PAPER I Sec 3 solutions for students

1 PAPER i Sec 3 Solutions for students Qn Solutions 1 3 3( 3 ) 2 84 22 39 2 ( 1 ) 231 ( 2 ) 3 Fr(2) : 69 3 ( 3 ) (1) (3) : 95 5 9 15 929 1 27 xy xy xy xy xy x x y y                    2(a) 43 216 24 6 463 36(6) 6(4) 6 23(6) 6 2 6 6 3 261 8 2 3 216 63 50 6 3         

2 2(b)               21 2 21 2 2 2 85 5 25 . 85 5 5 05 8 55 1 5 05 8 5 255 5 0 let 5 8 255 0 8 255 0 2550, 8 25550 ( . ) , 5 8 255lg 8 lg 5 2 . 1 5 ( 3 . ) xx x xx x xx xx x xx a aa aa aa rej x sf              3(i)     33 2 21 2 8 2 6 4 2 4 4 16 xx xx x      3(ii)            3 3 2 2 2 2 2 21 2 8 4 21 2 8 4 0 24 4 1 6 4 0 24 4 1 6 4 0 42 4 1 6 1 0 42 8 3 3 0 4, 2 8 33 0 86 4 4 ( 2 ) ( 3 3 ) ( ) 4 xx xx xx x x xx x x xx x xx x xx x x reject                           

3 4(a)    2 2 31 6 1 2 3 2 0 31 3 1 0 0 32 50 25 3 xx x xx xx xo r x           4(b) 2 2 ax byb a x y        22 2 2 22 2 2 22 2 2 22 2 2 22 2 2 22 2 22 2 2 452 0 0 , 45 2 0 0 ,2 245 2 0 0 4 16 5 2 80 0 16 5 10 5 80 0 discriminant, 4 0 100 4 16 5 5 80 0 100 4 80 1280 25 400 0 320 5120 xy ax bx ax a x b bx xa x a x b b a x abx b ba c ab a b ab b ab a b                        2 22 22 1600 0 320 5120 1600 16 5 (shown) a ba ba     5(a)     2 2 2 54 51 6 10 25 16 0 26 25 0 25 1 0 25, 1( ) xx xx xx x xx xx x x reject         

4 5(b)    2 3 3 3 3 3 2 2 33 2 log 2 log 3 5 log 32l og 2 5 log let log , 22 5 25 2 0 21 20 0.5, 2 log 0.5, log 2 3, 3 9 xx x x xa aa aa aa aa xx xx               6(i) 27x x  6(ii) 27 ( 1 ) - ------- ( 2 ) 27 37 7 3 yx yx xx x x         27 ( 1 ) (2) 27 7 yx yx xx x          

5  172 3x  Method 2 22 22 2 (2 7 ) () 42 8 4 9 32 8 4 9 0 7 or 73 xx x xx xx xx         172 3x  Method 3 27 37 7 3 27 7 x x x x x x x         172 3x  7(a)      2 2 2 2 91 8 At 3,0 : 0 (36) 18 1 2 1 91 82 1 18 81 182 14 5 922 ya x a a yx yx x xx               

6 Method 2 2 2 93 0 ( 1 ) 81 9 18 (2) 169 13 10 (3) solve simultaneous equations 14 5,9 ,22 14 5 922 ya x b x c ab c ab c ab c ab c yx x                   7(b) (i) 32 2 0 20 3 2 x r xr     7(b) (ii) Area of triangle 2 1 () s i n 6 02 3 4 xx x   Area of circle  2 2 20 3 2 20 3 4 x x        2232 0 3 4 x xA    

7 7(b) (iii)      22 22 2 2 22 2 32 0 3 4 3 400 120 9 4 1 9 3 120 4004 1 120 400934 9393 93 60 60 400 4 93 93 93 93 6 0 4 93 9 3 3600 4 (9 3 ) xxA xx x xx xx x x                                                  2 2 400 93 93 6 0 11.9944 93 x              Hence, minimum total area 212.0m (3s.f.) Method 2: 2 2 22 2 2 1.1492 9.5493 31.831 1.1492( 8.3095 27.698) 1.1492[( 4.1548) (4.1548) 27.698] 1.1492( 4.1548) 11.99 min. area = 12.0 m xx xx x x        8(i) coordinates of S = (x, y)   33 ,1 , 422 33 1, 422 1, 5 :1 , 5 xy xy xy S         

8 8(ii)      22 22 22 22 radius 4 3 1 3 5 units equation of circle is 1 4 5 21 81 6 5 281 2 0 xy xx yy xy xy          8(iii) ) 53 grad 12 2 54 11 1 2 121 2 , , w 90 . . (sho n RS grad QS gradRS gradQ RS is a tange e S Hence R nt to th ci R r SQ S S Q cle               8(iv)     2 A 1 re 30 2 1 31 32 3 532 6 0 10 3 0 4 3 152 1 a of quadrilateralP 782 7.5 unit R s TS             Method 2:  2 11(5)(1) 5 (2) A 2 rea of quadrilatera R 2 7. lPT 5u nt S is   9(i) f( ) 2 xx e Step 1: Reflect the graph of f( ) 2 xx e along x-axis to get ( ) 2 xhx e  Step 2: Translate 5 units in the positive y-direction, the graph of () 2 xhx e  to get the graph of   52 xg xe .

9 9(ii) 0 0 , 52 0 25 5 2 5ln 2 ln 2.5 0 , 52 3 x x x when y e e e x x when x ye          y-intercept: (0, 3) x-intercept: (ln2.5, 0) asymptote: y = 5 9(iii) 6ln 4 6 4 62 2 23 2 52 2 2 22 2 2 x x x x x x xe xe xe xe xy xy              

10 10a 23 43A  21 35B  23 2 133 43 3 5 69 2 1 12 9 3 5 41 0 15 4 AB                      10b (i) 290 240 200 235 362 117C  10b (ii) 240 95 110 D   10b (iii) 240290 240 200 95235 362 117 110 69600 22800 22000 56400 34390 12870 114400 103660 EC D        10b (iv) The elements in matrix E represents the total profit made for selling 290 air con units, 240 units of water heater and 200 units of humidifier in 2021 and 235 air con units, 362 units of water heater and 117 units of humidifier in 2020 respectively. 11(i) f( 2) 0 f(3) 185 82 8 24 0 8 2 24 (1) or 4 12 (1) 27 63 3 4 185 27 3 126 (2) or 9 43 (2) (1) (2) : 5 30 6 12 ab ab a b ab ab a b a a b                       