AS Mathematics MPC1 Unit: Pure Core 1 Mark scheme June 017 Version: 1.0 Final
FINAL MARK SCHEME AS MATHEMATICS MPC1 JUNE 017 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk
FINAL MARK SCHEME AS MATHEMATICS MCP1 JUNE 017 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A,1 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, the principal examiner may suggest that we award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Otherwise we require evidence of a correct method for any marks to be awarded.
FINAL MARK SCHEME AS MATHEMATICS MPC1 JUNE 017 Q1 Solution Mark Total Comment NO MISREADS ALLOWED IN THIS QUESTION 1 7 5 7 5 7 5 7 (Numerator = ) 5 0 7 7 56 at least this far (Denominator = 5 7 7 8 ) = B1 must be seen as denominator 58 7 Value = = 17 6 7 cso condone 6 7 17 x 9 5 " their "6 5 " their " 5 attempt to write each term as k 5 with either 5 6 5 or 80 5 (b) x 9 5 6 5 5 OE must have equation 1 x or x 1 or x 1. must be simplified to one of these Total 7 5 7 Condone multiplication by 5 7 instead of for only if subsequent working shows 5 7 multiplication by both numerator and denominator otherwise M0 For first each term must be evaluated correctly but may be seen in a grid An error in the denominator such as 5 5 7 5 7 8 should be given B0 and it would then automatically lose the final cso May use 1 7 7 5 5 7 7 5 ; numerator = 5 0 7 7 56 etc Alternative: 1 7 mn 7 5 7 leading to 5m 1n 1; m 5n (one correct) (both correct) either m 17 or n 6 ; answer = 17 6 7 cso (expression must be explicitly seen) (b) Alternative 1 Multiply or divide each term by 5 obtaining integer terms (possibly with one error) correct integer terms; eg x(5 0) 0 ; x Alternative 80 9 5 5 x 9 5 5 9 5 5 00 OE (integer/integer) 5 Squaring: Do not allow final unless justification for rejecting negative value May earn for 5 16 x k 80 and for x OE 9
FINAL MARK SCHEME AS MATHEMATICS MCP1 JUNE 017 Q Solution Mark Total Comment (b) d y two terms correct 0 x 6x dx all correct ( 6 x)( x) ( 0) OE correct factors or correct use of formula for correct quadratic possibly multiplied by 1 or divided by (other stationary point when) 5 0 x OE eg, 1, 1.6 0 but not cso 1 1 d y 1x dx d y when x, dx 0 therefore minimum (point) E1ft B1 B1ft Sub x into their correctly FT their value of reason d y d x d y and evaluate d x but must have (c) Cubic graph through origin with one max & one min on either side of y-axis may be reflection of given graph in x-axis for Graph roughly as shown in all quadrants Total 9 dy If candidate multiplies by 1 before differentiating and writes 6x x 0 this scores M0 dx For second, may have (5 x)( x) or (x5)( x ) or (5 x)( x) etc Second is earned for formula as far as 1 or etc 1 6 5 If both values given x, then allow cso Withhold cso if no =0 seen or incorrect equating of expressions. (b) Allow E1ft for their + > 0 so min etc or d y >0 (provided they have a value earlier) minimum etc d x (c) Allow if the curve does not actually cross the x-axis times For ignore any numbers on graph 5
FINAL MARK SCHEME AS MATHEMATICS MPC1 JUNE 017 Q Solution Mark Total Comment b clear attempt at p( ) c( ) 8 b c 0 b c8 0 AG must see powers of simplified correctly and = 0 appearing before last line (b) b c 0 clear attempt at p() and = 0 7 9b c 0 b c 7 0 ACF terms need not be collected but powers of must be evaluated No ISW - mark their final equation (c) Correctly eliminating b or c from b c8 0 and an equation from (b) PI by one correct answer b 7 or c 6 b 7 and c 6 Total 7 Condone poor use of brackets if recovered on next line for both and Note that = 0 must appear before last line; Example p( ) 8 b c b c8 0 scores A0 may also be earned for a full long division attempt by (x+) as far as the remainder in terms of b and c. also for p( x) ( x )( x dx 1) b d; c d 1 with for completion. Terms must be exactly as printed answer but accept 0 b c 8 for. (b) Do NOT treat use of 0 instead of 0 as a misread. may also be earned for a full long division attempt by (x ) as far as the remainder and equating the remainder (in terms of b and c) to 0. Ignore trailing equals sign for in both parts and (b). 6
