ACJC 2023 JC1 H1 CA Solutions for students

2023 JC1 H1 Math Solutions (Students) Qn Solutions (1) 2 12x x k+ − +  2 30x x k+ − +  For 2 30x x k+ − +  for all real values of x, Discriminant = 21 4( 3 ) 0 k− − +  4 13k  13 3.254k = 2 99log log log22=  − =x xyy   ---------(1) 52log 5 5log 2log 5=  + =x y x y   ----------(2) (1)+(2)x2: we have 7 log 14 log 2 = = x x   3 (i) (ii) (iii) 024 72(0.9) 96 C =+ = o T T , 0.9 0 24 → → → tt T T approaches/tends to 24 Co for large values of t.

(iv) Add the line 28 to curve in (iii), =T use GC to find the intersection point: t=27.4 minutes. 4 (i) (ii) 11ln 10 111 ln 1 10 11Add 1 10 1 10 −= − −+ = + − −=+ − = − xx x xx x xy x x (5)(i) 13 18 40 2830R J W+ + = ……..(1) 18 (0.8) 30 (0.75) 40 (0.9) 2934R J W+ + = 14.4 22.5 36 2934R J W+ + = ……..(2) 4 10 165RW− + = ……(3) From GC 50, 40, 36.5RJW= = = The manufacturing cost for a pair of walkers is $36.50. (ii) The manufacturing cost of 10 pairs of walkers is $165 more than the manufacturing cost of 4 pairs of runners. x=0 (0.368,0) or (e-1,0) x=10 0

(iii) From GC maximum point is x=349.99994, P = 12248 Since x is an integer, The number of pairs of shoes the manufacturer needs to manufacture is 350 and maximum profit is $12248. Method 1 When P = 0, 2 0 70 210 x x=− + − x = 0.04286 or x = 699.957 The maximum number of shoes the manufacturer can manufacture so that the investment is profitable is 699 pairs of shoes Method 2 From GC When P  0 0.0285726 699.9714x Largest value of x is 699 The maximum number of shoes the manufacturer can manufacture so that the investment is profitable is 699 pairs of shoes