Mark Scheme (Results) Summer GCE Further Pure FP3 (6669) Paper 1

Transcription

1 Mark (Results) Summer 1 GCE Further Pure FP3 (6669) Paper 1

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3 June Further Pure Maths FP3 Mark Question 1. (a) Uses formula to obtain 5 e = 4 A1 Uses ae formula (3) a Uses other formula e 16 Obtains both Foci are ( ± 5,) and Directrices are x = ± 5 (needs both method marks) A1 cso () (5 marks) a1: Uses b = a ( e 1) to get e > 1 a1a1: cao a: Uses ae b1: Uses a e b1a1: cso for both foci and both directrices. Must have both of the previous M marks may be implicit.

4 . Question dy sinh 3x dx = B1 so s = 1+ sinh 3 d x x s = cosh 3xdx A1 ln a 1 = sinh 3x = ln -3ln 3 sinh 3ln 6[e a a a = e ] D = 1 ( a ) (so k = 1/6) a A1 (6 marks) 1B1: cao dy 1: Use of arc length formula, need both and dx. 1A1: cosh 3xdx cao : Attempt to integrate, getting a hyperbolic function o.e. 3: depends on previous M mark. Correct use of lna and as limits. Must see some exponentials. A1: cao

5 Question 3. (a) uuur uuur AC = 3i + 6j+ k, BC = 3i + 4j+ 3k uuuv uuuv AC BC = 1i 15j+ 3k B1, B1 (4) Area of triangle ABC = 1 1 i 15 j+ 3 k = 1 15 = 17.5 () (c) Equation of plane is 1x 15y+ 3z = or x 3y+ 6z = 4 So r. (i 3j + 6k) = -4 or correct multiple A1 () ( 8 marks) uuur a1b1: AC = 3i + 6j+ k cao, any form uuur ab1: BC = 3i + 4j+ 3k cao, any form a1: Attempt to find cross product, modulus of one term correct. a1a1: cao, any form. b1: modulus of their answer to (a) condone missing ½ here. To finding area of triangle by correct method. b1a1: cao. c1: [Using their answer to (a) to] find equation of plane. Look for a.n or b.n or c.n for p. c1a1: cao

6 Question 4(a) n 1 π π 4 4 n 1 In = x ( cos x) nx cosxdx so n 1 π 4 1 n π π n 4 4 In = x ( cos x) + nx sin x n( n 1) x sinxdx i.e. n 1 π n 1 1 In = n n ( n 1) I * A1cso (5) π 4 π 1 sin d [ cos ] 4 = 1 I = xx = x I 1 π 1 = I π 1, so I = 8 4 (4) (c) 3 π 1 I = 4 3 I π π 1 = = 1 3 ( π 4 π + 48) * cso 64 () n a1: Use of integration by parts, integrating sin x, differentiating x. a1a1: cao a: Second application of integration by parts, integrating cos x, differentiating aa1: cao a3a1: cso Including correct use of 4 π and as limits. n 1 x. b1: Integrating to find I or setting up parts to find I. b1a1: cao ( Accept I = ½ here for both marks) b: Finding I in terms of π. If n s left in M ba1: cao c1: Finding I 4 in terms of I then in terms of π. If n s left in M c1a1: cso

7 Question 5. (a) ar sinh x, + x 1+ 4x arsinhxdx = [ ar sinh ] x x x dx 1+ 4x 1 xar sinh x (1 4 x ) + arsinh [ ] = [ ] 3 1 = = ln(3+ ) 1 1 A1, A1 (3) 1 1A1ft 3D 4 3A1 (7) (1 marks) a1: Differentiating getting an arsinh term and a term of the form a1a1: cao ar sinh x x aa1: cao x b1: rearranging their answer to (a). OR setting up parts b1a1: ft from their (a) OR setting up parts correctly b: Integrating getting an arsinh or arcosh term and a ( 1 ax ) ba1: cao b3d: depends on previous M, correct use of b4: converting to log form. b3a1: cao depends on all previous M marks. and as limits. px 1± qx 1 ± term o.e..

