Mark Scheme (Results) Summer International GCSE Further Pure Mathematics Paper 1 (4PM0/01)

Transcription

1 Mark Scheme (Results) Summer 0 International GCSE Further Pure Mathematics Paper (4PM0/0)

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3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Types of mark o M marks: method marks o A marks: accuracy marks. Can only be awarded if the relevant method mark(s) has (have) been gained. o B marks: unconditional accuracy marks (independent of M marks) Abbreviations o cao correct answer only o ft follow through o isw ignore subsequent working o SC special case o oe or equivalent (and appropriate) o dep dependent o indep independent o eeoo each error or omission No working If no working is shown then correct answers may score full marks. If no working is shown then incorrect (even though nearly correct) answers score no marks.

4 With working If there is a wrong answer indicated always check the working and award any marks appropriate from the mark scheme. If it is clear from the working that the correct answer has been obtained from incorrect working, award 0 marks. Any case of suspected misread which does not significantly simplify the question loses two A (or B) marks on that question, but can still gain all the M marks. Mark all work on follow through but enter A0 (or B0) for the first two A or B marks gained. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If there are multiple attempts shown, then all attempts should be marked and the highest score on a single attempt should be awarded. In some instances, the mark distributions (e.g., B and ) printed on the candidate s response may differ from the final mark scheme Follow through marks Follow through marks which involve a single stage of calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given. Ignore subsequent working It is appropriate to ignore subsequent working when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent working when the additional work essentially shows that the candidate did not understand the demand of the question. Linear equations Full marks can be gained if the solution alone is given, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.

5 General Principles for Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation ( x + bx+ c) = ( x+ p)( x+ q), where pq = c, leading to x = ( ax bx c) ( mx p)( nx q), where pq c and mn a, leading to x + + = + + = = =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square b Solving 0 :, 0, leading to x =... x + bx+ c= x± ± q± c q Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by.. Integration: Power of at least one term increased by. Use of a formula: Generally, the method mark is gained by either quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values or, where the formula is not quoted, the method mark can be gained by implication from the substitution of correct values and then proceeding to a solution. Answers without working: The rubric states "Without sufficient working, correct answers may be awarded no marks". General policy is that if it could be done "in your head" detailed working would not be required. (Mark schemes may override this eg in a case of "prove or show...") Exact answers: When a question demands an exact answer, all the working must also be exact. Once a candidate loses exactness by resorting to decimals the exactness cannot be regained. Rounding answers (where accuracy is specified in the question) Penalise only once per question for failing to round as instructed - ie giving more digits in the answers. Answers with fewer digits are automatically incorrect, but the isw rule may allow the mark to be awarded before the final answer is given.

6 Question Number. Scheme Marks (a) (b) (c) θ 6 = θ or π = θ = = = l θ 00.7 = = 44π l = 4 l = l = π = = (cm) Method (d) in Notes (4) Notes Question Method (a) and (c) for an expression in either degrees or radians using A=6 to find angleθ for a fully correct expression with correct numerical values for an expression in either degrees or radians with their θ to find arc length AB AB = (cm) cso Method (b) for a correct formula rl for correct substitution of the value of r, (=) for equating their formula to 6 cm = (cm) cso Method (d) for an area of a circle divided by 6 for using r = for the length of the circumference of the circle divided by their value of the scale factor using a value for r of only. for (cm) cso Note: Correct solution only seen award full marks Allow.0 (cm)

7 Question Number. ( ) 9 x + x+ < x x 7x 6 0 Scheme Marks + < (x )( x+ ) < 0 < x < (4) Notes Question for obtaining a TQ equation or expression (=0 not required for this mark) for attempting to find their critical values as far as x = (We are treating this as an M mark) for choosing the inside region for their critical values. cao for x <. Accept x and x < and x x <. Do not accept x or x <, or x, x <.These are all A0 Use of loses the final A mark

8 Question Number Scheme Marks. (a) a = b= B B (b) at (,0) c 0 = + c = c = at (0, d) d = + d = (6) Notes Question (a) B for either a or b B for both a and b for substituting in y = 0 and x = into the equation of the curve. a need not be substituted for this mark (b) for c = cso for substituting x = 0 and y = d into the equation of the curve to find d. Neither c nor a need to be substituted for this mark. d = cso.

