GCE. Mathematics. Mark Scheme for June Advanced Subsidiary GCE Unit 4721: Core Mathematics 1. Oxford Cambridge and RSA Examinations

Transcription

1 GCE Mathematics Advanced Subsidiary GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

2 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

3 7 Mark Scheme June 0 (i) ( x 6x) x ( x ) 9 p ( x ) seen or q q or q (their q) r Reasonably correct curve for quadrants only y in st and rd x If p, q, r found correctly, then ISW slips in format. (x ) + M0 A0 (x ) (BOD) (x x) B0 A0 (x ) B0 A0 (x + ) B0 (BOD) x (x ) B0 N.B. Ignore feathering now that answers are scanned. Reasonably correct shape, not touching axes more than twice. Very good curves for y in st and rd quadrants x SC If 0, very good single curve in either st or rd quadrant and nothing in other three quadrants. Correct shape, not touching axes, asymptotes clearly the axes. Allow slight movement away from asymptote at one end but not more. Not finite. (ii) (i) Translation units parallel to y axis 6x x x x Must be translation/translated not shift, move etc. 0 Or x For parallel to the y axis allow vertically, up, in the (positive) y direction. Do not accept in/on/across/up/along the y axis (ii) 6 x 5 6 or or 6 in final answer 6 seen 6 x (Allow x ) in final answer is M0 6 ± is A0 6

4 7 Mark Scheme June 0 x x 8x 8 6 x 5x 8( 0) (x 9)( x )( 0) 9 x, x 5 y, y 5 (i) Attempt to eliminate x or y Correct term quadratic (not necessarily all in one side) Correct method to solve quadratic x values correct y values correct SR If A0 A0, one correct pair of values, spotted or from correct factorisation www Attempt to express both surds in terms of One term correct Must be a clear attempt to reduce to one variable. Condone poor algebra for first mark. If x eliminated: 6 y y ( ) Leading to y 89y e.g. x00 x6 (y 5)(y ) =0 etc. 6 Fully correct (not 6 ) (ii) 5(5 0) Multiply numerator and denominator by 5 or - 5 or attempt to express both terms of numerator in terms of 5 (e.g. dividing both terms by 5 ) One of a, b correctly obtained Both a = and b= correctly obtained Check both numerator and denominator have been multiplied

5 7 Mark Scheme June 0 6 k x k 8k 0 (k )( k ) 0 k or k = x or x = 6 x or x = 6 8 If candidates use k x and rearrange: k - 8 k + = 0 8 k =k + 6k = 9k + k + 6 9k 0k + 6 = 0 (9k )(k )=0 k or k = 9 9 x or x = 6 x or x = (i) 6x 5 (ii) 0 < 7 5 x 6 x x ( x 6)( x ) x > 6, x < - * D * D Use a substitution to obtain a quadratic or factorise into brackets each containing x Correct method to solve a quadratic Attempt to calculate k Substitute, rearrange and square both sides Correct method to solve quadratic Attempt to calculate k equations or inequalities both dealing with all terms resulting in a 6 x b, a 9, b 0 - and -5 seen www Accept as two separate inequalities provided not linked by or (must be ) Rearrange to collect all terms on one side Correct method to find roots 6, - seen Correct method to solve quadratic inequality i.e. x > their higher root, x < their lower root (not wrapped, strict inequalities, no and ) No marks unless evidence of substitution (quadratic seen or square rooting or squaring of roots found). = 0 may be implied. Allow x x as a substitution. No marks if straight to quadratic formula to get x = x = and no further working No marks if k x then k 8k 0 SC If M0 Spotted solutions www each Justifies solutions exactly B Do not ISW after correct answer if contradictory inequality seen. 5 Allow x 6 6 Do not ISW after correct answer if contradictory inequality seen. e.g. for last two marks, - > x > 6 scores A0

6 7 Mark Scheme June 0 8 (i) dy 6x 6x dx 6 6x 0 x x y 7 (ii) d y 6 x dx d y When x =, dx > 0 so minimum pt ft 5 ft 7 Attempt to differentiate (one non-zero term correct) Completely correct Sets their d y 0 dx Correct value for x - www Correct value of y for their value of x Correct method e.g. substitutes their x from (i) into d y their (must involve x) and considers sign. dx ft from their dy differentiated correctly and correct dx substitution of their value of x and consistent final conclusion NB If second derivate evaluated, it must be correct (8 for x = ). If more than one value of x used, max A0 NB x = (and therefore possibly y = 7) can be found from equating the incorrect differential dy = 6x + 6 to 0. This could score A0 A0 ft dx If more than one value of x found, allow ft for one correct value of y dy Allow comparing signs of their either side of their dx, comparing values of y to their 7 d y SC = a constant correctly obtained from their dx dy and correct conclusion (ft) dx

7 7 Mark Scheme June 0 9 (i) Gradient of AB = 7 9 Gradient of AC = Vertex A OR: Length of AB = (7 ) ( ) 0 AC = ( ) ( 9 ) 60 BC = ( 7) ( 9 ) 00 Shows that AB + AC = BC Vertex A * D * D 5 y y Uses for any points x x One correct gradient (may be for gradient of BC =) Gradients for both AB and AC found correctly Attempts to show that m m oe, accept negative reciprocal Correct use of Pythagoras, square rooting not needed Any length or length squared correct All three correct Correct use of Pythagoras to show AB + AC = BC Do not allow final mark if vertex A found from wrong working. (Dependent on st M ) Accept BÂC etc for vertex A or between AB and AC Allow if marked on diagram. i.e must add squares of shorter two lengths 9 (ii) Midpoint of BC is 7 9, Length of BC = ( 7) ( 9 ) = (, -) Radius = 5 ( x ) ( y ) (5 ) ( x ) ( y ) 50 x y x 8y * ** D* D** 7 Uses x x y, y AC ( out of subs correct) Correct centre (cao) o.e. for BC, AB or Correct method to find d or r or d or r o.e. for BC, AB or AC (must be consistent with their midpoint if found) (x a) + (y b) seen for their centre (x a) + (y b) = their r Correct equation Correct equation in required form Substitution method (into x + y + ax + by + c = 0) Substitutes all points to get equations in a,b,c At least equations correct Correct method to find one variable One of a, b, c correct Correct method to find other values All values correct Correct equation in required form Alternative markscheme for last marks with f,g, c method: x x y 8y for their centre D* c = (±) + 50 D** c = 0 Correct equation in required form Ends of diameter method (p, q) to (c, d): Attempts to use (x p) (x c) + (y q) (y d) = 0 for BC,AC or AB M (x 7) (x + ) + (y ) (y + 9) = 0 A for both x brackets correct, A for both y brackets correct x + y x + 8y 0 = 0 SC If M A0 A0 then if both x brackets correct and if both y brackets correct for AC or AB 5

