PhysicsAndMathsTutor.com

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32 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme. (a) a = 7, d = 7, n = 4 n = 4 B n( n + ) Sn = n(a + b) or n(a + (n )d) or 7 (use of correct formula) = ( ) or ( ) or 7 = 7 07 A () 4 4 r= 4 (7r + ) = 7r + r= 4 r= split 4 r= = 4 4 (7r + ) = = 7 55 A () r= (6 marks). (a) B X 60 sin 60 = r or r = x, 4x = x +, x = r 0 r = 6 or r = A () A Area = r θ c θ or πr =, 60 6 π = π (cm ), A () (c) Arc = r θ c θ or πr =, 60 6 π Perimeter = Arc + r =, π + = (π + 6) (cm) ( ), A cso () (7 marks)

33 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number. (a) f(x) = 0 x + 7x 5 = 0 Scheme (x )(x + 5) = 0 attempt to solve f(x) = 0 points are (, 0), ( 5, 0) ; (0, 5) y 5 shape A (both); B () B vertex in correct quadrant B ft () 5.5 x (c) Symmetry: x = ( 5 +.5) or Calculus: 7 4x = 0 or Algebra: [(x + 47 ) k] 7 x =, y = 4 8 A, A () 4. (a) (x + k) 7 k = 0 (x + k) (LHS) A (8 marks) (x + k) = 7 + k = 0 x + k = (±) 7 + k (no need for ±) x = k ± 7 + k A (both) (4) 7 + k > 0 (or discriminant > 0) roots are real and distinct A () (c) k = x = ± 7 + x = + or A (both) () (8 marks)

34 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number 5. (a) f(x) x Scheme shape B 60, 0, 80 on x-axis B 5, 5 on y-axis (may be implied by part ) B () (0, 5); (50, 5); (90, 5) one x-coordinate B all x-coordinates B all correct B () (c) f(x) =.5 sin x = x = 0 (50, 90, 50) x = (α), 80 α, 60 +α, (540 α) x = 0, 50, 0, 70 one correct value B, A (ignore extras out of range) (4) (0 marks) 6. (a) x x = 0 x = x = x = α =.447 =.4 ( sf) A cao () f(x) = 4x + 9x + 5 A = 4 B = 4x 9 + x 7 B = 9, C = 7 B, B () (c) 4 f ( x ) dx = [ ] 9 x x 7x x n x n + A ft (candidate s A B, C) ( eeoo) = ( 4 9 4) ( 9 7) [] [] (use of limits) = 8 or.75 A (5) ( marks)

35 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 7. (a) l = (50 x) w = (40 x) B V = x(50 x)(40 x) V = xlw V = x(000 80x 00x + 4x ) = 4x(x 45x + 500) (*) A cso () 0 < x < 0 (accept ) B () (c) dv dx = x 60x (accept 4), A dv dx = 0 x 90x = 0 x = 90 ± (dv/dx = 0 & attempt to solve) x = (.6), required x = 7.6 or 7.4 or 7.6 A (4) (d) Vmax = 4 7.6(7.6 ), = 6564 or 6560 or 6600, A () (e) e.g. V = 4x 60 x = 7.6 (= 8 ) < 0, maximum full method A full accuracy () ( marks) 8. (a) Mid-point of AB = [ ( + 8), ( + 4)], = ( 5, ), A () MAB = 4 ( ) 6, = 8 ( ), A Equation of AB: y 4 = 6 (x 8) y 44 = 6x 48, 6x y 4 = 0 (or equivalent) A (4) (c) Gradient of tangent = Equation: y 4 = B ft 6 (x 8) (or 6y + x = 0) A () 6 (d) Equation of l: y = x B 88 Substitute into part (c): x 4 = 6 x x = 7 57, y = 4 45 A, A (4) ( marks) 4

36 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number n( n ). (a) + n(x), + (x) +! (c) n( n )( n ) 7 = 0 6 Scheme n( n )( n ) (x) B, B ()! n( n ) 9 n = A () n( n )( n )( n ) (x) 4 coefficient: A () 4! (6 marks). (a) x( x + ) + x 4 ( x + )( x ) B B = = ( x ) + x( x 4) x( x + )( x ) ( x )( x + ) x( x + )( x ) A A A (7) (7 marks). (a) 0, 9.05,.46 B B B () When t = 4.5, v =.76. Slower at t = 5 B () (c) s = (5)[ ( ) +.46] A Aft = (59.7, 60) A (4) (8 marks) 4. (a) Adding: sin (A + B) + sin (A B) = sin A cos B A + B = X A B = Y A = X + Y A = (X + Y), B = (X Y) A X + Y sin X + sin Y = sin cos sin 4θ + sin θ = sin θ cos θ X Y ( ) A (4) sin θ = 0 (or cos θ = 0) θ = 0, 60, 0, 80, 40, 00 4 correct: A 90, 70 6 correct: A 8 correct: A (5) (9 marks)

37 ( ) indicates final line given in the question paper; ft = follow-through mark

38 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 5. (a) log x = log x B Combine logs, e.g. log y = x y =, y = 8x ( ) A () x 4x = 8x (4x )(x ) = 0 Roots 4 and A () (c) log α = log 4 = log ( ) = ( ) B () (d) log.5 = k k =.5 k = log.5 log 6. (a) f(.) = 0 + ln 9. e. =. = A () (0 marks) f(.) = 0 + ln 9.6 e. = Sign change, so. < k <. A () f (x) = e x A (, 0) x () (c) f() = 0 + ln e B f (x) = e (i) y (0 + ln e) = ( e)(x ) (ii) x = 0: y = 0 + ln e + e B = 9 + ln A (5) (0 marks) ( ) indicates final line given in the question paper

39 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 7. (a) A: y = 6, B: y = B () y(x ) = 4, yx y = 4 x = y + 4 y ( ) () (c) x = 4 + = 9 + y y y A x dy = (9 + 4y + 6y ) dy 6 = 9y, + 4 ln y Aft, Aft y ln y 6y y = ( ln 6 ) (8 + 4 ln 8) V = π ( + 4 ln 8) A (7) (d) V ( ) A () ( marks) 8. (a) f(x) 4 B () Domain: x 4, range: f (x) B, B () (c) 4 Shape: B Above x-axis, right way round: B x-scale: 4 B y-intercept: B (4) (d) gf(x) = (x x ) 4 A () (e) x x 7 = 8: x x 5 = 0 (x 5)(x + ) = 0 x = 5, x = (reject) A Aft x x 7 = 8: x x + = 0 x = A (5) (4 marks) ( ) indicates final line given in the question paper; ft = follow-through mark 4

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41 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN June 00 Advanced Supplementary/ Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 number Scheme Paper no. P. (a) Complete attempt at remainder theorem, or long division Either f () = 7 + 9a+ b 0 = 4, Or complete attempt at long division by (x-) leading to equation. Either f ( ) = + a b 0 = 8 or long division by ( x+) leading to equation. Equation equivalent to 9a+ b= (a+ b= ) Equation equivalent to a b= 7 A A Solve two equations to get a =, b= 5, A (5) Either f () = = 0, Or complete division with no remainder. ( x ) is a factor. Or f ( x) = ( x )( x + 5), A ( A) (). (a) xsin x sin x xcos xx d = dx (integration in correct direction) A = xsin x cos x + ( + k ) (second integration) 4 A sin xcos x sin x x + ( + k) 4 (use of appropriate double angle formulae) (4) = = sin x(xcos x sin x) + + k for k sin x ( x cos x sin x ) + C A A c.a.o. ()

