January 2006 Question Number 1 6675 Pure Mathematics P5/FP2 Mark Scheme Scheme Marks x 2 2 x 17 ( x 1)2 16 4 I = 1 B1 dx = [ arsinh ( x - 1) ] or equiv. 4 = arsinh ¾ ( x 1) 2 16 [ M1 does not require limits; A1 f.t. on completing square, providing arsinh ] 1 3 Into ln form [ ln 4 9 1 ; 16 = M1 A1√ M1A1 ln2 [5] [ If straight to ln form : B1, ln ( x 1) ( x 1)2 16 Using limits correctly M1A1√ , ln2 2 (a) A1 ] Using b2 a 2 ( e2 1) ; [ 4 = 16 (e 2 1) ] e = 5 or equiv. (1.12) M1A1 2 (2) 5 ]; 2 = 4 5 (b) Distance between foci = 2 a e [ 2 x 4 x [ M1 M1A1√ (2) A1√ dependent on both Ms ] (c) Ellipse, centred on origin B1 Hyperbola, both branches B1 Totally correct, touching, with B1 correct intercepts (3) [7] 1 January 2006 3 6675 Pure Mathematics P5/FP2 Mark Scheme dx 1 cos t , dt s dy sin t , dt (1 cos t ) 2 (sin t ) 2 dt ; Use of “half-angle formula” B1 (both) = [ 4 cos t dt ] ; 2 Using limits correctly and surd form; M1A1 2 2 cos t dt ( t 2 s = 4 sin 2 ( 0) = 2 √2 (allow 4 ) 2 M1A1√ M1A1 [7] 4 (a) Using cosh x e x e 2 x and attempt to progress e3 x 3 e x 3e x e 3 x Correct intermediate step as far as 4 8 = M1 e x e x 3 2 e 3x e 3x cosh 3 x 2 A1 (3) M1 (b) Using part (a) to reduce to cosh2x = [ 2 ] Correct method to form ln x or find e X or e 2 x x = ln 2 1 , ln 2 1 A1 or equivalent or ½ ln 3 2 2 , ½ ln 3 2 2 , M1 A1 A1√ (4) ( after finding e 2 x = ……) [7] 2 Question Number 5 Scheme dy 1 . sec x tan x tan x ; dx sec x s 2 dy (1 dx dx c = B1B1 x M1 (1 tan xc2 dx ln sec x tan x sec x dx ; s 0, x Method to find “c”: Marks 3 ( c) 0 ln (2 3 ) c ln (2 3 ) A1A1 M1 c = ln (2 3 ) or ln tan s ln (sec tan ) or equivalent 5 or – 1.32 12 or equivalent A1 A1√ [8] 6 (a) dy 3 cosh 2 x sinh x dx B1 d2 y 3 cosh 3 x 6 cosh x sinh 2 x dx 2 or equivalent Application of formula for radius of curvature [ = M1A1 1 9 cosh 1 9 c (c 1) 3c {c 2(c 2 1)} (b) cosh x = 2 2 2 3 2 x sinh x ] 3 3 cosh x 6 cosh x sinh 2 x Use of sinh2 x = cosh 2 x 1 4 4 M1 M1 3 2 = 9c 9c 1 3c { 3c 2 2)} 6 4 3 2 5 4 AG A1* (6) B1 Using found value of cosh x in formula for ; = 4.8 M1A1 (3) [9] 3 4 7 (a) 3 3 4 2 n 2 n 1 2 I n x (4 x) n x (4 x) 2 dx 3 3 0 0 3 4 2 n x n 1 (4 x) 2 dx 3 0 = A1√ 1 4 1 4 2 2 n 4 x n 1 (4 x) 2 dx n x n (4 x) 2 dx 3 0 3 0 = In [ 2n 3I n 8 nI 3 8n I n 1 (b) Relating I 2 to I I2 = M1A1 16 8 . 7 5 I 0 n 1 ] 2 nI 3 I n n 8n I 2n 3 using result from (a) 0 = 2048 105 M1A1 53 19 105 n 1 AG A1* (6) M1 A1A1 (3) [9] 4 8 1 1 1 1 2 (a) ar tanh ln 2 1 1 2 2 = 1 2 1 ln x 2 2 1 = 1 ln 2 or equivalent M1 2 1 or equivalent 2 1 M1 2 2 1 ln 2 1 AG A1* (3) Alternative Approach 1 ar tanh (sin x) tanh y 2 and then use exponentials: If using y Progression as far as e y ....... or e2y ....... Converting to ln form Answer as given Note: 1 ln 3 2 2 2 [ or cosh2 y 2 ] M1 M1 A1* can earn M1M1 but for A1* there must be a convincing further step. (b) dy 1 . cos x ; dx 1 sin 2 x cos x sec x cos 2 x M1A1 (2) Note: If tanh y sin x is differentiated M1 requires dy f ( x) dx (c) Attempt at by parts and use of result in (b) = cos x ar tanh (sin x) M1 cos x sec x d dx A1 M1 = cos x ar tanh (sin x) x Using limits correctly : = 1 or exact equivalent ln 1 2 4 2 5 M1A1 (5) [10 ] 9 (a) Correct method for finding 1 P dy dx M1 Gradient of normal = – p A1 y 2ap ( p)( x ap 2 ) Equation of normal: y px 2ap ap3 M1 AG A1* (4) (b) Using both equations and eliminating x or y M1 ( p q) x 2a( p q) a ( p q ) 3 3 may be unsimplified A1 x 2a a ( p pq q ) 2 2 A1 Finding the other coordinate M1 y apq ( p q) A1 (5) (c) Using pq = 3 in both x and y [ x a ( p 2 q 2 5), (in any form) y 3 a ( p q) ] M1 Complete method for relating x and y, independent of p and q M1 A correct equation, in any form A1 [ e.g. : x 5a y 2 9 a 2 ( p 2 q 2 2 pq) = 9 a 2 6 ] a y 2 9 a ( x a) A1√ (4) [13] 6