TPJC_H2_MATHS_P1_ANS

ANNEX B TPJC H2 Math JC2 Preliminary Examination Paper 1 QN Topic Set Answers 1 Equations and Inequalities x < −3 2 Graphs and Transformation (i) 1 1 y − < ≤ (iii) Translation by 4 units in the positive x - direction, followed by -Stretch of factor 2 parallel to the x -axis. Alternative Answers: Stretch of factor 2 parallel to the x -axis, followed by Translation by 8 units in the positive x - direction 3 Functions (i) ( ) 1 1 f f ( ) D 0, x x k − − = − + = ∞ (ii) [ ] g R 1, 4 = − ( ) f D , k = −∞ Since 5 k > , g f R D ⊆ . Thus fg exists. (iii)(a) 2 fg( 1) f (0) k − = = (b) ( ) ( ) ( ) ( ) 2 2 fg 2 2 R 4 , 1 4 , 1 k k k k   = − − −     = − +   4 Complex numbers (i) smallest positive integer 5. n ∴ = (ii) 2 w = , ( ) 13 π arg 6 w = (iii) Hence Method: π π arg( ) π 6 12 z w   − = − − −     5 π 1 5 π π 6 2 6 3 π (exact) 4       = − − −             = − Otherwise Method: ( ) ( ) 1 3 1 3 i z w − = − − + − − π 3 π arg( ) π 4 4 z w   − = − − = −    

5 Differentiation & Applications 128 π 9 V = d d V t = 3 1 0.12 π cm s − 6 AP and GP (a)(i) 15 d = (ii) 20 S = 4150 cm (b)(i) 9 k = (ii) n = 6, Length = 235 cm 7 Sigma Notation and Method of Difference (ii) 1 1 4 2( 1)( 2) n n − + + (iii) As n → ∞ , 1 0. 2( 1)( 2) n n → + + 1 1 1 4 2( 1)( 2) 4 n n − → + + Sum to infinity 1 4 = (iv)13 8 Differential Equations (i) 3 3 e 1 e 2 kt kt x − = + (ii)1.45 hours (iii) 1 1 sin 2 2 x t t = − (iv) O The graph shows that as time increases, the drug concentration still continue to increase / the curve shows a strictly increasing function beyond the maximum level of drug concentration. 9 Application of Integration (i) 64 π (iv) The reflected light from the bulb produces a horizontal beam of light/ produces a beam of line parallel to x - axis.

(v) 2 4( 1) y x = − 10 Vectors (ii) 3 5 , 3, 2 2       (iii) ( ) 0, 3, 2 (iv) θ = 80.4 ° , 49.8 ° (v) 14 14 2 3 or 2 3 2 2 x y z x y z √ √ + − = − + − = (vi) 6 3.79 units cos 49.8 BD √ = = ° (vii) 60 °

H2 Mathematics 2017 Preliminary Exam Paper 1 Solutions 1 2 2 2 2 3 7 1 2 1 3 3 7 1 (2 1) 0 3 3 7 1 (2 1)( 3) 0 3 2 4 0 3 x x x x x x x x x x x x x x x x + + < − + + + − − < + + + − − + < + + + < + 2 ( 1) 3 0 3 x x + + < + Since 0 3 ) 1 ( 2 > + + x for all real x , the inequality reduces to: x + 3 < 0 ⇒ x < −3 2 2 2 2 2 2 1 Let , : 1 (1 ) 1 ( 1) ( 1) 0 x y x x y x x y x y − = ∈ + + = − + + − =

2 2 2 Discriminant 0 : 0 4( 1)( 1) 0