2020 A level H1 Math Solutions Students version (SAJC)

1 Questi on Suggested Solutions 1 Since ( ) 2 2 2 1 0x k x k+ − + +  , Discriminant < 0 ( ) ( )( ) ( ) 2 2 2 2 4 1 2 1 0 4 4 8 4 0 12 0 12 0 kk k k k kk kk − − +  − + − −  − − 0 12k Therefore the set of values of k is  | 0 12kk   2(ia) 3 2 3 2d 3d xxeex ++ = 2(ib) ( ) 2 3 3 2 3 d63ln 4 2 3d 4 2 9 2 xxxx x x +=  + = + k

2 2(ii) ( ) ( ) ( ) ( )( ) ( ) 2 2 1 5 d 5 3 4 d 34 345 14 5 , where C is an arbitrary constant4 3 4 x x x x x C Cx − − =− − −=+ −− =+ −  3(i) Let $x, $y and $z be the cost of mobile phone, a laptop and a television respectively. 2 2 0 (1) 3 2 8 24120 (2) 2 640 2 640 (3) Solving (1), (2) and (3) using GC, 1960 3920 1300 The cost of a mobile phone is $1960. y x x y x y z y x z x y z x y z =  − = − + + = − = + −  − + = − = = =  3(ii) Total amount Mia has to pay ( )( )( ) ( )( )( ) ( )( )( )$0.9 0.2 80 1960 0.2 80 3920 0.6 80 1300 $140832 = + + =

3 4(i) 4(ii) 32 2 32 d 36d y x x y xxx = − + =− ( ) ( ) ( ) ( ) 32 11 d 9When 0.5, 0.5 3 0.5 2 , 3 0.5 6 0.5 8 d 4 yxy x= = − + = = − =− Equation of tangent when 0.5x= , x 0 ( )0.732,0− y 32 32y x x= − +

4 11 9 1 8 4 2 11 9 9 8 4 8 8 11 18 9 8 18 20 0 4 9 10 0 or 4 9 10 0 yx yx yx yx y x y x − =− −  − =− + − =− + + − = + − = − − + = 4(iii) Area of region ( ) ( ) 4 32 3 443 3 44 33 2 3 2 d 3 243 43 4 2 4 3 2 344 8.75 units x x x xx x = − + = − +    = − + − − +       =  5(i) 6 atP ce −=+ When 0t= , 10P= 10 6 4 cc= +  = When 1t= , 8P=

5 ( ) 8 6 4 4 1 2 1ln 2 1ln 2 ln1 ln 2 ln 2 a a ec e a a − − = +  = = −= =− =− − = 5(ii) ( )ln 2 64 t Pe − =+ From GC, When t = 2, d d P t = − 0.693 ALT ( ) ( ) ( ) ln 2 ln 2 d 4 ln 2d 4ln 2 t t P et e − − =− =− When t = 2, ( ) ( ) ( ) 2 ln 2 ln 2 d 4 ln 2d 4ln 2 t P et e − − =− =−

6 5(iii) 5(iv) As ,6tP→ → since as ln 2 ln 2 1,0te e −→ = → . Therefore the approximate size of the population is 6 million. x 0 y

7 5(v) 11 22 13 22 13 22 1 1 2 2 1 2 5 dQ 1 5 d 2 2 dQWhen Q has a stationary point, 0d 15 022 15 0 2 15 10 5 1 5 when 5, 55 2 5 5 coordinates of stationary point is (5, 2 5). Q t t ttt t tt t tt tt t t t Q − −− −− =+ =− = −= −=    −= = = = = + =  5(vi) ( )ln 2 564 t P e t t − = + = + Using GC, 0.340, 9.16 milliont P Q= = =

8 6(i) Required probability = 80 1 1600 20= (ii) Require probability = ( )1600 256 64 4 1600 5 −+ = (iii) ( ) ( ) ( ) P student in year 1| student travels on fo ot P student in year 1 travels on foot P student travels on foot 400 400 112 25 32 = = + = (iv) Let A be the event that ‘the student travels by car’, and B be the event that ‘the student is in Year 1’. ( ) ( ) ( ) ( ) ( ) ( ) 432 27P 1600 100 432 144 1200 27PP 1600 1600 100 Since P P P , therefore events and are independent. AB AB A B A B AB  = = + =  =  =  7(i) ( ) Let be the random variable denoting num ber of oranges that is not ripe out of 12 oranges. ~ B(12,0.15) P( 1) 0.30122 0.301 to 3 s.f. X X X = = = (ii) P( 3) 1 P( 3) 0.092206 0.0922 (to 3 s.f.)XX = −  = =

9 (iii) Required probability ( ) =P( 3) P( 3) 0.0085020 0.00850 to 3 s.f. XX   = = (iv) ( ) Let be the random variable denoting the number of bags that were sold at the reduced price out of 20 bags. ~ B(20,0.092206) P( 4) P( 3) 0.89366 0.894 to 3 s.f. Y X YY =  = = 8(i) Number of ways 12 5 792 C= = (ii) Case 1: 5 Girls and no Boy chosen 75 50 21CC= Case 2: 4 Girls and 1 Boy chosen 75 41 175CC= Case 3: 3 Girls and 2 Boys chosen 75 32 350CC= Total number of ways 21 175 350 546 = + + =

10 (iii) 12 12 Group as 1 unit, then total permutation of 4 units = 4! Permutate the unit consisting of = 2! Number of ways = 4! 2! 48 BB BB   = 9(i) It means the probability of event A occurring given that event B does not occur. (ii) ( ) ( ) ( ) ( ) ( ) P | ' 0.8 P' 0.8P' P' 0.81 0.7 P ' 0.8 0.3 0.24 AB AB B AB AB =  =  =−  =  = (iii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P P P ' 0.4 0.24 0.16 Required probability P P P P 0.4 0.7 0.16 0.94 A B A A B AB A B A B  = −  =− =  = + −  = + − = _ _ _