Version.0 General Certificate of Education (A-level) January 0 Mathematics MPC (Specification 660) Pure Core Mark Scheme
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Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 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 eplanation 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, or (or 0) accuracy marks EE deduct 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, we must 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. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (a) = k( ) 5 d M Where k is an integer or function of 6 ( ) 5 = (ISW) A But note d y k ( ) 5 p d = + M0 u = y = u ( ) ( 6 ) d 6u 5 u = and d ( ) 5 du = M = 6 A Note = 6 ( ) 5 + c scores M A0 d (penalise + c in differential once only in paper) (b)(i) ln d =± ± M Product rule attempted and differential of ln = + ln (ISW) A Must have replaced ln e by (ii) ( = e) y = e PI B Condone y =.7 (AWRT) lne ( = ) d = + M Correct substitution into their d y d But must have scored M in (b)(i) ( ) y e= e or y = e OE, ISW A Must have replaced ln e by Total 7 4
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) f = 4 ln + 5 reverse (a) f.5 = 0.9 f.6 = 0.4 Attempt at evaluating both f (.5) and f (.6) f.5 = 0.9 M M f.6 = 0.4 But must see f = 5 ( 4) ln( + ) before A may be earned Condone f(.5 ) < 0 Only if f ( ) defined M f(.6 ) > 0 =.5 y= 4. ( < 5) M =.6 y= 5.4 ( > 5) < < OE A Either side of 5,.5 < α <.6 OE A Change of sign,.5 α.6 (b) ( 4) ln( + ) = 5 5 4= ln ( + ) 5 = 4 + ln =± ( + ) 4+ 5 ln ( + ) M AG A Either of these lines correct Condone poor use of brackets for M only Must have both middle lines and no errors seen (c) ( = ) 5 =.578 CAO B =.568 CAO B Total 6 Sight of AWRT.58 or.57 scores B B0 ±.578 or ±.568 scores B B0 =.578, =.568 scores BB0 5
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) (a)(i) Where k is an integer d ksec ( y ) M Condone omission of d But = ksec ( y+ ) d scores M A0 = sec y + ISW A Alternative methods y= ( tan ) d k( ) = + M ( ( y )) = + tan + A sin = cos ( y + ) ( y + ) kcos ( y ) ksin ( y ) ( y+ ) d ± + ± + = cos = ( y + ) cos M A (ii) d y = sec + M Substitution of y = into their d d or = sec 0 d BUT must have scored M in (a)(i) = d CSO A Condone 0. or better = d sec (y+ ) = As above sec 0 = (b) M A Total 6 Appro correct shape with no turning points, through (0,0) and only curve π Asymptotic at both ± and both values shown Condone ± 90 (degrees) Condone y= tan also drawn but clearly identified, otherwise M0 6
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 4(a) f M, < f < < f <, < y< f <, < f A Allow y, f (b)(i) y = cos y = cos y cos = M cos = y = cos Either order y= cos M Swap and y f ( ) = cos A (ii) If incorrect in (b)(i) BUT answer = cos M in form p cos ( q ) (condone p, q =) Then q = cos M or = f() M p = cos ISW A = cos A (c)(i) gf = cos B (ii) π π Modulus graph in st quadrant, starting from a +ve y-intercept, at least M continuous parts, first descending, then second increasing IGNORE CURVE OUTSIDE RANGE A Correct curvature, curves reaching -ais, condone multiple curves (no turning points at ais) A Approimately symmetrical graph with, π, π indicated (must have scored previous marks) Condone y = cos also drawn but clearly identified, otherwise M0 (d) STRETCH + direction M Either in -direction or y-direction s.f., parallel to y-ais A Either order s.f., parallel to -ais A Total 4 7
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 5(a)(i) d + = kln + M Where k is a rational number = ln ( + ) + c A if substitution u= +, du= d du = = kln u u M = ln ( + ) + c A (b) u= dv= sin M d sin ( d) = kcos, = d where k is a constant du = v= cos A All correct Correct substitution of their terms into = cos cos ( d) m parts formula (watch signs carefully) = cos + 4sin + c A 4 CAO Total 6 8
