ACJC JC1 H1 Maths Rev A-3 Complete Solution Eqns and Inequalities

2023 JC1 H1 REVISION SET A-3 COMPLETE SOLUTIONS Qn EQUATIONS & INEQUALITIES 1 2 2 2 26 6 60 10 ...(1)x y x y+ =  + = 12 12 48 4 ...(2)x y x y+ =  + = Substitute (2) into (1): ( ) 2 224 10 2 8 6 0y y y y− + =  − + = 2 4 3 0 ( 3)( 1) 0 3 1 since > y y y y y x y x  − + =  − − =  =  = 2 The equations are 1.2 (1) 4 2 34.4 (2) 9 3 11.1 (3) c a b c a b c = + + = − + =− From GC, 2.5, 11.6, 1.2a b c= = = 3 Let the price of 1 litre of A, B and C be a, b and c respectively. Given that 29 3.50 2.5 2 2 2 2.5 2 0 a b c bc b c a a b c + + = += + =  − − = Using GC, $4, $2, $1.50a b c= = = 4 2 11 3x kx x k+ + = + ( ) ( ) 2 3 11 0x k x k + − + − = b2 − 4ac > 0 ( ) ( ) 2 3 4 11 0kk − − −  2 2 35 0kk − −   k < −5 or k > 7 5 2 2 222 2 2 0 kk x kx x kx x kx k − =  − =  − + = Since there is no intersection, discriminant < 0 24 4(2)( ) 0 ( 2) 0k k k k−   −  0 2 02 k   6 (a) 6 (b) 25 2 3xx−  − 2 3 3 0xx − −  Let 2 3 3 0xx− − = 3 9 4(1)( 3) 2x  − −= 3 21 2 =  3 21 3 21 or 22xx −+ (i) 24 24 39xx−+ ( ) 24 6 39xx= − + 3 21 2 − 3 21 2 +

Qn EQUATIONS & INEQUALITIES ( ) 2 4 3 9 39x= − − + ( ) 2 4 3 3x= − + Since ( ) 2 30x− for all real values of x, we have ( ) 2 4 3 0x−  ( ) 2 4 3 3 3 0x− +    24 24 39 0xx− +  for all real values of x. (ii) ( )( ) 24 24 39 021 xx xx −+ +− From (i), 24 24 39 0xx− +  for all real values of x, so ( )( )2 1 0xx+ −  giving 21 x−   7 x2 + 2k = 4 – kx  x2 + kx + (2k − 4) = 0 Discriminant = k2 – 4(1)(2k – 4) = (k – 4) 2 0 Hence x2 + 2k = 4 – kx has real roots for all real values of k. 8 x2 + x + 7  2x2 + 1  x2 – x – 6  0  (x – 3)(x + 2)  0 giving x  −2 or x  3 (i) Replace x by ln x : ln x  −2 or ln x  3 giving 20 xe − or 3xe (ii) Replace x by ex : ex  −2 (reject or ex  3 because ex > 0 ) ln 3x 9 Intersection points at Therefore, solution set is { : , 4.70 0.822  or  0.518}x x x x −  −  22 77 551 1 ( ) xxxx+   − ++ + − Replace x by −x: 4.70 0.822  or  0.518 0.822 4.70  or  0.518 xx xx − − − −    −

Qn EQUATIONS & INEQUALITIES 10 2 2 42 3 4 0 kx x k x kx x k + +  − + + −  For quadratic curve to be positive, it must not cut the x-axis, i.e. no real solution, discriminant is negative ( 2 40b ac− ) and the curve must lie above the x-axis, so coefficient of x2 is positive 2 2 3 4( )( 4) 0 4 16 9 0 91 or 22 kk kk k k − −  − + +   − and k > 0 Overall, 9 2k  11 280 (1)A B C+