MI H2 Pre-U 1 Promo 2023 (Solutions)

1 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 2023 MI H2 Math Pre-U 1 Exam Solutions Qn Solution 1(i) [2] When 2n ,   1 22 22 41 4 1 42 1 4 4 23 nnnuSS nn n n nn nn n n         When n = 1, 2 11 14 ( 1 ) 5 2(1) 3 uS   Thus, 1u follows the form of 2 3nun when n = 1. So, 2 3nun 1(ii) [2]  1 23 2 (1 ) 3 23 21 2 nnuu n n nn        Since 1 2nnuu  is a constant independent of n, the sequence is an arithmetic progression. 2(i) [3] Let $x, $y and $z be the price of a short, tall and grande cup of coffee respectively. 3 12 8 147.9 4 8 7 120.10 2547 0 Using GC, 4.50, 6.20, 7.50 xy z xyz xyz xyz      The price of a short, tall and grande cup of coffee is $4.50, $6.20 and $7.50 respectively. 2(ii) [2] The required amount 0.9 4.50 4 6.20 2(7.50) 41.37 Andrew pays $41.37.    

2 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 3(i) [3] 3(ii) [3]      2222 22 22 2 2 1 1 I: replace by /4 1 1 4 1 1 II: replace by 2 1 1144 2 2 1 1 III: repl4 xxy x x y xy x yy y x y                      2 2 2 2 4 ace by 2 2 2 1 14 23 1x xyy y y          O (െ√2, 0) x y (√2, 0)  1y x  1yx    22 11xy   (0, െ1)

3 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme   4 (ii) [3] 5(i) [2] 1 , and are collinear, is the common point (shown) OA OB OC OA OB OB OC OA OB OC OB BA BC AB C B                                   Qn Solution 4 (i) [3] x y 𝑦ൌ 3  ሺ4, 2ሻ ሺെ4, 2ሻ O x = 2𝑥ൌെ 2 f| | 1yx ሺ0, 2ሻ (4, 1) 2 𝑦ൌ 1 2  𝑥 𝑦 𝑥ൌ 1 𝑂   1 fy x (0, 1)

4 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 5 (ii) [3] Given 1 6  , 51 66OA OB OC       65 65              OA OB OC OC OA OB   Method 1: Sub and Eliminate OC    By Ratio Theorem, 4 5 65 4 5 2 01 2 2 1 02 1 5 2 OC OAOD OA OB OA OD OD OA OB OD OD                                       Method 2: Find OC   and Sub in 015 65 6 2 5 1 1 7 02 1 0       OC OA OB              By Ratio Theorem, 4 5 50 1 17 4 2 5 10 0 1 5 2          OC OAOD OD OD            

5 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 5 (iii) [3] 101 12 3 20 2 10 1 12 1 202 AB AE                           Area of the triangle ABE 2 1 2 11 1 31 2 22 8 1 42 2 4 2 1 21 units AB AE                       5 (iv) [2] Length of projection of OE    onto  AB  .  11 1 13 14 22 1 6 14 31 4 units 7 ABOE AB                 

6 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme Qn Solution 6(i) [4]          2 2 2 39 8 21 39 8 21 01 37 6 01 33 2 01 33 2 1 0 xx x xx x x xx x xx x xx x             23 or < 13x x  6(ii) [2] 239 8 21 xx x     Replace x with ex 23e 9e 8 2e1 xx x    2e 3 or < e 1 3 2no solution ln < ln13 2ln < 03 xx x x     –3 2 3 1

7 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme Qn Solution 6(iii) [3] 2 2 2 2 2 2 2 2 39 8 21 1Replace by 1139 8 21 1 39 8 1 2 39 8 21 39 8 2 So, 12 1 3 or < 13 13 0 or 1 32 xx x x x xx x xx x xx xx x xx xx xx xx xx                    7(i) [3] 21 4 22 5 13 2 43 4 57 5 21 2 4 54 5 2 n r r                          Let x y z   r , 4524 5xyz (shown)

8 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 7(ii) [3] Line BF: 24 05 , 42               r  24 05 f o r s o m e 42   OF                Since OF   lies on  ,     4 54 5 2 24 4 05 5 4 5 42 2 4 2 45 52 4 2 4 5 45 45 1     OF                               246 055 42 2 OF                7(iii) [2] ' 2 '2 62 25 0 24 10 10 0 OB OBOF OB OF OB                     B B’ F

9 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme 7(iv) [3] Method 1 23 1 07 7 41 3 AB               10 3 7 '1 0 7 3 01 1 AB           Let angle BAB’ ൌ𝜃 1 1 1 'cos ' 17 1cos 7 3 59 59 31 31cos 59 59 121.697 121.7 (1 d.p.)      AB AB AB AB                              Method 2 23 1 07 7 41 3 AB               Let angle BAB’ ൌ𝜃 1 1 1 2sin 14 12sin 7 5 59 45 32 452sin 59 45 121.697 121.7 (1 d.p.) n n     AB AB                              B B’ F A 𝜃

10 © Millennia Institute 9758/01/PU1/EOY/23 Mark Scheme Method 3 Let angle BAB’ ൌ𝜃 23 1 07 7 41 3 AB               63 3 57 2 21 1 AF                1 1 1 2c os 13 12c os 7 2 59 14 31 142c os 59 14 121.697 121.7 (1 d.p.)         AB AF AB AF                             8(i) [2] R g =  ,0 0 ,   . Df = ,  . Since Rg  Df , fg exists. 8 (ii) [2]    2 2 1 1 11 2511 fg f g 2 51 f 1 1 xx x xx xx             fg gDD , 11 ,   