9 Question Number 4. Scheme Marks 6( cos x) cos x 4 = 0 6cos x+ cosx = 0 (cos x+ )( cos x ) = 0 (cos x= ) or cos x= x = 60 or x = 60 (6) Notes Question 4 for using cos x + sin x = to achieve an equation in terms of cos x only. (=0 not required for this mark) for forming the correct TQ for solving their TQ as far as cos x =.(usual rules for an attempt) Their quadratic need not = 0 at this stage cos x =, (cos x = this need not be seen) for either value of x = 60, x = 60 for both values x = 60 x = 60 If other values are given, ignore if not in range. Deduct one A mark for each extra value that is in range, up to a maximum of the last two A marks.

10 Question Number Scheme Marks 5. V = 500 4h = 500 h = 5 dv = h d h dh dh dv = = 6 dt dv dt h 6 = = = 0. cm/s 5 5 (7) Question 5 Notes Note: Parts of the question can be found anywhere in their working on the page for V = 500 4h = 500 h = 5 cso for differentiating V = 4h (usual rules apply) for dv dh = h cso dh dh dv for applying chain rule to find an expression for = or any correct arrangement dt dv dt (expression is sufficient substitution of values is not required for this mark) for substituting values into their dh dt dh for dt = 5 ) oe - exact answer only.

11 Question Scheme Marks Number 6. (a) (i) α + β = p B α + pα + = 0 (ii) or β + pβ + = 0 α + β = ( α + β) αβ α + β + p( α + β) + = 0 (iii) = p ( α β) α α β αβ β α + β = p + = α + β = ( α + β) αβ( α + β) = ( p) ( p) alternatives = p p α + β = ( α + β)( α αβ + β ) = p p = p p ( ) α + pα + α = 0 β + pβ + β = 0 α + β + p( α + β ) + ( α + β) = 0 α + β + p( p ) p= 0 α + β = p p (b) x p p x ( ) + = 0 ft ft (8) Notes Question 6 (a) (i) B p for α + β = p or (Note αβ = ) (ii) for α + β = ( α + β) αβ and substituting in values for α + β, and αβ Or α + pα + = 0 for β + pβ + = 0 α + β + p( α + β) + = 0 for α + β = p oe (Simplification is not required for this mark) (iii) for expanding ( α + β) = α + α β + αβ + β (allow some slips in algebra for this mark). Do NOT accept ( α + β) = α + β for this mark leading to α + β = ( α + β) αβ( α + β) fully correct for α + β = p p oe (Simplification is not required for this mark) (b) Please refer to ms for alternative methods for using x their sum x + product (= 0 not needed for this mark) `ft for x ( p p ) x+ = 0 (follow through their values for this mark) Note: = 0 must be seen with a correct equation for this mark Simplification is not required for this mark

12 Question Number Scheme 7. (a) (i) t 58 = a+ 57d B (ii) ( ) B (b) a+ 57 d = (a+ d) a = d 4 d = a 7 4 (c) t76 = a+ 75d = a+ 75( 7 a) OR 4 S = (a+ 0 d) = a+ 0( 7a) = a 00a = 99a 4 S = (a+ 0 d) = a+ 0( 7a) OR 4 t = a+ 75d = a+ 75( a) 76 7 = a 0a= 99a= t 76 Marks (d) a+ ( r ) d = 5( a+ 8 d) ( r ) d = 4( 4d) + 40 d or ( r )( 7a) = 4a+ 40( 7a) r = or 4( r ) = r = 4 () Notes Question 7(a) (i) B for any correct expression for t 58 (simplification not required for this mark) (ii) B for any correct expression for S (simplification not required for this mark) (b) for their t 58 = their S 4 for collecting like terms on either side leading to d = a cso (c) This is a show question so all working must be seen clearly. for an expression for t 76 or S in either a or d Substitution must be for the given value of d 69 for t 76 = a 00a= 99a or t 76 = for an expression for t 76 or S in either a or d Substitution must be for the given value of d 7 4 d cso OR S = a 0a = 99a for t 76 = a 00a= 99a 69 OR S = a 0a= 99a= t76 or t 76 = S = 4 d cso with a conclusion Alternative for t 76 = S using their expressions for correct unsimplified t 76 = S for 5d = 0aoe 4 for d = 7 a with a conclusion that must refer to part (b) (d) for equating expressions for t r and 5t 9 in r, a and d for an equation in r only (allow for slip ups in algebra for this mark) r = 4 cso

