Paper Reference(s) 6666/01 Edexcel GCE

1 Paper Reference(s). 6666/01 . Edexcel GCE. Core Mathematics C4. Gold Level (Harder) G2. Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Green) Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulas stored in them. Instructions to Candidates Write the name of the examining body ( Edexcel ), your centre number, candidate number, the unit title (Core Mathematics C4), the Paper reference (6666), your surname, initials and signature.

2 Information for Candidates A booklet Mathematical Formulae and Statistical Tables' is provided. Full marks may be obtained for answers to ALL questions. There are 8 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 gain no credit. Suggested grade boundaries for this Paper : A* A B C D E. 65 58 47 42 36 28. Gold 2 This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 2007 2013 Edexcel Limited.

3 1 A B C. 1. f(x) = = + + . x(3 x 1) 2. x (3 x 1) (3 x 1) 2. (a) Find the values of the constants A, B and C. (4).. (b) (i) Hence find f ( x) dx .. 2.. (ii) Find f ( x) dx , leaving your answer in the form a + ln b, where a and b are constants. 1. (6). June 2012. 2. The current, I amps, in an electric circuit at time t seconds is given by I = 16 16( )t, t 0. dI. Use differentiation to find the value of when t = 3 . dt Give your answer in the form ln a, where a is a constant. (5). January 2011. 3. (a) Use the binomial expansion to show that 1+ x 1 2. 1+ x + x , |x| < 1. 1 x 2. (6). 1. (b) Substitute x = into 26. 1+ x 1 2. =1 + x + x 1 x 2. to obtain an approximation to 3.

4 A Give your answer in the form where a and b are integers. b (3). June 2013. Gold 2: 10/12 2.. 4. Given that y = 2 at x = , solve the differential equation 4. dy 3. = . dx y cos 2 x (5). June 2012. 5. Figure 2. A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm, as shown in Figure 2. Water is flowing into the container. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm3. 4 h 3. (a) Show that V = . 27. (2). [The volume V of a right circular cone with vertical height h and base radius r is given by the 1.]

5 Formula V = r 2h .]. 3. Water flows into the container at a rate of 8 cm3 s 1. (b) Find, in terms of , the rate of change of h when h = 12. (5). January 2009. Gold 2 :10/12 3.. 6. (a) Find tan 2 x dx .. (2). 1. (b) Use integration by parts to find . 3 ln x dx . x (4). (c) Use the substitution u = 1 + ex to show that 3x e dx =. 1 2x x e e + ln (1 + ex) + k, . 1+ e x 2. where k is a constant. (7). January 2009. Gold 2: 10/12 4. 2. 7. (a) Express in partial fractions. 4 y2. (3). (b) Hence obtain the solution of dy 2 cot x = (4 y2). dx . for which y = 0 at x = , giving your answer in the form sec2 x = g( y). 3. (8). June 2008. Gold 2 :10/12 5. 8.

6 Figure 3. Figure 3 shows part of the curve C with parametric equations . x = tan , y = sin , 0 < . 2. 1 . The point P lies on C and has coordinates 3, 3 . 2 . (a) Find the value of at the point P. (2). The line l is a normal to C at P. The normal cuts the x-axis at the point Q. (b) Show that Q has coordinates (k 3, 0), giving the value of the constant k. (6). The finite shaded region S shown in Figure 3 is bounded by the curve C, the line x = 3 and the x-axis. This shaded region is rotated through 2 radians about the x-axis to form a solid of revolution. (c) Find the volume of the solid of revolution, giving your answer in the form p 3 + q 2, where p and q are constants.

