Mathematics HL - May 2001 - P2 \$

# Mathematics HL - May 2001 - P2 \$ - INTERNATIONAL...

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MARKSCHEME May 2001 MATHEMATICS Higher Level Paper 2 30 pages M01/510/H(2)M INTERNATIONAL BACCALAUREATE BACCALAUR&AT INTERNATIONAL BACHILLERATO INTERNACIONAL

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1. (a) Using integration by parts (M1)(A2) 11 cos3 d sin3 sin3 d 33 xx x x x =− ∫∫ (AG) ( not required) cos3 39 x C =+ + C [3 marks] (M1)(A1) (b) (i) Area 2 cos3 x 6 π 6 π  =  9  (A1) (ii) Area 4 cos3 x 6 6 π = 9 (A1) (iii) Area 6 cos3 x 6 6 π = 9 Note: Accept negative answers for part (b), as long as they are exact. Do not accept answers found using a calculator. [4 marks] (c) The above areas form an arithmetic sequence with (A1) 1 22 and ud ππ == 99 (M1)(A1) The required area 42 (1 ) 2 n n Sn + (A1) ) n n π 9 [4 marks] Total [11 marks] & 7 & M01/510/H(2)M
2. (a) Given the points A( 1, 2, 3), B(&1, 3, 5) and C(0, 1, 1), (A1) then 01 AB 1 , AC 3 22 →→       ==    (A1) and |AB| 5,|AC| 14 The size of the angle between the vectors is given by AB and AC (M1) AB AC 7 arccos arccos 51 4 |AB||AC| θ ⋅− (A1) radians 147 (3s.f.) or 2.56 = ! [4 marks] (M1) (b) Area 11 |AB||AC| sin or |AB AC| (A1) Area 2 2.29 units = 21 accept 2.28, 2.30, and 2 [2 marks] (c) (i) The parametric equations of are 12 and ll (A1) 1 :2 , 1, 2 lx y z λλ + = (A1) 2 :1 , 1 3 , 1 2 y z µµ µ =− + = − Note: At this stage accept answers with the same parameter for both lines. (ii) To test for a point of intersection we use the system of equations: 21 ! 3 λµ −+ =− " (M1) 2 =− # (A1) Then 3, 7 from and µλ !" (M1) Substituting into # gives RHS 14, LHS 5 Therefore the system of equations has no solution and the lines do not intersect. [5 marks] continued. .. & 8 & M01/510/H(2)M

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Question 2 continued (d) The shortest distance is given by where d and e are the position 12 () ( ) −⋅× × ed l l ll vectors for the points D and E and where are the direction vectors for the and lines . and (M1)(A1) Then 01 2 42 132 ×= = + − −− ijk i jk (M2) And ( )( 3 2) ( 4 21 −+ +⋅ + − = × i jk i jk (A1) or 1.96 9 21 = [5 marks] Total [16 marks] & 9 & M01/510/H(2)M
3. (a) ( ) 22 3 () ( 1 ) fx x x =− (A4) Notes: Award (A1) for the shape, including the two cusps (sharp points) at . 1 x (i) Award (A1) for the zeros at . 1and 0 xx = (ii) Award (A1) for the maximum at 1and the minimum at 1. = (iii) Award (A1) for the maximum at approx. , and the minimum at x =0.65 approx. x =−0.65 There are no asymptotes. The candidates are not required to draw a scale. [4 marks] (b) (i) Let 2 3 2 ( 1 ) fx xx (M1)(A2) Then 12 33 2 4 ( 1 ) ( 1 ) 3 fx x + 1 3 2 4 () ( 1 ) ( 1 ) 3 x x  +−   (or equivalent) 1 3 7 1 ) 1 3 x    (or equivalent) 1 3 2 2 73 3( 1) x x = (A1) The domain is (accept ) 1.4 1.4, 1 −≤ ± 1.4 1.4, 1 −< < ± (M1) (ii) For the maximum or minimum points let i.e. or use () 0 = 2 (7 3) 0 x −= the graph. (A1) (A1) Therefore, the x -coordinate of the maximum point is (or 0.655) and 3 7 x = the x -coordinate of the minimum point is (or 0.655). 3 7 x Notes: Candidates may do this using a GDC, in that case award (M1)(G2) .

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## This note was uploaded on 05/06/2010 for the course MATH 1102 taught by Professor Chang during the Spring '10 term at Savannah State.

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Mathematics HL - May 2001 - P2 \$ - INTERNATIONAL...

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