Mark Scheme (Provisional) Summer 2021 Pearson Edexcel International GCSE In Mathematics B (4MB1) Paper 02. Graded A+

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Question Working Answer Mark Notes 1(a) 56170 1 B1 (b) 1.368 10 2.144 10  −  9 7 or 1346560000 M1 for evidence of the correct subtraction (so M0 for 2.144 10 1.368 10  −  7 9 unle ... ss recovered later) or for a correct answer (to at least 3 significant figures) in non-standard form (e.g., 1346560000, 13.4656 10  8 , 1350000000, etc.). The correct answer implies this mark 1.34656 10  9 2 A1 allow answers which round to (awrt) 1.35 10  9 (c) 4 6 5.617 10 2.166 10   or 0.02593… M1 for evidence of division of the correct two values (condone for M1 6 4 2.166 10 5.617 10   ) or a correct answer (to at least 3 significant figures) in non-standard form (e.g., 0.0259, 0.259 10  −1 , 0.0259326, etc.) or for 2.59 10  −n where n is a positive integer 2.59 10  −2 2 A1 for awrt 2.59 10  −2 (e.g., 2.593259464 10  −2 scores both marks, but M1A0 for 2.6 10  −2 if more accurate answer not seen) Total 5 marks http://britishstudentroom.wordpress.com/2 d 3 2 2 y d x ax b x = + + M1 differentiating with at least 1 non-zero term correct. 3 2 2 2 9.8  +  + = ( )2 a b ( ) or 4 2.2 a b + = − oe M1 dep on 1st M mark substitute in x = 2 into their d y dx and equating to 9.8 (allow any equivalent, e.g., 12 4 9.8 + + = a b ) 6 8 4 2 8 = + + + a b or 2 5 a b + = − oe M1 substitute in x = 2 and y = 6 into y x ax bx = + + + 3 2 8 2 5 2.2 a = − or b = − + 10 2.2 M1 dep on 2nd and 3rd M marks. Correct method (but allow one sign slip) for eliminating a or b from their simultaneous equations Elimination method (oe with coefficients of either a or b the same) e.g. 2 5 (4 2 2.2 ( 5) ) ( ) 4 2.2 a b a b a b a b + = −  + − + = − − − + = − (so for this set of equations the candidate must be subtracting the two equations) or e.g. 4 2 10 (4 2 4 10 ( 2.2) ) ( ) 4 2.2 a b a b a b a b + = −  + − + = − − − + = − Substitution method e.g.b a a a = − −  + − − = − 5 2 4 5 2 2.2 ( ) or e.g. 1 5 ( 5 4 2.2 ) 2 2 b a b b = − −  + = −     − −   (or equivalent) This mark can be implied by either a correct value for a or for b. Allow by use of matrices. a = 1.4 b = −7.8 5 A1(oe e.g. a b = = − 7 39 5 5 , ) dependent on all four M marks Correct answers with no working scores no marks Total 5 marks http://britishstudentroom.wordpress.com/3 (a) (i) 4 18 24 160 x x 2 + + = oe M1 adding all the subsets together and equating to 160. Need not be simplified (but if all 7 terms not shown explicitly then need to see at least 4 18 24 160 x x 2 + + = ). Must see the 160 e.g. 4 18 24 160 0 x x 2 + + − = For reference (if fully un-simplified): ( ) ( ) ( ) ( ) ( ) ( ) 5 3 2 2 2 2 2 8 7 9 4 1 8 2 4 2 3 160 x x x x x x x x + + + + + − + + + + + + − = 2 9 68 0 x x 2 + − = A1 simplifying to the given 3 term quadratic (at least one intermediate line from initial line of working to given answer) – must include = 0 (allow 0 2 9 68 = + − x x 2 ) so all terms on one side equal to zero (ii) (2 17 4 [ 0] x x + − = )( ) oe M1 correct method for solving the given 3 term quadratic – either by formula, completing the square or factorising. By factorising: brackets must expand to give 2 out of 3 correct terms By formula: correct substitution into fully correct formula (allow 1 sign error) By completing the square: must see   2 9 2 ... 0 4 x     +  =   Either correct value of x 17 or 4 2 x x     = − =   can imply this mark NB anything appearing in square brackets [..] is not required x = 4 4 A1 (A0 if 17 2 x = − given as a final answer too) (b) 2 3 "4" 8 2 3 "4" 7.5 "4" 8  +  +  + oe M1 for either ( ) ( ) ( ) ( ) 3 2 3 2 2 2 8 8 4 1 2 4 2 3 x x x x x x + + + − + + + + − oe or for an equivalent expression with their value of x (which must be a positive integer) – if value for x substituted then numerator must be less than 160 7 43 2 A1 oe exact value (A0 if non [Show More]

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