RI Equations and Inequalities C3 Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________________________ Additional Practice C3: Equations and Inequalities Page 1 of 12 Additional Practice Questions for Chapter 3: Equations and Inequalities 1 9740/2007/01/Q1 Show that 22 22 21 9 4 2 1 132 32 xx x x xx xx     . [1] Hence, without using a calculator, solve the inequality 2 2 21 9 132 xx xx   . [4] Solution: 2 2 22 2 2 2 21 9LHS 1 32 21 9 3 2 32 42 1 RHS (Shown)32 xx xx xx x x xx xx xx          22 22 2 2 21 9 21 9 11 032 32 42 1 0 32 ( 3)( 7)( 2)( 1) 0, 2, 1. 3 or 2 1 or 7 xx xx xx xx xx xx xxxx x xx x                    2 Sketch the graphs of 42yx and  4yx x on a single diagram. [3] Hence solve 42 4 xx x  , giving your answer in exact form. [4] Solution: From the sketch 24 , 242 42 , 2 xxyx xx     At point A,   2 2 42 4 64 0 66 4 4 352 xx x xx x       Since 2x , 35x .

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 2 of 12 At point B,   2 2 24 4 24 0 22 4 4 152 xx x xx x        Since 2x , 15x . Hence the solution for the inequality 42 4 xx x  is 35 15 x   . OR : After finding the x co-ordinate at point A: We note the symmetry in the line 2x , so the graphs intersect at 223 5 1 5x    Hence the solution for the inequality 42 4 xx x  is 35 15 x   . 3 Without using a calculator, solve the inequality  2 25 1 1 xx x xx   . [4] Hence find the exact solution for each of the following: (i)  2e2 e 5 e e1e xx x xx    (ii) 2 2 25 1xx xx x   [6] Solution: 2 2 25 1 (1 ) 25 ( 1) 0(1 ) ( 1)( 2)( 3) 0, 0 and 1 xx xx x xx x xx xx x x x x           2x or 01 x or 3x (i)     2 2 e2 e 5 e e1e e2 e 5 1 ee1e xx x xx xx xxx       Replace x with ex ,

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 3 of 12 e 2 (no solution since e 0 ) or 0 e 1 or e 3 0 or ln 3 xx x x x xx          (ii)  2 2 2 25 1 25 1 1 xx xx x xx xxx     Replace x with x , 2x  (no solution since |x|0 x  ) or 01 x or 3x  3x or 10 x  or 01 x or 3x 4 (i) Solve     2 1 012 3 x xx   . [4] (ii) Hence find the exact solution of     2 1c o s 012 c o s c o s 3     , for 03  . [4] Solution: (i)           2 2 1 2 1 2 1 012 3 12 3 1 0 , , 3 or 1 or 3 x xx xx x x x xx x        (ii)         2 2 1c o s 012 c o s c o s 3 cos 1 012 c o s c o s 3         Replace x with cos 1 2 51 33 7 3 cos or cos 1 or cos 3 0, 2 (n.a. 1 cos 1 or 3                 571 33 30 or or 2 or 3         

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 4 of 12 5 ACJC Promo 9740/2012/Q5 Without the use of a graphing calculator, solve the inequality 2 2 27 6 12 xx xx   . [3] Deduce the range of values of x such that (a) 2 2 2(ln ) 7(ln ) 6 1(ln ) (ln ) 2 xx xx   , [2] (b) 2 2 27 6 112 xx xx   , [2] Solution: 22 2 2 2 2 27 6 ( 2 ) 02 68 02 (4 ) (2 ) 0(2 ) (1 ) (4 ) (2 ) (1 ) 0 , 2 , 1 xx x x xx xx xx xx xx xx x x             1 or 4xx  (a) 2 2 2(ln ) 7(ln ) 6 1(ln ) (ln ) 2 xx xx   Replace x with ln x, ln 1 or ln 4xx  140 e or e xx   Note: there is no equal sign in the inequality in part (a) (b) 2 2 2 2 27 6 1 12 1127 6 1 where 0 11 2 xx xx xx x xx            Replace x with 1 x , 11 1 or 4xx  11 0 or 0 4xx    Check 0x : 2 2 27 6 2 1 0 satisfy the inequality12 xx xxx   . 11 4x  

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 5 of 12 6 9758/2017/01/Q2 (i) On the same axes, sketch the graphs of 1y x a  and yb xa , where a and b are positive constants. [2] (ii) Hence, or otherwise, solve the inequality 1 bx axa  . [4] Solution: (i) (ii) Solve for point of intersection between the two curves.  211 ()) 1 ( 1 xax ab xa b xa xa bx a b        Hence, the solution for the inequality is 1or xa xa b   .

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 6 of 12 7 9758/2018/01/Q4 (i) Find the exact roots of the equation 223 2 2xx x  [4] (ii) On the same axes, sketch the curves with equations 223 2yxx  and 2.yx Hence solve exactly the inequality 223 2 2 .xx x   [4] Solution: (i)       2 121 2 , f o r 2 223 2 121 2 , f o r 2 2 xx x o r x xx xx x            Consider 2 122 3 2 , f o r 2 2xx x x o r x        2 2 2 22 3 2 22 0 22 4 2 2 22 3 2 1 3 valid solutions xx x xx x            Consider 2 122 3 2 , f o r 2 2xx x x x      2 2 22 3 2 22 0 0 or 1 valid solutions xx x xx x       Therefore the roots are 13 , 1 , 0 a n d 13   . (ii) For 223 2 2xx x   , 1 3 1 or 0 1 3.xx      

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 7 of 12 8 ACJCCommonTest9758/2021/2 Given that    32 2378 2 2 2xxx A x B xx     , where A and B are constants to be determined, solve algebraically the inequality 2 2031 0 1 8 . 1xx x   [4] Hence solve, exactly, the inequality  2 2031 1 0 1 1 8 . 11xx x    [2] Solution:    32 2378 2 2 2xxx A x B xx     Comparing coefficients of 3x , 3A . Comparing constant term, 1B . 22 20 203 10 18 3 10 18 0 11xx xx xx              2 32 2 2 31 0 1 8 1 2 0 01 378 2 01 31 2 2 01 31 1 2 20 , 1 xx x x xxx x xx x x xxx x x                  Since  22 22 11 1 0xx x    for all x  , hence   31 10 , 1xx x    . Therefore 1x or 1 3x .  2 2 2031 1 0 1 1 8 11 2031 1 01 1 8 11 xx x xx x         Replace x by 1x in the previous part, 11x  or 1421o r333xx x      (No solution) Hence, 42 or33xx  . 1 1 3

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ______________________________________________ Additional Practice C3: Equations and Inequalities Page 8 of 12 9 JPJC Promo 9758/2021/Q2(i) A curve C has equatio