2022 J1 RVHS End of Year Revision Package Solutions Ch1-4

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 1 RIVER VALLEY HIGH SCHOOL 2022 JC 1 H2 Mathematics (9758) End-of-Year Revision Package (Solutions)

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 2 1 Inequalities & Equations Basic Skills 1. Find the discriminant of the following quadratic expressions. (a) 2 1x x  (b) 2 4 5x x  (c) 2 1x x   (d) 22 4 3x x   (e) 2 5 2 9 x x  (f) 2 63 xx   (a) 2 1x x   21 4 1 1 3 D   (b) 2 4 5x x   24 4 1 5 36 D    (c) 2 1x x    21 4 1 1 3 D     (d) 22 4 3x x    24 4 2 3 40 D     (e) 2 5 2 9 x x    2 1 51 4 2 9 19 9 D           (f) 2 63 xx      21 4 1 63 1249 D        2. Find the exact roots of the following quadratic equations. (a) 2 3 2 0x x   (b) 2 2 3 0x x    (c) 2 2 1 02 x x    (d) 22 1 03 4x x   (e) 2 1x x     (f) 2 6 3 02 x x   (a) 2 3 2 0x x   2 1 0 1or 2 x x x      (b) 2 2 3 0x x    1 3 0 1or 3 x x x      (c) 2 2 1 02 x x     12 4 4 1 2 12 2 2 2 2 21 x                 (d) 22 1 03 4x x   2 11 1 4 3 4 22 3 5 51 3 33 3 4 4 3 3 15 4 x                       (e) 2 1x x          2 2 2 1 0 1 1 4 1 2 1 1 4 4 2 1 2 1 2 1 2 1 1 2 1or2 2 11or x x x                                      (f) 2 6 3 02 x x    16 6 4 32 12 2 6 x            

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 3 Tutorial Review Tutorial 1 Questions 4 and 7. Revision Questions 1. NYJC Promo 9758/2020/Q2 (i) The first four terms of a sequence nu are given by 132.1u, 217u, 30.7u and 47.8u. Given that nu is a cubic polynomial in n, find nu in terms of n. [3] (ii) Find the least value of n for which nu is greater than 555. [2] (i) Let 3 2nu an bn cn d    132.1u a b c d     28 4 2 17u a b c d     327 9 3 0.7u a b c d     464 16 4 7.8u a b c d     Using GC, 1.5a, 9.6b, 3.2c, 37d 3 21.5 9.6 3.2 37nu n n n     (ii) 3 2555 1.5 9.6 3.2 518 0nu n n n      Method 1 From GC graphing, 9.7871n  least value of n10 Method 2 From GC Table,  least value of n10

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 4 2. YIJC Promo 9758/2020/Q1 Four families, namely Chan, Lee, Tan and Wong, purchase masks, hand sanitisers and thermometers at a pharmacy. The Tan family made their purchase using a voucher, which entitled them to a 20% discount off the total amount paid. The quantity of each item purchased and the total amounts paid are shown in the following table. Masks (in boxes) Hand sanitisers Thermometers Total Amount paid Chan 4 6 5 $89.30 Lee 2 4 8 $83.30 Tan 3 5 3 $50.28 Wong 5 2 4 Calculate the total amount the Wong family paid. [4] Solution Let x, y and z be the selling price of a box of mask, hand sanitiser and thermometer respectively before discount. Amount that Tan family would have paid before discount (50.28)10080 $62.85    4 6 5 89.3 (1)x y z    2 4 8 83.3 (2)x y z    3 5 3 62.85 (3)x y z    From GC, 11.25, 1.80, 6.70x y z   Amount paid by Wong family 5 $11.25 2 $1.80 4 $6.70 $86.65       

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 5 3. 2017/Prelim/NJC/P2/Q1 There are 3 bike-sharing companies in the current market. For each ride, - bike charges a certain amount per 5 min block or part thereof, - bike charges a certain amount per 10 min block or part thereof and - bike charges a certain amount per 15 min block or part thereof. Rebecca rode each of the bike-sharing companies’ bikes once in each month. The table below shows the amount of time Rebecca clocked for each ride and her total spending for each month. In celebration of the company’s first anniversary, the pricings in February and March 2017 of - bikes are a 5% discount off the immediate previous month’s pricing. Determine which bike-sharing company offers the cheapest rate (without any discount) for a 40-min ride. Justify your answer clearly. [4] January 2017 February 2017 March 2017 - bike 25 min 17 min 36 min - bike 30 min 10 min 39 min - bike 15 min 44 min 33 min Total spending $5.70 $5.72 $9.71 Let , and    be the original amount charged per 5 min, 10 min and 15 min block for each ride by - bike, - bike and - bike respectively.   2 5 3 5.7 ---------- (1) 4 3 0.95 5.72 ---------- (2) 8 4 3 0.95 9.71 ---------- (3)                   Solving the above 3 equations simultaneously by GC, $0.4079329609, $0.8402234637, $1.139664804     $0.41, $0.84, $1.14     . Original pricing per 40-min block: Using calculator values - bike: $0.4079329609 8 $3.26  - bike: $0.84 4 $3.36  - bike: $1.14 3 $3.42  Thus, - bike offers the cheapest rate for a 40-min ride.

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 6 4. 2016/Promo/RI/3 Do not use a calculator in answering this question. Solve the inequality 241.2 3xx x   [4] 2412 3xx x   224 2 302 3x x xx x        2101 3x xx x   ,       221 1 3 0, 1, 31 1 3 0x x x x x xx x x x            Since 2 2221 11 12 21 30 2 4x x xx x                       Or considering 21 0,x x  discriminant = 1 4= 3 < 0 and coefficient of x2 is positive, so 21 0 .x x x     The inequality is reduced to   1 3 0x x  , 1, 3x x   1 or 3 x x   Deduce the solution of (i) 24 241,2 3xx x   [2] Method 1: Given 24 241,2 3xx x       222 2412 3xx x   Method 2: Given 241,2 3xx x   we replace x with 2xto obtain 24 241.2 3xx x   2 21 or 3 x x   3 1 x

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 End-of-Year Revision Package (Solutions) 7 By replacing x with 2x in 2 4 1,2 3 x x x     2 2 1 or 3 x x  Since 2 0 ,x x   2 1 x  has no solution. 2 3 3 or 3x x x     Since 2 0 ,x x   2 1 x  has no solution. 2 3 3 or 3x x x     (ii) 2 4 1.2 3 x x x    [2] Method 1: Given 2 4 1,2 3 x x x     2 4 1,2 3 x x x           2 4 1,2 3 x x x        By replacing x with x in 2 4 1,2 3 x x x     1 or 3 1 or 3 x x x x       Method 2: Given 2 4 1,2 3 x x x     we replace x with x to obtain    2 4 12 3 x x x         2 4 12 3 x x x        2 4 12 3 x x x    1 or 3 1 or 3 x x x x       5. 2017/Prelim/HCI/P1/Q3 (i) By first expressing 23 4x x  in completed square form, show that 23 4x x  is always negative for all real values of x. [2] 2 2 2 2 3 4 3 4 3 7 2 4 3 7 2 4 x x x x x x                      Since 23 02x     for all x, 23 02x      Hence 2 2 3 7 73 4 0 2 4 4x x x           23 4x x   is always negative for all values of x .

H2 Mathematics (9758) JC 1 Ri