JJC H2 MATH P2 Soln

1 Jurong Junior College 2015 JC2 H2 Mathematics Preliminary Examination Paper 2 Solutions Qn Solution Mark Remarks 1(a)(i) (ii)     arg i arg i arg i i 2 ab ba ab         1(b) i2 0 ( 1 )  zw *3 3 * ( 2 )     zw z w  Sub (2) into (1) i3 * 2 0 3i i * 2 0 Let i 3 i i ( i) 2 ( i) 0 (2 ) i ( 3 2 ) 0          ww ww wx y xy xy yx xy Comparing real and imaginary parts, 20 ( 3 ) 32 0 ( 4 ) Sub (3) into (4) 34 0 1 2 12 i 312 i 4 2 i                yx xy xx x y w z 2 Given   d d  t      25  ,   d d  t =  25k 1 dd25 kt      ln 25    = kt c 25    =  ekt c     25    = e e  ck t  =   e2 5 ktA , where A = e c When t = 0,  = 32 32 =   0e2 5A   A = 7 When t = 1,  = 30 30 = 7e 25  k  5ln 7k  =  ln 5 77e 25  t  = 572 57 t   (Shown) When  = 37, t =   12 7 5 7 ln 1.60ln   (3 sf)

3(i) 3(ii) The time o f r = 32r  ff r r = 323rr  = 323rr  = 323r r = 323rr    2 1  n r r   = = = = =   Let nP be for all n LHS of 1P RHS of 1P 1P is Assume P i.e.,  1 4 k r r   of death is 1 3 23rr 1r 2 2r r   2 2r r   2 32r rr  23 2rr =  1 1 f6   n r r 1 f f6 n   31 236 n n  21 26 nn   1 16 nn  the stateme  . 1P = 43  1P =  11  true. kP is true for  32 3r rr 10.24pm 313r r 32 33r r r 3 266rr  23rr =  f1r r f 0   2n n 31n  21n ent “  3 1 4 n r r   18   3 18  r some k  =  1kk  21 r   213r r 223r r  = 26r     3 23rr   ,  3 . 1    2 21r  63rr  3 1nn  ”  1r   1   ”

3 We want to prove that 1kP is true, i.e.,  32 1 43 k r rr r   =  3 11 1kk    =   3 12kk 1LHS of kP =    3232 1 43 41 31 1 k r rr r k k k     =         332 14 13 1 1kk k k k      22 11 4 1 3 1 1kk k k k      =    2212 1 4 2 1 3 3 1kk k k k k k        =   32 212 4 8 4 3 3 1kk k k k k k       =   3216 1 2 8kk k k   =   21( 2 ) 4 4kk k k   =   3 12kk = 1RHS of kP Since 1P is true, and kP is true  1kP is true,  by mathematical induction, nP is true for all n  .   3(iii)  32 1 43 n r rr r   =  3 1nn  32 11 1 43 nn n rr r rr r     =  3 1nn  3 1 4 n r r   =  3 2 11 13 nn rr nn r r     =      3 113 1 2 1 162 nnn nn n n      =    2 1111 2 1 22nn n n    =   212 1 1nn n n n    =    21nn n n  =    11nn n n =  2 1nn    3 1 n r r   =  21 14 nn    or  2 1 12 nn   

4(a) (i) (ii)   OR  = 23 5 pq d = OR OP OP   

 = 23 5 pq p p

= 23 5 pp qp p = 2 23 5 pp q p

(Shown)     4(b)(i)  Method 1 1l : r 1 01 31 k         ,  2l : 341 2 zxy      134 11 2 yzx      2l : r 41 11 32              ,  When 1l and 2l intersect, 1 01 31 k         = 41 11 32              1 3 k      = 4 1 32      k = 3 (1)   = 1 (2) 2  = 0 (3) Solving (2) and (3),  = 2 3 ,  = 1 3 Sub. into (1), 21 33k = 3 2 3 k = 10 3 k = 5 (Shown)    O P QR 3 2 d p q

5 4(b)(i)  Method 2 1l : r = 1 01 31 k         ,     x y z = 1 3       k x = 1 k , ,3yz    (1) 2l : 4x = 31 2 zy  (2) Sub. (1) into (2), 14  k = 1 = 33 2    3 k = 1 = 2  1 = 2   3 12   = 2 3 and  k = 4 (3) Sub.  = 2 3 into (3), 22 33 k = 4 2 3 k = 10 3 k = 5 [Shown]

