Core Mathematics C1 Advanced Subsidiary

Transcription

1 Paper Reference(s) 666/0 Edexcel GCE Core Mathematics C Advanced Subsidiary Monday 0 January 0 Morning Time: hour 0 minutes Materials required for examination Mathematical Formulae (Pink) Items included with question papers Nil Calculators may NOT be used in this examination. Instructions to Candidates Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics C), the paper reference (666), your surname, initials and signature. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for the parts of questions are shown in round brackets, e.g. (). There are questions in this question paper. The total mark for this paper is 75. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may not gain full credit. H50A This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 0 Edexcel Limited.

2 . (a) Find the value of 6. () (b) Simplify x x. (). Find 5 ( x x x ) dx, giving each term in its simplest form. (5). Simplify 5, giving your answer in the form p + q, where p and q are rational numbers. (). A sequence a, a, a,... is defined by where c is a constant. n a =, a = an c (a) Find an expression for a in terms of c. () Given that a i = 0, i (b) find the value of c. () H50A

3 5. Figure Figure shows a sketch of the curve with equation y = f(x) where f(x) = x x, x. The curve passes through the origin and has two asymptotes, with equations y = and x =, as shown in Figure. (a) In the space below, sketch the curve with equation y = f(x ) and state the equations of the asymptotes of this curve. () (b) Find the coordinates of the points where the curve with equation y = f(x ) crosses the coordinate axes. () 6. An arithmetic sequence has first term a and common difference d. The sum of the first 0 terms of the sequence is 6. (a) Show that 0a + 5d = 6. Given also that the sixth term of the sequence is 7, (b) write down a second equation in a and d, (c) find the value of a and the value of d. () () () H50A Turn over

4 7. The curve with equation y = f(x) passes through the point (, 0). Given that find f(x). f (x) = x 8x +, (5) 8. The equation x + (k )x + ( k) = 0, where k is a constant, has two distinct real roots. (a) Show that k satisfies (b) Find the set of possible values of k. k + k > 0. () () 9. The line L has equation y x k = 0, where k is a constant. Given that the point A(, ) lies on L, find (a) the value of k, (b) the gradient of L. () () The line L passes through A and is perpendicular to L. (c) Find an equation of L giving your answer in the form ax + by + c = 0, where a, b and c are integers. () The line L crosses the x-axis at the point B. (d) Find the coordinates of B. (e) Find the exact length of AB. () () H50A

5 0. (a) Sketch the graphs of (i) y = x(x + )( x), (ii) y = x. showing clearly the coordinates of all the points where the curves cross the coordinate axes. (6) (b) Using your sketch state, giving a reason, the number of real solutions to the equation x(x + )( x) + x = 0. (). The curve C has equation y = x 9x + x 8 + 0, x > 0. d y (a) Find. d x (b) Show that the point P(, 8) lies on C. (c) Find an equation of the normal to C at the point P, giving your answer in the form ax + by + c = 0, where a, b and c are integers. (6) () () END TOTAL FOR PAPER: 75 MARKS H50A 5

6 January 0 Core Mathematics C 666 Mark Question. (a) 6 = or or better 6 6 = or 0.5 (ignore ± ) A () (b) x = x or x x = x or equivalent or 6 A cao () (a) for a correct statement dealing with the or the power This may be awarded if is seen or for reciprocal of their 6 s.c ¼ is A0, also is A0 ± is not penalised so A (b) for correct use of the power on both the and the x terms A for cancelling the x and simplifying to one of these two forms. Correct answers with no working get full marks GCE Core Mathematics C (666) January 0

7 6 x x x 6. ( = ),, +,( +c) = 6 x x + x + c A A,A,A 5 for some attempt to integrate: a non zero constant st 6 x A for or better 6 nd x A for or better rd A for x or better n n x x + i.e 6 ax or ax or ax or ax, where a is th A for each term correct and simplified and the +c occurring in the final answer GCE Core Mathematics C (666) January 0

8 . ( + ) ( ) = denominator of Numerator = A So 5 = + + ( ) = 5, and form simultaneous equations in p and q -p + q = 5 and p - q = - A Solve simultaneous equations to give p = and q =. A st for multiplying numerator and denominator by same correct expression st A for a correct denominator as a single number (NB depends on M mark) Alternative: ( p q ) nd for an attempt to multiply the numerator by ( ± ) and get terms with at least correct. nd A for the answer as written or p = and q =. Allow 0.5 and.5. (Apply isw if correct answer seen, then slip writing p =, q = ) Answer only (very unlikely) is full marks if correct no part marks A GCE Core Mathematics C (666) January 0

9 (a) ( a ) = c 6 B () (b) a = (their a) c (= 8 - c) a+ a + a = + "(6 c)" + "(8 c)" 6 5c = 0 Aft So c = 5. A o.a.e (b) st for attempting a. Can follow through their answer to (a) but it must be an expression in c. nd for an attempt to find the sum a+ a + a must see evidence of sum st Aft for their sum put equal to 0. Follow through their values but answer must be in the form p + qc = 0 A accept any correct equivalent answer () 5 GCE Core Mathematics C (666) January 0

