ACJC NJC RVHS SAJC 2026 FM Prelim P2 (Ans)

1 Qn Suggested Solution 1(a) Let 111 123 23 k  = A 111 det( ) 1 2 3 23 1( 2 9 ) 1( 1 6 ) 1( 3 4 ) 29 (6 ) 1 4 k kk kk k = =⋅ − −⋅ ⋅ − +⋅ − =− − − − = − A det( ) 0=A when k = 4. Hence nonzero solutions exist when k = 4. (b) When k = 4, 10 1 RREF of 0 1 2 . 00 0 − =  A 3 1 :2 , 1 k xx Sy y tt zz        =∈ = − ∈           (c) 1 0 0 T   ∈ 3 11 :0 2 , 01 xx Ty y t t zz          =∈ = + − ∈              Since 1 20 0 T   ∉ , T is not closed under scalar multiplication, hence T is not a vector subspace of 3 . Or Since 0 0 0 T   ∉ , hence T is not a vector subspace of 3 . 2(a) ()d 11d d1 1 d yxy xx yx yx xx +− = −+=  Integrating factor: () 1 d lnIe e e x x x xxxx x− −−== = . Thus,

2 () ()d1 eed ee d e e e e1 . xx xx x x x x yx xxx yx x c cy x cy x −− −− − − − = = =− + −+= −=  (b) E.g., 0c = : E.g. 1c =− : (c) The curve e1xy x−= does not have any vertical asymptote (despite having a denominator of x). This is because the Maclaurin series gives 2 2 3 e 1 ... 2! 3! . 1 1 1 1. . 2! 3! x y x x x xx x x −= =− + ++ + +  =+ + + y x O y x O

3 As 0x → , 2e1 00 1 ... 12! 3!xx− →+ + + = which is a finite value. 3(a) Taking partial derivatives, we have 24T xyx∂ =+ −∂ 25T x yy∂ =+ −∂ At P , () ()21 1 4 1Tx∂ =+ − = −∂ and () ()12 15 2Ty∂ =+ − = −∂ , i.e., 1 2T −∇= − . Maximum rate of increase is 5T∇= . 3(b) Since the directional derivative of T at P in the direction of u is zero, 0T∇⋅= u . Taking cos sinu θθ=  , we have cos 2sin 0T θθ∇⋅= − − =u Solving, 1tan 2θ =− . Thus, 1 1π tan 2θ − =−  . 3(c) At ()1,1, 6− , () ()00 0 xyzz Txx Tyy−= − + − , we have () ()61 2 1zxy+= − − − − . Thus, 23xy z++ = − . 3(d) Normal vector for S at ()1,1, 6− is 1 2 1    . Thus, 11 12 61 λ      =+    −  r , λ ∈  .

4 4(a) AO1 () () () f 0.75 0.21191 0 f1 1 0 f 1.25 0.65527 0 => =− < => As f is a continuous function, th e sign change between 0.75 and 1 implies that there is at least 1 root between 0.75 and 1. Likewise, at least 1 root between 1 and 1.25. 4(b) AO1 () () 52 1 4 f 38 4 f ' 15 16 n nn nn n nn n x xxxx x xx x+ −+=− =− − 0.670 (to 3 dp)γ∴= − 4(c) AO3 The value of x = 1 is near to a turning point of the curve where ()f' 0x ≈ , which may cause iterates to diverge or converge to another root. 4(d) AO2 As n →∞ , if nxL→ , then 1nxL+ → . Thus, 42 5 5 231 83 4 3 84 082LL LL L L L=+  =+  −+ = From the given info, or or L αβγ= . However, if 0nx > , then 41 31 082nn nxx x+ =+ > . So, L cannot be negative, and or L αβ= . 4(e) AO2 Let () 431g 82nn nxx x=+ , so () 3 231g' 22nn nxx x=− . Use the GC to plot the graph of ()g' nyx= for ()0, 75,1.25nx ∈

5 Thus, ()'1 ngx < for 0.75 1 nx<< around the neighbourhood of α . Hence, the iteration would converge to α . However, ()'1 ngx > for 11 . 2 5nx<< around the neighbourhood of β . Hence, the iteration would not converge to β . Using an initial value of 0 0.9x = (for example), 0.780 (to 3 dp)α = 4(f) AO1 + AO3 Method 1: Bisection 11 . 2 5 1.1252β +≈= Method 2: Linear interpolation

