SAJC Check your Understanding Inequalities tutor

Check your Understanding (Inequalities) Section 1: Type of roots of Quadratic Equation 1. RVHS JC2 Prelim 8865/2019/Q1 Find the exact range of values of the constant k for which the equation 2 20kx x k    has 2 distinct real roots. [4] RVHS JC2 Prelim 8865/2019/Q1 (Solutions) Since the equation has 2 real distinct roots,      2 2 2 Discriminant 0 1 4 2 0 1 4 8 0 4 8 1 0 kk kk kk           551 1 022kk                            5511 22 k     2. CJC JC2 Prelim 8865/2019/Q2 Find the set of values of k for which the equation   2 3 3 0kx k x k     has real roots. Without carrying out further calculations, state the set of values of k for which   2 3 3 0kx k x k     for all real values of x. [4] CJC JC2 Prelim 8865/2019/Q2 (Solutions) 2 Since the equation has real roots, Discriminant 0        2 2 3 4 3 0 3 6 9 0 3 1 3 0 13 k k k kk kk k              The set of values is  : 1 3kk    . Side working: Let 24 8 1 0kk   .   8 64 4 4 1 8 8 80 8 51 2 k        x

If   2 3 3 0kx k x k     for all real values of x, 0k  and Discriminant 0    0 1 and 1 or 3 2k k k    Combining (1) and (2), 3k  The set of values is  :3kk . Section 2: Conditions for Quadratic Equation to be always positive or negative 3. ASRJC JC2 Prelim 8865/2019/Q1 Find algebraically the exact set of values of k for which 2 (3 1) (4 4 ) 0kx k x k     for all real values of x. [5] ASRJC JC2 Prelim 8865/2019/Q1 (Solutions) For 2 (3 1) (4 4 ) 0kx k x k     for all real values of x, 2 conditions need to be satisfied: (i) the coefficient of x2 must be positive 0k --- (1) and (ii) Discriminant < 0 2 22 2 2 (3 1) 4 (4 4 ) 0 9 6 1 16 16 0 7 10 1 0 7 10 1 0 k k k k k k k kk kk                 5 4 2 5 4 2or77kk     ---(2) Combining (1) and (2), 5 4 2 7k  The set of values of k is 5 4 2: 7kk  . 4. EJC JC2 Prelim 8865/2019/Q1 Find algebraically the set of values of k for which   2 20kx k x k    for all real values of .x [4] EJC JC2 Prelim 8865/2019/Q1 (Solutions) For   2 20kx k x k    , Discriminant 0 and 2coefficients of x > 0 Side working: Consider 27 10 1 0kk   210 10 4(7)( 1) 2(7) 10 128 5 4 2 14 7 k        

  2 2 4( )( ) 0k k k   and  02k  22 4 4 4 0k k k    23 4 4 0kk   (3 2)( 2) 0kk    22 or 1 3kk  Combining (1) and (2), the set of values is 2{ : } 3kk 5. NYJC JC2 Prelim 8865/2019/Q1 Find the exact range of values of k for which   212 k x x k   is non positive for all values of x. [4] NYJC JC2 Prelim 8865/2019/Q1 (Solutions) For   21 2 0k x x k    ,      2 2 2 2 2 coefficients of 0 and Discriminan t0 1 2 0 & 1 4 1 2 0 1 & 1 4 8 02 1 1 1 &02 2 8 1 1 3 & 02 4 16 x k k k k k k k k k kk                         1 1 3 1 31 & or 22 4 4 4 4k k k     Combining (1) and (2), 13 44k  k x