42 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN June 00 Advanced Supplementary/ Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 number Scheme Paper no. P. (a) Either completion of square, ( x 4) + ( y 8) = Or use of formulae, ( f, g), r = ( f + g c) Centre is (4,8), radius 7. A,A () dy dy Either x+ y 8 6 = 0 d x d x dy dy 4 x (y 6) = 8 x, so =. dx dx y 8 y 8 Or gradient of CP is x 4. Tangent is perpendicular to CP and so has gradient 4 x y 8., A A B, A () () (c) At (,8) the tangent is vertical, so equation is x =., A () 4. (a) x A x x B x C x + ( + )( ) + ( ) + ( + ) ; A = - =4B, B= ; 0 = 4C, C= 5 5 ( + + )dx ( + x) ( x) 5 = x+ ln( + x) ln( x) B A; A (4) A A 0 f ( x ) = ( + ln ) ( ln ) = -+ln 5 A (5)

43 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN June 00 Advanced Supplementary/ Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 number Scheme Paper no. P 5. (a) (c) (d) 5 4 ( + 5) = ( ), = ( 4) = ; = sin 60 (5 x) (5 x) x( ) + ( )( ) + ( )( )( ) = x, + x, x =, +,... =0.86(459..) sin 60 - (Ans) 0.006,A;A (), B, A, A (4) A A () () 6. (a) 5cost = 0 : t =, y = (0, ) t =, y = 6 (0, 6) π π A π 5 5π 5 5 4sin t = : t = 6, x= t = 6, x= ( ±,0) A (4) shape position B B (c) d x d d 4cos 5sin, y 4cos, y = t = t = t dt dt dx 5sin t π 5 tan 6 5 y = ( x ), i.e. 8 y = 0x 5 4 A, A () (4)

44 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN June 00 Advanced Supplementary/ Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 number Scheme Paper no. P 7. (a) dv dt = kv B d V = k d t, ln V kt V = kt lnv = kt + C V = Ae t = 0, V = 0 000: 0000 = A t =, V = 000 : 000 = Ae k A B, () (c) e k t = V = e = = k ; 700 k = ln 0.55 k 0.99() ( allow0.) kt kt 500 = 0000e e = t = ln 0.05 t 8.5 (8.44) accept8or9 yrs A ;A A A (5) ()

45 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN June 00 Advanced Supplementary/ Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 number Scheme Paper no. P 8. (a) r = (9i j+ k) + λ( i+ 4j+ 5 k ) ( or any correct alternative),a () Uses their line equation, or recognises B is mid point of AC or merely writes down p=6, q=, A () (c) Calculates OC AB OC AB Uses cosα = to obtain α. OC AB 70 cos α =, α = 9.8 (accept 9.79 or 40) A (d) Let OD be (9i j+ k) + t( i+ 4j+ 5 k ) Use scalar product OD. AB=0 Obtains equation in t and solves to obtain t = 0.6 ( or equivalent) Uses their t, to obtain 7. i j + 4k. A, A () (6)

46 EDEXCEL PURE MATHEMATICS P (667) NOVEMBER 00 PROVISIONAL MARK SCHEME Number Scheme. (a) x x > + 8 x >, A () x 5x 4 > 0 (x 7)(x + ) > 0 x = 7, B x < or x > 7, A ft () (5 marks). (a) f( ) = = 0 (x + ) is factor A () (x + )(x 6x + 8) A (x + )(x )(x 4) A (4) (6 marks). (i) Divide: + x A Differentiate: 6x + x x A (,0) (5) (ii) x x + 4 AA [ ] 4 [ ] = 4 = A (5) (0 marks) 4. (a) S = a + (a +d) + (a + d) + + [a +(n )d] B S = [a + (n )d] + [a + (n )d] + + a Add: S = n[a + (n )d] S = n[a + (n )d] A (4) a = and n = 9 B 6900 = 9 ( d) Aft d = 700 A (4) (c) a + (n )d = a + 0d = d = 9000 A () (d) ar n = (ft their n) Aft = (or 97000) A () ( marks)

47 EDEXCEL PURE MATHEMATICS P (667) NOVEMBER 00 PROVISIONAL MARK SCHEME Number Scheme 5. (i) arcsin 0.6 = 6.9 (awrt) α B x + 50 = 6.87, x = = x , x = = 9. x = 46.6, 7.4 A A (7) (ii) (a) sin 60 = BC 8, = sin 60 B, BC = 6 cos θ = sin θ = 9 BC = = 4 (*) A (4) sin θ = 8 9 = 8 = A () ( marks) 6. (a) 9 B () Shape B 5 Position of max. B 5 on y-axis B 5 and 5 on x-axis A (5) (c) Gradient: 8 ( 7) ( ) A y 8 = gradient (x ) y = x A (4) (d) Where y = 0, x = Aft () (e) Mid point:, =, k = A () (4 marks)

48 EDEXCEL PURE MATHEMATICS P (667) NOVEMBER 00 PROVISIONAL MARK SCHEME Number Scheme 7. (a) Integrate: y = x 0x + 9x (+C) A 6 = C C = 0 (y = x 0x + 9x 0) A (4) Substitute x = 4: = 0 A () (c) At x =, (d) d y = = B dx Tangent: y 6 = x (y = x + 4) A () d y = dx x 0x + 8 = 0 (x 4)(x ) = 0 A 4 x = A (5) (4 marks)

49 EDEXCEL PURE MATHEMATICS P (667) NOVEMBER 00 PROVISIONAL MARK SCHEME Number. y + y + ( y + )( y + ) ( y + ) ( y + ) ( ) ( y + 6y + 9 y + y + ) ( y + )( y + )( y + ) 4 ( y + ) ( y + )( y + )( y + ) Scheme ( y + ) ( y + ) ( y + )( y + )( y + ) 4y + 8 ( y + )( y + )( y + ) ( y + )( y + ) 4 or y 4 + 4y + A, A (5 marks). (a) x x ( ) 4 = = u or ( x + ) =. u 5 = ( u 5)( u + ) x = u, u u 5( = 0), A c.s.o () u, A u = 5 x log 5 = 5 x =, =., A (4) log [Ignore any other solution] (6 marks). (a).5 sin x + cos x = R sin (x + α) = R[sin x cos α + cos x sin α] R =.5 + =. 5 Full method for R or R, A tan α =, α = Full method for tan α, sin α, cos α, A (4) sin x cos x =.5 sin x 4 cos x = [ cos x] = (cos x + ) sin x cos x + 4 cos x =.5 sin x + cos x + A () (c) Maximum value of.5 sin x + cos x = R Maximum value of sin x cos x + 4 cos x = R + or 4.5 A ft () (8 marks)

50 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 4. (a) u = p + 5 B u = p(p + 5) = p + 5p + 5 or p + 5p = 0 (p ) (p + ) = 0 P =, or A, B cso (5) log ( ) = log = B () (c) log p = log p q log q Use of log a log b = log p log q Use of log a n = t accept log p t A ft () 5. (a) (0, ) on C = p + q Use of (0, ) equation in p and q dy = qe dx x (9 marks), at p 5 = q, A () Solving q =.5, p = 0.5 (or q) A, A (5) Gradient of normal at P is 5 B Equation of normal at P is: y (p + q) = (x ln ) 5 at L y = 0 at M x = 0 xl =.5 + ln or 5(p + q) + ln or.9. ym = ln or p +q + 5 ln or Area of triangle OLM is : xl ym = , A cso (5) (0 marks)

51 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 6. (a) y Shape with vertex on +ve x-axis B O ½ x (, 0) and (0, ½) B () x = α given by: e -x = ( x ) Use of ( ) x A e -x = x +, i.e. x + e -x = 0 A cso () (c) f(x) = x + e x : f(0) = = correct value to.s.f f( ) = 4 + e =.4. Change of sign root in < α < 0 Both correct and comment A () (d) x = 0.69(.), x = 0.6(..) B, B () (e) f( 0.575) = f( 0.585) = Change of sign so root is 0.58 to dp. A () ( marks) ALT (e) x = , x4 = , x5 = , x6 = , A (x7 = )

