2022 Ch1 Inequalities APQ Soln (RVHS)

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 1 Chapter 1: Equations & Inequalities Additional Practice Questions Solutions 1 2 2121 x xx     2 2 1 2 1 021 x x x x      243 021 xx x        22 1 4 3 0x x x       2 1 4 3 1 0x x x    31 42 x   or 1x (ans) 22 22 2 1 2 2 xx xx  2 2 2 12 2 12 1 x x x     2 3 1 1 42 x    or 2 1 1 x   2 110 2x  or 2 1 1 x   2 2x  or 2 1x   22xx  or or 1 1, 0xx    2. Sketch ( 3)( 3) 16 2 10 21( 3)( 2) ( 3)( 2) x x x xyx x x x x            1 1/2 -3/4

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 2 Find 0, From GC, 4 3 or 2 or 1 y x x x       Hence, Let 2 , 4 2 3(n.a.) 2 2 1 2 1 0 1 or 0 x x x x y x x xx              3. (a) 4 3 2 3, 4 3 3 2 11From GC, 2 or 453 xx xx xx          (b)             2 2 2 2 2 3 2,1 1 3 2 1 0 1 2 1 0 1 1 0 1 0,since 1 0 1, 1 x x x x x x x x xx xx xx                    

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 3 2 3For 2,let 1 1 1, 1 x yxx x xx        4.  2 22211 xa ay xx     Horizontal asymptote: y = 2; vertical asymptote: x = 1 Therefore, solution set:  : , 1 or 3x x x x   . 5. (i) 32y ax bx cx d    When x = 0, y = −8, we get d = −8 (ii) So, 32 8y ax bx cx    When x = 4, y = 0, we get 64 16 4 8 16 4 2a b c a b c       ---- (1) When x = 3, y = −8 − 2(3) = −14, we get

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 4 27 9 3 8 14 9 3 2a b c a b c        ---- (2) 2d 32d y ax bx cx    When x = 2, d 2d y x  , we get 12 4 2a b c   ---- (3) Putting equations (1), (2) and (3) together and using the GC to solve, We get a = 1, b = −3 and c = −2. Therefore, equation of curve is 32 3 2 8y x x x    . 6. 7. (a) Let x = 1, 2, 3, 4 (eg. x = 1 for year 1920), y represents the population in millions. Let the cubic polynomial be 32y ax bx cx d    Sub. (1, 106), (2, 123), (3, 132), (4, 151) into the expression, 106 8 4 2 123 27 9 3 132 64 16 4 151 a b c d a b c d a b c d a b c d                 By using GC, 3, 22, 62, 63a b c d    323 22 62 63 When 5 (year 1925), 198 millions y x x x xy      (b) My estimate is higher than the actual value. This value of x (for year 1960) is outside the data range for us to make a reliable prediction. 8. Let x, y and z be the cost of a ticket for “under 16 years”, “between 16 and 65 years”, and “over 65 years” categories respectively. 9x + 6y + 4z = 162.03 7x + 5y + 3z = 128.36 10x + 4y + 5z = 158.50 For “under 16”, ticket costs $7.65. For “between 16 and 65 years”, ticket costs $9.85. For “over 65 years”, ticket costs $8.52.

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 5 Let x, y and z be the prices per gram of swordfish, salmon and tuna respectively. 4.6 2.4 64.60 8.6 3.6 102.20 4.8 2.8 68.00 x y Tz x y Tz x y Tz          By using GC: 8.05882x , 4.47059y , 16.8Tz Given y z x , so 4.47059 8.05882 11 4.47059 16.8 8.05882 16.8 16.8 4.47059 8.05882 3.75789 2.08467 z T T T     Since T is an integer, so 3T  and 5.6z Ans: 8.06x , 4.47y , 5.6z , 3T  9.

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 6 10. [2020 ASRJC J1 MYE Q1] Using an algebraic method, solve the inequality [4] Hence solve the inequality 2 2 e 12e 3 e e 6x xx x    , leaving your answers in exact form. [2] 2 2 123 6 xx xx    2 2 023 16 x x x x    2 2 2 02) 6 (36x x x x xx     2 2 026 6 x x x x    2 0( 1) 5 ( 3)( 2) x xx    Since 2( 1) 5 0x   ( 3)( 2) 0xx   3 or 2xx  For 2 2 e 12e 3 e e 6x xx x    , replace by .exx e 3 or e 2xx 

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 7 For e 3 ,x  there is no real solution. For 2e 2 lnx x  

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 8 11. [2019 EJC Promo Q1] At the start of the year, Mr Toh investe d in three types of savings bonds, namely “Ucare”, “Ushare” and “Ugain”. The amount invested in “Ugain” is equal to the sum of the amounts invested in the other 2 bonds. In addition, the sum of the amount invested in “Ushare” and twice the amount invested in “Ugain” is 8 times the amount invested in “Ucare”. At the end of the year, “Ucare”, “Ushare” and “Ugain” paid out interest at a rate of 2.5%, 1.75% and 3% respectively. Mr Toh received a total of $657.60 in interest. Express this information as 3 linear equations and hence find the amount invested in savings bond “Ushare”. [4] Let c, s and g be the amount of investments in saving bonds ‘Ucare’, ‘Ushare’ and ‘Ugain’ respectively. Amount invested in bond ‘Ugain’ equals the sum of the amounts invest ed in the other 2 bonds: c s g 0c s g   ----- (1) The sum of the amount invested in “Ushare” and twice the amount invested in “Ugain” is 8 times the amount invested in “Ucare”. 28s g c 8 2 0c s g   -----(2) Total interest of $657.60: 0.025 0.0175 0.03 657.60c s g   ----(3) Solving simultaneously, $4384, $8768 and $13152c s g   Total amount of investments for Ushare is $8768

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 9 12. [2020 TJC J1 MYE Q3] (i) Solve algebraically the inequality 2 1a xa  , where a is a positive constant. [3] (ii) Hence solve the inequality   1 3 2 3ln x  , leaving your answers in exact form. [4] 2 1a xa  2 10a xa    2 0a x a xa   3 0ax xa     3 0 and a x x a x a     or 3x a x a     1 3 2 3ln x      1 3 1 3 2 1ln x  Replacing x by ln x and let 1 3a , we have 1ln or ln 13xx 1 30 e or exx   