NYJC 2026 FM Prelim P2 (Ans)

1 NYJC 2026 JC2 Preliminary Examination 9649/01 2026 Preliminary Examination Further Mathematics Paper 2 Solutions 1 The linear transformation 33:T →  is defined by : xx T y y zz                  M  , 3 1 1 where 1 3 1 1 1 3 − = − − −M . (a) Find the dimension of the null space of T. [2] (b) Given that 1 2 3 1 0 1 1 1 0 01 ,, 1                       === −     v v v . Determine with a reason, whether 23( ), ( ), ( )T T T1v v v form a basis for ℝ3. [2] (c) For matrix M , 𝑠 tate the eieenvalue correspondine to the eieen vector , 1 0 1   −=u and determine an eieenvector correspondine to another eieenvalue. [3] By GC, 3 1 1 1 1 3 1 3 1 0 1 1 1 1 3 0 0 10 −−       −−      −−    → Rank(M) = 3 Dim(K) = dim( 3  ) – rank(M) = 3-3=0 Since 1 2 3 0− − =v v v 1 2 3,,v v v are linearly dependent. By linearity of ,T 23( ) ( ) ( ) 0T T T++=1v v v 23( ), ( ), ( )T T T1v v v are linearly dependent and cannot span ℝ3. 23( ), ( ), ( )T T T1v v v do not form a basis for ℝ3. =Mu u 3 1 1 1 2 1 1 3 1 0 0 2 0 1 1 3 1 2 1 −             − − = =            − − − −       The correspondine eieen value is 2

From GC, other eieenvalue is 5. Alternative: Solvine for another eieen value, 3 1 1 0, i.e. 1 3 1 0 1 1 3    −− − = − − − = −− MI 2(2 ) (5 ) 0 2 or 5   − − = == For eieenvector, we solve ( 5 ) 0−=M I w 2 1 1 0 1 2 1 0 1 1 2 0 x y z −−         − − − =        −−     20 20 20 x y z x y z x y z − − + = − − − = − − = Eieenvector is 1 1 1  − . 2 Anand and Bene set up fish farms on two identical, adjacent plots of land such that the fish populations in both farms may be modelled by the same differential equation d 21d 25PP PHt = − − , where P is the population of fish (in thousands) and H is a constant (in thousands) representine the chanee in fish population due to harvestine. One unit of t represents one harvest season. Anand started his farm with 4000 fish, but the fish population in his farm died out after several years. Bene started his farm with 16500 fish, and the fish population in his farm erew and stabilised at 20000 after many years. (a) Find the number of fish harvested each season. [2] (b) On the same eraph, illustrate how the populations of fish in Anand’s and Bene’s farm vary over time. [2] (c) Find the smallest number of fish Bene could have started his farm with to reach the same population size after many years. [2] 2 (a) Given that d 0dPt = when P = 20, 200 2(20) 1 8 25 HH= − −  = . Hence 8000 fish are harvested each season.

(b) (c) setd 2 1 8 0 5 or 20d 25PP PPt = − − =  = . If Bene starts with 5001 (least positive inteeer > 5000) fish, the model predicts that the fish population will increase to 20000 in the lone run. 3 Amy wishes to find the root of the equation 5 1.0093 0xx− − = ------ (*) First, she tries usine a Newton-Raphson iteration method with 0 0x = . (a) Find the values of 1 2 3,,x x x and 4x , and explain whether Amy will be able to find a solution for the equation. [3] (b) Sueeest what A my should have done before choosine an appropriate initial value for h er iterative process. [1] (c) Without usine a eraphine calculator, sketch two suitable curves to show that (*) has exactly one real positive root, justifyine your answer. [1] Next, Amy decides to find the root of (*) usine a fixed-point iteration formula 1 4 1.0093n n n xx x + += with 0 1x = . (d) Sketch a eraph showine how 1x and 2x are obtained usine this method, and explain why Amy will not be able to find a solution to this equation. [2] (e) Rewrite the equation (*) in a suitable form and hence use the iterative formula ( )1 gnnxx+ = to find the solution of (*) correct to 3 sienificant fieures usine 0 1x = . [3] (a) Let ( ) 5f 1.0093x x x= − − . Then ( ) 4f 5 1xx =− . 20 P t 5 16.5 4 Bene’s farm Anand’s farm

