Paper D Sec 3 IP solutions for students

1 Paper D Sec 3 IP Exam Solutions for students Qn Solutions 1(a)    1 1 1 3 1 3 1 3 1 3 1 3 1 3      2 113  1(b) 2 2 2 339 3 ( 3)( 3 9) xxx x x x x      = 22 23 ( 3) ( 3 9) ( 3)( 3 9) x x x x x x x       = 2 5 ( 3)( 3 9) x x x x     1(c) Sum of roots =    20 17 20 17 40 Product of roots =       20 17 20 17 400 17 383 The equation is 2 40 383 0xx   Or 2 2 ( 20 17)( 20 17) 0 20 17 20 400 20 17 17 20 17 17 0 40 383 0 xx x x x x x xx                  2(a) 2( ) 1 3 1x y x y x y       ……….(1) 221 2 11x y x y    ……….(2) Sub (1) into (2): 22(3 1) 1 2(3 1) 11y y y y      27 16 4 0yy   22 7y or y   15 7x or x   125, 77x , y = 2 or x y    2 (b) y x m meets the curve 2 7y mx x 0 2 60mx x m   226 4 0Meet m   , 2 9m 33 m   3(i) ( ) ( 4)( 2)g x x x   or   2 19x   42 12a  

2 3(ii) 3(iii)   22 2 8 1 9Let y x x x      .   2 91yx    19xy    9 1,xy    1 accept 9 1x x y     1 : 9 1g x x   , 9x 4(i) p(1) 0 5ab   ……….(1) p( 2) 0 2 4 0ab    ……….(2) Sub (1) into (2), 3, 2ab 4(ii) 4 3 2 ( ) 3 2 2 4let p x x x x x     2( 1)( 2)( 2)x x x Ax     Comparing coefficient of x, 2 ( 2) ( 2 ) 2 A     2A Third factor is 2( 2 2)xx 4(iii)   22( 2 2) 1 1 1 0x x x       Hence, this factor is positive for all value of x. Alternative mtd: Discriminant =-4 < 0 4(iv) For ( ) 0, ( 1)(x 2) 0 2 or 1p x x x x       5(a) 55log ( 1)( 2) log 6xx   22 3 2 6 3 4 0x x x x       ( 1)( 4) 0 1( ), 4x x x rejected x      4x 5(b) log log 3 log 6 2,a a axx   log (3 ) log 6 2aaxx  23log 2 6 a x

3 22 1 2ax 222ax 2xa  Since 0a , 0x , 2xa 6(a) | 2 | 32 2 3 or 2 322 3 1 522 2 -10 (reject)3 x x xx xx xx xx                6(b)   28 3 2 2 3x x x     322 3 2 2 3 0x x x     Let 2x be y .   32 3 3 0y y y     Let 32( ) 3 3f y =y y y   . (3) 27 27 3 3 0f=      3y is a factor of   32 3 3 0 y y y    Let 2 3 2( ) ( 3)( 1) 3 3f y y y Ay y y y        . Comparing coeff of 2y , 0A 2( ) ( 3)( 1) ( 3)( 1)( 1)f y y y y y y        3, 1, 1y y y    2 3, 2 1, 2 1x x x     ln 3 0ln 2x or x or na   1.58 0x or  7(i) 7(ii) 7(iii) 4 8(a) RHS = sec² x cosec² x = (tan² x+1)(1+ cot² x) = (tan² x+1)+(1+ cot² x) == sec² x+ cosec² x or LHS= sec² x+ cosec² x = 22 11 cos sinxx 22 22 2 2 2 2 sin cos 1 sec coscos sin cos sin xx x ec xx x x x   

4 8(b) 3 sin x tan x + 8 = 0 sin3sin 8 cos xx x  23sin 8cosxx  23(1 cos ) 8cosxx   23cos 8cos 3 0xx    (3cos 1)( cos 3) 0xx     1cos or cos 33xx   (na) 109.5 or 250.5x 9(i) Angle CAD1 = 30° (alt angle) Angle D1AT = 102°−30° = 72° Bearing of T from A = 180°−72° = 108 9(ii) 150 sin 41 sin 37 AC 137.6AC m 9(iii) 58tan 70.5 20 h 58 20 tan 70.5 1.52h    10(i) By Pythagoras theorem, Adjacent side = 2 1p  x 1 p N N C T 37° 30° 102° A D D1

5 2 1 cos( ) cos p p x x     10(ii)     11sec 90 sincos 90 x xx      = p 11(i) 11(ii) 11(iii) 11(iv) (i) 0 8 12 10 8 0 4 7 12 4 0 9 10 7 9 0       (ii) 1 1 1 1       (iii) 30 19 25 26       school B (iv) 3460 1750 2330 2860       school B 12(i) 12(ii) (a) general formula, to find solution of 20 0.43 1.712 5.183xx   . ie. 21.712 (1.712) 4( 0.43)(5.183) 2( 0.43)x      (b) checking of discriminant, if   0, then it will have intercepts. Or (b) Graphing using Desmo, Visually see if there is an x intercepts or alternatives 13(i) Gradient of AD = 70 192   2yx or 2yx   13(ii) Radius = 22(9 5) (7 11) 32 or 4 2    units 22( 5) (y 11) 32x    or 22 10 22 114 0x y x y     13(iii) Area of triangle CAB= 5 9 15 51 11 7 18 112 = 11(35 162 165 99 105 90) (68) 3422        unit2 13(iv) 226 11 157AB   units or 12.53 units 1 ˆ( A)(AB)sin AB 342 CC  34 2ˆsin AB 32 157 C  ˆAB 73.6C

6 Alternative mtd : Cosine rule Using 149BC , 157BA , 32AC  BC = 12.21 units, BA =12.53 units, AC = 5.65 units 14 3 ( 3)( 1)( 2) 76 y x x x xx        15(i) 2y ax bx c   1712y ax bx    15(ii) Refer to graph Paper Line of best fit Points plotted, Label Axes Table of Points 15(iii) 4.2 64yx  Accept 4.2 0.4a  and 64 5b Note: Deduct only 1M (iii)&(iv) for inaccurate range due to poor line of best fit if correct method use for these three answers 15(iv) Accept 1585 40

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9 (i) Correct translation of 2 units to the left 1M Correct translation of 3 units up 1M (ii) Correct reflection about the x-axis 1M Correct stretch along x axis by factor ½ or correct compression along x-axis by factor 1/2 1M