MARKSCHEME May 2011 MATHEMATICS Standard Level Paper 2

Transcription

1 M/5/MATME/SP/ENG/TZ/XX/M MARKSCHEME May 0 MATHEMATICS Standard Level Paper 7 pages

2 7 M/5/MATME/SP/ENG/TZ/XX/M SECTION A. (a) attempt to form composite f (x 5) h( x) 6x 5 N [ marks] (b) interchanging x and y evidence of correct manipulation x 5 y 5 6x, y 6 x 5 h ( x) 6 N3 Total [5 marks]. (a) Note: Award for approximately correct shape, for left end point in circle, for local maximum in circle, for right end point in circle. N4 [4 marks] (b) attempting to solve g ( x) marking coordinate on graph, xsin x 0 x 3.7 N [ marks] Total [6 marks]

3 8 M/5/MATME/SP/ENG/TZ/XX/M 3. (a) terms N [ mark] (b) evidence of binomial expansion n n r r a b, an attempt to expand, Pascal s triangle r evidence of choosing correct term 0 th 9 term, r 9,, ( x ) () 9 correct working 9 ( x ) (), x N [4 marks] Total [5 marks] 4. (a) 6 3 M 4 4, 5 N 5 A N3 (b) evidence of appropriate approach (M) X M N, attempting to solve a system of three equations 5 X 0 N3 (c) x 5, y 0, z N [ mark] Total [7 marks]

4 9 M/5/MATME/SP/ENG/TZ/XX/M 5. (a) Note: Award for label, for labelling [AU] 5 metres, for drawing [TU]. N3 (b) TAU 86 evidence of choosing cosine rule correct substitution x 5 36 (5)(36) cos86 x 4.4 N3 [4 marks] Total [7 marks]

5 0 M/5/MATME/SP/ENG/TZ/XX/M 6. (a) symmetry of normal curve P( X 5) 0.5 P( X 7) 0. N [ marks] (b) METHOD finding standardized value 7 5 evidence of complement p, P( X 7), 0.8 finding z-score z 0.84 attempt to set up equation involving the standardized value 7 5 X 0.84, 0.84 M.38 N3 [5 marks] METHOD set up using normal CDF function and probability P(5 X 7) 0.3, P( X 7) 0.8 correct equation P(5 X 7) 0.3, P( X 7) 0. attempt to solve the equation using GDC solver, graph, trial and error (more than two trials must be shown) A.38 N3 [5 marks] Total [7 marks]

6 M/5/MATME/SP/ENG/TZ/XX/M 7. METHOD evidence of antidifferentiation x (0e 5) dx y 5e x 5x C A Note: Award A for 5e x, for 5x. If C is omitted, award no further marks. substituting (0, 8) 8 5 C C 3 ( y x 5e 5x 3) substituting x y 34.9 ( 5e ) N4 [8 marks] METHOD evidence of definite integral function expression x f ( t )d t f ( x ) f ( a ), a x (0e x 5) 0 initial condition in definite integral function expression x t (0e 5)dt y 8 0, x x (0e 5)dx 8 0 correct definite integral expression for y when x x (0e 5)dx 8 0 (M) (A) (A) y 34.9 ( 5e ) A N4 [8 marks]

7 M/5/MATME/SP/ENG/TZ/XX/M SECTION B 8. (a) appropriate approach AO OB, B A AB N [ marks] (b) any correct equation in the form r a tb A N where b is a scalar multiple of r t, 4 t r t, r i j 5 k t ( i j k ) 5 t [ marks] (c) choosing correct direction vectors, 3 finding scalar product and magnitudes scalar product 3 ( 4) magnitudes ( ) (.73 ), 4 9 ( 3.74 ) substitution into uv u v accept uv, but not sin u v uv u v M cos 3 ( ) 3, cos N5 [7 marks] continued

8 3 M/5/MATME/SP/ENG/TZ/XX/M Question 8 continued (d) METHOD from r t 4 appropriate approach p r, t 4 s, two correct equations t s, t 4 s, 4 t 7 3s attempt to solve one correct parameter t 3, s C is (,, ) N3 [6 marks] METHOD from r t 5 appropriate approach p r, t 4 s, two correct equations t s, t 4 s, 5 t 7 3s attempt to solve one correct parameter t 4, s C is (,, ) N3 [6 marks] Total [7 marks]

9 4 M/5/MATME/SP/ENG/TZ/XX/M 9. (a) three correct pairs N3 (, 4), (3, 3), (4, ), RG4, R3G3, R4G (b) p, 6 q, 6 r 6 N3 (c) let X be the number of times the sum of the dice is 5 evidence of valid approach X B( n, p), tree diagram, 5 sets of outcomes produce a win one correct parameter n 4, p 0.5, q 0.75 Fred wins prize is P( X 3) appropriate approach to find probability complement, summing probabilities, using a CDF function correct substitution ,, , M probability of winning N3 [6 marks] Total [ marks]

10 5 M/5/MATME/SP/ENG/TZ/XX/M 0. (a) evidence of finding height, h h sin, sin evidence of finding base of triangle, b b cos, cos attempt to substitute valid values into a formula for the area of the window two triangles plus rectangle, trapezium area formula correct expression (must be in terms of ) cos sin sin, (sin )( 4cos ) attempt to replace sin cos by sin 4sin (sin cos ) M y 4sin sin AG N0 [5 marks] (b) correct equation y 5, 4sin sin 5 evidence of attempt to solve a sketch, 4sin sin (49.0 ),.5 (7.4 ) N3 [4 marks] (c) recognition that lower area value occurs at finding value of area at 4sin sin, draw square A 4 recognition that maximum value of y is needed A A 5.0 (accept 4 A 5.9 ) A N5 [7 marks] Total [6 marks]