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 4( 1) 0 1 0 y y y y ≥ − + − ≥ − − ≥ − ≤ 2 1 1 1 Since 1 is an asymptote, 1 1 y y y y ≤ − ≤ ≤ = − − < ≤ Alternative Method: 2 2 2 2 2 2 2 1 Let , : 1 (1 ) 1 ( 1) ( 1) 0 1 , 1 1 1 Since 0 , 0 1 1 1 x y x x y x x y x y y x y y y x x y y − = ∈ + + = − + + − = − = ≠ − + − ≥ ∀ ∈ ≥ + ∴− < ≤

– 1 1 x + – +

2 (ii) 2 2 2 2 1 ( ) p( ) 1 ( ) 1 1 p( ) for all (shown) x x x x x x x − − − = + − − = + = ∈

2(iii) Graph of 1 q( ) p 4 , 2 x x x   = − ∈    

is obtained from the graph of ( ) p x by: - Translation by 4 units in the positive x -direction, followed by Stretch of factor 2 parallel to the x -axis. 3(i) ( ) ( ) ( ) 1 2 1 f Let f ( ) D 0, y x k x k y x y k x k x x k − − = − − = ± = − + < = − + = ∞ Q 3(ii) [ ] g R 1, 4 = − ( ) f D , k = −∞ Since 5 k > , g f R D ⊆ . Thus fg exists. 3(iii) 2 fg( 1) f (0) k − = = Using [ ] g R 1, 4 = − , and the fact that f is a strictly decreasing function in the given domain, ( ) ( ) ( ) ( ) 2 2 fg 2 2 R 4 , 1 4 , 1 k k k k   = − − −     = − +   4(i) 2 2 1 3 2 z = + = 1 3 2 π arg π tan 1 3 z −     = − − = −             2 π i 3 2e z   −     = ( ) π 2 π i i 2 3 2 4 π i 2 3 π 2 π 4 π i 2 2 3 3 (8 ) π i 2 6 i e 2 e 2 e 2 e 2 e n n n n n n n n n z z     −           −       − +   −   −     −   = = = ( ) 2 i n z z is purely imaginary: (8 ) π cos 0 6 n −   =     (8 ) π π (2 1) , 6 2 5 6 , n k k n k k − = + ∈ = − ∈

Note: You may also have alternative form: (8 ) π π (2 1) , 6 2 11 6 , n k k n k k − = − ∈ = − ∈

smallest positive integer 5. n ∴ = Alternative Method: arg(i ) 2arg( ) arg(i) arg( ) 2arg( ) π 2 π 4 π 2 3 3 (8 ) π 6 n z z n n z z n n n − = + − = − + − = 4 (ii) ( ) ( ) ( ) 2 4 2 4 2 * 5 π arg 6 5 π arg 2 arg 6 5 π 2 π arg 2 6 3 13 π 6 π Since π arg( ) π , arg( ) (exact). 6 wz w w w z w z w w w = = =   = −     − − = −   = − −     = − < ≤ = 4(iii) 2 A O arg( ) z w − 2 B 1 π π π π π π 2 2 3 2 6 12 OAB       ∠ = − − + + =             Hence Method: π π arg( ) π 6 12 z w   − = − − −     5 π 1 5 π π 6 2 6 3 π (exact) 4       = − − −             = − w π 3 π 6 π 3 1 π 6 3 • • 1 − 3 − Re( ) z z

Otherwise Method: ( ) ( ) 1 3 1 3 i z w − = − − + − − π 3 π arg( ) π 4 4 z w   − = − − = −     5 Using similar triangles: 6 4 6 r h − = ( ) 2 6 3 r h = − 2 π V r h = ( ) ( ) ( ) 2 2 2 3 2 π 6 3 4 π 36 12 9 4 π 36 12 (shown) 9 h h h h h h h h   = −     = − + = − + For maximum V , d 0 d V h = : ( ) 2 4 π 36 24 3 0 9 h h − + = Using GC: 2 h = or 6 h = (Rejected as h = 6 is height of cone) Method 1 (1st derivative sign test) h 2 − 2 2 + Sign of dV dh + 0 - slope Thus, maximum volume 128 π 9 V = when h = 2 cm. Method 2 (2nd derivative test) ( ) 2 2 2 2 d 4 π 24 6 d 9 d 16 π When 2 : 0 d 3 V h h V h h = − + = = − < Thus, maximum volume 128 π 9 V = . ( ) ( ) 2 3 1 3 1 d d d . d d d 4 π 36 24(1.5) 3 1.5 (0.04) 9 0.12 π cm s (Ac cept: 0.377 cm s ) V V h t h t − − = = − + = 6(a)(i) 20 ( 1) 350 65 19 15 u a n d d d = + − = + = 6(a)(ii) 20 20 (65 350) 2 4150 cm (Accept: 41.5 m) S = + =

6(b)(i) 8 1 9 9 integer 9. a S a k ∞ = − = ∴ = 6 (i) Method 1: Number of ways 14 3! 2184 3   = × =     Method 2: Number of ways 14 13 12 2184 = × × = 6(b)(ii) ( ) ( ) ( ) ( ) ( ) 2000 8 423 1 9 2000 8 1 9 2000 8 1 9 3807 1807 8 9 3807 1807 ln 3807 8 ln 9 6.3267 Largest integer 6. n n n n S n n n ≤   −     ≤ − − ≤ ≥ ≤ ≤ ∴ = Length of shortest plank is 6 1 6 8 423 9 235 cm (3 s.f.) u −   =     = 7(i) 2 1 1 1 1 2( 1) 2( 1) r r r = − − − + 2 1 1 1 1 ( 1) 2( 1) 2( 1) 1 1 1 2 ( 1) ( 1) r r r r r r r r r   = −   − − +     = −   − +   7 (ii) 1 1 1 ... ( th term) 2 3 3 8 4 15 n S n = + + + + × × ×

1 2 2 1 2 1 ( 1) 1 1 1 2 ( 1) ( 1) 1 1 1 2 2 1 2 3 1 1 3 2 3 4 1 1 4 3 4 5 1 1 ( 1) ( 2) ( 1) 1 1 ( ) ( 1) ( 1) 1 1