6 () () () () 1 f 1.25 1.25 f 1 f1 . 2 5 f1 0.65527 1.25 0.65527 1 1.15 (to 3 sf) β +≈ + += + = 5(ai) AO1 () 1 2 2 1 0 0 0 1 100 11 100 100 ... 1 1 1 ... 1100 100 100 11 1001 100 100 100 1001 100 nn n nn n n n CC k Ck k Ck Ck kkC α αα αα α αα α α αα − − − =+ −  =+ −+ −   =   =+ − ++ + ++       +−=+ − ×   =− + +     5(aii) AO2 Since 11 100α+> , nC →∞ as n →∞ , unless 0 100 0kC α−≤ . 0 1000 kCα∴< ≤ 5(bi) AO2 10 0 00 0 1 100 1 100 200 1 200 CC S CC C α αα α =+ − =+ − =+ For 1n ≥ , 11 11 11 1 100 200 200 1 200 200 10 200 200 nn n n nn n nn n CC C C CC C CC C αα α αα αα +− +− +−   =+ − +    =+ − −+ + = 5(bii) AO2 Auxiliary equation: 2 10 200 200ααλλ −+ + =

7 () 10 200αλλ −− = Thus, 1λ = or 200αλ = . We have repeated roots when 1 200200α α=  = . When 200α = , then nCA B n=+ . Let 00:nC A== Let ()10 0 001: 1 1nC C B B C C C== +  =+ − = ()00 0 1nCC C n C n∴=+ = + For 200α ≠ : 200 n nCA B α=+  Let 00:nC A B== + --- (1) Let 11: 200nC A B α== +  --- (2) () ( ) 10 0 0 21 : 1 200 200 200 200 200 CC B CB BC α αα α α −− = −  − =  = − From (1): 000 200200 200 CAC C ααα=− = −− 00200 200 200 200 n n CCC α ααα ∴= + −−  If 200α > , 0B > and 1200α > , so { }nC diverges to infinity. For 0 200α<< , 01 200α<< , so { }nC converges to 0200200 Cα− . Therefore, 0 0 200 3 200 400 600 3 0200 2 3C C ααα ≤  ≤−  <≤−

8 5(biii) AO3 Criticism 1: The idealised technological model seems to assert that ever more anthropogenic carbon can be stored, especially if { }nC is increasing. This is unrealistic as one would expect a limit to the capacity of the technology. Improvement 1: Set a cap on nS , e.g. 1min ,200 200nn nSC C k αα −=+  Criticism 2: The idealised technological model operates on the basis that a constant %α of anthropogenic carbon is generated per year when it may decrease due to improvements in carbon capture technology OR increase due to continued development. Improvement 2: Allow α to depend on n or prior terms in the recurrence. 6(a) Let X and Y be the anxiety scores of students on the dietary intervention programme and the standard exerci se programme respectively. Let DX Y=− be the difference in anxiety scores, and Dm be the population median difference. Sign Test: To test 0H : 0Dm = against 1H : 0Dm < Pair 1 2 3 4 5 6 7 8 9 d –3 +4 –5 –7 –9 0 +2 –6 –10 sign – + – – – --- + – – (Pair 6 is excluded as d = 0, leaving n = 8 valid pairs.) Let S+ be the number of + signs out of 8. Under 0H , S+ ~ B(8, 0.5) From the table, 6s− = and 2s+ = . p-value = P( 2)S+ ≤ = 0.1445 > 0.05 We do not reject 0H .

9 There is insufficient evidence at 5% significance level that the dietary intervention programme is more effective than the standard exercise programme in reducing anxiety. 6(b) Wilcoxon Matched-Pair Signed Rank Test: Pair 1 2 3 4 5 6 7 8 9 d –3 +4 –5 –7 –9 0 +2 –6 –10 rank 2 3 4 6 7 --- 1 5 8 To test 0H : 0Dm = against 1H : 0Dm < Sum of positive ranks, P = 1 + 3 = 4 Sum of negative ranks, Q = 2 + 4 + 5 + 6 + 7 + 8 = 32 T = min(P, Q) = 4 From table, since 5T ≤ , we reject 0H . There is sufficient evidence at 5% significance level that the dietary intervention programme is more effective than the standard exercise programme in reducing anxiety. The conclusions of the two non-para metric tests are different. The Wilcoxon test makes use of the magnitude of the differences between matched pairs of the anxiety scores and is hence a better test compared to the sign test as it incorporates more information about the data. Hence the conclusion of the Wilcoxon matched-pair signed rank test is more reliable. 7(a) Test H0: Preferred content genre is independent of subscriber type. Against H1: Preferred content genre is dependent on subscriber type. Using a 2χ test for independence. Expected frequencies: Action Romance Documentary Comedy Horror Anime Student 85.52 52.96 35.86 55.72 39.72 50.21 Working adult 69.48 43.03 29.14 45.28 32.28 40.79 Degree of freedom (2 1)(6 1) 5=− − = 2 16.265χ= p-value 0.00613= < 0.01, we reject H0. There is sufficient evidence at 1% significance level that genre preference is dependent on the type of subscribers.

10 7(b) Contributions of each cell to the test statistic: Action Romance Documentar y Comedy Horror Anime Student 0.0270 0.0