presbyterian high school 2023 A math paper 1 answer key

Name: Index No.: Class: PRESBYTERIAN HIGH SCHOOL ADDITIONAL MATHEMATICS 4049/01 Paper 1 18 August 2023 Friday 2 hours 15 min PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL 2023 SECONDARY FOUR EXPRESS / FIVE NORMAL (ACADEMIC) PRELIMINARY EXAMINATIONS MARK SCHEME Cynthia – Q1 to 10 Sabrina – Q11 to 13

2 1 The line 2 15yx=+ intersects the curve 2 63y x x= + + at points A and B. Find the value of p for which the distance AB can be expressed as 5.p [5] 2 A curve is such that 2 2 2 d 12e e . d xxy x −=+ The curve intersects the y-axis at (0, 5)P and the tangent to the curve at P is parallel to 4 3.yx=+ Find the equation of the curve. [6] 2 6 3 2 15x x x+ + = + M1 (equate curve to line) 2 4 12 0xx+ − = ( )( )2 6 0xx− + = M1 (factorise) 26x or x= =− 19 3y or y== M1 (find y) ( ) ( ) 22 2 ( 6) 19 3AB= − − + − M1 (apply distance formula) 320AB= 85AB= p = 8 A1 ( ) 22 1 d 12e e d 6e ed x x x xy xcx −−= + = − + M1 (any 2 correct terms) At ( )0,5 , d 4d y x = M1 (seen gradient at P = 4) 2(0) (0) 16e e 4 c−− + = M1 (sub. gradient at x = 0, attempt to find c1) 1 1c =− ( ) 226e e 1 d 3e ex x x xy x x c −−= − − = + − + M1 (any 2 correct terms) At ( )0,5 , 2(0) (0)3e e 0 5 c−+ − + = M1 (sub. x = 0 & y = 5, attempt to find c) 1c= 23e e 1xxyx − = + − + A1

3 3 A function is defined by 2f ( ) 2 2 3x x kx k= + + + for all real values of x, where k is a constant. (a) Find the discriminant of f ( )x in terms of k. [2] (b) Show that the discriminant of f ( )x in part (a) can be expressed in the form ( ) 2 4, k a b−− where a and b are integers. [2] (c) Find the range of values of k for which f ( ) 0x = has no real roots. [3] For 2f ( ) 2 2 3x x kx k= + + + , 22 4 (2 ) 4(1)(2 3)b ac k k− = − + M1 (apply discriminant) 24 8 12kk= − − A1 2 2 2 24 8 12 4 2 1 1 12k k k k − − = − + − − ( ) 2 4 1 1 12k= − − − M1 (completing the square) ( ) 2 4 1 16k= − − A1 2 40b ac− ( ) 2 4 1 16 0k− −  M1 (apply discriminant < 0) ( ) 2 21 2 0k− −  ( 1 2)( 1 2