Paper 2

1 2018 H2 Mathematics 9758 Paper 2 Suggested Solutions Section A (Pure Math) 1(i) 1 3d1 15d3 y yx =− 1 31 15 d 1 d3 y y x −  −= 2 31 153 21 33 y xc  − =+       , where c is an arbitrary constant 2 391 1523 y x c − = + When x = 0, y = 69 ( ) 2 391 69 1523 c −= c = 18 2 391 15 1823 yx − = + 2 3 3 2 3 2 3 2 12 15 439 12 15 439 12 4 1539 23 4 459 yx yx yx yx  − = + − = + = + + = + + 1(ii) Since d 4d y x = , 1 31 15 43 1 15 643 237 y y y  −= −= = 3 223 4 45 2379 x + + =

2 From GC, x = 54 Coordinates of the point is (54, 237) 2(a) Since the coefficients of the equation are real, the complex roots occur in conjugate pairs. Therefore since 2 – 3i is a root, then 2 + 3i is also a root. So we have ( )( )( ) ( )( ) 4 3 2 2 22 4 20 56 2 3 2 3 4 4 13 4 , x x sx x t x i x i x ax b x x x ax b − + − + = − + − − + + = − + + + where ,ab  Comparing coefficients of 3x , 20 16 4 a a − = − =− Comparing coefficients of x, 56 4 13 4 56 13( 4) 1 ba b b − =− + = + − = Comparing constants, 13 13 tb= = Comparing coefficients of 2x , 4 52 1 4( 4) 52 69 s b a= − + = − − + =  we have ( )( ) ( )( ) 4 3 2 2 2 22 4 20 69 56 13 4 13 4 4 1 4 13 2 1 x x x x x x x x x x x − + − + = − + − + = − + − The other roots are 2 + 3i and 1 2 . Note: 1 2x= is a repeated real root. (bi) 3 3 27 27 0 w w = −= Since 3w= , ( )( ) 32 27 3w w w aw b− = − + + , where ,ab 

3 Comparing constants, 27 3 9 b b − =− = Comparing coefficients of w, 03 3 3 ba ba =− = =  we have ( )( ) 32 27 3 3 9w w w w− = − + + To find the other roots, let 2 3 9 0ww+ + = 3 9 4(1)(9) 2 3 27 2 3 3 3 2 w i −  −= −  −= −= The other roots are 3 3 3 22 i− + and 3 3 3 22 i− − (bii) Let 1w = 3, 2 3 3 3 22 iw −=+ and 3 3 3 3 22 iw −=− 1 3w = 22 2 3 3 3 9 27 32 2 4 4w −= + = + =   22 3 3 3 3 9 27 32 2 4 4w −−= + = + =   1arg 0w = 1 2 33 22arg tan 3 3 2 w  −= − = − 1 3 33 22arg tan 3 3 2 w  − − =− + =− −

4 ( )1 3 cos0 sin 0wi = + , 2 223 cos sin33wi =+ and 3 223 cos sin33 223 cos sin33 wi i   −−=+ =− (biii) 22 33 1 2 3 3 3 3 27 ii w w w e e  −  ==      1 2 3 2 2 2 23 3 cos sin 3 cos sin3 3 3 3 1 3 1 33 3 3 2 2 2 2 0 w w w i i ii       + + = + + + −          −−= + + + −          = 3(i) 5 5 5 4 4 4 2 0 1 5 4 3 AD BC OD OA OC OB OD OC OB OA OD = − = − = − + −           = − + −                − =− ( )5, 4,3D − − 3 Re(w) Im(w) 3 3

5 (ii) 5 0 5 4 0 4 2 10 8 EC −−           = − =           −      5 0 5 4 0 4 0 10 10 EB            = − =           −      5 5 8 4 4 90 8 10 40 EC EB −−            =  = −          − − −      Normal vector of plane BCE is 4 2 45 20  −  Therefore the equation of plane BCE is 4 4 0 45 45 0 200 20 20 10 r            ==                 Cartesian equation of plane BCE is 4x + 45y + 20z = 200 (iii) 5 5 10 4 4 0 2 0 2 BC −−           = − =                5 5 0 4 4 8 0 1 1 AB            = − − =           −      10 0 16 8 0 8 10 2 5 2 1 80 40 BC AB −−                =  = − =−              −−        Normal vector of base of the pyramid is 8 5 40    Let the angle between the planes be  0 84 5 45 40 20 1057cos 1689 2441 1689 2441 58.6                    == =

