GCE Examinations Advanced Subsidiary / Advanced Level Further Pure Mathematics Module FP2 Paper H MARKING GUIDE This guide is intended to be as helpful as possible to teachers by providing concise solutions and indicating how marks should be awarded. There are obviously alternative methods that would also gain full marks. Method marks (M) are awarded for knowing and using a method. Accuracy marks (A) can only be awarded when a correct method has been used. (B) marks are independent of method marks. Written by Rosemary Smith & Shaun Armstrong Solomon Press These sheets may be copied for use solely by the purchaser’s institute. FP2 Paper H – Marking Guide 1. 2x2 + y2 = 4, 4x + 2y dy dy = 0 or 2x + y =0 dx dx d2 y dy + ( )2 = 0 2 dx dx dy dy at (1, −√2) 2 − √2 =0 ∴ = √2 dx dx d2 y d2 y 2 − √2 2 + 2 = 0 ∴ = 2√ 2 dx dx 2 2+y 3 ρ= 2. (a) (1 + ( 2) 2 ) 2 2 2 1 2 1 2 ln(xy) = arcosh ( 53 ) = ln ( (e−x + ex) = f(x) ∴ cosh x is even 5 3 + 25 9 ) − 1 = ln 3 ∴ xy = 3 cosh (3x − y) = 1 ∴ 3x − y = 0 3. ∫ A1 A1 M1 A1 M1 A1 A1 B1 ∴ 3x = y = 3x , x2 = 1 M1 x > 0 ∴ x = 1, y = 3 A1 1 dx = 13cosh x − 5sinh x = ∫ ∫ ∫ 13 2 1 dx (e + e ) − 52 (e x − e − x ) x −x 1 dx 4e + 9e − x x ex dx 4e2 x + 9 du u = ex, = ex dx 1 = ∫ du 4u 2 + 9 1 = 14 ∫ du u 2 + 94 = (8) M1 A1 M1 M1 A1 M1 = 1 4 × 23 arctan ( 23u ) + c A1 = 1 6 arctan ( 23 ex) + c A1 Solomon Press FP2H MARKS page 2 (8) (ex + e−x) (e−x + e−(−x)) = 1 2 M1 A1 M1 A1 4 2 2 f(x) = cosh x = f(−x) = (b) = 3 3 = 3 6 M1 A1 (8) 4. (a) let y = arcsin (2x − 1) ∴ sin y = 2x − 1 dy =2 ∴ cos y dx dy 2 2 = = 2 dx 1 − sin y 1 − (2 x − 1) 2 2 = (b) x= 3 4 dy = dx 5. (a) 1 − (4 x − 4 x + 1) , y = arcsin 1 − 9 16 eqn. is y − π 6 3 4 = 2 1 2 = 4 3 3 4 (x − π 6 = y= π 6 − √3 4 3 2 4x − 4x 2 = 1 x − x2 A1 B1 4 3 3 16 x=0 ∴y− M1 A1 π 6 = 1 = = M1 M1 ) M1 × (− 34 ) M1 A1 dy dy = 4a ∴ = 2a y dx dx dy a at P, = 22at = 1t dx 2y (9) M1 A1 eqn. is y − 2at = 1t (x − at2) giving yt = x + at2 A1 (b) at Q, y = 0 ∴ x = −at2 ; at R, x = 0 ∴ y = at ∴ M is (− 12 at2, 12 at) M1 A1 A1 (c) grad of OM = 1 at − 0 2 − 12 at 2 −0 M1 = − 1t ; grad of OP = 2 at2 −0 = 2t at −0 OM perp. OP ∴ 2t × − 1t = −1 ∴ t2 = 2 M1 A1 M1 A1 (11) Solomon Press FP2H MARKS page 3 6. (a) In − In−2 = ∫ = ∫ = ∫ ∴ In = (b) cos nθ − cos(n − 2)θ dθ sin θ −2sin(n − 1)θ × sin θ dθ sin θ −2 sin (n − 1)θ dθ = n2−1 cos (n − 1)θ M1 2 n −1 A1 cos (n − 1)θ + In−2 4 ∫ π cos θ dθ = [ln | sin θ | ] 2π 4 sin θ π 2 π 4 1 2 = ln 1 − ln = −ln 2 − 12 1 2 = π 4 π I5 = [ 24 cos 4θ ] 2π + I3 = 1 2 4 (b) ae = √3, ae = 1 2 M1 A1 ln 2 = − (− 12 ) + 1 2 1 2 ln 2 − 1 ln 2 − 1 = 1 2 M1 A1 ln 2 M1 A1 ∴ ae × ae = a2 = 4 M1 a>0 ∴a=2 b2 = a2(1 − e2) = a2 − (ae)2 = 4 − 3 = 1 b>0 ∴b=1 A1 M1 A1 x2 4 A= (c) M1 A1 ln 2 I3 = [ 22 cos 2θ ] 2π + I1 = −1 − 0 + (a) M1 A1 π In = [ n2−1 cos (n − 1)θ ] 2π + In−2 I1 = 7. M1 4 3 + y2 = 1, ∫ 2 1 2 x + 2y 2πy 1 + −2 2 = 2π ∫ −2 = 2π ∫ −2 = π 2 −2 = π 2 ∫ ∫ y 4y 2 2 dx 2 16(1 − x4 ) + x 2 dx M1 16 − 3x 2 dx 3x2 = 16 sin2θ, x = π 3 A= π 2 ∫− = π 2 ∫− = 8π 3 = 4π 3 ∫− = 4π 3 [θ + = 4π 3 [ 4 3 sin θ, A1 dx = dθ 16 − 16sin 2 θ × π 3 π 3 4 cos θ × π 3 π 3 π 3 π 3 π 3 M1 A1 16 − 4x 2 + x 2 dx −2 ∫− M1 A1 x2 dx 16 y 2 16 y 2 + x 2 16 y 2 y 2 dy dy =0 ∴ = −4 yx dx dx π 3 + 4 3 4 3 4 3 cos θ M1 A1 cos θ dθ M1 cos θ dθ cos2 θ dθ A1 1 + cos 2θ dθ M1 1 2 π sin 2θ ] −3 π A1 3 3 4 − (− π 3 − 3 4 )] = 4π 3 ( 2π 3 + 3 2 )= Solomon Press FP2H MARKS page 4 (13) 8 9 π2√3 + 2π M1 A1 (18) Total (75) Performance Record – FP2 Paper H Question no. Topic(s) Marks 1 rad. of curv. 8 2 3 4 5 6 7 eqn. in hyp. fns. integr. hyp. fns diff. inv. trig. parabola, tangent reduction formula 8 8 9 11 13 Total ellipse, surface area 18 75 Student Solomon Press FP2H MARKS page 5
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