January Core Mathematics C1 Mark Scheme

Transcription

1 January Core Mathematics C Mark Scheme Question Scheme Mark. 4 k or k (k a non-zero constant) M, +..., ( 0) A, A, B (4) 4 Accept equivalent alternatives to, e.g. 0.5,,. M: 4 differentiated to give k, or differentiated to give k (but not for just 0 ). st A: (Do not allow just 4 ) nd A: or equivalent. (Do not allow just, but allow or ). B: differentiated to give zero (or disappearing ). Can be given provided that at least one of the other terms has been changed. Adding an etra term, e.g. + C, is B0.

2 . (a) 6 (a = 6) B () (b) Epanding 7, 4 ( ) to get or 4 separate terms M ( = 7, c = 4) (a) ± 6 also scores B. (b) M: The or 4 terms may be wrong. st A for 7, nd A for 4. b A, A () Correct answer 7 4 with no working scores all marks with or without working scores M A A0. Other wrong answers with no working score no marks. 4

3 8 y. (a) Shape of f() B Moved up M Asymptotes: y = B = 0 (Allow y-ais ) B (4) ( y is B0, 0 is B0). 4 (b) + = 0 No variations accepted. M = (or 0. ) Decimal answer requires at least d.p. A () 6 (a) B: Shape requires both branches and no obvious overlap with the asymptotes (see below), but otherwise this mark is awarded generously. The curve may, e.g., bend away from the asymptote a little at the end. Sufficient curve must be seen to suggest the asymptotic behaviour, both horizontal and vertical. M: Evidence of an upward translation parallel to the y-ais. The shape of the graph can be wrong, but the complete graph (both branches if they have branches) must be translated upwards. This mark can be awarded generously by implication where the graph drawn is an upward translation of another standard curve (but not a straight line). The B marks for asymptote equations are independent of the graph. Ignore etra asymptote equations, if seen. (b) Correct answer with no working scores both marks. The answer may be seen on the sketch in part (a). Ignore any attempts to find an intersection with the y-ais. e.g. (a) This scores B0 (clear overlap with horiz. asymp.) M (Upward translation bod that both branches have been translated). B0 M B0 M B0 M0 No marks unless the original curve is seen, to show upward translation.

4 4. ( ) = or ( y + ) = y + 4y + 4 M: or 4 terms M ( ) + = 0 or y + ( y + ) = 0 M: Substitute M 4 6 = 0 or y + 4y 6 = 0 Correct terms A ( )( + ) = 0, =... or ( y + )( y ) = 0, y =... M (The above factorisations may also appear as ( 6)( + ) or equivalent). = = or y = y = A y = y = or = = M A (7) 6 (Allow equivalent fractions such as: = for = ). 7 st M: Squaring a bracket, needs or 4 terms, one of which must be an or y term. nd M: Substituting to get an equation in one variable (awarded generously). st A: Accept equivalent forms, e.g. 4 = 6. rd M: Attempting to solve a -term quadratic, to get solutions. 4 th M: Attempting at least one y value (or value). If y solutions are given as values, or vice-versa, penalise at the end, so that it is possible to score M MA M A M0 A0. Strict pairing of values at the end is not required. Non-algebraic solutions: No working, and only one correct solution pair found (e.g. =, y = ): M0 M0 A0 M0 A0 M A0 No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M A M A Both correct solution pairs found, and demonstrated, perhaps in a table of values: Full marks Squaring individual terms: e.g. y = + 4 M = 0 M A0 (Eqn. in one variable) = M0 A0 (Not solving -term quad.) y = + 4 = 7 y = 7 M A0 (Attempting one y value) 4

