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 of the line,  1200 3 300 3 900 1200 3 2100 yS yS S           2 Total Selling Price Total Cost 240000 3 2100 240000 2100 3 240000 P yS SS SS          (v) d 2100 6d P SS  When P is stationary, d 0d 2100 6 0 350 P S S S    2 2 d 60d P S   Therefore when S is $350, P will be maximum. 6i Let X be the random variable the number of people who have blood group A, out of 6. ~ B(6,0.4)X P( 2) 0.31104 0.311 (3 s.f.)X    ii P(at least 4 people do not have blood group A) = P( 2) 0.54432 0.544 (3 s.f.)X    7i No of selections = 8 4 6 3 1 2 3360C C C   ii Case 1: no guitarist chosen No of selections = 5 3 10C  Case 2: 1 guitarist chosen No of selections = 53 21 30CC Total number of selections = 30 + 10 = 40 iii P( band contain none of the three guitarists) = 5 3 8 3 5 28 C C  8i P( ) P( ) P( ) P( )A B A B A B    

0.8 2 P( ) P( ) 3 0.8 p p A B A B p        Given P( )P( | ) P( ) ABAB B  Then 3 0.80.3 2 p p  0.6 3 0.8pp 2.4 0.8 1 3 p p   ii P( ' )AB refers to the probability of event B occurring and A does not occur at all. P( ' ) P( ) P( )A B A B A    170.8 3 15   9i 112 28P( ) ( student studies Business) = 300 75BP ii P( ) ( student is female or studies Busine ss) 140 112 40 53 = 300 300 300 75 F B P    iii P( ) ( student is male and studies Art) 54 9 = 300 50 M A P  iv 160P( ) ( student is male) = 300MP 90P( ) ( student studies Art) = 300AP 160 90 4P( ) P( ) 300 300 25AM    From (iii), 9P( ) 50AM Since P( ) P( ) P( )A M A M   Events M and A are not independent. v 90 210 21 300 3 ( exactly 2 students study Art out of 3) = 0.189 90 89 210Or 3 0.189 (3 d.p.)300 299 298 CCP C        A B

10i ii Product moment correlation coefficient is ‒0.976 (3 s.f.), which is close to ‒1, suggesting a strong negative linear correlation between the times (in seconds) to run 100 metres and 10000 metres. iii Using GC, 2.2181 66.261yx  2.22 66.3yx  (to 3 s.f) (see (10i) for line on the scatter diagram) iv When 13.1,x 2.2181(13.1) 66.261 37.204 37.2 y   The required time taken is 37.2 minutes. v Since 13.1,x is within data range and the product moment correlation coefficient is close to ‒1, this estimate is reliable. 11i Unbiased estimate of the population mean, 15 30 30.15100x   Unbiased estimate of the population variance, 2 2 1 15 298299 100 36s    ii Let X be the length, in cm, of a randomly chosen adult fish in a particular species of fish and  be the population mean. Test : 30oH   against 1 : 30H   Under oH , since sample size = 100 is large, by Central Limit Theorem, 293630, 100XN    approximately. Use a one-tailed test at 2.5% level of significance. Using GC, with 30.15x , 29 36s , n =50, value 1.6712z  value 0.047335 0.025p  We do reject oH and conclude that there is insufficient evidence at 2.5% significance level to conclude that mean length of an adult fish in a particular species of fish is greater than 30cm. 42.2 10.8 14.8 15.6 33.1 32.6 x (secs) y (secs)

iii Not necessary since the sample size of 100 is large. By Central Limit theorem, the sample mean lengths of adult fish will be approximately normal. iv Test : 30oH   against 1 : 30H   Under oH , since sample size = 100 is large, by Central Limit Theorem 0.930, approximately.100XN   To support the scientist’s claim that the mean length of the fish is greater than 30cm at 10% means sample mean lies in critical region. Using GC, 30.12158m . Or use critical value method, 0 ~ (0,1)XZN n    30 1.2816 0.9 100 calc mz  0.930 1.2816 100m   0.12158 30 30.12158 30.2 m m m    12i Let A be the random variable “mass of Type A component in grams”   2N 250,3A Let B be the random variable “mass of Type B component in grams”   2N 240, 4B P( 1.02(250)) ( 255) 0.047790 0.0478A P A     ii     22N 10,3 4 N 10, 25 AB AB   P( ) ( 0) 0.97725 0.977A B P A B      z 1.2816 0.1

iii       1 2 6 1 2 3Let ... ( ) (6 250) 3 240 2220 Var( ) (6 9) 3 16 102 ~ N 2220,102 T A A A B B B ET T T                 P(2190 2230) 0.83746 0.837T    iv       1 2 10 1 2 10 22 Let 0.03( ... ) 0.02( ... ) ( ) 0.03(10 240) 0.02 10 250 22 Var( ) 0.03 (10 16) 0.02 10 9 0.18 ~ N 22,0.18 C B B B A A A EC C C                 P( 22.50) 0.119296 0.119C   