FINAL MARK SCHEME AS MATHEMATICS MCP1 JUNE 017 Q Solution Mark Total Comment 6 5 grad AB 56 or 8 8 y 5 " their " ( x ) or y 6 " their " ( x 8) x y 8 correct unsimplified; PI by ft their gradient but must use A or B coordinates correctly integer coefficients with x and y terms on one side and integer on other side (b) k k 7 grad AC or grad BC B1 correct simplified expression eg k k 8 k k7 1 OE condone one error in one term k k8 k k 8 k 6k 16 k k 0 k k ( k )(k) ( 0) d attempt at factorisation or correct use of formula for their quadratic 1 k, k 5 OE 5 Total 9 May earn second for y " their " x c and attempt to find c using x, y 5 or x 8, y 6 For, condone any equivalent equation of the correct form such as 8 y x or x 0y 56 but not x y 8 0 or any equation with non-integer coefficients (b) Alternative AC ( k ) ( k ) or BC ( k 8) ( k 7) B1 (may be under square root) Pythagoras ( k ) ( k ) ( k 8) ( k 7) (condone one error in one term) k k k k k k k k 8 16 16 6 1 9 1 Allow d for factors that would multiply out to give their k 6k88 0 etc k and constant terms. To earn d using formula then it must be correct for their quadratic and discriminant evaluated correctly ; if correct quadratic used, you must see 61 OE to award d 7
FINAL MARK SCHEME AS MATHEMATICS MPC1 JUNE 017 Q5 Solution Mark Total Comment d y dx 8x 9x one term correct d y dx correct d y dx 8( 1) 9( 1) 17 d correct substitution of x 1 and correct evaluation for their d y dx (Grad of normal =) 1 17 ft FT their negative reciprocal provided is earned 1 1 y x 9 17 17 5 1 15 or y x 17 17 equation must be in this form (b)(i) 5 x x x 5 5 5 5 ( 1) ( 1) ( 1) 5 16 8 5 5 19 = 1 or 1.95 or 79 0 0 d two terms correct all correct (may have +c) correct substitution of and 1 to find their F() F( 1) correct with powers of and ( 1) and minus signs handled correctly 5 correct single equivalent fraction or decimal (ii) Area of trapezium = 1 (9 1) = 1.5 or 6 OE B1 or 1 ( x ) dx =1.5 their trapezium answer from (b)(i) must be trapezium (Area of shaded region =) 1.5 1.95 = 17.55 Total 1 OE 17, 51 0 0 1 15 Do not accept y x c,... c for final ; equation must be written in full on one line. 17 17 x 15 x 15 Allow y for final but not y. 17 17 17 (b)(i) Use of F( 1) F() scores dm0 (ii) For condone use of their (b)(i) their trapezium. Be generous in awarding this provided you are convinced they are considering the area of a trapezium. 8
FINAL MARK SCHEME AS MATHEMATICS MCP1 JUNE 017 Q6 Solution Mark Total Comment ( x) ( y 7) = one of these terms correct LHS correct ignoring any extra constants or ( x ) ( y 7) = ( x ) ( y 7) (or =0) or ( x ) ( y 7) (or =0) (b) ( y 7) putting x=0 in their equation ( y 7) 0 y 7 and attempt to solve for y Repeated root means circle touches y-axis E1 completely correct working and both parts of the conclusion ( x ) 7 0 putting y = 0 in their equation ( x ) 51 x 51 and attempt to solve for x Two roots so circle crosses x axis twice E1 completely correct working and both parts of the conclusion (c)(i) ( x ) ( kx 5) 0 x 0x 0 k x kx 5 0 sub y kx into their circle equation and attempt to multiply out brackets (1 k ) x ( k) x 5 0 cso AG be convinced - must have terms exactly as printed answer condone 0 (1 k ) x ( k) x 5 (ii) ( k) 5(1 k ) correct discriminant unsimplified 00 00k 0k 0 0k 0 multiplying out correctly k cso must see =0 before final answer Total 1 (b) May have their y 1y 9 0 and attempt at formula or discriminant for first with E1 awarded for being completely correct with both parts of conclusion such as only one value of y / root/ solution so touches y axis or discriminant = 0 therefore repeated root so touches y axis Second can