5 5. Use of b 4ac, perhaps implicit (e.g. in quadratic formula) M ( ) ( 9 + 8( k + ) < 0) 4 ( k + ) < 0 A 8k < 7 (Manipulate to get pk < q, or pk > q, or pk = q ) M 7 k < Or equiv : k < or k < Acso (4) 4 st M: Could also be, for eample, comparing or equating b and 4 ac. Must be considering the given quadratic equation. There must not be terms in the epression, but there must be a k term. st A: Correct epression (need not be simplified) and correct inequality sign. Allow also 4 ( k + ) < 0. nd M: Condone sign or bracketing mistakes in manipulation. Not dependent on st M, but should not be given for irrelevant work. M0 M could be scored: e.g. where b + 4ac is used instead of b 4 ac. Special cases:. Where there are terms in the discriminant epression, but then division by gives an inequality/equation in k. (This could score M0 A0 M A).. Use of instead of < loses one A mark only, at first occurrence, so an 7 otherwise correct solution leading to k scores M A0 M A. 8 N.B. Use of b = instead of b = implies no A marks. 5

6 6. (a) ( 4 )(4 + ) k 0. M (The k value need not be eplicitly stated see below) , or k = 4 Acso () + seen, or a numerical value of k seen, ( ) (b) 6 c or k c or 9 ( ) d = C, c M A, Aft () 5 (a) e.g. ( 4 + )(4 + ) alone scores M A0, (but not e.g scores M A0. ( 4 + ) alone). k = 4 or ,with no further evidence, scores full marks M A. Correct solution only (cso): any wrong working seen loses the A mark. 9 9 (b) A: C. Allow 4.5 or 4 as equivalent to. k Aft: (candidate s value of k, or general k). 48 For this final mark, allow for eample as equivalent to 6, but do not allow unsimplified double fractions such as ( ) 4, and do not allow unsimplified products such as 4. A single term is required, e.g is not enough. An otherwise correct solution with, say, C missing, followed by an incorrect solution including + C can be awarded full marks (isw, but allowing the C to appear at any stage). 6

7 7. (a) c or 6 c or 8 c M 8 8 f ( ) = 6 ( + C) 6 + A A Substitute = and y = into a changed function to form an equation in C. M = C C = Acso (5) 8 (b) 6 M = 4 A Eqn. of tangent: y = 4( ) M y = 4 7 (Must be in this form) A (4) (a) First A marks: + C is not required, and coefficients need not be simplified, but powers must be simplified. All terms correct: A A Two terms correct: A A0 Only one term correct: A0 A0 Allow the M A for finding C to be scored either in part (a) or in part (b). (b) st M: Substituting = into 6 (must be this function). nd M: Awarded generously for attempting the equation of a straight line through (, ) or (, ) with any value of m, however found. nd M: Alternative is to use (, ) or (, ) in y = m + c to find a value for c. 8 9 If calculation for the gradient value is seen in part (a), it must be used in part (b) to score the first M A in (b). Using (, ) instead of (, ): Loses the nd method mark in (a). Gains the nd method mark in (b). 7

8 8. (a) 4 k or k or k M d y 9 = A A () d (b) For = 4, y = ( 4 4) + ( 4 4) ( 6) = = 8 (*) B () dy (c) = = d Gradient of normal = M: Evaluate their dy at = 4 M d Aft Equation of normal: y 8 = ( 4), y = + 0 (*) M, A (4) (d) = 0 : =... ( 0) y and use PQ = or ( y M ) + ( y ) PQ = Follow through from (k, 0) Aft and 8 May also be scored with ( ) ( ) 4. = 8 0 A () (a) For the A marks coefficients need not be simplified, but powers must be simplified. For eample, is acceptable. All terms correct: A A Two terms correct: A A0 Only one term correct: A0 A0 (b) There must be some evidence of the 4 value. (c) In this part, beware working backwards from the given answer. d y Aft: Follow through is just from the candidate s value of. d nd M: Is not given if an m value appears from nowhere. nd M: Must be an attempt at a normal equation, not a tangent. nd M: Alternative is to use (4, 8) in y = m + c to find a value for c. (d) M: Using the normal equation to attempt coordinates of Q, (even if using = 0 instead of y = 0), and using Pythagoras to attempt PQ or PQ. Follow through from (k, 0), but not from (0, k) A common wrong answer is to use = 0 to give 0. This scores M A0 A0. For final answer, accept other simplifications of 640, e.g. 60 or