52 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 7. (a) x: y: correct ys R, [ + { } + 6] B, [ A ft] 7 or 6.75 A 4 Since curve bends under straight line overestimate (c) V = 4 π = ( x + x 4) y dx dx π y 0 B π, y = ( ) 5 x = π x + x 4 0 x n x n+ A = ( ) π Use of correct limits 5 π 5 = [ ] 76 = π 5 (or 5 5 or 5.π ) A (6) ( marks)

53 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number 8. (a) Scheme x y = y(x ) = x x y x = f (x) = y yx x = y x(y ) = y Collect x and factorise x = f(x) A cso () x ff(k) = f f(k), = k A () (c) g( ) = 5 B (d) f( 5) = 5, = 8 y 6 =, A () 8 g (x) 5 6 x shape (0, ) and (, 0) Domain: 5 x 6 B B B () 6 g(x ) x Translation + (lines join at (0,0)) B y 5 h(x) Stretch B x Range: 0 h(x) B () 0 (4 marks)

54 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number. (a) d y = x 0 = 5x d x A 5 7x + 4x + C A(,0). (a) Scales (, and 60) B 60 Shape, position B (0, 0.5) (50, 0) (0, 0) B B B (c) ( ) x + 0 = 0 or 0 One of these B x = 80, 00 M: Subtract 0, A: Both A. (a) x ( y = ) x = (y ) (*) A (y ) = y + 7, y 8y = 0, A (y + )(y ) = 0, y = (or correct substitution in formula) y =, y = A x = 8, x = 4 Aft

55 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number 4. (a) a r = = A r 960 ( r) = 00 r = 4 (*) A T 9 = 00 (- 0.5) 8 (or T 0 ) Difference = T 9 T 0 = ( ) (c) S n = n ( (. ) ) (. ) = 0.0 (or 0.0) A A (d) Since n is odd, ( 0.5) n is negative, so S n = 960 ( n ) (*) A

56 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number 5. (a) d C d v v = 60v + 00 v 60v + = 0 00 A v = v = 0 A d C d v = 0v + 50 > 0, therefore minimum A (c) v = 0 : C = = Bft 00 Cost = 50 = 0 A

57 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number 6. (a) P: x = 0 y = B Mid-point: ( 0 + 5) ( ) 5 5, =, Aft Gradient of l is, so gradient of l is B l : y ( ) = (x 5) Aft x + y = A (c) Solving: x y = 4 x + y = x = 4 y = 5 A Aft 4

58 EDEXCEL FOUNDATION Stewart House Russell Square London WCB 5DN January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number 7. (a) BM = (7 + 4 ) = 5 (*) B 7 tan α = or equiv. and BMC =α, or cosine rule A 4 BMC = radians (*) A (c) ABM : Sector: (4 4) (= 68 mm ) (or other appropriate ) B ( ) A Total: = 5 mm (or 54, or 50) A (d) Volume = 5 85 mm (M requires unit conversion) = 44 cm A 5

59 January 00 Advanced Subsidiary / Advanced Level General Certificate of Education Subject PURE MATHEMATICS 667 Paper No. P Scheme number 8. (a) A: y = B: y = 4 B d d y = = 5 where x = 5 A x x 5 Tangent: y = 5 (x 5) (5y = x 5) A (c) x = 5y B B (d) Integrate: y 0y = 5 Aft [ ] 4 [ ] = , = (,.) A, A Alternative for (d): Integrate: x 75 A Area = (0 4) (5 ) , = (,.) A, A In both (d) schemes, final M is scored using candidate s 4 and. 6

60 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 number Scheme. x 9 = (x )(x + ) seen B Attempt at forming single fraction x( x ) + ( x + )( x + ) ; = ( x + )( x + )( x ) x + 0x + ( x + )( x + )( x ) ; A Factorising numerator = ( x + )( x + ) ( x + )( x + )( x ) or equivalent = ( x + ) ( x + )( x ) A (6) (6 marks). ( + px) n + npx, + n( n ) p x + B, B Comparing coefficients: np = 8, n( n ) = 6, A Solving n(n ) = 7 to give n = 9; p = A; A ft (7) (7 marks). (a) y V graph with vertex on x-axis a a O x y { a, (0)} and {(0), a} seen A () Correct graph (could be separate) B () O (c) Meet where x = x + a x x + a = 0; only one meet B () (d) x + x B Attempt to solve; x = (no other value) ; A () (7 marks)

61 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 number Scheme 4. Volume = π 4 + dx x + d x = + + dx B x x 4x Using limits correctly = x + x + ln x A Aft 4 Volume = π 8 + ln A = π 5 + ln A (8) (8 marks) 5. (a) Distance from one side (m) Height (m) y = 7.80 when x = 4 or 6 B Symmetry B ft () Estimate area = [0 + ( )] B Aft = 55.7 m A (4) (c) 40 = 84. m A ft () (d) Over-estimate; B reason, e.g. area under curve is under-estimate (due to curvature) B () (9 marks)

62 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 number 6. (a) y Scheme Shape B p = or {, 0} seen B () O Gradient of tangent at Q = q B Gradient of normal = q Attempt at equation of OQ [y = qx] and substituting x = q, y = ln q or attempt at equation of tangent [y ln q = q(x q)] with x = 0, y = 0 or equating gradient of normal to (ln q)/q q + ln q = 0 (*) A (4) (c) ln x = x x = e x ; x = x e (d) x = ; x = , x = 0.07, x4 = ; A () ; A Root = 0.04 ( decimal places) A () 7. (a) sin x + cos x = R sin (x + α) R cos α =, R sin α = = R (sin x cos α + cos x sin α) Method for R or α, e.g. R = ( + ) or tan α = A ( marks) Both R = and α = 60 A (4) sec x + cosec x = 4 cos x sin x + cos x = 4 sin x cos x + = 4 B sin x = sin x (*) () (c) Clearly producing sin x = sin (x + 60) A () (d) sin x sin (x + 60) = 0 cos cos sin x + 60 x 60 x + 60 sin x 60 = 0 = 0 x = 40, 60 A A ft = 0 x = 60 B (5) ( marks)

63 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Scheme number y 8. (a) shape B (0, d) intersections with axes (c, 0), (0, d) B () O (c, 0) x y (0, c) O ( d, 0) x shape B x intersection ( d, 0) B y intersection (0, c) B () (c)(i) c = (ii) < f(x) (candidate s) c value B B ft () (d) ( x ) = x = and take logs; x = ln ln log x x (e) fg(x) = f[log x] = [ ( ) ]; = [ ( ) ] or B ; A d (or x) =.585 ( decimal places) A () log log x ; A = x A () (4 marks) 4

64 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number (a) Scheme ( x+ ) A B +, and correct method for finding A or B ( x+ )( x ) x+ x A =, B = f ( x) = ( x+ ) ( x ) Argument for negative, including statement that square terms are positive for all values of x. (f.t. on wrong values of A and B) A, A () A Aft () (a) a= 4, b= 5 (both are required) B () ( x 4) + ( y 5) = 5 Aft () (c) Finding the distance between centre and ( 8, 7), [(8 a) + (7 b) Complete method to find PT, i.e. use Pythagoras theorem and subtraction, PT =.6 A ()

