ASRJC 9649 2025 Prelim P2 Solutions

1 Consider 21nnnUUU−− += 1 22 1 nn nn UU UU − −− += 11 2 12 1 n nn n nn U UU U UU −− − −− +=   ( )111 n nnR RR−−+= If sequence converges, 1nRL− → , nRL→ when n→∞ where L is the limit of the sequence 21 LL∴+ = 2 10LL−−= ( )( )1 1 41 1 15 22L ±− − ±= = 15 2 += since all the Fibonacci numbers are positive, where a = 1 and b = 5 2 (a) [ ] 2 3 f () f () 0 f ()0 o r f ()0 2f( ) nn nn n xx xx x ′′ ′′= ⇒= = ′ This implies that the function f is a linear function, i.e. it is of the form f( )x ax b= + . (b) 1f () l n,f() l n 1 ,f()x xx x x x x ′ ′′= = += n nx (Newton-Raphson) nx (Chebyshev) 1 5 5 2 1.916121 1.551663 3 1.161072 1.027829 4 1.010204 1.000013 5 1.000051 1.000000 At each iteration, Chebyshev’s results are markedly closer to the actual root, demonstrating a better convergence rate compared to Newton-Raphson method. (c) Using Chebyshev’s method, 2 2.5277868 0x = −< , which is outside the domain of f. We cannot continue to compute 3x as 2ln x is undefined. 3 (a) ( ) ( ) 22f 4 11 2 7x xx y x y= +− + + − ( ) ( ) 22f 2 11 4 7y x y yx y= +− + + − ( ) ( ) ( ) ( ) ( )( ) 22 22 41 2 1 1 21 2 7 36 9f 1, 2 432 82 1 2 11 4 2 1 2 7  +− + + − −  ∇= == −  − +− + + −   

For steepest descent, the unit vector is in the opposite direction of the gradient vector, hence unit vector is 91 8145   . ( )f 1,2 4 145uD =− (b) ( ) 22f 4 11 8 2xx xy x= +− + + ( ) 22f 28 4 7yy y xy= + + +− f 44xy xy= + At ( )( )3 ,0 ,f 3 ,0 , ( )f 3, 0 20= , f 32x =− , f4y =− , f 66xx = , f 14yy =− , f 12xy = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 21, f 3, 0 3 f 3, 0 f 3, 0 3 f 3, 0 3 f 3, 0 2 x y xx xyQxy x y x yx≈ +− + + − + − ( ) 21 f 3, 02 yyy+ ( ) ( ) ( ) ( ) 2 2, 20 32 3 4 33 3 12 3 7Q x y xy x y xy≈− −− + − + −− For saddle point, ( ) ( ) ( ), 0 32 66 3 12 0 33 6 115 1xQ xy x y x y= ⇒− + − + = ⇒ + = −−−−− ( ) ( ) ( ), 0 4 12 3 14 0 6 7 20 2yQ xy x y x y= ⇒− + − − = ⇒ − = −−−−− Solving gives 925 103.46, 0.112267 89xy= ≈= ≈ 266 0, 14, 12 1068 0xx yy xy xx yy xyQ Q Q QQ Q= > = − = ⇒ − = − <⇒ saddle point, shown Coordinates are ( )3.46,0,112,13.7 . 4 (a) ( ) ( ) ( )det 1 7 12 2 0 4 3 0 1 0= − − −+ −=E Since det 0 =E , the word vectors are linearly dependent. (b) Set containing any 2 columns in the matrix E. (c) 13 13 123 6 123 6 1236 014 9 014 9 0149 1 3 7 18 0 1 4 12 0 0 0 3 RR RR−+ −+         → →          The system is inconsistent (last row is not a whole row of zeros). The vector v cannot be represented in the model’s current embedding space. The model lacks the capacity to understand or process this particular word/concept, as it lies outside the span of the stored word vectors. (d) 12310 1 0 1 40140 21 151 5137 − −−   = =       AE . Note that the two rows are linearly independent. So rank(AE) = 2. Using rank-nullity theorem, nullity( ) 3 2 1=−=AE .

Solving 014 0 1 5 15 0 x y z −−    =       : 1 12 1 25014 014 1 05 151 5 10 5 014 R RR R R − +↔−− −− −    → →   −    50 5 40 4 xz x z yz y z − =⇒= + =⇒= − 5 4, 1 x yz z z      = −∈        . A basis of the null space is 5 4 1  − . 5 (a) Let ieur θ= . Then 6 6 i6 eur θ= . So 66Re( ) cos6 0zr θ= < . Solving cos6 0θ < , we get ( 1 4) ( 3 4),12 12 kkππθ ++∈ for 3, 2, 1,0,1,2k = −−− . (b) 3 3 1arg 12 v v π − =− + 3 3 33 3 3 1 i, where 01 1 i( 1) (1 i) 1 i 1i 1i v kkv v kv vk k kv k − = <+ −= + −= + += − 2 3 2 1i 1 111i 1 k kvv k k + += = = ⇒=− + 3arg( ) arg(1 i) arg(1 i) 2arg(1 i)v kk k= +− −= + . Since k can be any negative real number, arg(1 i) 02 kπ−< + < . This gives (2 1) 22 3arg( ) 2 arg( ) fo r 1,0,133 kkk vk v kππππ π −−< < ⇒ < < = − . The locus consists of three arcs of distance 1 unit from the origin. Total length of arcs 3 1 units3 π π= ×= . Re Im