8 Question 6(a) x y dy dy xb bcosθ + = and so = = a b dx dx ya asinθ bcosθ y bsinθ = ( x acos θ ) asinθ xcosθ ysinθ Uses cos θ + sin θ = 1 to give + = 1 a b A1cso cosθ Gradient of circle is and equation of tangent is sinθ y asinθ = cos θ ( x acos θ ) or sets a = b in previous answer sinθ So ysinθ + xcosθ = a A1 (4) () (c) a b Eliminate x or y to give y sin θ ( 1) = or x cos θ ( 1) = b a b a a l1 and l meet at ( cos θ, ) A1, B1 (3) (d) The locus of R is part of the line y =, such that x a and x -a Or clearly labelled sketch. Accept real axis B1, B1 () (11 marks) a1: Finding gradient in terms of θ. Must use calculus. a1a1: cao a: Finding equation of tangent aa1: cso (answer given). Need to get cos θ + sin θ on the same side. b1: Finding gradient and equation of tangent, or setting a = b. b1a1: cao need not be simplified. c1: As scheme a c1a1: x =, need not be simplified. cosθ c1b1: y =, need not be simplified. d1b1: Identifying locus as y = or real/ x axis. db1: Depends on previous B mark, identifies correct parts of y =. Condone use of strict inequalities.

9 Question 7(a) 1 x x 1 x x f(x) = 5cosh x 4sinh x = 5 ( e + e ) 4 ( e e ) (c) = 1 ( x 9 x ) e + e A1cso 1 x 9 x ) e + e x x = 5 e 1e + 9 = x So e = 9 or 1 and x = ln9 or x Integral may be written e dx x e + 9 B1 () (4) This is x e arctan Uses limits to give 3arctan1 3arctan( ) 3 = π π = 18 π * Dcso (5) (11 marks) x x a1: Replacing both coshx and sinhx by terms in e and e condone sign errors here. a1a1: cso (answer given) x b1: Getting a three term quadratic in e b1a1: cao b: solving to x = ba1: cao need ln9 (o.e) and (not ln1) c1b1: cao getting into suitable form, may substitute first. c1: Integrating to give term in arctan c1a1: cao c: Depends on previous M mark. Correct use of ln3 and ½ ln3 as limits. ca1: cso must see them subtracting two terms in π.

10 Question 8. (a) λ 1 1 λ 1 4 λ = ( λ)( λ)(4 λ) (4 λ) = (4 λ) = verifies λ = 4 is an eigenvalue (can be seen anywhere) (4 λ) { 4 4λ+ λ 1} = λ { λ λ } (4 ) = A1 (4 λ)( λ 1)( λ 3) = and 3 and 1 are the other two eigenvalues 1 x x Set 1 y = 4 y 1 4 z z or 1 x 1 y = 1 z (5) Solve x+ y = and x y = and x= to obtain x =, y =, z = k Obtain eigenvector as k (or multiple) A1 (3) (c) 3+ λ l B1 1 has equation which may be written λ + λ 1 3+ λ So l is given by r = 1 λ λ 8 + λ i.e. r = 7 λ 11+ 7λ So (r c) d = where c = 8i + 7j - 11k and d =i - j + 7k A1ft (5) (13 marks) a1: Condone missing =. (They might expand the determinant using any row or column) a: Shows λ = 4 is an eigenvalue. Some working needed need to see = at some stage. a1a1: Three term quadratic factor cao, may be implicit (this A depends on 1 st M only) a: Attempt at factorisation (usual rules), solving to λ =. aa1: cao. If they state λ = 1 and 3 please give the marks. b1: Using Ax = 4x o.e. b: Getting a pair of correct equations. b1a1: cao c1b1: Using a and b. c1: Using r = M x their matrix in a and b. c: Getting an expression for l with at least one component correct. c1a1: cao all three components correct ca1ft: ft their vector, must have r = or (r-c)x d = need both equation and r.

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