65 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number Scheme (a) Using f ( ± ) = Showing that p = 6, with no wrong working seen. S.C. If p = 6 used and the remainder is shown to be award B A () Attempt to find quotient when dividing ( n + ) into f(n) or attempting to equate coefficients. Quotient = n + 4n+, or finding either q = or r = A (c) Finding both q = and r = The product of three consecutive numbers must be divisible by Complete argument A () A () 4. (a) ( )( ) ( )( )( 4)!! =, 6x, + 7x...( 08x ) ( + x) = + ( )( x) + ( x) + ( x) +... B, A, A (4) Using (a) to expand ( x+ 4)(+ x) or complete method to find coefficients [e.g. Maclaurin or (+ x) + (+ x) ]. = 4 x, + 0 x, 405x = 4, x, + 0x...( 405x ) A,Aft,Aft (4)

66 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number 6(a) Scheme r = i+ j k± λ(4i 5j+ k ) or r = 5i j± λ(4i 5j+ k ) (or any equivalent vector equation) Show that µ = - A () B () (c) Using cos θ = (4i 5j+ k).( i j+ k) (4 5 ) ( ) (d) 0 4 = = ( ft on 4i 5j + k) num, denom. 5 θ = 9.5 ( allow 9 or 0 if no wrong working is seen) Shortest distance = AC sin θ Aft Aft A (4) AC = (( a ) ( b ) ) ( = ) A Shortest distance = unit A (4) Alternatives Since X = (+ 4 λ, 5 λ, + λ) CX = ( + 4 λ) i+ ( 5 λ) j+ ( + λ) k Use Scalar product CX. (4i 5j + k ) = 0, OR differentiate and equate to zero, CX or CX to obtain λ = 0.4 and thus CX = A A (4)

67 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number Scheme 5. (a) dv = 0 V dt 5 dv 5 = V, no wrong working seen dt 5 Separating the variables d d V 450 V = t 5 5 Integrating to obtain ln V 450 = t OR ln V 5 = t 5 5 Using limits correctly or finding c ( ln550 OR ln 775) V 450 ln = t, or equivalent A A* () d A A Rearranging to give V t 5 e = da (7) (c) V = 5 B ()

68 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number 7(a) Scheme x dy x e = e x + dx x Putting d y = 0 and attempting to solve dx x = 4 Volume x 4x = π ( xe )dx = π xe dx 0 0 4x 4x 4x xe d x = xe + e dx 4 4 4x 4x = xe e π 4 Volume = π[ e e ] [ ] = [ 5 e ] A A d A (5) A A A ft A (7)

69 EDEXCEL PURE MATHEMATICS P PROVISIONAL MARK SCHEME JANUARY 00 Number Scheme 8 (a) cos( ) cos sin A+ A = A A = = cos A ( cos A) cos A π π [ x=, θ = ; x= 6, θ = ] 4 dx x = sin θ, = cosθ dθ Using x x= θ θ θ = θ θ 8 d cos cos d 8cos d cos θ cos θ = to give 4( + cos θ)dθ = 4θ + sin θ A () B B A d Aft Substituting limits to give π + or given result A (7) (c) dy sin θ = dθ + cos θ dx Using the chain rule, with secθ tanθ dθ = to give d y ( = cos θ ) dx π Gradient at the point where θ = is. Equation of tangent is y+ ln = ( x ) (o.a.e.) B Aft A (5)

70 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme. (a) y = 5x x + C A (,0) 7 = 5 + C, C = A ft x = : y = 0 + = A. (a) 6 x x < + 7 x < A (7 marks) (x )(x 5) Critical values and 5 A (c) < x < 5 A ft < x < B ft. (a)(i) a + (n )d = 80 + (5 5) = 455 A (ii) n [a + (n )d ] = 8 [560 + (5 5)] = 0 A ft 8 [560 + (5 d)] = A (7 marks) d = 0.98 x = (allow.0 or 0.98 or 0.99 or ) (ft = follow-through mark) A (8 marks)

71 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 4. (a) r θ = r.5 = 5 A r = 0 = (4 5) r = 5 (*) A rθ + r = = 7 5 cm (or 5.7, or a.w.r.t 5.65.) A (c) OAB: r sinθ = 0 sin.5 (= ) Segment area = 5 OAB = 5.05 cm A (8 marks) cos θ cos θ = cos θ cos θ cos θ = 0 A ( cos θ + )(cos θ ) = 0 cos θ = or A θ = 0 θ =.8 B A θ = (60.8 ) = 8. A ft (8 marks) 6. (a) 6 m = = 4 A y 6 = (their m)( x 4) x + y = 6 A y = 4x B 6 (c) x + ( 4x) = 6 7x = 6 x = 7 y = 7 64 A (4, 6), C 6 64 x + x y + y 6 5, 7 7 :,, 7 7 (ft = follow-through mark) A A ft A ft (0 marks)

72 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 7. (a) x x + = 9 x x x 6 = 0 (x + )(x ) = 0 x =, A y =, 6 A ft x A ( x x + ) dx = x + x x x + x = 8 ( ) 4 6 = A Trapezium: Area = Alternative: ( 6 + x x ) ( + 6) 5 = 4 B ft 5 4 = 0 A 6 ( 9 x) ( x x + ) = 6 + x x A x x dx = 6x + A ft x x x + = , = 0 A, A 6 (ft = follow-through mark) ( marks)

73 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme dy 8. (a) = 4x 6x A dx 4x 6x = 0 4x ( x 4) = 0 x = 0,, A (, 0) d y (c) = x 6 dx y =,, x = 0 Max. One of these, ft x = Min. x = Min. All three A Aft A (d) x = : y = 8 + = 4 B dy At x =, = 4 6 = dx (m) B ft Gradient of normal = = m y ( 4) = ( x ) x y 49 = 0 A (ft = follow-through mark) (5 marks)

74 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme. (a) x + 4x + x + x = ( x + )( x + ) x( x + ) Attempt to factorise numerator or denominator = x + x or + x A () x LHS = log + 4x + x + x Use of log a log b RHS = 4 or 6 B x + = 6x Linear or quadratic equation in x x = 5 or 5 or 0. A (4) (6 marks). (a) y x x + = (x ) + Full method to establish min. f f(4) = + = f A x f B () f() = ; 6 = gf() 6 = λ + M for using their f() for eqn B; λ = 5 ft their genuine f() A ft () (6 marks). ( px) 6 = ( px) + 4 ( px) n Coeff. of x or x okay r = ( px); ( px) n No A; A r 5 6p = 5 p 9 = 6 or p = 4 (only), A 6.p = A (ft = follow-through mark) A = 44 A ft (their p (> 0)) (7 marks)

75 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number 4. (a) y (, ) Scheme Translation in or Points correct B B,, 0 ( eeoo) () x y ( 4, ) x < including points x > correct reflection B B cusp at (, 0) (not ) B () O x (c) y ( 4, ) ( 4, ) O x 5. (a) y y = x y = e x x correct shape x 0 symmetry in y-axis correct maxima correct x intercepts y = x: starting (0,0) y = e x : shape & int. on + y-axis correct relative posns B B B B (4) (0 marks) B B B () Where curves meet is solution to f(x) = 0; only one intersection B () (c) f() = 0.8 f(4) = 0.08 one correct value to sf change of sign root in interval both correct ( sf) + comment () (d) x0 = 4 x = ( e 4 ) =.9707 expression or x to dp dp x =.958 x, x to 4 x =.95 carry on to A x 4 x4 =.9(9) to dp Approx. solution =.9 ( dp) A cao (4) (0 marks) ( eooo = minus mark for each error or omission; cao = correct answer only)