7 (7). June 2011. TOTAL FOR Paper : 75 MARKS. END. Gold 2: 10/12 6. Question Scheme Marks Number 1 A ( 3 x 1) + Bx ( 3 x 1) + Cx 2. 1. (a) = B1. x 0 (1 = A) M1. x 13 1 = 13 C C = 3 any two constants A1. correct Coefficients of x 2. 0= 9 A + 3B B =. 3 all three constants A1 (4). correct 1 3 3 . (b) (i) + dx x 3 x 1 ( 3 x 1)2 .. 3 3. = ln x ln ( 3x 1) + ( 3x 1) ( +C ). 1. M1 A1ft A1ft 3 ( 1) 3. 1. = ln x ln ( 3 x 1) ( +C ) . 3x 1 . 2. 2 1 . (ii) f ( x ) d=x 1 ln x ln ( 3 x 1) 3 x 1 . 1. 1 1 . = ln 2 ln 5 ln1 ln 2 M1. 5 2 . 2 2. = ln + .. M1. 5. 3 4 . = + ln A1 (6). 10 5 . [10]. dI. 2. = 16 ln ( ) M1 A1. dt dI. At t = 3 = 16 ln ( ) M1. dt = 2 ln = ln 4 M1 A1.

8 [5]. Gold 2 :10/12 7. Question Scheme Marks Number 1 + x . 1 1. 1 1. 3. (a) =(1 + x ) 2. (1 x ) 2. (1 + x) (1 x). 2.. 2 B1. 1 x . 1 ( 1 )( 12 ) 2 1 ( 1 )( 32 ) . M1 A1. = 1 + x + 2 x + .. 1 + ( x) + 2 ( x) 2 + .. A1. 2 2! 2 2! . 1 1 1 3 . = 1 + x x 2 + .. 1 + x + x 2 + .. 2 8 2 8 . 1 3 2 1 1 2 1 2. =1 + x + x + x + x x + .. M1. 2 8 2 4 8. 1 Answer is given = 1 + x + x2 A1 *. 2 in the question. (6). 1 + ( 261 ) 1 1 1 . 2. (b) =1+ + M1. 1 ( 26 ) . 1 . 26 2 26 . 3 3 1405. ie: = B1. 5 1352. 7025 7025. so, 3= A1 cao 4056 4056. (3). [9]. 3. 4. y d y = cos 2. x dx Can be implied. Ignore integral signs B1. = 3sec 2 x dx 1 2. = y 3 tan x ( +C ) M1 A1.

9 2.. y 2,=. = x 4. 1 2 . = 2 3 tan + C M1. 2 4. Leading to C = 1. 1 2. = y 3 tan x 1 or equivalent A1 (5). 2. [5]. Gold 2: 10/12 8. Question Scheme Marks Number r 16 2h Uses similar triangles, ratios or 5 (a) Similar triangles = r= trigonometry to find either one of M1. h 24 3. these two expressions oe. 1 2 1 2h . 2. 4 h3 Substitutes r = 23h into the formula for =V = r h = h AG A1. 3 3 3 27 the volume of water V. (2). dV dV. (b) From the question, =8 =8 B1. dt dt dV 12 h 2 4 h 2 dV 12 h 2 4 h2. = = = or B1. dh 27 9 dh 27 9. dV dV. Candidate's ; M1;. dt dh dh dV dV 9 18. = 8 . = =2 12 h 2 9 18. dt dt dh 4 h 2. h 8 or 8 or 27 4 h 2. h2 A1. oe 1.

10 Dh 18 1 18 A1 oe When = h 12, = = or 8 . dt 144 8 144 isw Note the answer must be a one term exact (5). value. Note, also you can ignore subsequent 18. working after . 144 . [7]. Gold 2 :10/12 9. Question Scheme Marks Number tan 2. 6 (a) x dx NB : sec 2 A = sec 2 A 1 . 1 + tan 2 A gives tan 2 A = The correct underlined identity. M1 oe . sec 2. = x 1 dx = tan x x ( + c ) Correct integration A1. with/without + c (2). 1. (b) x ln x dx 3. u= ln x ddux= 1x . dv . v =x 2 =2 x12. 3 2. dx =x . 1 1 1 Use of integration by = . 2x 2 . ln x 2 . dx 2x x parts' formula in the correct direction. M1. Correct direction means that u = ln x . Correct expression.