8 0(i) (ii) (iii) (iv) 7 Mark Scheme June 0 0,, 0,0 x +5x, x + x, x x + (x +5x )(x ) x + x 8x + dy 6x 6x 8 dx When x =, gradient = Gradient of l = On curve, when x = -, y = 5 y 5 = (x + ) y = x + Attempt to find gradient of curve when x = - 6( ) 6( ) 8 So line is a tangent 6 6 +ve cubic with distinct roots (0, ) labelled or indicated on y-axis (-, 0), (, 0) and (, 0) labelled or indicated on x- axis and no other x- intercepts Obtain one quadratic factor (can be unsimplified) Attempt to multiply a quadratic by a linear factor Attempt to differentiate (one non-zero term correct) Fully correct expression www Confirms gradient = at x = **AG May be embedded in equation of line Correct y coordinate Correct equation of line using their values Correct answer in correct form Substitute x = - into their d y dx Obtain gradient of CWO Correct conclusion Substitution method into (x p) + (y q) = their r Correct method to find d or r or d or r * Substitutes all points to get equations in p,q D At least equations correct Correct method to find one variable One of p, q correct Correct equation [ ( x ) ( y ) 50 ] Correct equation in required form [ x y x 8y 0 0 ] For first, left end of curve must finish below x axis and right end must end above x axis. Allow slight wrong curvature at one end but not both ends. No cusp at either turning point. No straight lines drawn with a ruler. Condone (0, ) as maximum point. To gain second and third B marks, there must be an attempt at a curve, not just points on axes. Final can be awarded for a negative cubic. Alternative for first marks: Attempt to expand all brackets with an appropriate number of terms (including an x term) Expansion with at most incorrect term Correct, answer (can be unsimplified) Allow if done in part(i) please check. M mark is for any equation of line with any non-zero numerical gradient through (-, their evaluated y) Alternatives ) Equates equation of l to equation of curve and attempts to divide resulting cubic by (x + ) Obtains (x + ) (x 5) (=0) Concludes repeated root implies tangent at x = - ) Equates their gradient function to and uses correct method to solve the resulting quadratic Obtains (x + )( x ) = 0 oe Correctly concludes gradient = when x = - 6

9 7 Mark Scheme June 0 Allocation of method mark for solving a quadratic e.g. x 5x 8 0 ) If the candidate attempts to solve by factorisation, their attempt when expanded must produce the correct quadratic term and one other correct term (with correct sign): ( x )( x 9) 0 x and 8 obtained from expansion ( x )( x ) 0 x and 5x obtained from expansion ( x 9)( x ) 0 M0 only x term correct ) If the candidate attempts to solve by using the formula a) If the formula is quoted incorrectly then M0. b) If the formula is quoted correctly then one sign slip is permitted. Substituting the wrong numerical value for a or b or c scores M0 5 ( 5) 8 earns (minus sign incorrect at start of formula) 5 ( 5) 8 earns (8 for c instead of 8) 5 ( 5) 8 M0 ( sign errors: initial sign and c incorrect) 5 ( 5) 8 5 M0 (b on the denominator) Notes for equations such as x 5x 8 0, then b = 5 would be condoned in the discriminant and would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0. c) If the formula is not quoted at all, substitution must be completely correct to earn the ) If the candidate attempts to complete the square, they must get to the square root stage involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions! x 5x x x x x 5 x This is where the is awarded arithmetical errors may be condoned 5 provided x seen or implied If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last full attempt. 7

10 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

11 GCE Mathematics Advanced Subsidiary GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

12 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

13 7 Mark Scheme June 0 (i) BC = x 9 x 7 x cos0 o BC =.6 cm Attempt use of correct cosine rule Must be correct formula seen or implied, but allow a slip when evaluating eg omission of, or incorrect use of an additional big bracket. Allow even if subsequently evaluated in radian mode (.96). Allow if expression is not square rooted, as long as it is clear that correct formula was used ie either BC = or even just a = if the power disappears from BC. Obtain.6, or better Actual answer is.69 so allow more accurate answer as long as it rounds to.6 (ii) area = 9 7 sin 0 = 9. cm Attempt triangle area using (½)absinC, or equiv Condone omission of ½ from this formula, but no other errors allowed. If using right-angled triangle, must use ½bh with reasonable attempt at perpendicular sides. Allow if subsequently evaluated in radian mode (57.00). If using 0 o, must be using sides of 9 and 7, not.6 from (i). If using another angle, can still get as long as sides used are consistent with this angle. Obtain 9., or better Actual answer is 9.75 so allow more accurate answer as long as it rounds to 9.7 Must come from correct working only. (iii) BD = sin 0 9 sin 6 BD = 6.9 cm Attempt use of correct sine rule, or equiv, to find length BD No further rearrangement required. Could have both fractions the other way up. Must be angles of 0 o and 6 o if finding BD directly. Must be attempting BD, so using 77 o to find AD is M0 unless attempt is then made to find BD by any valid method. Placing D on BC is M0. Obtain correct unsimplified expression involving BD as the only unknown Can still get even if evaluated in radians (0.07). If using a multi-step method (eg use 77 0 to find AD and then use cosine rule to find BD) then this A mark is only given when a correct (unsimplified) expression involving BD as the only unknown is obtained. 7 Obtain 6.9, or better Actual answer is so allow more accurate answer as long as it rounds to 6.9 Must come from correct working only not eg sin 7.

14 7 Mark Scheme June 0 (i) x 6 dx = x x + c Obtain kx Obtain x Obtain x (don t penalise lack of + c) Any k, as long as numerical. Allow both and for equiv eg x x Allow for unsimplified coefficient as well (ie 6 /.5 ). Allow -x. Maximum of marks if or dx still present in final answer. Maximum of marks if not given as one expression eg the two terms are integrated separately and never combined. (ii) y = x x + c 7 = + c c = hence y = x x * State or imply y = their integral from (i) Must have come from integration attempt ie increase in power by for at least one term, but allow if - disappeared in part (i) ie at least one of the and the must have been awarded in part (i). Can still get this if no + c. The y does not have to be explicit it could be implied by eg 7 = F(). M0 if they start with y = their integral from (i), but then attempt to use y 7 = m(x ). This is a re-start and gains no credit. d* Attempt to find c using (, 7) M0 if no + c. M0 if using x = 7, y =. Obtain y = x x Coefficients now need to be simplified, so -x is A0. Allow for equiv eg x x Must be an equation ie y =, so A0 for equation = or f(x) = 6