(c) (i) (ii) max 3 10 3z MC CD+= + = + (iii) Point A has coordinates 4 3 862 ,2 ,5 5 55     −= −        . 3 65 9tan 3 8 5 23AKC −∠= = + . Also, 3sin 7EKC∠= . Range of values of arg( 3 3i)z−− is 11 93, tan sin ,23 7ππ π π −−  −−+ ∪ −     . (iv) Consider the amount of anticlockwise rotation θ about the origin such that the rotated point B would have y-coordinate = 1− . 8OB= . 1 3tan 4MOB −∠= . Required angle 111131sin tan sin ( 0.769 rad)848MOBθ −−−= ∠+ = + ≈ 6 (a) Width of 95% CI ˆˆ(1 )2(1.96) pp n −= The maximum value that can be attained by ˆˆ(1 )pp − occurs when 1ˆ 2p= . x x K(3, 3) O M(–3, 0) C A B (–4, 3) D E 3 Re Im

Therefore, the maximum width 0.5(1 0.5) 1.96 22(1.96) n nn −= = < . (b) Sample proportion, 0.27369 0.42631ˆ 0.352p += = 0.35(0.65)2(1.96) 0.42631 0.27369 150 n n = − = (c) 14.125x = , 2 0.50931818s = Degree of freedom = 12 – 1 = 11. 3.105807t = 0.713714.1525 3.1058 12 14.1525 0.6399 13.5126,14.7924 sxt n ±⋅ = ± × = ± = 99% confidence interval (13.51,14.79)= (d) 99% of intervals calculated in a similar manner would contain the true population mean. 7 (a) H0: Choice of habitat where Anna’s Hummingbird forage is independent of the time of the day. H1: Choice of habitat where Anna’s Hummingbird forage is dependent on the time of the day. Under H0, Expected frequencies Ei = ( )( ) total grand totalcolumntotalrow iO iE Morning Midday Evening Total Urban Gardens 148158.6 54 47.58 10398.82 305 Chaparral (Mediterranean Shrubland) 6359.8 1017.94 4237.26 115 Oak Woodlands 49 41.6 1412.48 17 25.92 80 Total 260 78 162 500 Test Statistic: ( ) ( ) 2 22 ~4ii calc i OE Eχχ −=∑ degree of freedom ( )( )3131 4ν =− −=

( ) 2 2 10.611ii calc i OE Eχ −= =∑ value 0.0313p−= Since p-value = 0.0313 < 0.05, the null hypothesis is rejected, so there is significant evidence at 5% level of significance that the choice of habitat where the Anna’s Hummingbird forage is dependent on the time of the day. (b) Contributions to test statistic iO iE Morning Midday Evening Urban Gardens 0.7084 0.8663 0.1768 Chaparral (Mediterranean Shrubland) 0.1712 3.5141 0.6030 Oak Woodlands 1.3163 0.1851 3.0697 The largest contributor to the test statistic is Chaparral at midday and the observation is less than expected. Hence, the conservationist can conclude that the hummingbird prefers to forage in other habitats more than in Chaparral during midday. 8 (a) 1 3 01 41 d d1 b xx xbb +=∫∫ d1ln d yyx xx= ⇒= ( ) ( ) 0 ln 3 0 41e ed ed 1 b yy yyybb−∞ +=∫∫ ( ) 44e , 0 1h e , 0 ln 0, otherwise y y yb y yb b  ≤ = <<    (b) 0 04441 1ed eyy ybb b −∞ −∞ = =∫ Since 1120 2b b>⇒<< 0 11 3 ed 4 y y ybb∴+ =∫

0 11 3 e 4 y y bb +=  11 1 3e 4 y bb b+ −= 13 3e ln44 y byb =⇒= . The upper quartile of Y is 3ln 4 b . (c) 441ed 4 y y yb−∞ =∫ 411 e 4 yy b −∞  = 411 1e ln4 44 y byb =⇒= Interquartile range: 3 1 375ln ln ln444 128 2 bb −=  1 4 3 3754ln ln 128 2 4 b b =  1 4 3 3754 128 2 4 b b =  3 43 375 2 2 128 2 b = 43 4 125 5 64 4bb = ⇒=  (d) 1Geo 4V    ( ) 2 11 4Var 1 2 1 4 V − = =  9 (a) The average rate of occurrence of earthquakes with magnitudes 7.0 and above is constant throughout the year. The occurrences of earthquakes with magnitudes 7.0 and above are independent. (b) Let X be the random variable denoting the number of earthquakes with magnitude 7.0 and above in a