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 6(a) y B Using 4 correct -values, PI 0.05 cos.5 = 0.4780 0.5 cos.45 = 0.585 0.5 cos.75 = 0.454 0.5 cos.05 = 0.86 M At least correct y-values, (condone unsimplified correct epressions), correct values rounded to s.f. or truncated to s.f. 0. Σ y m Used and must be working in radians = 0. CAO A 4 Must be s.f. (b) du d = M du= d OE u ± = u kdu m = ± All in terms of u, with k = or Condone omission of du u u ( du) 9 m p u ± u d u 5 u u = 5 9 5 = 4 4 9 5 5 A m (must have scored first marks) OE Must have earned all previous method marks and then correct substitution, into their integral, of, 4 for u or 0, for and subtracting 6 = ISW A 6 equivalent fraction 5 Total 0 9
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 7(a) cos = 0. M tan =± 4 =.77, 4.5 AWRT A One correct value A Second correct value and no etra values in interval 0 to 6.8 Ignore answers outside interval SC =.8, 4.5 with or without working M A A0 SC (using degrees) 0.54, 8.54 M A A0 0.5, 8.5 M A0 A0 SC No working shown correct answers / correct answer / (b) LHS cosec ( cosec ) cosec ( + cosec ) Correctly combining fractions but = M condone poor use, or omission, of ( + cosec )( cosec ) brackets cosec cosec cosec cosec = A Allow recovery from incorrect brackets cosec cosec ( + cot ) Correct use of relevant trig identity = or m cot cot eg cosec = + cot sec = 50 All correct with no errors seen sec = 5 AG A 4 INCLUDING correct brackets on st line cosec cosec = 50 + cosec cosec cosec cosec cosec + cosec = 50( + cosec )( cosec ) cosec cosec cosec cosec 48cosec = 50 sin = 50 cosec (M) (A) 4 = cos = (m) 5 5 sec = 5 AG (A) Correctly eliminating fractions but condone poor use, or omission, of brackets Allow recovery from incorrect brackets Correct use of relevant trig identity eg sin = cos All correct with no errors seen INCLUDING correct brackets on st line 0
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 7(c) sec =± 5 M cos =± 0. tan =± 4 =.77, 4.5,.7, 4.9 (AWRT) A correct A 4 correct and no other answers in interval Ignore answers outside interval SC.8, 4.5,.4, 4.9 With or without working M A SC their answers from (a) +.7, 4.9 (AWRT) / SC For this part, if in degrees ma mark is M A0 Total 0 SC No working shown 4 correct answers / correct answers / 0,, correct answers 0/
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 8(a) e = 4 = ln 4 M ln4 = ln 4 ISW A OE, eg ln, ln, (b)(i) ( y = ) B Condone ( 0,) but not (,0 ) (ii) y = 0 4 4e e = 0 4e = 0 M a ± e ± b =0 e = or e = 4 A 4 = ln ISW A OE, eg ln4, ln, ln 4 and no etra solutions 4 4e = e ln 4 = 4 (M) = ln4 (A) OE = ln 4 (A) OE (iii) 4 y = 8e + 4e B 4e = 8e 4 e = 0 or e = 0 or ln 4 4= ln8 or e = or e = ln M Equating d y = 0 and getting d ± ae ± b =0 from pe qe d = + ln 4 ln8 and no etra solutions = ISW A OE, eg ( ) 4
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) 8(b)(iv) Must be completely correct including d ln seen on this line or net line 4 V = π ( 4e e ) d B 0 Limits, brackets and π PI from later working 4 8 6 = ( π) 6e + e 8e ( d) B Correct epansion, PI from later working 6 ln 4 8 4e 6 = ( π ) 4e e + B e 4 OE 8 ( 0) 4 B 8 e OE 8 B 8 e 6 OE may be two separate terms 6 Correct substitution of = ln and 0 into 4ln 8ln 4 6ln = ( π ) 4e e + e their integrated epression 8 4 6 8 M ( must be of form ae + be + ce ) 0 0 4 0 and subtracting. 4e e + e 8 PI 547 = π A 7 OE eact fraction eg 5856 π 048 9804 Total 6 TOTAL 75