15 7 Mark Scheme June 0 (i) perimeter = r + rθ 6 + 8θ =. 8θ = 7. θ = 0.9 rads * d* State or imply that arc length 8θ, or equiv in degrees ie θ / 60 x 6π Equate attempt at perimeter to. and attempt to solve for θ Allow by implication for. / 8 or equivalent in degrees. Need to get as far as attempting θ. Must include radii and correct expression for arc length, either in radians or degrees. M0 if using chord length. Obtain θ = 0.9 rads Obtaining 0.9 and then giving final answer as 0.9π is A0 do not isw as this shows lack of understanding. Finding θ in degrees (5.6 o ) and then converting to radians can get as long as final answer is 0.9 (and not eg from premature approximation). (ii) ½ x 8 x 0.9 = Attempt area of sector using (½) r θ Obtain 8.8 Condone omission of ½, but no other error. Allow if incorrect angle from part (i), as long as clearly intended to be in radians. Allow equivalent method using fractions of the area. Allow working in degrees as long as it is a valid method. Allow if using 0.9π (even if 0.9 was answer to (i)), as long as clearly attempting (½) r θ with error on angle rather then (½) πr θ. Or any exact equiv. If 0.9 obtained incorrectly in part (i), full credit can still be gained in part (ii). Condone minor inaccuracies from working in degrees, as long as final answer is given as 8.8 exactly.

16 7 Mark Scheme June 0 (i) x + = (y + ) x + = y + y + x = y + y A.G. Attempt to make x the subject Allow for x = (y ± ) ± only. Allow if (y + ) becomes y +, but only if clearly attempting to square the entire bracket squaring term by term is M0. Must be from correct algebra, so M0 if eg (x + ) = x + is used. Verify x = y + y Need to see an extra step from (y + ) to given answer ie explicit expansion of bracket. No errors seen. (ii) ( y + y ) dy = [ ] y + y y = ( ) (⅓ + ) State or imply that the required area is given by ( y + y ) dy SR for verification, using y = + ( y + y + ), and confirming relationship convincingly, or for rearranging x = f(y) to obtain given y = f(x). No further work required beyond stating this. Allow if x appears in integral. Any further consideration of other areas is B0. = (9) ( ⅔) = 0⅔ Attempt integration Increase in power of y by for at least two of the three terms. Can still get if the - disappears, or becomes x. Allow for integrating a function of y that is no longer the given one, eg subtracted from, or using their incorrect rearrangement from part (i). ft Obtain at least two correct terms Allow for unsimplified coefficients. Allow follow-through on any function of y as long as at least terms and related to the area required. Condone, dy or + c present. Attempt F() F() for their integral Must be correct order and subtraction. This is independent of first so can be given for substituting into any expression other than y + y, including y +. If last term is x allow for using and throughout integral, but M0 if x value is used instead. 5 7 Obtain 0⅔ aef Must be an exact equiv so 0.6 is fine (but 9 5 / is A0). 0.7, 0.66 or 0 / + c are A0. Must come from correct integral, so A0 if from x. Must be given as final answer, so further work eg subtracting another area is A0 rather than ISW. Answer only is 0/5, as no evidence is provided of integration. SR Finding the shaded area by direct integration with respect to x (ie a C technique) can have 5 if done correctly, if non-exact decimal given as final answer but no other partial credit.

17 7 Mark Scheme June 0 5 Throughout this question, candidates may do valid work in the incorrect answer space. This can be marked and given credit wherever it occurs, as long as it does not contradict the working and final answer given in the designated space. (i) State, or 5 B0 if other terms still present eg 5 C 0 or 0. Could be part of a longer expansion, in which case ignore all other terms unless also solely numerical. (ii) nd term = 5 x x (kx) = 05kx rd term = 0 x x (kx) = 70k x 05k = 70k k =.5 Obtain 05k as coeff of x Either stated, or written as 05kx. Allow unsimplified expression ie 5 x x k or 5 x x (kx), even if subsequently incorrectly evaluated. B0 if still 5 C unless later clearly used as 5. Attempt coeff of x Needs to be an attempt at a product involving the relevant binomial coefficient (not just 5 C unless later seen as 0), and an intention to square the final term (but allow for kx ). 67.5k is M0 (from 5 / x ). Obtain 70k Allow unsimplified ie 0 x x k or 0 x x (kx) even if subsequently incorrectly evaluated. Allow 70k following 0 x x kx ie an invisible bracket was used. Equate coefficients and attempt to solve for k Must be one linear and one quadratic term in k, and must be appropriate method to solve this two term quadratic eg factorise or cancel common factor of k. Condone powers of x still present when equated, as long as not actually used in solution method. Could still gain if incorrect, or no, binomial coefficients used each term must be product of powers of (poss incorrect), correct powers of k and any binomial coefficient used. 5 Obtain k =.5 (ignore any mention of k = 0) Any exact equivalent, including unsimplified fraction. Could be implied by writing ( +.5x) 5. NB If expansion is given as 05kx + 70kx, and candidate then concludes that k = 05 / 70 this is only as k never seen. 5

18 7 Mark Scheme June 0 (iii) 0 x x.5 = Attempt 0 x x k Obtain 0.75 (allow 0.75x ) Need to see 0 so just 5 C is not good enough for. Need to see correct powers intended, even if incorrectly evaluated. This includes a clear intention to cube.5 (or their k), so 0 x 9 x.5x = 5x is M0. Can get if using their incorrect k, including 0, but M0 if the value of k used is different to that obtained in (ii). For incorrect numerical answer (following incorrect k), we need to see evidence of method it cannot be implied by answer only. If k = -, 0 or we still need to see evidence of cubing. If 90k is seen in part (ii) (or even (i)) then this is sufficient for unless contradicted by their work in part (iii). Allow if still k rather than numerical. Or any exact equivalent. Ignore if subsequently rounded eg to 0 as long as exact value seen. If.5 obtained incorrectly in part (ii), full credit can still be gained in part (iii). 6