9 9. (a) Recognising arithmetic series with first term 4 and common difference. B (If not scored here, this mark may be given if seen elsewhere in the solution). a + ( n ) d = 4 + ( n ) = n + M A () ( ) n 0 S n, M A, A () (b) = { a + ( n ) d} = { 8 + (0 ) }, = 75 k (c) S k < 750 : { 8 + ( k ) } < 750 or > 750 : { 8 + k} > ( or k + 49 > 0) k + S k M k + 5k 500 < 0 k M A (Allow equivalent -term versions such as k + 5k = 500). ( k 00)( k + 5) < 0 Requires use of correct inequality throughout.(*) Acso (4) (d) or equiv. seen or, k = (and no other values) M, A () (a) B: Usually identified by a = 4 and d =. M: Attempted use of term formula for arithmetic series, or answer in the form (n + constant), where the constant is a non-zero value. Answer for (a) does not require simplification, and a correct answer without working scores all marks. (b) M: Use of correct sum formula with n = 9, 0 or. A: Correct, perhaps unsimplified, numerical version. A: 75 Alternative: (Listing and summing terms). M: Summing 9, 0 or terms. (At least st, nd and last terms must be seen). A: Correct terms (perhaps implied by last term ). A: 75 Alternative: (Listing all sums) M: Listing 9, 0 or sums. (At least 4, 7,.., last ). A: Correct sums, correct finishing value 75. A: 75 Alternative: (Using last term). n M: Using S n = ( a + l) with T 9, T 0 or T as the last term. 0 A: Correct numerical version (4 + ). A: 75 Correct answer with no working scores mark:,0,0. (c) For the first marks, allow any inequality sign, or equals. st M: Use of correct sum formula to form inequality or equation in k, with the 750. nd M: (Dependent on st M). Form -term quadratic in k. st A: Correct terms. Allow credit for part (c) if valid work is seen in part (d). (d) Allow both marks for k = seen without working. Working for part (d) must be seen in part (d), not part (c). 9

10 0 y 0. (a) (i) Shape or or B Ma. at (0, 0). B (, 0), (or shown on -ais). B () (ii) Shape B (It need not go below -ais) Through origin. B 0 (6, 0), (or 6 shown on -ais). B () (b) ( ) = (6 ) M 6 = 0 Epand to form -term cubic (or -term quadratic if divided by ), with all terms on one side. The = 0 M may be implied. ( )( + ) = 0 =... Factor (or divide by ), and solve quadratic. M = and = A = : y = 6 Attempt y value for a non-zero value by M substituting back into ( ) or ( 6 ). = : y = 9 Both y values are needed for A. A (, 6) and (,9) (0, 0) This can just be written down. Ignore any method shown. (But must be seen in part (b)). B (7) (a) (i) For the third shape shown above, where a section of the graph coincides with the -ais, the B for (, 0) can still be awarded if the is shown on the -ais. For the final B in (i), and similarly for (6, 0) in (ii): There must be a sketch. If, for eample (, 0) is written separately from the sketch, the sketch must not clearly contradict this. If (0, ) instead of (, 0) is shown on the sketch, allow the mark. Ignore etra intersections with the -ais. (ii) nd B is dependent on st B. Separate sketches can score all marks. (b) Note the dependence of the first three M marks. A common wrong solution is (-, 0), (, 0), (0, 0), which scores M0 A0 B as the last marks. A solution using no algebra (e.g. trial and error), can score up to marks: M0 M0 M0 A0 M A B. (The final A requires both y values). Also, if the cubic is found but not solved algebraically, up to 5 marks: M M M0 A0 M A B. (The final A requires both y values). 0