76 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number dy c 6. (a) dx x Scheme = Attempt dy dx c When x = p 4 = c = 4p (*) A cso () p 5 = + p c and solve with c = 4p 5 = + 4p p = c = 4 (*) A cso () (c) y 8 6 = + x + x y = ; terms correct 6 y dx = x + 8ln x x some correct ; all correct ; A 6 y dx = ( + 8 ln ) ( + 8 ln 6) Use of correct limits V = π y dx B V = π(9 + 8ln ) k = 9; q = 8 A; A (7) 7. (a) M is (0, 7) B d y = e x dx Attempt gradient of normal is ft their y (0) dy dx ( marks) equation of normal is y 7 = (x 0) or x + y = 4 A (4) y = 0, x = 4 N is (4, 0) (*) B cso () (c) (e x + 5) dx = [e x + 5x] some correct R T ln 4 R = 0 (e x + 5) dx = ( ln 4) ( + 0) limits used = ln 4 A T = (4 ln 4) Area of T B T = (7 ln) ; R = ln Use of ln 4 = ln B R = T + R, R = 97 ln, A (7) (marks)

77 EDEXCEL PURE MATHEMATICS P (667) JUNE 00 PROVISIONAL MARK SCHEME Number Scheme 8. (i) cos x cos 0 sin x sin 0 = (cos x cos 0 + sin x sin 0) Use of cos(x ± 0) cos x sin x = cos x + sin x i.e. 4 sin x = cos x tan x = Sub. for sin 0 etc decimals, surds A sin x (*) Use tan x = cos x, A, Acso (5) (ii) (a) LHS = ( sin θ ) sinθ cosθ Use of cos A or sin A; both correct ; A = sinθ = tan θ (*) A cso () cosθ Verifying: 0 = (since sin 60 = 0, cos 60 = ) B cso (c) Equation = ( cos θ ) sin θ Rearrange to form cos θ sin θ tan θ = A i.e. θ = 6.6 or 06.6 (Accept 7, 07 ) (any acc.) Alt (c) sin θ cos θ = ( sin θ) Use of cos A and sin A 0 = sin θ( sin θ cos θ) (sin θ = 0) tan θ = etc, as in scheme A A (both) (4) ( marks) Alt (c) cos θ + sin θ = cos (θ α) = α =.6 (or 7) 5 A θ = α, 60, 60 + α θ = α or 80 + α θ = α, 80 + α i.e. θ = 7 or 07 ( or dp) A both ((*) indicates final line is given on the paper; cso = correct solution only; ft = follow-through mark) 4

78 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS number Scheme Paper no. P. Attempt to use correctly stated double angle formula cos t = cos t, or complete method using other double angle formula for cost with cos t+ sin t = to eliminate t and obtain y = x y = ( ) or any correct equivalent.(even y =cos(cos x ( )) A () shape position including restricted domain - < x < B B (). (a) 6 p q= p+ 6 + q p = p = 0 Remainder = p + q + 8 = p + (=), A (4) B ft on p ()

79 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS number Scheme Paper no. P. (a) Centre is at (,-4) B radius = ( + ( 4) 75) = 0 A () st circle nd circle Circles touching B B B At ( 9, 4 ) B (4) 4. (a) dy dy 4 x + (48x 48 y) 4y 0 dx + dx = dy Substitutes = into derived expression to obtain dx x+ x+ 48y y = 0 (B) A 50x+ 500y = 0 x+ y = 0 Eliminates one variable to obtain, for example, 7( y) + 48( yy ) 7y + 75 = 0 and obtains y (or x) Substitutes y to obtain x (or y) A (5) Obtains coordinates (-,) and (,-) A, A (4)

80 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS number Scheme Paper no. P 5. (a) dy x = sin x + (sin x) cos x dx,a At A x sin x + (sin x) cos x = 0 d x sin x+ cos x= 0 (essential to see intermediate line before given answer) tan x+ x= 0 A (4) V = π y dx = π x sin xdx = π + x cos x x cos xdx = π + x cos x xsin x sin xdx = π + + x cos x x sin x cos x = π π = π π 4 ] ] ] π 0 ] π 0 ] π 0 A A A (7)

81 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS 667 number 6. (a) Scheme AB = (i j+ k), CB = ( i+ j+ k), ( or BA, BC, or AB, BC, stated in above form or column vector form. Paper no. P A CB AB 4 cos ABC = = CB AB 9 A (4) (c) Area of ABC = sinb 6 65 sin B = ( 8) = 9 Area = AC = DC = or given in alternative form with attempt at scalar product A () AC DC = 0, therefore the lines are perpendicular. A () (d) 4 AD = DB = and AD:DB = :- (allow :) 4, A ()

82 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS 667 number Scheme Paper no. P 7. (a) (c) dv dv =± c V or V dt dt dv As V = Ah, = A or V h dh dh c c use chain rule to obtain = V = h = k h dt A A dh = h h = A kt A kt h = h ( A Bt) kdt = t = 0, h= : A= t = 5, h= 0.5: 0.5 = ( 5 B) A, A A B () (4) ( 0.5 ) B= ( B= ) 5 A 5 h= 0, t = = = 7.min B 0.5 B B () (d) A h = = 0.5m 4 A ()

83 EDEXCEL Mathematics June 00 Provisional Mark Scheme Subject PURE MATHEMATICS 667 number Scheme Paper no. P 8. (a) Method using either A B C A Dx + E + + or + ( x) (x+ ) (x+ ) x (x+ ) A = C=0, B= or D=4 and E = 6 x x+ A Dx + E dx x (x+ ) [ + + 0(x + ) ] dx or + B A, A (4) (c) ln x + ln x+ 5(x+ ) ( + c) or ln x ln x (x 8)(x ) ( c) Either ( x) + ( + x) + 0( + x) = + x+ x +... x 4x + ( +...) 9 0 x ( )( ) x = x+ x Or ( ( )( ) ( )...) ] ] 5 (9 + x+ 4 x )( x) = 5 (9 + x 8x 4 x ) 5 x 8x 4x x 8x 4x x = ( ) + (..) A A A (5) A A A A A A A (7) = x+ x A (7)

84 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme. (a) B B d B () () (c) S50 n a n d 75 ( ) 5 ( 77) (49 ) A A () 6. (a) 4 xx ( ) or x(4x ) (or use of quadratic formula) x 0 x A A () Using b ac c c A (x )(x ) 0 x (or quadratic formula) x A (4) 7. x y (y ) y(y ) y y y 0 0 A ( y 5)( y ) 0 y 5 y A x 4 x 7 A (7) 7

85 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 4. (a) 4x 9, x 4 x x 9 dx x 8 x 9 x ( dep. on terms) B, B A () (c) 8 (8 ) 8 ( 8 9) = 7 6 A (5) 7 5. (a) Shape B Position B () 5 0,,,0,,0 4 4 B B B () (c) x 4 B Other value 5 Subtract 7 x, x 4 A (4) 9

86 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 6. (a) V r h 500, A rh r B A r r r r r * A (4) da r 000r dr A 500 r r r A (4) (c) d A 000 r, 0 therefore minimum A dr () (d) 000 A r 77 (nearest integer) r A ()

87 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 7. (a) 5 ( ) 4 8 A () M x x y y :, (5,) A Gradient of CM is 4 B Equation of CM: y ( x 5) 4 A (5) (4y x 9) (c) When x=4, 7 y 4 A () (d) Radius = (4 ) 4 7 A A * (4) 4

88 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 8. (a) x x 6x 5 x 5 x x A () and 5 B () (c) dy x x 5 dx A dy At x. 5 4 dx A () (d) x 6x 5x 6 5 d 4 x x x x 4 A R A Evaluating at 5: A To find S: 4 4 Total Area = 4 4 A (7) 4 5

89 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme. (a) x x ( x 7) ( x 7) x 7 ( x )( x 7) ( x )( x 7) A x 7 x 4x x x 48 ( x 8)( x 6) 0 x 8, 6 A 6. (a) y 0 x Shape domain, intercept B B cao A (c).04 x ln x 8 (years) ln.04 A 7