19 7 Mark Scheme June 0 6(i) f() = f( ) = f() = 0, hence (x ) is a factor Attempt use of factor theorem at least once Just substituting at least one value for x is enough for, showing either the working or the result, or both. Just stating f(a) = k is enough don t need to see term by term evaluation. If result is inconsistent with the f(a) being attempted, then we do need to see evidence of method used. No conclusion required. M0 A0 for division attempts, even if considering remainder. Obtain factor of (x ) (ii) f(x) = (x )(x + x 5) Attempt complete division by a linear factor, or equivalent ie inspection or ± 9 coefficient matching x = or x = Allow for sight of (x ), even if x = also present. No words required, but penalise if used incorrectly ie A0 if explicitly labelled as root. A0 if (x ) not seen in this part, even if subsequently used in (ii). SR for (x ) stated with no justification, and no incorrect terminology. Need linear factor of form (x ± a), a 0. Allow if factor different to their answer to (i), inc no answer to (i). Must be complete attempt at all three terms. If long division then need to be subtracting lower line; if coefficient matching then need to be considering all possible terms from their expansion to equate to relevant coefficient from cubic; if inspection then expansion must give at least one correct coefficient for the two middle terms in the cubic. 6 8 Obtain x + x + c or x + bx 5 Obtain x + x 5 Attempt to solve quadratic equation Obtain ½ ( ± 9) State as root, at any point 7 Obtain x and one other correct term. Just having two correct terms does not imply need to look at method used for third term. If coefficient matching allow for stating values eg a = etc. If quadratic factor given with minimal working in (ii), there may be more evidence of method shown in (i). Could appear as quotient in long division, or as part of product if inspection. If coefficient matching, must now be explicitly stated rather than just a =, b =, c = -5. Using quadratic formula or completing the square see extra guidance sheet. Quadratic must come from division attempt, even if this was not good enough for first. Or / ± 9 / from completing the square. Ignore terminology and ignore if subsequently given as factors, as long as seen fully simplified as roots. Must be stated in this part, not just in part (i). Ignore terminology.

20 7 Mark Scheme June 0 7(a) (i) u 9 = 7 x ( ) 8 = 79 Attempt u 9 using ar 8 Allow for 7 x - 8. Using r = will be marked as a misread. Obtain 79 Condone brackets not being shown explicitly in working. SR B for listing terms, as long as signs change. Need to stop at u 9 or draw attention to it in a longer list. (ii) S 5 = 5 ( ( ) ) ( ) 7 = 76,6 Attempt sum of GP using correct formula Must be using correct formula, so denominator of is M0 unless r clearly seen previously. If n = used, then only mark as misread if no contradictory evidence seen starting with S 5 = implies error in using in formula so M0. Obtain 76,6 Condone brackets not being shown explicitly in working. SR B for listing terms and then manually adding them. (b) N / ( x 7 + (N ) x ) = 900 N (6 N) = N 8N 900 = 0 (N 58)(N + 50) = 0 N = 58 State correct unsimplified S N Equate attempt at S N to 900 and rearrange to f(n) = 0 If (n )d is written as (N ), then give benefit of doubt and allow, even if misused in subsequent work (eg becomes N ), unless there is clearly an error in the formula used. Must be attempt at S N for an AP, so using u N or GP formulae will be M0. M0 if (N ) becomes N. To give at least one of the two terms in the bracket must have been multiplied by -. Can still get if incorrect formula as long as recognisable, and is quadratic in N. Expand brackets and collect all terms on one side of equation. Allow slips eg not dividing all terms in bracket by. Obtain N 8N 900 = 0 Any equivalent form as long as f(n) = 0 (but condone 0 not being explicit). Attempt to solve term quadratic Any valid method as long as it has come from equating an attempt at S N of an AP to Obtain 58 only 58 must clearly be intended as only final answer could be through underlining, circling or deleting other value for N. No need to see other value for N if seen, allow slips as long as factorisation / substitution into formula is correct. 58 from answer only or trial and improvement can get 5/5. 58 and -50 with no working is /5. 8

21 7 Mark Scheme June 0 8(i) translation of units in negative y-direction State translation State or imply units in negative y- direction Not shift, move etc. Independent of first. Statement needs to clearly intend a vertical downwards move of, without ambiguity or contradiction, such as down, - in the y direction etc or vector notation. B0 if direction unclear, such as in the y-axis (could be along or towards) or along the y-axis (unless direction made clear). Allow or units but not places, spaces, squares, coordinates or mention of (scale) factor of. If both a valid statement and an ambiguous statement are made eg units down on the y-axis then still award. Ignore irrelevant statements, such as where the y-intercept is, whether correct or incorrect. Give BOD on double negatives eg down the y-axis by units unless clearly wrong or contradictory eg negative y-direction by. 0 (ii) y = State or imply y = Just stating - is enough. B0 for final answer of 0 or. (-, 0) is B0 unless - already seen or implied as y-coordinate. (iii) x = x = log Attempt to solve x = 0 Rearrange to x =, introduce logarithms (could be no base or any base as long as consistent) and then attempt expression for x. M0 for x = log. A0 for alternative, correct, log expressions such as log / log or /log. Decimal equivalent of.58 can get A0. x = log (y + ) is M0 (unless y then becomes 0). State log Doesn t need to be x = Change of base is not on the specification, but is a valid method and can gain both marks. Allow if base not initially specified, but then both logs become base. NB x log = 0 leading to correct answer, can get full marks as there is no incorrect statement seen. 9

22 7 Mark Scheme June 0 (iv) p = 65 log p = log 65 plog = log 65 p = 6.0 * d* Rearrange equation and introduce logs (or log ) Drop power and attempt to solve Must first rearrange to p = k, with k from attempt at 6 ±, before introducing logs. Can use logs to any base, as long as consistent, or equiv with log. Dependent on first. p = log k will gain both M marks in one step. If taking logs to any other base, or no base, or log on both sides then need to drop power of p and attempt to solve using a sound algebraic method ie p = log k / log. Obtain 6.0, or better Decimal answer reqd, if more than sf it must be in range [6.0, 6.0]. Answer only, or trial and improvement, is 0/ as no evidence of using logs as requested. (v) { ( ) } Attempt y-values at x =,.5, = 8.66 M0 if other y-values also found (unless not used in trap rule). Allow for using incorrect function as long as still clearly y- values that are intended to be the original function eg x or (x ).. Attempt correct trapezium rule Obtain 8.66, or better Must be correct structure ie 0.5 x 0.5 x (y 0 + y + y ). Must be finding area from to, so using eg x = 0, 0.5, is M0. Allow if still in terms of y 0 etc as long as these have been clearly defined elsewhere. Using x-values in trapezium rule is M0, even if labelled y-values. Allow a different number of strips (except ) as long as their h is consistent with this, and the limits are still and. If final answer given to more than sf, allow answers in range [8.655, 8.657]. Exact answer from integrating e xln is 0/. Answer only is 0/. Attempting integration before using trapezium rule is 0/. Using two separate trapezia is fine. 0

