2021 A level H1 Math Solutions (SAJC)

Appendix 2 1 by: Questi on Suggested Solutions 1 Since ( ) 2 1 2 0x k x k+ − + +  , Discriminant < 0 ( ) ( )( ) ( )( ) 2 2 2 1 4 1 2 0 2 1 4 8 0 6 7 0 7 1 0 kk k k k kk kk − − +  − + − −  − −  − +  17 k−   Therefore the set of values of k is  | 1 7kk −   k

Appendix 2 2 2(a) 21 1 2 3 3 2 3 3 3 1 32 3 2 3 d5 2 5 2 2 5 6d3 52 2 5 6 3 x x x x x xx x x x x −    − = − −           = − −   2(b) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 d 5 4 d 54 54 1 42 1 5 4 , where C is an arbitrary constant2 x x x x x C xC − =− − −=+ − =− − +  3(a) 2 2 1614 d 16 2d yx x y xxx = − + =− At stationary points, d 0d y x = . 2 2 3 1620 162 8 2 x x x x x x −= = = =

Appendix 2 3 When 2x= , 2 162 14 2 2y= − + =− Therefore stationary point is ( )2, 2− . (b) (c) 2 22 2 2 2 162 20 0 1614 6 0 1614 6 x x xx x x x x x − + = − + + − = − + + = − The equation of the curve is 26yx=− . (d) From GC, 0.865 or 2.64xx== . x y

Appendix 2 4 4(a) When 0x= , 2ye=− When 0y= , 13 2 1 3 ln 2 1 ln 2 3 xe x x − = −= −= 4(b) 13 13 2 d 3d x x ye y ex − − =− = 1 3 2 22 1 d 3When 1, 2 2 , 3 d yx y e e e x e −−= = − = − = = Equation of tangent when 1x= , x 0 ( )0,2 e− y

Appendix 2 5 ( ) ( ) 22 22 22 1321 31 12 34 2 yx ee yx ee yx ee − − = − = − + − = − + 4(c) ( ) 1 13 1 3 111 3 1 3 11 33 1 3 0 22 2 d 22 33 1 2 121 3 3 3 11 3 x xx A e x eexx ee qpee − −− − =−    = − = +   −       = + − +       = + = +  Comparing 11, 3pq== 5(a) Let $x, $y and $z be the amount spent on internet advertising, television advertising and posters advertising respectively.

Appendix 2 6 220000 (1) 2 2 0 (2) 11000 3 3 11000 (3) Solving (1), (2) and (3) using GC, $61000 $122000 $37000 x y z y x x y y z y z x y z + + = − =  − = − − =  − = − = = = (b) When 4t= , ( )1.5 0.8ln 2 4 1 1.5 0.8ln 9 3.26 S = + +  =+ = 3.26 represents the annual sales in millions of dollars in the 4th year. (c) t 0 S

Appendix 2 7 (d) ( ) ( ) 1.5 0.8ln 2 1 d2 0.8d 2 1 8 5 2 1 S t S tt t = ++ =  + = + (e) For 1t , d 0d S t  . Since d 0d S t  , S increase when t increases. Therefore the sales in the future will increase. As t increases, d d S t decreases, so the sales in the future will increase at a decreasing rate in the future. 6 Let be the random variable that denotes the height of a sunflower plant Given that P( 1.72) 0.3X = and P( 1.52) 0.1X = , X ( ) 2~ N ,X 

Appendix 2 8 P( 1.72) 0.3 1.72P 0.3 1.72P 0.3, ~ (0,1) 1.72 0.52440 1.72 0.52440 0.52440 1.72...(1) X X ZZ         = −− = − = − = −= += P( 1.52) 0.1 1.52P 0.1 1.52P 0.1, ~ (0,1) 1.52 1.2816 1.52 1.2816 1.2816 1.52...(2) X X ZZ         = −− = − = − =− − =− −= Solving equati