Practice Paoer C Sec 3 Math Marking Scheme

3 Practice Paper C Sec 3 Math Marking Scheme Q1(a) 1 1 1 1 3 3 3 3( )( )x y x y  2 1 1 2 2 1 1 2 3 3 3 3 3 3 3 3( )( )x x y y x x y y    M1 = (x – y)(x +y) A1 = 22xy Q1(b) x a x a x a x a x a x a x a x a             ( ) 2 ( ) ( ) 2 ( ) ( ) ( ) x a x a x a x a x a x a x a x a x a x a                 M1 2 2 2 2 2 x a x a a    M1 2x a A1 Award M1 for rationalising y Q1(c) 5 5 6log 1 logx x M1 2 55(log ) log 6 0xx    M1 55(log 3)(log 2) 0xx   55log 3 or log 2xx  1 or 25125xx A1 Q2 Given that 3tan is in the 1st & 3rd Q4AA 4sin is in the 3rd & 4th Q5BB  M1 When A is in the 1st Q, 3sin 5A , When A is in the 3rd Q, 3sin 5A . M1 When B is in the 3rd Q, 4tan 3B , When B is in the 4th Q, 4tan 3B . M1 When A and B are in different quadrants : Case (i) A is in the 1st Q and B is in the 3rd Q, 3 4 18 20 22sin tan 2( ) 5 3 15 15AB      A1 Case (ii) A is in the 1st Q and B is in the 4th Q, 3 4 18 20 82sin tan 2( ) ( ) 2 5 3 15 15AB       A1 Case (iii) A is in the 3rd Q and B is in the 4th Q, 3 4 18 20 22sin tan 2( ) ( ) 5 3 15 15AB        A1 -1 for additional case Q3(i) 11 1 1.5 1 4 and 1 2 1 5            AB B2 (ii) (ii) 1 1.5 4 0 1 4 1 2 6 2 1 5                 M M1FT 5 3 1 4 8 4 1 5              M1 FT 25 4 12   A1 Q4(a) 22 7 6 0x h hx x hx x h       As the line meets the curve i.e. 2 40b ac M1 22 6 4 0 h   2 9h  M1 3 3, 0hh    A1 No M1 for 2 40b ac No M1 for 2 40b ac (b) As y is always positive and 0a 2 4 0b ac   where 1, 4 and 1a b k c    2( 4) 4 0k   M1 2 8 12 0kk   ( 2)( 6) 0kk   M1 2 6 k   A1 No marks for this question if student considers 2 40b ac Q5 As 2 6 ( 2)( 3)x x x x     M1 Let x = 2 and subs into the eqn. M1 13 (1)ab    Let 3x and subs into the eqn. 4 67 (2)ab    M1 Solve the above two equations, a = 16 and b = 3 A1 Method marks are awarded for substituting 2x and 3x Q6(a) ( )( ) ( )( ) ( )( ) ( )( )( ) a b a b b c b c c a c a a b b c a c            M1 2 2 2 2 2 2 ( )( )( ) a b b c c a a b b c a c         M1 0 ( )( )( )a b b c a c    = 0 A1 Students should simplify to obtain 0 as the answer

4 (b) 2 2 2( 3 2) ( 5)( 4) ( 1)( 5) 3( 2 8) x x x x xx xx      2( 1)( 2) ( 5)( 4) ( 1)( 5) 3( 2)( 4) x x x x x x x x         M1 M1 2 3 A1 If constants “2” and “3” are missing from factorisation, award 2 method marks Q7 [G1] for correct period [G1] for correct amplitude [G1] for smoothness and correct shape of the graph Q8(a) sin 3 sin(90 2 )xx  M1 3 90 2 , 180 (90 2 ), x x x     M1 360 (90 2 ), 720 (90 2 )xx    18 , 90 , 90 , 162x      Hence 18 , 90 , 162x    A1 Award 2 method marks if students obtained 18x (b) Let tan(2 40 ) 2.718 69.800 (B.A.)xx      M1 Hence 2 40 180 69.80 ,x    360 69.80  M1  75.1 , 165.1x  A1 Deduct A1 for additional x values Q9(a) 2 2 2 2LHS=sin (sin cos ) cosx x x x  M1 22sin cos 1 RHSxx    M1 Alternatively 2 2 2 2 2LHS=(sin ) (1 cos )cos cosx x x x   2 2 2 2 2(1 cos ) (1 cos )cos cosx x x x     M1 242 42 1 2cos cos cos cos cos xxx xx      M1 Alternatively 2 22 22 22 22 1 2sin 12 sin sinLHS= 1 2cos 1 2sin cos sinsin sin sin cos sin 2sin cos cos x xx x x x xxx xx x x x x      M2 2 (sin cos )(sin cos ) sin cos sin cos(sin cos ) RHS x x x x x x xxxx      M2 Q10(a) 22 x kxzz yy   M1 When x = 2 and y = 4 then 6z , 2262 4 k k 3k A1 Hence 23xz y  . Therefore 2325 9 x M1FT 5 or 5 (rej)xx   A1 (b) Let 11 2 , 4 yx x y 22 1 (2 ) 8then 8 4 k x kxzz yy    M1FT Acceptable answers: Hence z is increased to 800% Hence z is increased by 700% % change is 700% A1 Q11 Given 2 3 2xx 22 3 2 or 3 2x x x x     M1 Case 1 : 2 32xx or 2 2 3 0xx   ( 3)( 1) 0xx   10 8 6 4 2 2 4 6 8 10 7π 3 2π 5π 3 4π 3 π 2π 3 π 3 π 3 2π 3 π 4π 3 5π 3 2π 7π 3 f x( ) = 1 4∙cos 2∙x( )

5 1 RHS (b) 2 2 2 (1 cot )LHS=1 cot 2cot x xx   M1 2 2 1 cot (1 cot ) x x  