23 7 Mark Scheme June 0 9(a) (i) π radians State π Allow. radians or B0 for 0 x π. (ii) ( π /, ) State x = π / State y = Allow.57 radians, or better. Allow A = π /. B0 for Allow cos A =. SR Award for (-, π / ) (iii) cosx = 0.5 x = π /, 5π / x = π / 6, 5π / 6 hence π / 6 x 5π / 6 Attempt correct solution method Obtain π / 6 (allow 0.5 or 0 o ) Obtain 5π / 6 (allow.6 or 50 o ) Inverse cos and then divide by, to find at least one angle. Just mark angle, ignore any (in)equality signs. Needs to be single term so π - π / 6 is A0. Just mark angle, ignore any (in)equality signs. A0 if any other angles in range 0 x π. Obtain π / 6 x 5π / 6 (exact radians only) Allow two separate inequalities as long as both correct and linked by and (not or, a comma or no link). Mark final answer and condone incorrect inequality signs elsewhere in solution. SR If alternative methods (eg double angle formulae) or inspection are used (or no method shown at all) then mark as B Obtain one correct angle (degrees or radians) Obtain second correct angle and no others Obtain correct inequality (exact radians only)

24 7 Mark Scheme June 0 (b) tan x = / x = π / 6, 7π / 6 x = π /, 7π / Obtain tan x = / Attempt correct solution method of tan x = k Allow for decimal equiv ie tan x = Allow for tan x =. Inverse tan and then divide by, to find at least one angle. Could follow error eg tan x =, even if tan x = cos x / sin x clearly used. Obtain one correct angle Could be exact ( π / or 7π / ), decimals (0.6 or.8) or degrees (5 o or 05 o ). Must come from correct working only. Obtain both correct angles Must now both be in exact radians. A0 if any other angles in range 0 x π. OR cos x = sin x sin x = cos x = sin x = ± / cos x = ± / x = π / 6, 5π / 6, 7π / 6, π / 6 x = π /, 5π /, 7π /, π / Obtain correct equation in either sin x or cos x Attempt correct solution of sin x = k or cos x = k Square both sides and use sin x + cos x = to obtain sin x = or cos x =, or any equiv, including unsimplified eg sin x = sin x. Inverse sin or cos and then divide by, to find at least one angle. x = π /, 7π / Obtain one correct angle Could be exact ( π / or 7π / ), decimals (0.6 or.8) or degrees (5 o or 05 o ). Must come from correct working only. Obtain both correct angles Must now both be in exact radians. A0 if any other angles in range 0 x π. SR If using alternative methods, such as more advanced trig identities, or no method at all shown, then mark as B Obtain one correct angle, (degrees or radians)with no errors seen Obtain second correct angle, now both in radians

25 7 Mark Scheme June 0 Guidance for marking C Accuracy Allow answers to sf or better, unless an integer is specified or clearly required. Answers to sf are penalised, unless stated otherwise in the mark scheme. sf is sometimes explicitly specified in a question - this is telling candidates that a decimal is required rather than an exact answer eg in logs, and more than sf should not be penalised unless stated in mark scheme. If more than sf is given, allow the marks for an answer that falls within the guidance given in the mark scheme, with no obvious errors. Extra solutions Candidates will usually be penalised if an extra, incorrect, solution is given. However, in trigonometry questions only look at solutions in the given range and ignore any others, correct or incorrect. Solving equations With simultaneous equations, the method mark is given for eliminating one variable allowing sign errors, addition / subtraction confusion or incorrect order of operations. Any valid method is allowed ie balancing or substitution for two linear equations, substitution only if at least one is non-linear. Solving quadratic equations Factorising candidates must get as far as factorising into two brackets which, on expansion, would give the correct coefficient of x and at least one of the other two coefficients. This method is only credited if it is possible to factorise the quadratic if the roots are surds then candidates are expected to use either the quadratic formula or complete the square. Completing the square candidates must get as far as (x + p) = ± q, with reasonable attempts at p and q. Using the formula candidates need to substitute values into the formula and do at least one further step. Sign slips are allowed on b and ac, but all other aspects of the formula must be seen correct, either algebraic or numerical. If the algebraic formula is quoted then candidates are allowed to make one slip when substituting their values. Condone not dividing by a as long as it has been seen earlier.

26 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

27 GCE Mathematics Advanced GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

28 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

29 7 Mark Scheme June 0 (i) Obtain integral of form ke x any non-zero constant k different from 6; using substitution u xto obtain ke u earns (but answer to be in terms of x ) Obtain correct e x 6 or equiv such as (ii) Obtain integral of form ln( ) any non-zero constant k ; allow if brackets absent; ln (after sub n) earns 0 Obtain correct 5l n(x ) or equiv such as ln( ) brackets rather than modulus signs but brackets or modulus signs must be present (so that 5lnx earns A0) Include + c at least once 5 anywhere in the whole of question ; this mark available even if no marks awarded for integration 5 Apply one of the transformations correctly to their equation Obtain correct ln x ln or equiv Show at least one logarithm property correctly applied to their equation of resulting curve (even if errors have been made earlier) Obtain y ln( x ) or equiv of required form; ln x earns ; correct answer only earns /; condone absence of y (a) State sin cos sin or unsimplified equiv such as 7(sin cos ) sin Attempt to find value of cos by valid process; may be implied Obtain exact answer required; ignore subsequent work to find angle (b) Attempt use of identity for cos of form cos ; initial use of cos sin needs attempt to express e x sin in terms of cos to earn Obtain 6cos 9cos 0 or unsimplified equiv or equiv involving sec Attempt solution of -term quadratic eqn for cos or (after adjustment) for sec Use sec at some stage cos or equiv Obtain 5 or equiv; and (finally) no other answer 8

30 7 Mark Scheme June 0 (i) Draw sketch of y ( x) * touching positive x-axis and extending at least as far as the y-axis; no need for or 6 to be marked; ignore wrong intercepts Draw straight line with positive gradient * at least in first quadrant and reaching positive y-axis; assess the two graphs independently of each other Indicate two roots AG; dep *B *B and two correct graphs which meet on the y-axis; indicated in words or by marks on sketch [SC: Draw sketch of y ( x) x6 and indicate the two roots : (i.e. max mark)] (ii) State 0 or x 0 not merely for coordinates (0, 6) (iii) Obtain correct first iterate to at least dp; any starting value (> 6) Show correct iteration process producing at least iterates in all; may be implied by plausible converging values Obtain at least correct iterates allowing recovery after error; iterates given to only d.p. acceptable; values may be rounded or truncated Obtain.8 answer required to exactly dp; A0 here if number of iterates is not enough to justify.8; attempt consisting of answer only earns 0/ [ ; ; ; ; ; ] 8 5 Attempt use of product rule * to produce ln( ) kx kx x x form Obtain xln(x ) x Obtain x or equiv Attempt second use of product rule * Attempt use of quotient (or product) rule * allow numerator the wrong way round Obtain 8x 8 x(x) 6x ln(x ) x (x ) or equiv Substitute into attempt at second deriv dep *M *M *M 96 Obtain ln5 5 8 or exact equiv consisting of two terms 8