13 Question Number Scheme Marks 8. (a) 5 + x x = 0 (5 x)( + x) = 0 x= 5, x= 5 (5 + )d (b) x x x 5 = 5x x x + 5 = ( ) ( ) 85 = (c) x + 9= 5+ x x x x 6= 0 ( x )( x+ ) = 0 x=, x= (d) = { + + } = 85 { 6 + x x } dx M 85 (5 x x ) ( x 9) dx x x x = 85 [6 + ] 8 = 85 {( ) ( + + ) } 5 = 85 0 = 64 (4) 6 Alternative 5 (d) M = (5 + x x )d x+ (7 + )5 + (5 + x x )dx = [5 x + x x ] + + [5 x+ x x ] = ( ) ( ) ( ) ( ) = = 64

14 Question 8 (a) for setting 5 + x x = 0 for solving the quadratic as far as x = for x= 5, x= (b) Ignore limits for first and for an attempt at for an area = 5 5 x + x x dx Notes (Usual rules) ft their values of x in (a) for a fully correct integrated expression for an evaluation of their integrated expression with their limits 85 or 56 or awrt 85. (with a minimum of dp) cso. (c) for equating line l with curve C ( x + 9= 5+ x x ) for forming a TQ and attempting to solve as far as x = for x=, x= (d) for forming a COMPLETE expression of the area, either from, M = 85 (or their area in part (b)) { + + } 5 (5 )d (7 )5 (5 )d (5 x x ) ( x 9) d x or, M = + x x x x x x using their limits found in (c) for correct integration of their expression for the area d for evaluating their integrated expression for the area 5 either, = = 64, or = = 64 oe exact answer only NOTE: If they do not form a complete expression for the area, then M0 A0 dm0 A0

15 Question Number Scheme Marks 9. (a) ABC = 90 B BC BC cos 0 = or sin 60 = BC = cos0 = 6 cm or BC = sin 60 = 6 cm BP (b) sin 0 = 6 BP = 6 sin0= 6 = cm (c) BF tan 5 = or tan 65 = BF BF = or BF = tan 65 tan 5 BF =. cm (SF) (d) BD = ( tan 65) + (6 ) or DP = ( tan 65) + ( tan 60) BD =.7 = 5.4 or DP = 05. = 4. sin BDP = or 5.4 BDP = 9.9 tan BDP = 4. (e) Volume = ( tan 65) 6 tan cm (SF) = = (4)

16 Notes Please note the stipulations on exact answers and the rounding required. Please refer to General Principles. Question 9 (a) B for ABC = 90, can be implied from working for any acceptable trigonometry using a complete method to find BC for the value 6 only. Do not accept any decimal value for this mark (b) for using any acceptable trigonometry using a complete method to find BP for the value of only (this is a show question, all working must be correct) (c) for using any acceptable trigonometry using a complete method involving angles 5º or 65º for a correct expression for BF for BF =. (cm) correct to sf for this mark (d) for an attempt at an expression for BD or DP, please refer to the ms for examples - ft their values for BC and BF, but must use for BP for BD =.7 = 5.4 or DP = 05. = 4. for using an expression of any acceptable trigonometry to find BDP for BDP = correct to dp (e) for an expression of the volume using the given AC (=), BP = only, and their BF for 47 cm (correct to sf) Lengths of line in the prism for examiners AC = DE = AB = EF = 6 BP = BF = CD = AE =.4. AD = CE = 6.7. CP = 9 AP = BC = DF = 6 DP = 4.. BD = 5.4

