Paper Reference(s) 6666/0 Edexcel GCE Core Mathematics C4 Advanced Level Thursday June 0 Afternoon Time: hour 0 minutes Materials required for examination Mathematical Formulae (Pink) Items included with question papers 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 formulae 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. 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 not gain full credit. P4484A This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 0 Edexcel Limited.
. f(x) = x(x ) A B = + x ( x ) C ( x ) +. (a) Find the values of the constants A, B and C. (b) (i) Hence find f ( x). (4) (ii) Find f ( x ), leaving your answer in the form a + ln b, where a and b are constants. (6). Figure Figure shows a metal cube which is expanding uniformly as it is heated. At time t seconds, the length of each edge of the cube is x cm, and the volume of the cube is V cm. dv (a) Show that d x = x. () Given that the volume, V cm, increases at a constant rate of 0.048 cm s, (b) find when x = 8, dt (c) find the rate of increase of the total surface area of the cube, in cm s, when x = 8. () () P4484A
. f(x) = 6 9, x <. (9 4x) 4 (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x. Give each coefficient in its simplest form. (6) Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x, of (b) g(x) = (c) h(x) = 6, (9 4x) 9 x <, 4 6, (9 8x) 9 x <. 8 () () 4. Given that y = at x = 4, solve the differential equation d y =. y cos x (5) 5. The curve C has equation 6y + 9x y 54x = 0. dy (a) Find in terms of x and y. d x (b) Find the coordinates of the points on C where d y = 0. (5) (7) P4484A
6. Figure Figure shows a sketch of the curve C with parametric equations x = sin t, y = 4 cos t, 0 t. d y (a) Show that d x = k tan t, where k is a constant to be determined. (5) (b) Find an equation of the tangent to C at the point where t =. Give your answer in the form y = ax + b, where a and b are constants. (c) Find a cartesian equation of C. (4) () P4484A 4
7. Figure Figure shows a sketch of part of the curve with equation y = x ln x. The finite region R, shown shaded in Figure, is bounded by the curve, the x-axis and the lines x = and x = 4. (a) Use the trapezium rule, with strips of equal width, to find an estimate for the area of R, giving your answer to decimal places. (4) (b) Find x ln x. (c) Hence find the exact area of R, giving your answer in the form a ln + b, where a and b are exact constants. () (4) P4484A 5
8. Relative to a fixed origin O, the point A has position vector (0i + j + k), and the point B has position vector (8i + j + 4k). The line l passes through the points A and B. (a) Find the vector AB. (b) Find a vector equation for the line l. () () The point C has position vector (i + j + k). The point P lies on l. Given that the vector CP is perpendicular to l, (c) find the position vector of the point P. (6) END TOTAL FOR PAPER: 75 MARKS P4484A 6
June 0 6666 Core Mathematics C4 Mark Question. (a) ( ) ( ) x 0 ( A) = A x + Bx x + Cx B = x = C C = any two constants correct A Coefficients of x 0= 9A+ B B = all three constants correct A (4) (b)(i) + x x ( x ) = ln x ln ( x ) + x + C ( ) = ln x ln x + C x ( ) ( ) ( ) ( ) Aft Aft (ii) f( x) lnx ln( x ) = x = ln ln 5 ln ln 5 = ln +... 5 4 = + ln 0 5 A (6) [0]
Question. (a) dv V x x = = cso B () (b) dv 0.048 = = dt dv dt x At x = 8 0.048 = = 0.0005 ( cms ).5 0 4 dt 8 A () (c) ( ) ds S = 6x = x B ds ds 0.048 = = x dt dt x At x = 8 ds = 0.04 ( cm s ) dt A () [6]
Question. (a) ( ) ( ) f x =......... x 6 = 6 9 (... ) 9, 6, or equivalent B 5 ( )( ) ( )( )( ) ( ) ( ) =... + ( )( kx) ; + kx + kx +... ; Aft!... 9 x 4 = + + or + x 9 A 4 4 40 = + x+ x + x +... 9 7 79 A (6) 4 4 40 g x = x+ x x +... Bft () 9 7 79 (b) ( ) h 4 4 40 = + + + +... 9 7 79 A () 8 6 0 = + x+ x + x +... 9 7 79 [9] (c) ( x) ( x) ( x) ( x)
Question 4. ydy = Can be implied. Ignore integral signs B cos x = sec x ( ) tan y x C = + A π y =, x = 4 π = tan + C 4 Leading to C = tan y = x or equivalent A (5) [5]
Question 5. (a) Differentiating implicitly to obtain ± ay y y and/or ± bx dy 48 y... 54... A dy 9x y 9x + 8xy or equivalent B dy ( 48y + 9x ) + 8xy 54 = 0 dy 54 8xy 8 6xy = = 48y + 9x 6y + x (b) 8 6xy = 0 Using x = or y = y x 6y + 9 y 54 = 0 y y Leading to or 4 6y + 8 6 = 0 or 4 8 y = 6 or d d 6 + 9x 54x= 0 x x 4 4 6 x x 0 A (5) + = 4 x = 6 y =, or x =, A A Substituting either of their values into xy = to obtain a value of the other variable.,,, both A (7) []
Question 6. (a) cost dt B dy 8costsint dt A dy 8costsint = cost 4sint = cost dy tan t k = A (5) (b) When π t = x =, y = can be implied B π m = tan ( = ) y = x y = x A (4) (c) x = sin t = sin tcost ( ) x = sin tcos t = cos t cos t x Alternative to (c) y y = 4 4 sin t+ cos t = ( y ) or equivalent A () y = cost+ [] x + = A () 4
Question 7. (a) x 4 y ln ln4 ln6 ln8 0.69.9605.04 4.589 Area = (...) B ( ( ) )... 0.69+.9605 +.04 + 4.589 4.97989... 7.49 7.49 cao A (4) (b) (c) x = x ln x x 4 = x ln x x 9 ( + C) A (4) x ln x= x ln x x A 4 4 4 4 x ln x x = 4 ln 8 4 ln 9 9 9 ( 6ln...)... n = Using or implying ln = n ln 46 8 = ln 9 A () []
Question 8. (a) (b) (c) 8 0 uuur AB = = 4 0 r = + t 0 t 7 t uur CP = + t = t 0 + t t A () 8 r = + t Aft () 4 A 7 t t 0. = 4 + 4t+ t 0 + t = 0 t Leading to t = 4 A 0 8 Position vector of P is + 4 = 6 + 4 7 A (6) [0] Alternative working for (c) 8 t 5 t uur CP = + t = t 9 4+ t t+ A 5 t t 9. = 0+ 4t+ t 9+ t+ = 0 t + Leading to t = A 8 6 Position vector of P is + = 6 4+ 7 A (6)