31 7 Mark Scheme June 0 6 Method : (Differentiation; assume value 0 ; eqn of tangent; through origin) Differentiate to obtain k(x 5) any constant k Obtain ( x 5) or equiv Attempt to find equation of tangent at P and attempt to show tangent passing through origin assuming value 0 ; or equiv Obtain y x and confirm that 5 tangent passes through O AG; necessary detail needed Method : (Differentiation; equate y change change x to deriv; solve for x) Differentiate to obtain k(x 5) any constant k Obtain Equate Obtain y change change ( x 5) or equiv x to deriv and attempt solution x5 x obtain 0 only (x 5) and solve to Method : (Differentiation; find x from y f( x) x and y x 5 ) Differentiate to obtain k(x 5) any constant k Obtain ( x 5) or equiv State ( 5) x, y x 5, eliminate y and attempt solution condone this attempt at eqn of tangent Obtain 0 only Method : (No differentiation; general line through origin to meet curve at one point only) Eliminate y from equations y kx and y x 5 and attempt formation of quadratic eqn Obtain k x x50 or equiv Equate discriminant to zero to find k 0 Obtain k or equiv and confirm x 5 Method 5: (No differentiation; use coords of P to find eqn of OP; confirm meets curve once) 0 5 Use coordinates (, 5) to obtain y x 0 or equiv as equation of OP Eliminate y from this eqn and eqn of curve and attempt quadratic eqn should be 9x 60x000or equiv Attempt solution or attempt discriminant Confirm 0 only or discriminant = 0

32 7 Mark Scheme June 0 Either: Integrate to obtain Obtain correct 9 k(x 5) * any constant k ( x 5) Apply limits 5 and 0 dep *M; the right way round Make sound attempt at triangle area and calculate (triangle area) minus (their area under curve) or equiv 0 0 Obtain and hence or exact equiv involving single term Or: Arrange to x = and integrate to obtain ky ky form * 5 Obtain y y 9 Apply limits 0 and 5 dep *M; the right way round Make sound attempt at triangle area and calculate (their area from integration) minus (triangle area) Obtain and hence 5 5 (9) or exact equiv involving single term (i) Either: Attempt solution of at least one linear eq n of form ax b Obtain A and (finally) no other answer Or: Attempt solution of -term quadratic eq n obtained by squaring attempt at g( x ) on LHS and squaring or on RHS Obtain A () and (finally) no other answer (ii) Either: Obtain (x 5) 5 for h Attempt to find inverse function of function of form ax b Obtain ( x 0) 9 or equiv in terms of x Or: State or imply g is ( x 5) g g Attempt composition of with 5 Obtain 9 () or more simplified equiv in terms of x (iii) State x 0 B give for answer x 0 8

33 7 Mark Scheme June 0 0.0t k 8 (i) Differentiate to obtain form e any constant k different from t 0.0t Obtain 5.6e or 5.6e or (unsimplified) equiv Obtain.9 or.9 or.87 or.87 but not greater accuracy; allow if final statement seems contradictory; answer only earns 0/ differentiation is needed (ii) Either: State or imply M 75e kt or equiv Attempt to find formula for M 0.07t ( ln 8 ) t Obtain M 75e 0 or equiv such as 75e 5 Equate masses and attempt rearrangement as far as equation with e appearing once 0.06t 6 Obtain e 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii 0.t Or: State or imply M 75 r for positive value r 0. Obtain 75.6 t Attempt to find M in terms of e Equate masses and attempt rearrangement 0.06t 6 Obtain e 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii (iii) Attempt solution involving logarithm of any equation of form e mt c whether the conclusion of part ii or not Obtain 7. or greater accuracy 7. ; correct answer only earns both marks 0 5

34 7 Mark Scheme June 0 9 (i) Use at least one identity correctly angle-sum or angle-difference identity Attempt use of relevant identities in single rational expression not earned if identities used in expression where step equiv to A BC A B C D E F D E F or similar has been carried out; condone (for ) if signs of identities apparently switched (so that, for example, denominator appears as cos cos sinsin cos cos cos sinsin ) sin cos sin Obtain cos cos cos or equiv but with the other two terms from each of num r and den r absent Attempt factorisation of num r and den r Obtain sin and hence tan cos 5 AG; necessary detail needed (ii) State or imply form k tan50 obtained without any wrong method seen State or imply sin50 tan50 or equiv such as 9cos50 Obtain or exact equiv (such as ); correct 9 answer only earns / (iii) State or imply tan 6 k State k 6 Attempt second value of using 6 tan k (multiple of 80) Obtain 6 tan k 0 and no other value 6

35 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

36 GCE Mathematics Advanced GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

37 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

38 7 Mark Scheme June 0 Attempt to factorise both numerator & denominator completely or partially Num = e.g. x x 9 or x x x x or x x x x Denominator = e.g. x x x 5x or x x x 5x x x 5 or 6 WWW ISW but not if any further cancellation x 5 Alternative start, attempting long division Expand denom as quartic & attempt to divide numerator denominato but not divide denominator numeratorr Obtain quotient = & remainder = Final available as before 6x 6x 5x 5 soi e.g. 5 or 5 Allow 5 May be implied by 5 or /5 in final answer 5 or AEF FT their 5. Accept or 5 5 (i) The words quotient and remainder need not be explicit Long division For leading term x in quotient Suff evidence of div process ( x, mult back, attempt sub) (Quotient) = x (Remainder) = x AG No wrong working, partic on penult line Identity x x 0x Qx R * Q ax b,r cx d & attempt at least operations dep* If a, this operation a, b c, d 0 No wrong working anywhere B or state quotient x Inspection x x 0x x x x Clear demonstration of LHS = RHS B (ii) x Change integrand to their (i) quotient + x Correct FT integration of their (i) quotient x x dx ln x Exact value of integral = ln ln AEF ISW Answer as decimal value (only) A0 8

39 7 Mark Scheme June 0 Indefinite integral Attempt to connect dx and dθ Incl Denominator 9x becomes cos d x d, d, dx... dθ ; not dθ x θ dx dθ Reduce original integral to cos θ dθ Change cos d to May be implied, seen only as sec d tan Ignore at this stage Use appropriate limits for (allow degrees) or x Integration need not be accurate AEF, exact answer required, ISW (i) Attempt to set up equations of type s, 6 s t, s t s,t, or, or ( 0 0, ) * or s & or t two of,, Show clear contradiction e.g.,, 6 dep* Allow unsimpl contradictions. No ISW. 0 SC If s found from nd & rd eqns and contradiction shown in st eqn, all marks may be awarded. 0 (ii) Work with and Clear method for scalar product of any vectors Clear method for modulus of any vector () 79. or better ( ).8 (rad) ( ) ISW (From. ) (iii) Use s 6 s. 0 s Obtain s from 9s s s 0 A is or i + j + k final answer B Accept,, 0