6. TMJC JC2 Prelim 8865/2019/Q1 The equation of a curve is       22 2 4 2 1y k x k x k      where k is a real constant. Find the range of values of k for which the curve lies completely above the x-axis. [4] TMJC JC2 Prelim 8865/2019/Q1 (Solutions) For       22 2 4 2 1 0k x k x k      Discriminant 0 and coefficient of 2x > 0   2 0 2 2kk    .        2 22 2 2 4 4 2 2 1 0 4 16 16 4 2 5 2 0 4 4 8 0 k k k k k k k kk                   24 4 8 0 2 1 0 kk kk          1 or 2 1kk   Combining (1) and (2), 2k  Section 3: Show Questions 7. ACJC JC2 Prelim 8865/2019/Q1 Show that there are no real values of k for which     22 1 1x x k x    is always positive. [4] ACJC JC2 Prelim 8865/2019/Q1 (Solutions) Let     222 1 1 2 ( 2) 1y x x k x x k x k          Assume 0y for all real values of x. Since 0y for all real values of x, Discriminant 0 . Method 1: 2 2 2 2 Discriminant ( 2) 4(2)( 1) 4 4 8 8 4 12 ( 2) 8 0 for all real values o f kk k k k kk kk                  since 22( 2) 0 ( 2) 8 0 8 0kk        for all real values of .k Discriminant can never be negative for all real values of k      22 1 1 0x x k x     will always have 2 real and distinct roots, i.e.  There are no real values of k for which     22 1 1x x k x    is always positive. Method 2: x

Solving Discriminant 0 , 2 2 2 2 ( 2) 4(2)( 1) 0 4 4 8 8 0 4 12 0 ( 2) 8 0 kk k k k kk k                 But 2( 2) 0k for all real values of k so 2 ( 2) 8 0k   .  There are no real values of k for which     22 1 1x x k x    is always positive. 8. MI PU2 Prelim 8865/2019/Q2 Show that there are no real values of k for which   22 2 1 1x k x k    is always positive. [4] MI PU2 Prelim 8865/2019/Q2 (Solutions) Assume that that there are real values of k such that   22 2 1 1x k x k    is always positive. Since   22 2 1 1x k x k    is always positive, Discriminant 0        2 2 2 2 Discriminant 2 1 4 2 1 4 4 1 8 8 4 12 9 2 3 0 for all real values of kk k k k kk kk                 This contradicts with the original assumption. Hence there are no real values of k such that   22 2 1 1x k x k    is always positive. Section 4: Intersection of a curve and a line 9. SAJC JC2 Prelim 8865/2019/Q2 The curve C has equation 225y x kx   and the line L has equation 3y x k . Find the exact range of values of such that C intersects L. [4] SAJC JC2 Prelim 8865/2019/Q2 (Solutions) When C intersects L, 22 5 3x kx x k    22 ( 3) (5 ) 0x k x k     Since C intersects L, Discriminant 0              2 2 2 22 ( 3) 4 2 (5 ) 0 14 31 0 7 80 0 7 80 0 7 80 7 80 0 kk kk k k kk                    k

7 80 or 7 80 7 4 5 or 7 4 5 kk kk         10. NJC JC2 Prelim 8865/2019/Q1 Find the range of values of m such that the line 2y x m and the curve   22 2 4y mx m x    intersect at two distinct points. [4] NJC JC2 Prelim 8865/2019/Q1 (Solutions) When the line intersects the line,                     2 2 2 2 2 2 4 2 2 4 4 0 Since the line and the curve intersect at 2 dis discriminant 0tinct points, 4 2 4 0 4 8 4 0 4 4 8 0 4 9 4 0 4 mx m x x m mx m x m mm m m m m m m m mm                             44 or 9mm  When m = 0, the curve becomes a line y = 2x  4 and the other line will have equation y = 2x. These 2 lines intersect at a single point. Hence, for the lines to intersect at two distinct points, 44 or , 9 0m m m   . 11. TJC JC2 Prelim 8865/2019/Q1 Find algebraically the range of values of k for which the curve 24 2( 1) 9y x k x    intersects the x-axis. [4] TJC JC2 Prelim 8865/2019/Q1 (Solutions) When the curve intersects the the x-axis, 24 2( 1) 9 0x k x     Since the curve must intersect the x-axis (either at one or two points), Discriminant 0 24( 1) 4( 4)( 9) 0k     x

2 2 35 0 ( 5)( 7) 0 kk kk         5 o