90 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme. (a) nn nn n nax, ax ax 6 accept!,! B, B nn na 8, a 0 both a a a n nn 0, 0 either n 6, a A, A 4 (c) A 8 4. (a) 8 x 0 x 8 x A x x x 6x x x (or 4) terms A 5 4 x 64 x 6x 64x dx 8x 5 x A 5 x 64 8x x 5 5 A ft 7 5 Volume is (units ) A 7 9

91 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 5. (a) sin sin tan cos or cos or equivalent tan sin sec cos cos sin cos cos sin cos * cso A 4, 8 cos t t t cso A 5 9 Alternative to 5 * t t t t 0 tan t * t cso A 5

92 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number Scheme 6. (a) x 0 6, 0 B y ,.4 B, B NB. Not giving d.p. loses a maximum of one mark I ft their ys A ft.8 accept.77 A 4 B (c) underestimates diagram or explanation B B 9 NB. Exact answer is e.905 4

93 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number 7. (a) Scheme V shape right way up vertex in first quadrant g a eeoo; a, a, - 4 B B B B (, 0) 5 y a a a 0 4 x 4x a a x a 5x a, x 5 9a y 5 a both correct A (c) x 4x a a a 4x a fg A (d) 4x a a 4x a A a a x, A, A 5

94 EDEXCEL PURE MATHEMATICS P (667) PROVISIONAL MARK SCHEME NOVEMBER 00 Number 8. (a) f ' x Scheme A x x 0 x x 0 x x x y n n k A 4 A (c) x y B f ' m ' y x y x A 4 (d) i x n x x leading to 6n x x 0* cso x A ii g(0.)=0.7 g(0.4)=-0.70 Both, accept one d.p. Sign change (and continuity) root 0., 0.4 A 4 4 6

95 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME Mark Scheme. (a) f ( ) = ( ) (9 ) 0 M: Evaluate f ( ) or f () f ( ) = 0, so (x + ) is a factor Alternative: ( x 9x 0) ( x ) ( x ax b), a 0, b 0 [] A () ( x x 5), so (x + ) is a factor [A] ( x 9x 0) ( x )( x x 5) = (x + )(x + )(x 5) A A (4) (6). (a) r (a.w.r.t. if changed to degrees) A () x, A sin 0.4, x = 6.5 sin 0.4, (where x is half of AB) 6.5 (n.b. 0.8 rad = 45.8 ) AB = x = 5.06 (a.w.r.t.) (*) A () Alternative: AB cos 0. 8 [] AB cos0.8 [A] AB = 5.06 [A] (c) r 5.06 ( ) (a.w.r.t) (or 0.) A ().(a) ( 5 p 8) p (p 8) (5 p 8) Solve, showing steps, to get p = 4, or verify that p = 4. (*) A c.s.o. () Alternative: Using p = 4, finding terms (4,, 0), and indicating differences.[] Equal differences + conclusion (or common difference = 8 ). [A] (7) a = 4 and d = 8 (stated or implied here or elsewhere). B T 40 a ( n ) d 4 (9 8) 6 A () (c) S n n[a ( n ) d] n[8 8( n )] Aft 4n (n) A () (8)

96 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME

97 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME Mark Scheme 4.(a) b 4ac ( k) 6 k 6 A Or, (completing the square), x k k 9 4 Or, if b and 4ac are compared directly, [] for finding both [A] for k and 6. No real solutions: k 6 0, 6 < k < 6 (ft their 6 ), Aft (4) x 4x 9 ( x )... (p = ) Ignore statement p = if otherwise correct. x 4x 9 ( x ) 4 9 ( x ) 5 (q = 5) M: Attempting ( x a) b 9, a 0, b 0. B A () (c) Min value 5 (or just q), occurs where x = (or just p) Bft, Bft () Alternative: f ( x ) x 4 (Min occurs where) x = [B] Where x =, f ( x ) 5 [Bft] 5.(a) 8 = seen or used somewhere (possibly implied). B (9) 8 or Direct statement, e.g. (no indication of method) is M0. At x = 8, d y = 8 + = 6 + = 9 (*) A () dx 8 x Integrating: x ( C) (C not required) A A At (4, 0), 4 4 C 0 (C required) f( ) x 4x, 4 x A, A (6) (9)

98 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME Mark Scheme 6.(a) (, 0) (or x =, y = 0) B () y x y 4 or 4( x ) y 6y 6 0 or x x 54 0 (or equiv. terms) A ( y )( y 8) 0, y = or ( x )( x 8) 0, x = ( term quad.) y =, y = 8 or x =, x = 8 A x =, x = 8 or y =, y = 8 (attempt one for M Aft mark) (6) (Aft requires both values) (c) Grad. of AQ = 8 0, Grad. of AP = 8 0 ( ) (attempt one for M mark) Aft m m, so PAQ is a right angle (A is c.s.o.) A (4) Alternative: Pythagoras: Find lengths [] AQ = 0, AP = 5, PQ = 5 (O.K. unsimplified) [Aft] (if decimal values only are given, with no working shown, require at least d.p. accuracy for (implied) A) AQ AP PQ, so PAQ is a right angle [, A] () requires attempt to use Pythag. for right angle at A, and A requires correct exact working + conclusion. 4

99 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME Mark Scheme 7.(a) Solve x x 0 to find p = 6, or verify: (*) B () 4 4 dy dx x x 4 A m = 9, y 0 9( x 6) (Any correct form) A (4) (c) x x 0, x = 4, Aft 4 () (d) 4 x x x x dx 4 (Allow unsimplified versions) 6 A M: Need 6 and 0 as limits. A (4) 8.(a) 0 = 5 = 5 (cos( 0) cos cos0, etc, is B0) B () 0 = 45 = 55 M: Using 60 5 (can be implied) A () Stating = 45 scores A0 (Other methods: for complete method, A for 5 and A for 55) =.8 ( ) (At least d.p.) (Could be implied by a correct B ). = + 80 or = + 60 or = (One more solution) = 0.9, 00.9, 90.9, 80.9 (: divide by ) Aft A (Aft: correct, ft their ) (A: all 4 correct cao, at least d.p.) (5) (c) sin sin, sin cos cos, A ( cos ) cos cos cos 0 ( cos )(cos ) 0 cos (M: solve term quadratic A up to cos = or x = ) = 60, = 00 A (6) 5 (4)

100 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME 6

101 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number Mark Scheme (a) + x + ( x + ) +.. = = x + x + 7 x + or ( x + ) + x + B () y = + x + OR y = x + 7 x + y = x + y (x + ) = x + 7 yx x = 7 y x + = x = y y - x (y ) = 7 y 7 y x = y f ( x ) = x f -- 7 x ( x ) = o.e A x () (c) Domain of f (x) x εr, x [NB x + ] Notes B () (5) A y = f (x) and st step towards x =. One step from x =. y or f - (x) = in terms of x.

102 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number Mark scheme (a) u = 0 + = =. 86 u =.9000 =. 90 A c.a.o u4 = =. 89 A c.a.o S.C. [If u = AWRT.90 and u4 = AWRT.89 penalise once only] () a (i) = + or 9 = a + a = 9 or a = 9 (ii) (If u = u, then u = u..) u5 = Notes (a) Correct expression or AWRT.86 a =.5 A B () () (7) (i) A correct equation for a, with or without. Attempt correct manipulation to ka =, ( k > 0).

103 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number (a) log log ( 6 x) = log 6 + log x 4 = 4 + a 4 = log log x = 4 log x log Mark Scheme x A c.a.o () = 4a (accept 4 log x - ) (c) = 4+ a ( 4a ) a = A x x log = = x = 8 or or or ( ) A Notes (a) Correct use of log ( ab) = log a + log b a Correct use of log = b Use of log x n = n log x A () (4) (9) (c) Use their ( a )& ( b) to form equ in a a Out of logs: x = A Must write x in surd form, follow through their rational a.

