2018 H1 Mathematics A Level Paper solutions (SAJC)

2018 H1 Mathematics A Level Paper 1 For   23 6 2 0kx k x    to have two distinct roots, Discriminant > 0        2 2 2 2 6 4 3 2 0 12 36 24 0 12 36 0 60 ,6 kk k k k kk k kk               2(i)   3 2 3 2 3 Let 3ln 4 2 d 12 3d 4 2 36 42 yx yx xx x x       (ii) 312 22 51 22 5 2 32 32 3 25 2 1 2 6 45 x dx x x dx x xx C x x C          where C is an arbitrary constant 3(i) Let x, y and z be the cost of training one accountant, one financial advisor and one consultant respectively 5 12 8 294100 9 7 3 270100 3 6 0 122700 x y z x y z x y z          Using GC to solve, 18500, 11200, 8400x y z   Therefore the cost of training one financial advisor is $11200 last year. (ii) Let n be the number of accounts that company S will be able to train this year.      1.1 18500 1.1 11200 200000 32670 200000 6.12 nn n n    The greatest possible number of accountants that company S will be able train is 6.

4 12 12d 12d x x y x e y ex     At stationary point, d 0d y x    12 12 12 1 2 0 21 1 2 11 2 ln ln ln1 ln 22 1 2 ln 2 1 ln 2 2 x x x e e e xe x x               When 1 ln 2 2x  , 1 ln 212 21 ln 2 2ye   1 1 ln 2 ln 2 1 ln 2 2 1 ln 2 2 1 ln 2 1 22 ln 21 2 e e       The exact coordinates of the turning point of C are 1 ln 2 ln 2,122   . (ii) (iii) When 2x , 32ye  , 3d 12d y ex  y x (0, e)

Equation of tangent when 2x ,          33 3 3 3 33 2 1 2 2 1 2 2 4 2 1 2 5 0.900 0.249 y e e x y e x e e y e x e yx                     Alternatively: ( better method since the question did not ask for Exact form) Using GC, gradient of tangent when 2x is 0.900 0.249yx  Sketch 32ye   Select Draw  followed by Tangent(  Enter 2x and press Enter

(iv) 2 1 2 12 2 1 2 d 22 22 x x x xex e x C xe C             where C is an arbitrary constant Required Area 12 1 2 0 1 2 22 1 02 2 2 11 units2 2 2 xxe ee e e            5(i) Number of times that the manager needs to order in a year 1200 x Total ordering cost 1200 6000050 xx     Storage Cost Purchase Cost Ordering Cost 600006 200 1200 C x x       (ii) 2 d 600006d C xx When C is stationary, 2 2 d 0d 6000060 10000 100 or 100 (reject since 0) C x x x xx       2 23 2 2 d 120000 d d 0 for all 0d C xx C xx  

Therefore minimum value of C   600006 100 240000 $241200100    (iii) This is not a reasonable model as various components of the cost differs depending on external factors such as the economical climate and real estate climate. For example the rental cost of warehouse will differs from year to year. (iv) Let y be the number of television sets sold. Gradient of the line 1200 0 3300 700    Equation