17 Question Number Scheme 0. dy (a) dx x x x dy at R, 4 4 dx l has equation ( ) [ ] = (b) 4x x 4x+ = 4x x 4x+ = 0 4( x )( x x ) = 0 4( x )( x+ )( x ) = 0 x=, x=, x= At P, x=, y = = 5 so P(, 5) At Q, x=, y = = so Q(, ) 5+ (c) Gradient of PQ = = Equation of l is y+ = ( x ) [ y = x 4] or y+ 5 = ( x+ ) Marks (d) Gradient of l = gradient of C at P = gradient of C at Q [= ] [Since l passes through P and Q with the same gradient as the curve at these points, it must be a tangent to C at P and at Q.] B (e) Normal at R has equation y = ( x ) At intersection with l, ( x 4) = ( x ) or y = ( y+ 4 ) x= 8 or y = 0 x= 9 and y = 5 RS = ( 5) + ( 9) RS = = 8 (f) PQ = ( ) + ( 5 + ) = = 4 Area PQR = 8 4 = (8) alternative (f) Area PQR = = [ ( + 9 5) ( 5 ) ] 5 5 = (5 9)

18 (a) Notes for an attempt at differentiation (usual rules reducing the power of at least one term, the disappearance the constant is insufficient for this mark) for a complete correct differentiated expression for finding and using a numerical value of the gradient, derived only from using dy into either y dx (their m) (x ), or by applying y = mx+ cincluding finding a value for c for any correct equation y (x ) [ y x, y x 0] (b) = + = etc for setting their dy = and re-arranging to give a cubic equation (=0) dx for factorising their equation leading to three values of x for either of the correct coordinates (, 5) or (, ) ( x =, y = 5 or x =, y for both (, 5) and (, ) correct, ( x =, y = 5 and x =, y (c) for finding the numerical gradient of l using their coordinates of P and Q, and attempting to form an equation using their gradient and the points P or Q for a correct equation eg y+ 5= ( x+ ) or y+ = ( x ) [ y = x 4] (d) B please refer to ms (e) for forming the equation of the normal at R. They must use a numerical gradient derived from their gradient of the tangent in part (a) using the rule m t m n =, and use the given coordinate of R. y = ( x ) oe ( y = x 4) for finding the point of intersection of the Normal at R and l, by any acceptable method eg., simultaneous equations for the point of intersection of S, either x = 9 and y = 5, or gives coords (9, 5) for using Pythagoras with point R and their S for 8, 8 oe exact answer only (f) for any method to find the area of triangle PQR ft their P and Q for area PQR = (units )

19 Question Number. (a) AB = p q oe Scheme (b) BC = 6p 4 q ( p q) or AC = 6p 4 q ( p+ q) = p q = 5p 5q AB = BC or AB BC or AB = 5 AC or AB AC ABC,, are collinear Marks B (c) AB : BC = : oe B (d) CD = AC or AD = AC or BD= 0 AC = (( p q ) + ( p q )) = (5p 5 q ) = (5p 5 q ) = (5p 5 q ) 0 OD = (6p 4 q) + (5p 5 q) or ( p+ q) + (5p 5 q) or ( p q) + (5p 5 q) 0 = 8 p 6 q (8) alternative OA + OD OA + OC = OC or OD = + + OD ( ) (6 4 ) = 6p 4q or OD = + 7p q OD = 7p q or OD = OD = 8 p 6 q

20 Notes Question (a) B for AB = p q or any equivalent expression (b) for finding a vector for BC ( = p q) or AC ( = 5p 5q) for AB = BC or AB BC or AB = 5 AC or AB AC So A, B,C are collinear cso there must be two correct vectors (c) B for AB : BC = : (oe) (d) First Method for forming a vector equation for either CD, AD, or BD for k ( 5p 5q) where k is either, or for CD, AD, or BD respectively 0 for finding an expression for OD (alternatives in ms) for OD = 8 p 6 q oe Second method for the ratio of AC: CD for AC:CD = : for either component of p or q correct, ie., 8 p OR 6 q 7p q for a complete correct expression for, OD = 8 p 6 q, OD =, oe

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