104 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number 4(a) y Mark Scheme x>0 B -4a 4a x x<0 ( 4a,0) & ( 4a,0) and shape at ( 0,0) B B () ( ) ( ) ( ) f f a = a 4a a = 4a 8a = 4a ( a) [ = f ( a) ( even function) ] = 4a (c) a = and f ( x) = = x x ( x > 0) Notes 0 = x x 45 ( x 5)( ) 0 = x + x = 5 ( or ) Solutions are x = ± 5 only 4 B their f ( a) B B A A () (4) (9) (c) A A Attempt TQ in x Attempt to solve At least x = 5 can ignore x = To get final A must make clear only answers are ± 5. 4

105 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number 5(a)(i) (ii) Mark Scheme y x = a B In both sides of (i) i.e ln x ln a y y ln a = ln x y ln x = ln a or ( y =) loga x = = ln a Bcso () () ALT. y lna = or x ln x, dy = lna dx dy, dx, = lna dy dx (c) 0 A is ( 0, ) log 0 from ( b) = x xlna = y equ of target y = m( x 0) m= or or (or better) 0lna 0ln0 A =, Acso () B B (d) 5(a) or y= x + (o.e) 0ln0 0ln0 ln0 i.e y = ( x 0) x y = 0in 0ln0 ln0 ln0 Notes B B ( c) 0 = + x, = 0ln0 x = 0 0ln0 or 0( ln0) or 0ln0( ) ln0 ylna x = e is BO y Must see ln a or use of change of base formula., Acso needs some correct attempt at differentiating. A A (4) () (0) (c) B Allow either their y A and m (d) Attempt to solve correct equation. Allow if a not = 0. 5

106 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number Mark Scheme 6(a) f ( ) = 0 for maximum f (.48). = e f (.49) = = 0.00 x (or stationary point or turning point) change of sign root / max imum in range at y = e x x ( + c) 0 ( 0,5) 5 = e 0 + c B A A () (c) c = 4 x y = e x + 4 Area x = e x + 4 dx 0 ( c = 4) A (4) = e x x4 + 4x 0 A 6 = e + 8 ( e ö 0 + 0) 6 6(a) Notes A = e + 4 or e + May be if maximum mentioned at A One value correct to S.F. Both correct and comment Some correct A cao (4) () A e x x Attempt to use (0,5) No + c is M0 (c) Some correct other than ex e x. A [ ] their ( 0) c. Attempt both limits 6

107 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number Mark Scheme 7(a) 4, 4.84, 7.06 B//0 () B I 0.5 [ ( ) ] [ A ] = 0.5 [ 9.44] = 4.9 or 4.9 (AWRT 4.9 or just 4.9) A (c) x dx = lnx x 0.5 x A (4) = ln ln = ln+.55 or ln+ 80 (d) [4.9 ( c) ] 00, =.5% ( i.e < % ) ( c ) AWRT.5% Notes 7 B 0.5 A (4), A () () A [ ] (c) Some correct A 5 lnx + x 5 Use of limits 7

108 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 MARK SCHEME number Mark Scheme 8(a)(i) cosθ 5sin θ = R cosθcosα R sin θsin α. ( θ +.6) cos = R α 4 5 R = 5 +, R =, A tan α = 5, α =.6 ( AWRT.6) or 0.9 c (AWRT 0.9 c ), A θ +.6 = 7., (4) (ii) θ = 49.5 (only) A 8 tan θ = tan θ () i.e. 0 = tan θ + tanθ 8 0 = ( tan θ 4)( tan θ+ ) 4 tanθ = or 4 A tan θ = θ = 5. [ ignoreθ not in range e. g. θ = 6. 6 ] A Notes 8(a)(i), A for correct expression for R or R, A for correct trig expression for α 4 cos ( θ + α) = R θ + α = their R (ii) Use of cot θ = tanθ TQ in tan θ = 0 Attempt to solve TQ = 0 A For Final A mark must deal with tan θ = (5) () 8

109 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme. Either for f.t. either on gradient Obtains centre ( 0, 6.5) a Finds radius or diameter by Pythagoras Theorem, to obtain r =.5 or r = 6.5 B, A Or x + ( y 6.5) =.5 or x y y = 0 B (4) y 8 y 5 = x+ x Gradients multiplied and put = to - B A x y y Or Obtains centre ( 0, 6.5) = 0 B B (4) x + ( y 6.5) = r or substitutes either (, 5) or (-, 8 ) x + ( y 6.5) =.5 or x y y c + + = 0 x y y = 0 B A (4). (a) nn ( ) na = 6, a = 7 nn ( )6 Attempts solution by eliminating variable e.g. = 54 or n 6 6 a a = ( ) a 54 B, B n = -, a = A, A (5) ( )( )( 4) 6 = 08 for allow a instead of a A () (c) x < or < x < B f.t. ()

110 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme. (a) 0, ( dy dy x + y + x ), 6y 0 dx dx =,(B),A At (, ) 0 + (4 + dy dy ) 0 dx dx = dy 4 7 = =.4 or or 5 dx 0 5 A (5) 5 The gradient of the normal is 7 5 Its equation is y = ( x ) (allow tangent) y = x+ or y = x Acao () 4. (a) Uses the remainder theorem with x =, or long division, and puts remainder = 0 To obtain p+ q= 5 or any correct equivalent (allow more than terms) A Uses the remainder theorem with x =, or long division, and puts remainder = ± 7 To obtain p+ q= or any correct equivalent (allow more than terms) A Solves simultaneous equations to give p = -7, and q = - 4 Then 6x 7x 4x+ 8 = (x ) ( x x 8) A A ft (6) So f(x) = (x )(x + 4)(x ) B ()

111 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme 5. (a) Area of triangle = ( = 444.) Accept 440 or π B () Either Area shaded = π π 4 0sin t.tdt = 480tcos t+ 480cos t 480tcos t+ 40sin t π π = [ ] 4 π π 4 A A A ft = 40( π ) A (7) or π 4 60cos.(6 - π )d π t t t A = t π t t t+ t (0sin ( 6 ) 480 cos 480cos 480tcos t+ 40sin t π π = [ ] 4 π π 4 A A ft = 40( π ) A (7) (c) Percentage error = 40( π ) estimate 00=.6% (Accept answers in the 40( π ) range.4% to 4.4%) A ()

112 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme 6. (a) A B Uses + ( x ) ( x+ ) Considers x + = A (x + ) + B ( x ) and substitutes x = - or x =.5, or compares coefficients and solves simultaneous equations To obtain A = 4 and B = -. A, A (4) Separates variables 4 dy = dx y x x+ ln y = ln(x ) ln( x+ ) + C A, B ft Substitutes to give ln 4 = ln ln + C and finds C (ln 08) ln y = ln(x ) ln( x+ ) ( + ln08) C(x ) = ln ( ) x + 08(x ) y = ( x + ) A A cso (7) Or y ln(x ) ln( x++ ) ln08 = special case A e 4

113 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme 7. (a) dy x x x = dx (4 + x ) (4 + ) ( ) Need numerical answers for or (from product rule) (4 + x ) x (4 + x ) Solve dy =0 to obtain (, ¼ ), and (-, -¼ ) or ( and A, full solution A) dx A A, A (5) d y When x =, = <0 thus maximum dx d y When x = -, =0.065 >0 thus minimum. dx B B () (c) Shape for - x B Shape for x > B Shape for x < B () 5