10 5. (a) y y= y= Correct shape with a single crossing of each axis y = labelled or stated B B x= x= x x = labelled or stated B () (b) (b) Horizontal translation so crosses the x-axis at (, 0) x ± New equation is ( ) ( ) When x = 0 y = y = x ± B = A B for point (,0) identified - this may be marked on the sketch as on x axis. Accept x =. st for attempt at new equation and either numerator or denominator correct nd for setting x = 0 in their new equation and solving as far as y = A for or exact equivalent. Must see y = or (0, ) or point marked on y-axis. Alternative f( ) = = scores A0 unless x =0 is seen or they write the point as (0, ) or give y = / Answers only: x =, y = / is full marks as is (,0) (0, /) Just and / is B0 A0 Special case : Translates unit to left (a) B0, B, B0 (b) Mark (b) as before May score B0 A0 so /7 or may ignore sketch and start again scoring full marks for this part. () 7 GCE Core Mathematics C (666) January 0 5

11 6. (a) 0 S0 = [ a+ 9 d] or S0 = a+ a+ d+ a+ d+ a+ d+ a+ d+ a+ 5da+ 6d+ a+ 7d+ a+ 8d+ a+ 9d 6 = 0a + 5d * Acso () u = a+ ( n ) d 7= a+ 5d B (b) ( ) n () 0 ( b) gives 0a+ 50d = 70 (a) is 0a + 5d = 6 Subtract 5d = 8 so d =.6 o.e. A Solving for a a = 7-5d so a = 9 A (a) for use of S n with n = 0 () 7 (b) st for an attempt to eliminate a or d from their two linear equations nd for using their value of a or d to find the other value. GCE Core Mathematics C (666) January 0 6

12 7. ( f( x) ) x( c) x 8x = + + A A ( f( ) = 0 ) 0= ( ) + c c = 9 A f( x) = x x + x st n n for an attempt to integrate x x + st A for at least terms in x correct - needn t be simplified, ignore +c nd A for all the terms in x correct but they need not be simplified. No need for +c nd for using x = - and y =0 to form a linear equation in c. No +c gets M0A0 rd A for c = 9. Final form of f(x) is not required. (a) b ac= ( k ) ( k) ( ) ( ) k 6k+ 9 k > 0 or k + 8k > 0 or better k + k > 0 * Acso (b) ( k )( k )[ 0] + = Critical values are k = or - A (choosing outside region) k > or k < - A cao (a) st for attempt to find b acwith one of b or c correct nd for a correct inequality symbol and an attempt to expand. Acso no incorrect working seen (b) st for an attempt to factorize or solve leading to k = ( values) nd for a method that leads them to choose the outside region. Can follow through their critical values. nd A Allow, instead of or > loses the final A < k < - scores A0 unless a correct version is seen before or after this one. () () 7 GCE Core Mathematics C (666) January 0 7

13 9. (a) ( 8 k 0) = so k = 5 B (b) y = x+ k y = x+... and so m= o.e. A (c) Perpendicular gradient = Equation of line is: y ( ) Bft = x Aft y+ x = 0 o.e. A () () () c (d) y = 0, B(7,0) or x = 7 x = 7 or a Aft () (e) ( 7 ) ( 0) AB = + AB = 5 or A (b) for an attempt to rearrange to y = A for clear statement that gradient is.5, can be m =.5 o.e. (c) Bft for using the perpendicular gradient rule correctly on their.5 for an attempt at finding the equation of the line through A using their gradient. Allow a sign slip st Aft for a correct equation of the line follow through their changed gradient () nd A as printed or equivalent with integer coefficients allow y+ x= or y = x (d) for use of y = 0 to find x = in their equation c Aft for x = 7 or a (e) A for an attempt to find AB or AB for any correct surd form- need not be simplified GCE Core Mathematics C (666) January 0 8

14 0. (a) y (i) correct shape ( -ve cubic) B Crossing at (-, 0) B Through the origin B Crossing at (,0) B - O x (ii) branches in correct quadrants not crossing axes One intersection with cubic on each branch B B (6) (b) solutions Since only intersections (b) Bft for a value that is compatible with their sketch dbft This mark is dependent on the value being compatible with their sketch. For a comment relating the number of solutions to the number of intersections. Bft dbft () 8. (a) (b) (c) [ Only allow 0, or ] dy 7 = x x 8x dx AAA 8 x= y = = = -8 * Acso x= 7 8 y = 6 7 A = 7 = 7 Gradient of the normal = " " Aft Equation of normal: y 8= ( x ) 7 7y x+ 6 = 0 A () () (6) GCE Core Mathematics C (666) January 0 9

15 (a) st n n for an attempt to differentiate x x st A for one correct term in x nd A for terms in x correct rd A for all correct x terms. No 0 term and no +c. (b) A (c) st A for substituting x = into y = and attempting note this is a printed answer Substitute x = into y (allow slips) Obtains.5 or equivalent nd for correct use of the perpendicular gradient rule using their gradient. (May be slip doing the division) Their gradient must have come from y rd for an attempt at equation of tangent or normal at P nd Aft for correct use of their changed gradient to find normal at P. Depends on st, nd and rd Ms rd A for any equivalent form with integer coefficients GCE Core Mathematics C (666) January 0 0