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,