ACJC_EJC_NJC_RVHS_9649_2024_Prelim_P1_Solutions

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q1 (a) A is a 3 4 matrix. Hence rank( ) nullity( ) 4+=AA . As 3rank( ) A , 1nullity( ) A . As the dimension of the null space is at least 1, it is not the zero vector space. (b) 21 1 1 2 1 2 2 1 3 3 3 3 2 2 33 2 3 4 8 1 1 1 3 1 1 1 3 2 3 4 8 3 6 9 15 3 6 9 15 1 1 1 3 1 0 1 1 0 1 2 2 0 1 2 2 0 3 6 6 0 0 0 0 RR R R R R R R R R R R R R  → − → − → + → − −−       = − ⎯⎯⎯→ −       −−    − − −       ⎯⎯⎯⎯⎯ → − ⎯⎯⎯⎯⎯ → −       −    A 1 0 1 1 0 0 1 2 2 2 2 0 0 0 0 0 0 0 x x z wy y z wz w − − − −            − = + − =                   11 22 10 01 x y zwz w            −     =+                  11 22ker( ) , , 10 01 T a b a b        −   = +          

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q2 (a) Using GC, 3 2 1 e sin d 100.96xI x x== (2 d.p) (b) Using Simpson’s Rule,  0 1 2 1 43I h y y y= + + ( )1 3 1 6.217676 4 49.645957 56.93187532 −=  + +   87.24446008 87.24== (2 d.p.) Percentage error 100.96 87.245 100% 13.6%100.96 −  = (3 s.f.) (c) From diagram, we observe that there is a large discrepancy in terms of area between the actual curve 2e sinxyx= and the approximated quadratic curve for the interval 23 x , which is caused by the existence of both a turning point and a point of inflexion . Hence Simpson’s Rule using 3 ordinates will not provide a good approximation in this case. Increasing the number of intervals/ordinates will give a better approximation. x O y 1 2 3 (2.68, 94.7) (3, 56.9) (2, 49.6) (1, 6.22)

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q3 (a) From the GC, k 1 2 … 50 51 52 21kM − 1 2 … 2.0288 2.0288 2.0288 2kM 2 2.25 … 4.1686 4.1686 4.1686 Since 2 4.1686kM → and 21 2.0288kM − → as k→ , The sequence  nM alternates between the 2 values of 2.0288 and 4.1686 as n→ . (b) Observe that n 1 2 3 … nM 2 2 2.25 … So the sequence in part (b) is the same as moving the terms in the sequence in part (a) up by 1 term, and it will have the same alternating behaviour but odd terms will approach 4.1686 and even terms will approach 2.0288 as n→ . (c) For a constant sequence, 12n n nM M M ++= = = for all n + . Then we have 2 2 1 2 1 2 2 22 9 4 nn n nn MMM MM    + + +  = +  = +  =

2024 H2 FM 9649 Prelim Paper 1 (ACJC_EJC_NJC_RVHS) Solution P1 Q4 (a) f ( ) 8sin 2 5 2 xxx = − + . f (4.5) 2.2246 0= and f (5) 0.21222 0=−  Since f (4.5) f (5) 0 and f is continuous in the interval  4,5 , the equation f ( ) 0x = has a root  in the interval  4.5,5 . 4.5 5   (Shown) (b) Note that 54sin 22 xx =+  is equivalent to 8sin 2 5 02 x x − + = . Let 5g( ) 4sin 22 xx