114 EDEXCEL 667 PURE MATHEMATICS P JANUARY 004 PROVISIONAL MARK SCHEME Number Scheme 8. (a) Any two of + λ = + Need µ two of these for + λ = + µ 5 λ = 4+ 4µ Solve simultaneous equations to obtain µ =, or λ = A intersect at (, 5, 4) A Check in the third equation or on second line + + ( ) 4 = 0 perpendicular B A (6) () (c) P is the point (, 7, ) [ i.e. λ = ] and R is the point ( 4, 6, 8) [i.e. µ = ] A ( ) 6 PQ = + + = RQ = + + = 4 A ft PR = 7 4 The area of the triangle = 6 = A (6) Or area = 6 7 sin P where sin P = 7 = 4 Or area = 7 sin R where sin R = = 4 (must be simplified) 6

115 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme number x + x y + y (a), =, = (,5) Gradient = y x y = m( x ) y x 8 = 5 A () A y = x + A (4) + Allow = x y or y = (x + ) 6. (a) 8 = 4 = (seen or implied) B ( 8)( 8) = = 7 A () , = = , A () Allow ( + ) or equiv. (in terms of ) 4 6. (a) x + ( x + 0) < 00 (Using x 0 is A0) A () x ( x + 0) > 4800 (Using x 0 is A0) A () (c) 65 (i.e. Allow wrong inequality sign or x = 65). Bft Solving term quadratic, (x + 80)(x 60) = 0 x = x > 60 (x < 80 may be included here, but there must be no other A wrong solution to the quadratic inequality such as x > 80) 60 < x < 65 A (4) 8

116 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme number 4. (a) C C : U shape B l C : Position l : Straight line through origin with negative gradient B () (, 0), (, 0), (0, 4) of these correct: B B All correct: B () (c) 5. (a) x 4 = x x + x 4 = 0 ( x + 4)( x ) = 0 x =... x = 4 x = A y = y = M: Attempt one y value A (4) 9 8 tan x = (or exact equivalent, or s.f. or better) B () 8 tan x = x = 69.4 (α), x = 49.4 (80 + α) A, Aft () (c) ( cos y ) 8cos y = 0 cos y + 8cos y = 0 A ( cos y )(cos y + ) = 0 cos y =..., (or ) A y = 70.5 (β), x = 89.5 (60 β) A Aft (6) 0

117 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme number 4 6. (a) ( x 6x + 9) ( 4 x x + x = x x + x (*) A () 6 9) 6 9 (c) (d) f ( x When 4 x ) = + 6x 7 First A: terms correct (unsimplified) A A () Second A: all correct (simplified) 6 7 x = ±, f ( x) = + f ( x ( ) ( ) = + = 0, Stationary A () 9 5 x ) = x + 08 M: Attempt to diff. f ( x), not g( x ) f ( x) ( ) = + ( ) ( ) f 5 ( = 4.69) = > 0, Minimum (not dependent on a numerical version of f ( x) ) Aft () 0 A 7. (a) ( n S = ) a + ar ar S = not required. Addition required. B ( rs = ) ar + ar n ar rs = not required (M: Multiply by r) S( r) = a( r n ) n a( r ) S = r (M: Subtract and factorise each side) (*) A (4) r = 0.9 B 0 0( 0.9 ) S 0 = = 87.8 A () 0.9 a 0 (c) Sum to infinity = = = 00 r 0.9 a r (d) = = 0 r r (ft only for r < ) Aft () (Put a = r in the formula from (c), and equate to 0) 0 r = 0 ( r) r =..., (or exact equivalent), A ()

118 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme number d y 8. (a) = x 4x + 5 dx x 4x + 5 = 0 ( x 5)( x ) = 0 x =..., (A requires correct quadratic factors)., A A () y = (Following from x = ) A (4) (c) P: x = y = B Same y-coord. as Q (or zero gradient ), so PQ is parallel to the x-axis B () 4 x 7x 5x A A 4 (First A: terms correct, Second A: all correct) (d) ( x 7x + 5x + ) dx = + + x 4 x 4 7x 5x x = = 5 4 ( ) = (or equiv. or s.f or better) A (6) 4 4

119 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Number Scheme. ( x )( x 5) x( x + ) ( x )( x + ) ( x 5) ( factorising) B B B = x x 5 B (4 marks). (i) A correct form of cos x used (ii) (a) 5 or 4 5 sec x = ; cos x = or 4 or 5 cos x + or sin x sin x 4 A (4) 7 tan Forming single fraction (or multiplying both sides by sinx) 7 5 A + x sin x Use of correct trig. formulae throughout and producing expression in terms of sin x and cos x cos x cos x Completion (cso) e.g. = = cot x (*) A (4) sin x cos x sin x 0. (a) ( x ) ;... + ( x ) ( ) + ( x ) ( ) +... B; x x [For, needs binomial coefficients, n C r form OK, at least as far as shown] n Correct values for s :, 66, 0 used (may be implied) C r B (8 marks) ( x ) ( x ) + 66( x ) + 0( x ) x x x... x 6 8 6x + x 55 4 x A(,0) (5) Term involving 9 ( ) ( ) x ; x coeff = 55 = ( ) A.. (or ) A () (8 marks)

120 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Number 4. (a) y Scheme x + x + 4x = = = x A x x x π y dx [dependent on attempt at squaring y ] B x + 4x + 4 x y dx = ( )dx; = + 4x + ln x x 4 ;A ft Correct use of limits: [ ] [ ] [ ] [M dependent on prev. ] 4 4 [ A must have ln x term] = 9 Volume = + 4 ln 4 π or equivalent exact A (7) Showing that y = at x = and x = 4 B () (c) Volume = answer to (a) ; = 69.5 cm 60 cm (*) ;A () [allow 69 60] (0 marks) 5. (a) Attempting to reach at least the stage x ( x + ) = 4x + Conclusion (no errors seen) 4x + x = (*) A () x + [Reverse process: need to square and clear fractions for ] 4 + x = =.58 + x =.68, x.70 AA () 4 = [Max. deduction of for more than d.p.] (c) Suitable interval; e.g. [.695,.705] (or tighter ) f(.695) = 0.07, f(.705) = Dep. Change of sign, no errors seen, so root =.70 (correct to d.p.) A () (d) x =, division by zero not possible, or equivalent B,B () or any number in interval < x < ¼, square root of neg. no. (0 marks)

121 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme Number 6. (a) log5 log y ; = log5 log y = a b A () x 5 x 5 log5 5 = or log5 y B log5 5 + log5 x + log y ; = + a + ½ b ;A () 5 (c) a b =, + a + ½ b = (must be in a and b) B ft () (d) Using both correct equations to show that a = 0.5 (*) b =.5 B () [Mark for (c) can be gained in (d)] (e) Using correct method to find a value for x or a value of y: x = = 0.669, y =. 5 = A A ft () [max. penalty for more than d.p.] ( marks) 7. (a) Differentiating; x e f ( x ) = + ;A () 5 A: 0, B 5 Attempt at y y f(0) = f (0) x ; 6 = x or equivalent one line termed equation A ft () 5 5 (c).4,.55,.86 B(,0) () (d) Estimate = 0.5 ; ( ) [( ) + ( )] B A ft.48 =.475,.4.49 A (4) ( marks)

122 EDEXCEL PURE MATHEMATICS P (667) JUNE 004 PROVISIONAL MARK SCHEME Scheme Number y 8. (a) y = ln( x 6) x 6 = e y e + 6 x = ; { f x + e 6 ( x )} = ;A () Domain: x R B Range: f ( x ) > B () e + 6 (c) Attempting to find f () [ = ] ; = 8.70 ;A () (d) ln curve passing through y = 0 B Symmetry in x = k, k > 0 All correct and asymptote at x = labelled A (e) Meets y-axis: (x = 0), y = ln 6 B Meets x-axis: x = 5 7, (0) ; x =, (0) BB () [May be seen on graph] ( marks) 4