4(b)(ii)  4(b)(iii)  12dd = = = A norm Since P 1 0 3        

[Note: O 2l , the p An equa Let N b Method AP  =    Perpend = AP n 

= 8 1 1        = 83 1  = 5 1 1         =  2 1 10 51     = 3 9 6    mal to  is n 1, 0, 3P  1 3 2      = 1 + Other points pt of intersec ation of i e the foot o d 1 Use dot p 17 01 32       dicular dista n   1 3 2 1 3 2            

2 4  = 7 14 1 1 2     1 1    1 33 2    1 3 2   n . lies on 1l ,  6 = 7 s in  that c ction of 1l a s 1 3 2    r f perpendic product 8 1 1        (re ance from A 4 or 14 2 o  P also lie can be used and 2l , 13 3R 7 .   cular from A ecall 1, 0,P A to  or 1.87 (3 sf s in  . are 4, 1Q  3 21 33,, 2  7,1, 2A   , 3 lies in f)  1,3 on , etc.]    to  . n  )

4(b)(iii)  Method 2 Vector equ Equation o Sub. (1) in    7  Sub. into ( Perpendic Find N by s uation of AN of  :    r nto (2), 7 13 22          

   39 4  (1), ular distanc solving eqn N: r =    1 3 2         = 7 1 3 2         = 7 4  = 7 14 = 7  = 1 2 ON  =     AN  =    ce from A to = A  = = ns of AN and 71 1 3 2 2       1 2  1 2 3 2 1 2 7 1 22         13 2 5 2 7 1 32            o    AN   2 13 22       7 2 or 14 2 d  3 2    ( ( 13 2 5 2 3   1 2 3 2 1      2 2 1  or 1.87 (3 (1) (2) sf)   77

5(a) Number all the sections from 1 to 90 and then generate 15 random numbers (from 1 to 90) to get a sample of 15 sections. Free from bias since every section has an equal chance of being selected/Easy to conduct 5(b) Systematic sampling. Teams are evenly spread out along the trial making supervision difficult. 6(a) Let X = Weight of a bag of oats. From GC, unbiased estimate of μ = 991.2x  unbiased estimate of σ2 = 22 23.967 574.42s  0H : 1000  1H : 1000  Since n = 10 is small, we need to assume that X has a normal distribution. Under H 0, Test statistic , 2 10 1000 ~( 9 ) S XTt   = 0.05 From GC, p-value = 0.13773 = 0.138 Since p-value = 0.138 >  = 0.05, we do not reject 0H at the 5% level of significance and co nclude there is insufficient evidence that Sheena’s suspicion is valid/the mean weight is less than 1 kg. Assumption: The population distribution of the we ight of a bag of oats is a normal distribution. 6(b) Because the sample mean is more than 1 kg, test statistic is positive and does not fall inside the critical region. The conclusion is unreliable because the sample is not random/may be biased/the manager may select relatively heavy bags. 7(a)(i) No. of numbers formed 4! 2! 48 7(a) (ii) No of numbers where I1 is first = 4! No of numbers where 5 is lastI = 4! No. of numbers where 1 is firstI 5and is lastI = 3! No. of numbers formed 4! 4! 3! 42

9 7(b) (i) P(Ivan improves on each subsequent jump) = 0.75 0.3 = 0.225 7(b) (ii) P(Ivan improves on his lst jump) = 0.75 0.25 0.4 or 0.75 0.3 0.75 0.7 0.25 0.4    = 0.85 [shown] 7(b) (iii) P(2nd jump is longest | Ivan improves on lst jump) = P(2nd jump is longest and improves on lst jump) P(Ivan improves on his lst jump) = P(2nd jump is the longest) P(Ivan improves on his lst jump) = 0.75 0.7 0.85  = 21 or 0.61834 8(a) Let T = Time taken by Augustine to complete the course. Then T ~ N(45.1, 1.92+ 1.52 + 1.32) = N(45.1, 7.55). P(T < 47) = 0.75537  0.755 8(b) (i) Let X = Time taken by Augustine to complete the r