Hence the Newton-Raphson iteration is eiven by 5 1 4 1.0093 51 nn nn n xxxx x+ −−=− − . From GC, x0 = 1, x1 = −1.0093, x2 = −0.75925, x3 = 0.00015346, x4 = −1.0093. As the iterations repeat themselves with a period of 3, Amy will not be able to find a solution. (b) Amy should have sketched suitable eraphs to determine the roueh location of the root and choose an initial value nearer to the root that is also away from turnine points. (c) Sketch y = x5 and y = x + 1.0093: From the sketch it is clear that the eraphs intersect exactly once at x > 0; therefore the equation (*) has exactly one real positive root. (d) Based on the sketch, successive iterations will eet further away from the intersection, so Amy will not be able to find a solution. (e) x5 – x – 1.0093 = 0 5 1.0093xx = + From GC, x0 = 1, x1 = 1.1498, x2 = 1.1664  1.17, x3 = 1.16821  1.17 Check: f(1.165) = −0.0283 < 0 and f(1.175) = 0.0554 > 0, hence the root is 1.17 (3sf) x y x0 x1 x2 y = x x y y = x + 1.0093 y = x5

4 The polar equations of two curves are eiven by 1 : 1 sinCr =+ and 2 : 1 cos2Cr =+ for   −   . (a) Find the exact polar coordinates of all point(s) of intersection of 1C and 2C . [3] (b) Find the exact area of the reeion of 1C that lies inside 2C . [4] Let P be the point of intersection of 1C and 2C in the first quadrant. A straieht line parallel to the initial line passes throueh P and intersects 2C at another point, Q, in the first quadrant. (c) Show that at Q, 13 1sin 4 −= . [4] (d) The reeion bounded by the line seements OP, OQ and the arc of the curve 2C between P and Q forms a polar sector. Find the total perimeter of this sector, eivine your answer correct to 4 decimal places. [2] 4(a) ( )( ) 2 2 1 sin 1 cos 2 sin cos 2 sin 1 2sin 2sin sin 1 0 2sin 1 sin 1 0 1sin or sin 12 5, 6 6 2          + = + = =− + − = − + = = =− = =− 53 1 sin 1 sin6 6 2r = + = + = 1 sin 0 2r = + − = Polar coordinates of points of intersection are 3 3 5, , , , 0,2 6 2 6 2        −           , (b)

Exact area ( ) ( ) 62 26 62 26 62 26 62 26 22 22 112 1 sin d 1 cos 2 d22 1 2sin sin d 1 2cos 2 cos 2 d 3 cos 2 3 cos 42sin d 2cos 2 d2 2 2 2 3 sin 2 3 sin 42cos sin 22 4 2 8 3 2                            − − − − =  + + + = + + + + + = + − + + +    = − − + + +       =    3 3 32 0 06 2 8 2 2 3 3 3 3 002 2 2 6 2 16 9 3 9 3 8 2 16 3 27 32 16          − − − − − −                + + + − + +              = − + −          =− (c) At point P 3 ,26 , 3sin 1 cos 2 sin 6 6 4yr   = = + =   ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 2 2 3 2 2 2 3sin 1 cos 2 sin 4 31 cos 2 sin 4 32 2sin sin 4 4 2 2sin sin 3 8sin 8sin 3 0 2sin 1 4sin 2sin 3 0 4sin 2sin 3 0 2 2 4 4 3 2 2 13 1 13sin 2 4 8 4 yr             = = + = += −= −= − + = − + − = + − = −  − − −  − = = = Since Q is in first quadrant, 13 1sin 0 4 −= (shown). (d) Total perimeter of this sector ( ) ( ) 1 6 1 13 1sin 224 13 11 cos 1 cos 2 sin34 1 cos 2 2sin 2 d 3.0793 (4dp)      − −  −   −= + + +    + + + −