40 7 Mark Scheme June 0 6 ax = ax.. +. ( ax), N.B. third term = 8 a x Change x x into k, where is likely to be k // / x, & work out expansion of x = x ( ) x, N.B. third term = 8 x x x x into l, where l is likely to be / / /,& work out expansion of - OR Change x x x (for all terms simplified) 8 8 k (with possibility of + + to follow) l (with no further marks available) Multiply ax by x or The required three terms (with/without x ) identified as a 6 56 a or 6 8a 56 SC for x x x x ; for 6 If result is x Ignore irrelevant products a AEF ISW + 8 for one correct term + for other two p qx rx, then to find p qx rx ; for multiplying ax award for... correct st & nd terms of expansion, correct rd term; p, by their x., as before, for correct answers. 8 7 Attempt to sep variables in format q py (dy) (dx x ) where constants p and/or q may be wrong Either y & ln( x ) or y & ln x + Accept lnx 6 for ln x & for ( ) If indefinite integrals are being used (most likely scenario) Substitute x, y into an eqn containing +const Sub y =.5 and their value of const & solve for x or q x or q =.97 only A [SC x or q =.970 or. 97 or or ] 7 If definite integrals are used (less likely scenario) Use...dy...d x where corresponds with... M &.5 corresp with q (at top/bottom or v.v.).5 q Then A or SC as above q Use...dy...d x where corresponds with q.. &.5 corresp with (at top/bottom or v.v.).5 Then for.97 only 7

41 7 Mark Scheme June 0 8 Cartesian equation may be used in parts (i) - (iii) and corresponding marks awarded (i) Sub parametric eqns into y x & produce t OR sub t into para eqs, obtain, & state y x OR other similar methods producing (or verifying) t Value of t at other point is t is sufficient for + (ii) Use (not just quote) dy dt dx dt dy dx t x = or or y x Attempt to use for gradient of normal dy dx Gradient normal = cao Subst t into the parametric eqns. to find pt at which normal is drawn Produce y x as equation of the normal WWW 6 A marks in (ii) are dep on prev A (iii) Substitute the parametric values into their eqn of normal Produce t = 0 as final answer cao This is dep on final in (ii) N.B. If y x is found fortuitously in (ii) (& given A0 in (ii)), you must award A0 here in (iii). (iv) Attempt to eliminate t from the parametric equations Produce any correct equation e.g. x y Produce y or x x y ISW Must be seen in (iv) x {N.B. Candidate producing only y is awarded both marks.} x

42 7 Mark Scheme June 0 9 (i) Treat x ln x as a product If ln x, use parts u = ln x, dv = Obtain Show x. ln x x ln x x d x = x ln x x x. ln x = ln x WWW AG And state given result x (ii)(a) Part (a) is mainly based on the indef integral ln x dx [A candidate stating e.g. ln x dx ln x dx or = ln x x dx is awarded 0 for (ii)(a)] Correct use of ln x dx x ln x x anywhere in this part Quoted from (i) or derived Use integ by parts on ln x dx with u ln x, dv ln x or u ln x, dv [For integration by parts, candidates must get to a st stage with format f x / gxdx ] x st stage = ln xx ln x x x ln x x dx soi xln x x nd stage = x ln x x ln x AEF (unsimplified) x. ln x dx x Value of definite integral between & e = e - cao Use limits on nd stage & produce cao Volume = e - ISW 6 Answer as decimal value (only) A0 Alternative method when subst. u = lnx used Attempt to connect dx and du Becomes u e u du First stage u e u u e Third stage ( u u ) e u u du Final available as before (b)indication that reqd vol = vol cylinder vol inner solid Clear demonstration of either vol of cylinder being (including reason for height about the y-axis (including upper limit of x dy e e y d y e ln e ) or rotation of x e y ln e ) Could appear as e dy or. or.8 or better May be from graphical calculator Possible helpful points. M is Method; does the candidate know what he/she should be doing? It does not ask how accurate it is.. d e.g. in Qu., a candidate saying x cos is awarded. d. When checking if decimal places are acceptable, accept both rounding & truncation.. In general we ISW unless otherwise stated.. The symbol is sometimes used to indicate follow-through in this scheme. 0 5

43 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

44 GCE Mathematics Advanced GCE Unit 75: Further Pure Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

45 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

46 75 Mark Scheme June 0 (i) a 0 B seen or implied elements correct Other elements correct, a.e.f., including brackets (ii) a a Sensible attempt at matrix multiplication for AB or BA Obtain correct answer 5 Establish result true for n = or * Add next term to given sum formula D Combine with correct denominator Obtain correct expression convincingly 5 Specific statement of induction conclusion, provided st marks earned 5 k 6 Obtain correct det Equate their det to 0 k Obtain correct answers Express as sum of two series n(n )(n ) n 6 Each term correct a.e.f. Attempt to factorise n (n ) A 6 Completely correct answer, ( if one factor not found ) 6 5 (i) a Correct modulus arg a 60,,.05 Correct argument (ii) Circle Centre (, ) Through origin, centre (, ) and another y intercept Vertical line * Through a or their centre, with +ve gradient D Correct half line 6 8

47 75 Mark Scheme June 0 6 Show correct expansion process for or multiplication of C and adjc Correct evaluation of any det C = 5a 5 Obtain correct answer Show correct process for adjoint entries a a a a Obtain at least correct entries in adjoint Obtain completely correct adjoint Divide their adjoint by their determinant (i) Obtain given answer correctly (ii) Express at least st two and last two terms using (i) st two terms correct Last two terms correct Show that correct terms cancel n ( n ) 5 Obtain correct answer, a.e.f. in terms of n (iii) ft Sum to infinity stated or implied or start at 000 as in (ii) S their (ii) with n = 999 or 000 or show correct cancelling Obtain correct answer, a.e.f. ( condone 0.00 ) 9 8 (i) ( 0, ) seen (, 0 ) seen Square with A B and C positioned correctly (ii) 0 0 or 0 0 * Reflection in y = x or y = -x D Correct matrix, dep on stating reflection 0 0 or 0 0 * Enlargement scale factor or s.f. - D Correct matrix, dep on stating enlargement S.C. B for a pair of transformations consistent with their diagram. 7