be earned for their x 0x 9 0 and attempt at formula or discriminant with E1 awarded for being completely correct with both parts of conclusion such as two (different) values of x / roots/ solutions so crosses x-axis in two places May use geometry : award first for stating x-coord of centre = their and radius = their ; then E1 for both stating that lengths are equal hence circle touches y-axis (or y-axis is a tangent to the circle) Second for stating y-coord of centre = their 7 and radius = their ; E1 for both explaining that yc r and concluding statement that the circle crosses x-axis at two distinct points. To award each of these E1 marks all working must be correct. (c)(i) Penalise trailing equals signs and any incorrect algebra even if recovered awarding A0cso All terms must be exactly as in the printed answer but 0= may be on the LHS for cso. May use x ( kx ) 0x 1( kx ) 9 0 and attempt to multiply out brackets for giving x k x kx x kx 0 1 8 9 0 leading to (1 k ) x ( k) x 5 0 for cso (ii) May write ( k) 5(1 k ) for then k k 1 k for Correct discriminant appearing within the quadratic equation formula can earn. 9
FINAL MARK SCHEME AS MATHEMATICS MPC1 JUNE 017 Q7 Solution Mark Total Comment (i) x x ( x y) ( x y) 15 OE ACF eg x ( x y) 15 1 y 15 x OE (ii) ( ) OE S x y y S x xy or x( x y) xy etc x x(15 x) d Sub their y into correct S expression 15x x S (5 x x ) AG be convinced (b)(i) A( x.5) must have both minus signs but A may be negative but not zero 6.5 ( x.5) (ii) (Max S =) " their " p (must be > 0) or = 18.75 Total 9 p ( x q) ; p6.5, q.5 OE 5 5 x (5 " their " q " their " q ) (i) Penalise candidates who are clearly working back from printed answer in part (ii) OE eg 75 withhold if marks not scored in (b)(i) (ii) Only allow d if their y is of the form a bx ; expression need not be simplified so S ( x 7.5 x) (7.5 x), for example, earns d. Must have S = for final but allow (5 x x ) S ; condone S m = but not S = Allow omission of S = on final line provided it appears on an earlier line with = on all subsequent lines. (b)(i) Example 1 5 ( x.5) scores (and ISW any incorrect rearrangement) 5 Example ( x.5) scores A0 16 Alternative for : x 5 x ( x.5).5 but must have both sides of this identity If M0 is scored then award SC B1 for 6.5 ( x.5) or 6.5 (.5 x) (ii) Allow SC B1 if 18.75 is clearly obtained via differentiation or NMS
FINAL MARK SCHEME AS MATHEMATICS MCP1 JUNE 017 Q8 Solution Mark Total Comment dh 1t 59t 7 dt OE dh t 1 59 7 dt d (8 177 + 7 =) two terms correct (may have x for t) all correct must have t substituting t = into their d h dt (b) (Decreasing) 1t 59t 7 0 B1ft FT their d h but must have < 0 dt (t9)(t 8) attempt at factors or correct use of formula 9 8 CVs are t, t + + use of sign diagram or sketch may have 9 8 t 9 5 fractions must be simplified & B1 earned for final 8 9 t AND t no ISW here 8 Total 9 (b) 59 5 For first, if using formula need to go as far as if inequality is correct and to a similar form if following through from their quadratic; if factorising quadratic (correct or incorrect) then factors should multiply out to give t and constant terms to earn first. For second, if critical values are correct then sign diagram or sketch must be correct with correct CVs marked in correct order. However, if CVs are not correct then second can be earned for attempt at sketch or sign diagram but their CVs MUST be marked correctly on the diagram or sketch. For final, inequality must have t and no other letter. If B1 is earned and correct quadratic inequality is seen, final answer of 9 t 8 (with or without working) scores final marks. 9 8 8 9 8 9 (A) x (B) t OR t (C) t, t (D) 9 8 5 6 t (E) t OE with or without working, each score marks (SC) If B1 is NOT earned, then only the next marks are available, namely,, and A0 even if final inequality is correct. 8 9 Example NMS t, t scores M0 (since both CVs are correct)