ACJC JC1 H1 Maths Rev A Complete Solution CA1

2023 JC1 H1 REVISION SET A COMPLETE SOLUTIONS for ACJC CA1 ACJC 2015 CA1 [24 April 2015] 1 4x2 + 3 = 2kx has two real distinct roots  4x2 − 2kx + 3 = 0 has real, distinct roots Discriminant = (−2k)2 − 4(4)(3) > 0  4k2 − 48 > 0  k2 − 12 > 0  (k − 12)(k + 12) > 0  k < −12 or k > 12 (or k < −23 or k > 23) 2 (a) Given lg IM S =  , let intensity in Alaska be IA. Then intensity in Iceland is 4IA. Magnitude in Iceland 4lg AI S =  lg 4 lg AI S =+  lg 4 8.3=+ = 8.9 (1 decimal place) 2 (b) In Italy, 7.1 lg I S =  In Alaska, 8.3 lg AI S =  lg lg lg A A II I S S S S I    − =          7.1 8.3 lg A I I −=  1.2 lg A I I −=  1.210 A I I −= Ratio is I : IA = 10−1.2 : 1 or 1 : 101.2 Method 2: In Italy, 7.1 lg I S =  In Alaska, 8.3 lg AI S =  I = ( )S1.710 IA = ( )S3.810 2.110 1= AI I I : IA = 1 : 101.2 3 13 2 x y =−  and 3 13 1 344 xy xx += = +++ 3 13 4 xy x += + 3 13 1 3042 3 13 1 342 x x x x x x + − + +   +   − +   x y O y = 3 x = −4 13 2 x y =−  (0, 2) (0, 3.25) (−4.33, 0) (−1.58, 0) or ( 14 3− , 0)

ACJC 2015 CA1 [24 April 2015] Intersection at (−4.05 9956, −13.67894). Range of values of x is x  −4.06 or x > −4. 4 y = x − 2a and y2 = ax 4 (i) At intersection, (x − 2a)2 = ax x2 − 4ax + 4a2 = ax x2 − 5ax + 4a2 = 0 (x − a)(x − 4a) = 0 x = a or x = 4a Points of intersection are (a, −a) and (4a, 2a). Length of AB = ( ) ( )22 42a a a a− + + 2 2 29 9 18a a a= + = 9 2 3 2aa=  =  AB = 3a 2 units 4 (ii) Coordinates of C are (2.5a, 0.5a) or 51 ,22aa Radius of circle is ( )11 3222AB a= Equation of circle is ( )22 25 1 9 2 2 2 4x a y a a   − + − =       22 25 1 9 2 2 2x a y a a    − + − =       x y O y = x − 2a y2 = ax (0, −2a) (2a, 0)

ACJC 2015 CA1 [24 April 2015] or any equivalent form, e.g. ( ) ( )22 22 5 2 18x a y a a− + − = 2016 JC1 H1 Mathematics CA1 [21 April 2016] 1 23 3 xy x −= + 2 ( )( ) 2 2 1 2 2 22 1 e 2e 3e 2e 3e 3e 2e 0 e 2e 0 e e or e 2e 1 or e 2 1 or 1 ln 2 i.e. 1 l n2 xx xx x uu u u u u x x x x + − + =  + =  − + =  − − =  = = == = − = = + 3 22 3 4 4 3 0kx k x kx x k+ +   − + +  No intersections with the x-axis so NO real roots and curve has a minimum point  discriminant < 0 and coefficient of x2 > 0 ( ) ( ) ( )( ) 2 2 2 0 & 0 4 4 3 0 4 3 0 3 4 0 4 1 0 4 or 1 & 0 1 Dk k k k k k k k k k k k k  − − +   − −   + −   + −   −    4 Let the cost of a ticket in each category be x, y, z. 10x + 4y + 5z = 320 9x + 6y + 4z = 352.5 7x + 5y + 3z = 282.5 By GC, x = 12.50, y = 30, z = 15 Total cost for Lim family = 5(12.5) + 10(30) + 5(15) = $437.50 5(i) 1.5e When 0, . tnA t n A = == x = −3 y = 2 y x O (3/2, 0) (0, −1)

5(ii) 1.5 1.5 When 50 , 50 e e 50 1.5 ln 50 2 ln 50 or 2 .61 (3 s.f.)3 t t n A A A t t == = = = 6 (i) 6 (ii) By GC, x = 9.38 (3 s.f.) 6 (iii) For ( ) ( )2 0.5 ln 2 x x−  − , x > 9.38 (3 sf) x = 2 y = 2 y x O (0, 1) (−1,0) (3, 0) y = 2 – (0.5)x y = ln(x – 2)

