (eactly five correct) C Blunders (-) B Binomial error B Answer not in decimal form Slips (-) S Arithmetic error S Answer not to decimal places Attempts ( marks) A 0 C 5 used or implied (b) (iii) 0 marks Att 9 (b) (iii) Probability (four, five or si) C C5 C Blunders (-) B Each term omitted B Binomial error B Answer not in decimal form B Rounding off too early Slips (-) S Arithmetic error Attempts ( marks) A Effort at probability of four or si correct Page 70 of 8

71 Part (c) 0 (5, 5, 5, 5) marks Att (,,, ) 9 (c) A bakery produces muffins. A random sample of 50 muffins is selected and weighed. The mean of the sample is 80 grams and the standard deviation is 6 grams. Form a 95% confidence interval for the mean weight of muffins produced by the bakery. (c) Correct S.E. 5 marks Att Two tailed 5 marks Att Mean ±.96 SE 5 marks Att Finish 5 marks Att 9 (c) 80, σ 6 and n 50. σ σ 6 6. n The 95% confidence interval is ( σ ) ( σ ) 96, , [ 78, 8 6 ] grams. 5 5 Blunders ( ) B Error in standard error of mean. B Error from tables. B Answer not simplified. Slips ( ) S Arithmetic error. Attempts (,,, marks) A Standard error of mean with some substitution. A Incomplete substitution. Page 7 of 8

72 QUESTION 0 Part (a) 0(5, 5) marks Att (, ) Part (b) 0(5,5,5,5,5,5,5,5) marks Att (,,,,,,,) Part (a) 0 (5, 5) marks Att (, ) 0 (a) The binary operation is defined by y y y, where, y \ { }. (i) Find the identity element. (ii) Epress, the inverse of, in terms of. (a) (i) 5 marks Att 0 (a) (i) e e e e 0 e 0. Blunders (-) B * e incorrect B e e 0 and stops Slips (-) S Arithmetic error Attempts ( marks) A * e A * e e -e ( ) (a) (ii) 5 marks Att 0 (a) (ii) e 0. ( ), (provided ). Blunders (-) B * incorrect Slips (-) S Arithmetic error Attempts ( marks) A * correct and stops A * 0 Page 7 of 8

73 Page 7 of 8 Part (b) 0(5, 5, 5, 5, 5, 5, 5, 5) marks Att (,,,,,,, ) 0 (b) G is the set of permutations of {,, } and the si elements of G are as follows: a b c d f. g ( ), G is a group, where denotes composition. (i) Write down b and, d the inverses of b and d respectively. (ii) Verify that ( ). b d d b (iii) Write down the subgroups of ( ), G of order. (iv) K is the subgroup of ( ), G of order. List the elements of K. (v) ( ), H is a group, where { },, w w H and. w Give an isomorphism φ from ( ), K to ( ), H, justifying fully that it is an isomorphism. (b) (i) b 5 marks Att d 5 marks Att 0 (b) (i)., d b Blunders (-) B Incorrect element (ma of ) Slips (-) S Arithmetic error Attempts ( marks) A Permutation incomplete A One element correct with another repeated (b) (ii) One composition correct 5 marks Att Finish 5 marks Att 0 (b) (ii) ( ) ( ).. d b b d d b Blunders (-) B Incorrect element (ma ) B d b correct instead of b d B Incorrect conclusion or no conclusion implied

74 Slips (-) S Arithmetic error Attempts (, marks) A Permutation incomplete A One element correct with another repeated (b) (iii) 5 marks Att 0 (b) (iii) {a, b}, {a, d}, {a, g}are subgroups of order two. Blunders (-) B Subgroup omitted B Incorrect subgroup Slips (-) S Arithmetic error Attempts ( marks) A One correct subgroup (b) (iv) 5 marks Att 0 (b) (iv) K {a, c, f} is a subgroup of order three. Blunders (-) B One incorrect element (other than identity) Slips (-) S Arithmetic error Attempts ( marks) A {a,b,d} Worthless (0) W { b,d,g} W No identity element Page 7 of 8

75 (b) (v) Establishing link 5marks Att Finish 5 marks Att 0 (b) (v) K φ H a.. c.. w f.. w a and are the identities of (, ) and ( H, ) φ( c c) φ( f ) w and φ( c) φ( c) w w w. φ( f f ) φ( c) w and φ( f ) φ( f ) w w w w. φ( c f ) φ( a) and φ( c) φ( f ) w w w. φ( f c) φ( a) and φ( f ) φ( c) w w w. K respectively. φ( a a) φ( a) and φ ( a) φ( a) φ( a c) φ( c) w and φ( a) φ( c) w φ( a φ( c a ) φ( c) φ( f ) φ( f ) w f a ) φ( f ) w Isomorphism. w and φ( c) φ( a) w and φ( a) φ( f ) w and φ( f ) φ( a) w Alternative Methods: K is a cyclic group with generator c. K: {a,f,c} { c, c, c } H is a cyclic group with generator w Isomorphism: c ( or w ), c w, c w Justification: K and H are both cyclic groups of same order (order ) K and H isomorphic, under any function that maps a generator to a generator and corresponding powers accordingly, as this one does. or (alternative justification) Theorem: Any cyclic group of order n is isomorphic to the group of comple nth roots of unity K is s cyclic group of order, and H is the group of the cubic roots of unity K and H isomorphic under this function. * Using alternative methods above, it is not sufficient to show the groups are isomorphic; an isomorphism must also be given. Blunders (-) B Cayley table but links not established B Incomplete justification Slips (-) S Arithmetic error Attempts (, marks) A Link identities only A States order of groups. Page 75 of 8

76 QUESTION Part (a) 0 marks Att Part (b) 0 (0, 0) marks Att (, ) Part (c) 0 (0, 0) marks Att (, ) Part (a) 0 marks Att (a) An ellipse with centre (, 0) is units. Find its equation. 0 has eccentricity 5 and the length of its major ais (a) 0 marks Att (a) 6 9 a a. b a ( e ) b b. 5 5 y 5y Ellipse:. a b 9 Blunders (-) B Incorrect a B b calculated, but equation not found B Error in forming equation Slips (-) S Arithmetic error Attempts ( marks) A a A Some substitution into b formula Part (b) 0(0, 0) marks Att (, ) (b) f is an affine transformation. The point M is the mid-point of the line segment [ AB ]. (i) Show that f ( M ) is the mid-point of the line segment [ f ( A) f ( B) ]. (ii) A triangle ABC has centroid G. f A f B f C has centroid f ( G). Show that the triangle ( ) ( ) ( ) Page 76 of 8