6 (iv) Let OM be the midpoint of AD OM = 55 44 013 42 2 −       − + −            =− ME = 0 0 0 0 4 4 10 2 8            − − =                Distance from M to the face BCE = length of projection of ME on 4 45 20    2 2 2 4 0 4 45 4 45 20 8 20 340 6.88 2441 24414 45 20 ME                            = = = = ++ 4(i) ( ) ( ) 224 1cos 1 2! 4! 2 ! r rxxxx r −= − + − + + ( ) ( ) ( ) 2 4 6 2 4 6 2 2 2cos 2 1 ... 2! 4! 6! 241 2 ... 3 45 x x xx x x x = − + − + = − + − + ( ) 123 1ln(1 ) 23 r rxxxxx r + −+ = − + − + +

7 2 4 6 23 2 4 6 2 4 6 2 4 6 466 6 2 4 6 2 4 6 4 6 6 24 24ln(cos 2 ) ln 1 2 3 45 2 4 2 42224 3 45 3 452 3 45 2 3 4442 4 8 332 3 45 2 3 2 4 4 822 3 45 3 3 4 642 3 x x x x x x x x x x x x x x x x xx x x x x x x x x xx  − + −    − + − − + −        − + − − +  −− −− + − − +    =− + − − + − =− − − 6 45 x Expansion will be not valid for cos 2 0x= since ln(0) will be undefined. cos 2 0x= 2 2 4 x x   = = Hence when 4x = , expansion will not be valid. (ii) ( ) 2 4 6 22 24 35 4 642ln cos 2 3 45dd 4 642d 3 45 4 642 9 225 x x xx xx xx x x x x x x c − − − = = − − − =− − − +   ( ) 0.50.5 35 2 0 0 ln cos 2 4 64d2 9 225 1.0644 x x x x x x = − − − −  (iii) Using GC, ( )0.5 2 0 ln cos 2 d 1.0670x x x − Section (B) Statistics 5(i) Since the time to failure does not follow normal distribution, in order for the mean time to failure to follow normal distribution, sample size must be at least 30 for the manager to apply Central Limit Theorem. The fans should be randomly chosen. That means each fan in the population must have an equal chance of being selected and the fans must be selected independently of one another.

8 (ii) Let X be the random variable “time to failure of a fan in hours” with population mean  . 0H : 65000= 1H : 65000 (iii) At the 5% level of significance. Under 0H , since n = 43 is sufficiently large, by Central Limit Thorem, 2 ~ N 65000, 43 sX   approximately Using a 1-tailed z-test, Critical value: z = − 1.64485 In order not to reject, test stat value 0 2 1.644853626calc xz s n −= − 64230 65000 1.64485 43 770 1.64485 43 468.13 43 3069.7 s s s s − − − −   2 2 9423264.967 943000 (3 significant figures) s s   6(i) Let X be the random variable that represents the number of steps to the left out of 8 steps. ~ B(8, )Xp 53 53 P(bug finished at ) P( 5) 8 5 56 (sho wn) DX pq pq == = = The bug has to make the decision to make the right or left turn at 8 different junctions, and at 5 of the 8 junctions, bug has to make a left turn, that’s why 8 5   .

9 (ii) Since P( 5)X = is the highest, X = 5 is the modal number. Therefore ( )P 5 P( 4)XX=  = and ( )P 5 P( 6)XX=  = 5 3 4 4 856 4p q p q  and 5 3 6 2 856 6p q p q  5 3 4 456 70p q p q and 5 3 6 256 28p q p q 5 4pq and 2pq ( )5 14pp  − and ( )21pp  − 5 9p and 2 3p Therefore the possible range of values of p is 52 93 p . (iii) Let Y be the random variable that represents the number of steps out of 8 that ends up at the black hole. ~ B(8,0.1)Y P(reaching an endpoint) P( 0) 0. 430 (3sf) Y== = Alterative ( ) 8 P(reaching an endpoint) P(not being swall owed by a black hole at all 8 junctions) 0. 9 0.430 = = = 7(i) ( ) ( )P P( ) P( ) P (sin ce , ar e independent) A B A B A B a b ab A B  = + −  = + − ( ) ( )P ' ' 1 P