2017 JC1 H1 Mathematics CA1 [21 April 2017] 1 kx2 + (k – 2)x + k > 0 D = discriminant = (k −2)2 – 4k2 < 0 and k > 0 −3k2 – 4k + 4 < 0 and k > 0 (−3k + 2)(k + 2) < 0 and k > 0 k < -2 or 3 2k and k > 0  3 2k 2 ( ) 03lnln 22 −+ xx Let u = lnx u2 + 2u – 3  0 (u + 3)(u -1)  0 3−u or 1u  3ln −x or 1ln x  3− ex or ex but x > 0  0< 3− ex or ex 3 −2x2 + 400x = 120x + q −2x2 + 280x – q = 0 If company cannot break even then equation has no real roots. D = discriminant = (280)2 – 4(−2)(−q) < 0 (280)2 < 8q q > 9800 4 (ii) Using GC alcohol content is a maximum at 1.85 h after drinking 56g of alcohol (iii) The period in which the 77kg man is legally drunk is 1.11 < x < 2.73 5 (i) →t , 4→m Mass of chemical in the long term is 4 g. (ii) m = 2.56 = 21.0 )2( te−− 6.12 1.0 =− − te 4.01.0 =− te or 6.31.0 =− te − 0.1 t = ln 0.4 or - 0.1 t = ln 3.6 t = − 10 ln 0.4 or t = −10 ln 3.6 (NA because t 0 ) t = 10 ln (2.5) (iii) When t = 0, m = 1 Asymptotes m = 4. Intersects with the y – axis at (0,1) 6 (a) 0.7 170 0.7100 100 xy xx= =− +−− Horizontal asymptote y = − 0.7 and x = 100 When x = 0, y = 0 . Curve passes through (0,0) -2 3 2 -3 1 x A(x) (0,0.009) (5,0.0324) (0,1) t m m = 4

(b) No. Because as x tends to 100 the cost y approaches infinity OR No. Because when x is 100, y is undefined OR NO. The cost to remove the pollutant is relatively low at first but skyrocketed as we get closer and closer to removing 100 percent of all the pollutants. 100 y = -0.7 x = 100 -0.7 (0,0)

2018 JC1 H1 Mathematics CA1 [26 April 2018] 1 ( ) ( )( ) 2 3 6 3 5 6Let 3 , 5 5 6 0 6 1 0 3 6 or 3 1 (rejected since 3 0) ln 3 ln 6 ln 6 ln 3 xx x x x x yy y yy yy yy x x −−= = − = − − = − + = = = = =−  = = 2 ( ) ( ) ( ) 22 2 5 4 2 1 1 0k x kx x k x k x+ + + = +  + + − + = ( ) ( )( ) ( )( ) 22 2 2 4 0 1 4 2 1 0 2 1 4 8 0 6 7 0 7 1 0 b ac k k k k k kk kk −   − − +  − + − −  − −  − +  1 or 7kk−  The curve has a minimum point 2 0 2kk +   − 2k − and 1k − or 7k  Hence –2 < k < −1 or k > 7 3 Let the number of Chocolate, Strawberry and Vanilla ice cream tubs be c, s, v respectively. 60 ----- (1) 16 14 12 860 ----- (2) 1 3 3 0 ----- (3)3 c s v c s v v s c c s v + + = + + = − =  + − = By GC, 30, 10, 20c s v= = = New amount with membership = ( ) ( ) ( )30 0.8 $16 10 0.9 $14 20 0.95 $12 $10 $748 +  +  + = Amount saved = $860 $748 $112−= 4 (i) 1 21 xy x += − 4 (ii) Suitable graph added is y = x. 1 2x 1 2y y x O (−1, 0) (0, −1) y = x

At points of intersection, 1 21 x xx + =− ( ) 2 2 1 2 1 2 2 1 0 2 ( 2) 4(2)( 1) 2 12 1 3 2(2) 4 2 x x x xx x + = − − − =  − − − = = = For 1 21 x xx + − , 1 3 1 22 x−  or 13 2x + 5 (i) When t = 0, 98 30 68 aa= +  = 5 (ii) 12 12 63, 12 63 30 68e 33 e 68 1 33 ln or 0.0603 (3sf)12 68 b b Tt b − − = =  = + =  =− 5 (iii) 5 (iv) The temperature of the pot of soup in the long run is 30◦C. T (◦C) t (min) (0, 98) 30 30 btT ae −=+ T = 30

2019 JC1 H1 Mathematics CA1 [9 May 2019] 1 ( ) ( ) ( ) ( ) ( ) 22 2 22 2 2 23 2 2 2 2log 3 log (1 ) 3 log 3 log (1 ) 3 3log 3 (1 ) 6 9 2 1 6 9 8 8 0 2 1 0 10 1 xx xx x x x x x x x x xx x x + − + = + − + = + =+ + + = + + + − − = − + = −= = 2 ( )( ) 2 2 2 3 4 0 0 4( 3)( 4) 0 48 0 48 48 0 48 48 x kx D k k kk k − + −   − − −  − + −  −   Greatest integer value of k = 6 3 Let x, y and z be the unit cost of craft paper, marker and glue stick. 3.21 4.28 5.35 26.75 --------(1) 5.136 4.28 1.712 26.12 --------(2) 2.889 1.926 0.963 12.91 --------(3) OR 26.753 4 5 --------(1)1.07 26.126 5 2 --------(1.07 0.8 x y z x y z xyz x y z x y z + + = + + = ++= + + = + + =  2) 12.913 2 --------(3)1.07 0.9 By GC, $1.65, $3.70, $1.05 x y z x y z + + =  = = =

4 ( )12 52 Asymptotes: 3 24 ax aya x b x b xb ya −= =− −++ =− = =− = 3b=− 2a=− 0.298 3 x or 6.70x 5    2 2 2 2 2 3 --------(1) 3 28 --------(2) Solving (1) & (2) 3 3 28 ( 3) 25 0 is tangential to , D 0 ( 3) 4(1)(25) 0 ( 3) 100 0 ( 3) 10 ( 3) 10 0 ( 7)( 1