Mathematics HL - May 2000 - P2 \$

# Mathematics HL - May 2000 - P2 \$ - INTERNATIONAL...

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MARKSCHEME May 2000 MATHEMATICS Higher Level Paper 2 M00/510/H(2)M INTERNATIONAL BACCALAUREATE BACCALAUR&AT INTERNATIONAL BACHILLERATO INTERNACIONAL 30 pages

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1. (a) Using the formula for the area of a triangle gives (M1) Ax x = 1 2 3s i n θ (A1) sin . = 442 3 2 x [2 marks] (b) Using the cosine rule gives (M1) cos 22 2 (3 ) ( 3) 23 xxx xx +− + = ×× (A1) = −− 32 3 2 2 2 x [2 marks] (M1) (c) (i) Substituting the answers from (a) and (b) into the identity cos sin 1 θθ =− gives (AG) 3 2 1 3 2 2 2 2 2 F H G I K J F H G I K J . (G1)(G1) (ii) (a) x = 124 294 ., . (M1) (b) = F H G I K J arccos 3 2 2 2 x (G1)(G1) 1.86 radians or 0.171 (accept 0.172) radians (3 s.f.) = = Notes: Some calculators may not produce answers that are as accurate as required, especially if they use &zoom and trace± to find the answers. Allow 0.02 ± difference in the value of x , with appropriate ft for . Award (M1)(G1)(G0) for correct answers given in degrees ( or ). 106 ! 9.84 ! Award (M1)(G1)(G0) if the answers are not given to 3 s.f. Award (M0)(G2) for correct answers without working. [6 marks] [Total: 10 marks] ² 6 ² M00/510/H(2)M
2. (a) Let g x ax bx cx d () =++ + 32 (A1) gd 04 4 =− ⇒ =− (M1) =+ + g x ax bx c () 3 2 2 (A1) ′ =⇒= gc 00 0 (2 ) 0 8 4 ga b −=⇒ −+ 4 = (M1) ) 0 1 2 4 b −=⇒ − 0 = 4 a = 4 (A1) a 1 = (A1) b = 3 (AG) Therefore, gx x x =+ − 34 [6 marks] (b) Under reflection in the y -axis, the graph of is mapped onto the graph of yx x =− + 3 (M1) x =−− + − () () 3 (A1) i.e. y . xx 3 Under translation , the graph of is mapped onto the graph of F H G I K J 1 1 yx x 3 (M1) yh x = (1 )3 )1 =+ + + − (A1) =+ ++ + ++ xxx 2 33 1 36 3 1 (A1) hx 69 3 [5 marks] (c) The graph of is mapped onto the graph of , with point x 3 A mapped onto point , using the following combination of transformations: A (A1) Reflection in the x -axis (A2) followed by the translation . F H G I K J 2 0 (or vice versa.) [3 marks] [Total: 14 marks] & 7 & M00/510/H(2)M

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(G2) 3. (a) (i) Award (G1) for correct shape, including three zeros, and (G1) for both asymptotes (G1)(G1)(G1) (ii) ( ) 0 for 0.599,1.35,1.51 fx x == [5 marks] (b) is undefined for () (M1) xx 52 30 −= 23 (A2) Therefore, 03 13 or [3 marks] (M1)(A1) (c) 43 4 56 ( ) or 33 x  −− =   (A1) ′= = x x ( ) is undefined at and [3 marks] (R1) (d) For the x -coordinate of the local maximum of , where 0 1.5 put ( ) 0 xf x << = (M1) 3 0 x (A1) 1 3 6 5 x = [3 marks] (e) The required area is (A2) 1.35 0.599 ()d Af x x = Note: Award (A1) for each correct limit [2 marks] [Total: 16 marks] & 8 & M00/510/H(2)M ln 3 yx x =− asymptote asymptote
4. Note: In all 3 parts, award (A2) for correct answers with no working. Award (M2)(A2) for correct answers with written evidence of the correct use of a GDC (see GDC examples ). (a) (i) Let X be the random variable &the weight of a bag of salt±. Then , where is the new standard deviation. 2 ~ N(110, ) X σ Given , let P( ) . X <= 108 0 07 Z X = 110 (M1) Then P Z < F H G I K J = 108 110 007 . (M1)(A1) Therefore, =− 2 1476 . (A1) Therefore, = 1355 . GDC Example: Graphing of normal c.d.f. with as the variable, and finding the intersection with p . 0.07 = [4 marks] (ii) Let the new mean be , then .

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

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Mathematics HL - May 2000 - P2 \$ - INTERNATIONAL...

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