48 75 Mark Scheme June 0 9 (i) 6 + 0i State correct value (ii) Use a = ( sum of roots ) a = Obtain correct answer Use b = product of roots b = 56 Obtain correct answer Substitute, expand and equate imag. parts Obtain a = - Equate real parts Obtain b = (iii) Attempt to equate real and imaginary parts of (p+iq) & 6 0i or root from (ii) p q 6 and pq 5 Obtain both results cao Obtain quadratic in p or q Solve to ob tain p () 5 or q () Obtain correct answers as complex nos Attempt at all roots (5 ± i) 7 State other two roots as complex nos 0 (i) 0 u u Use substitution correctly EITHER Rearrange Square 9 u u u Obtain correct equation u u 9u 0 5 Obtain given answer OR e. g. ( u u )( u u ) 0 M Multiply their equation in u by appropriate related expression A Obtain given answer (ii) Stated or imply that u x b Us e a Obtain correct answer c Use a 9 5 Obtain correct answer 0

49 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

50 GCE Mathematics Advanced GCE Unit 76: Further Pure Mathematics Mark Scheme for June 0

51 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

52 76 Mark Scheme June 0 x A Bx C ( x)( x 9) x x 9 For correct form seen anywhere with letters or values A 6 x A( x 9) ( BxC)( x) B, C 6 x 9 6( x ) 6( x 9) (i) Asymptote x = y x x asymptote y x (ii) METHOD x ( y6) x(y5) 0 b ac0y6 (y5) 0 y y56 0 y 5 0 : this is true y So y takes all values METHOD dy x x7 Obtain OR dx x x 5 For correct A (cover up or otherwise) For equating coefficients at least once.(or substituting values) into correct identity. For correct B and C For correct final statement cao, oe For correct equation For dividing out (remainder not required) For correct equation of asymptote (ignore any extras) N.B. answer given For forming quadratic in x For considering discriminant For correct simplified expression in y soi For completing square (or equivalent) and correct conclusion www dy For finding either by direct dx differentiation or dividing out first For correct expression oe. dy x, dx so y takes all values. For drawing a conclusion For correct conclusion www Alternate scheme: Sketching graph Graph correct approaching asymptotes from both side Graph completely correct Explanation about no turning values Correct conclusion A graph with no explanation can only score

53 76 Mark Scheme June 0 (i) x. x.0, x.8 (ii) e F ( ) 0.8 (0.86) e For correct x For correct x For dividing by For estimate of e e F( ) (iii) F ( ) 0.78 (0.78) For true F( ) obtained from d ( ln x ) dx TMDP anywhere in (i) (ii) deduct once (but answers must round to given values or A0) For y x and y F( x) drawn, crossing as shown For lines drawn to illustrate iteration (Min horizontal and vertical seen) Staircase For stating staircase

54 76 Mark Scheme June 0 (i) x rcos, y rsin a cos sin r cos sin for 0 (ii) f sin sin a cos cos a sin cos sin cos For substituting for x and y For correct equation oe (Must be r =..) For correct limits for θ (Condone <) N.B. answer given For replacing θ by in their f( ) For correct simplified form. (Must be convincing) f( ) f For correct reason for (iii) (iv) r a.. a For correct value of r.oe Closed curve in st quadrant only, symmetrical about Diagram showing 0, tangential at O

55 76 Mark Scheme June 0 5(i) dx x sin y cos y dy dy d x sin y x For implicit diffn to d y dx cosy oe For using sin ycos y to obtain N.B. Answer given + taken since sin x has positive gradient For justifying + sign (ii) f(0) 0, f (0) For correct values f ( x) f ( x) x x x x x x f ''(0) = 0, f '''(0) Use of chain rule to differentiate f (x) Use of quotient or product rule to differentiate f (0). For correct values www, soi (iii) 6 sin x x x Alternative Method: f(0) = 0, f'(0) = f'( x) x x x... x 8 f ''( x) x x... 9 f '''( x) x... f '(0),f ''(0) 0,f '''(0) sin x x x 6 sin x ln( x) x x x x x 6 5 For correct series (allow!) www For correct values Correct use of binomial Differentiate twice Correct values Correct series ft For terms in both series to at least x f.t. from their (ii) multiplied x x x together For multiplying terms to at least x For correct series up to x www For correct term in x www

56 76 Mark Scheme June 0 6(i) n I x ( x) dx n n n x ( x) n x ( x) d 5 5 x n In n x ( x) dx n n n n I n x x x dx n I ni ni 5 5 For integrating by parts (correct way round) For correct first stage For splitting 5 x suitably For obtaining correct relation between I n and In (ii) I I n n I n 5 n 5 0 x For correct result (N.B. answer given) For evaluating I 0 [OR I by parts] For using recurrence relation [OR ] times (may be combined together) I I I I For [OR ] correct fractions I 55 For correct exact result 5

57 76 Mark Scheme June 0 7(i) y = tanh - x Both curves of the correct shape (ignore overlaps) and labelled y = tanhx y = tanhx gradient = at x = 0 stated For asymptotes y = ± and x = ± (or on sketch) y = tanh - x Sketch all correct (ii) k 0 k x 0 tanh x dx ln(cosh ) ln(cosh k) For substituting limits into ln cosh x For correct answer (iii) Areas shown are equal: x = tanhk y = k For consideration of areas For sufficient justification tanh k 0 tanh x dx = rectangle ( k tanh k ) (ii) For subtraction from rectangle For correct answer N.B. answer given ktanh k ln(cosh k) Alternative: Otherwise by parts, x as tanh x OR ln x PTO for alternative schemes 6

58 76 Mark Scheme June 0 7(iii) Alternative method By parts: tanh k I tanh x dx 0 u tanh x dvdx du dx v x x tanh k tanh k x I xtanh x 0 x k k x k k k ktanh k ln(sech k) ktanh kln(sech k) Alternative method By substitution Let y tanh x x tanh y tanh k tanh ln( ) 0 tanh ln( tanh ) d sech d x y y When x 0, y 0 When x tanh k, y k tanh k 0 dx k tanh d sech d I x x y y y 0 0 I ytanh y tanh y dy ktanh kln cosh k k 0 k 0 u y dvsech y dy du dy v tanh y For integrating by parts (correct way round) For getting this far Dealing with the resulting integral For substitution to obtain equivalent integral Correct so far For integration by parts (correct way round) Final answer 7

59 76 Mark Scheme June 0 8(i) x cosh u du cosh usinh udu For correct result x cosh u dx coshusinhudu x sinhu cosh udu For substituting throughout for x For correct simplified u integral cosh u du sinh ucosh uu For attempt to integrate For correct integration cosh u (ii) x ( x) ln x ( x) ( c) ln( ) 7 For substituting for u For correct result oe as x x f( ) ln(g( )) (iii) x V ( ) d x ( ) xln x x For attempt to find x x dx For correct integration (ignore π) V For statement that volume is infinite (independent of M mark) 8

60 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0