NYJC_VJC_TJC_9649_2025_Prelim P1 Solutions

1 2025 JC2 FM Prelim Paper 1 Solution 1 Suggested Answers (a) ( ) ( )( ) ( ) 2 2 d1 180d 175 1 90 8100175 90 324 175 7 y y yhx yh y h = −− = −− + − −= − +− For d 0d y x ≥ for some values of y, 324 07 324 7 h h −≥ ≤ 3240 7h∴<≤ (b) ( ) 2 90d 324 25d 175 7 yy x −= − +− Let d 0d y x = ( ) ( ) 2 2 90 324 25 0175 7 90 3725 90 3725 y y y −− + −= −= = ± 90 3725− is the unstable equilibrium soln 90 3725+ is the stable equilibrium soln

2 2 Suggested Solution (a) dcos 2 2 sin 2 cos 2d yyx x x x xx= ⇒= − + At P, 11 112 sin 2 cos 2 2 tan 2 t an22xx x x x xx − = ⇒ =⇒=  111tan 022x x − ∴− =  (b) Let 111f ( ) tan 22xx x − = −  f(0.3) 0.21518 0= −< and f(0.5) 0.10730 0= > Since f is continuous in the interval [0.3, 0.5] and there is a sign change, there must be at least one root in the interval [0.3, 0.5]. (c) 2 3 4 0.51518 0.38521 0.45717 x x x = = = (d) O y x

3 3 Solutions (a) 0 0 f (4) 4 f (0) 5 4 f (4) + 1 f (0)x += = (b) 1 f( ) ,f( ) n nn n xxx x + = − ′ where 1f() 4 2 xx x ′ = − 1 2 3 4 5 6 7 9 8 24.38936 12.25320 6.22125 3.26743 1.89602 1.36799 1.25436 1.24848 1.248 (3 d.p) 1.24846 1.248 (3 d.p) x x x x x x x x x = = = = = = = = = = = Thus 1.248 (3 d.p.)β = Since f(1.2475) 0.00440 0, f(1.2485) 0.000141 0 => = −< and f is continuous over (1.2475, 1.2485), the result obtained is correct to 3 decimal places. (c) 4 f( ) f( 1 4 2 For 1 40 ) 0: 2 1 x x x x x x x ′ −= = ′ − = = Newton-Raphson method would fail if 0 1 4x ≤ y x O 0.25

4 0 4 0 f( ) fi (0) f( ) + f( 0 ) 1.e. kkx k ≤+= ( )f0 2 = , thus ( ) 2f2 2kkk=−− 2 2 2 1 422 1 0 (2 1 ) 2 2 42 2 42 kk k k k k k k k ≤ −+− − − − ≤ ≤ Since k β> , ( 1.248β = from part (ii)), 4.24k ≥ ) y x O

5 4 Suggested Solution (a) As n→∞ , *nCC→ , 1 *nCC− → , *nTT→ and 1 *nTT− → * 0.8 * 0.2 * 5 0.2 * 0.2 * 5 0C CT CT= ++ ⇒ −− = --------(1) * 0.1 * 0.7 * 3 0.1 * 0.3 * 3 0T CT CT= + +⇒ − += ---------(2) Using GC, * 52.5C = and * 27.5T = (b) Using * 52.5nn n nX CC C X=−⇒=+ * 27.5nn nnYT T TY=−⇒=+ Therefore ( ) ( )1152.5 0.8 52.5 0.2 27.5 5nn nXX Y −−+= + + + + 110.8 42 0.2 5.5 5 52.5nn nXX Y −−⇒ = + + + +− 110.8 0.2nn nXX Y −−⇒= + ----------- (3) Similarly ( ) ( )1127.5 0.1 52.5 0.7 27.5 3nn nYX Y −−+= + + + + 110.1 0.7nn nYX Y −−⇒= + ------------(4) From (3): 1 11 0.8 540.2 nn n n n XXY XX− −− −= = − ----------- (5) Sub (5) into (4): ( )11 15 4 0.1 0.7 5 4n n n nnX X X XX+− −⇒ −= + − 11 15 4 0.1 3.5 2.8nn n n nXX X X X+− −⇒ −= + − 115 7.5 2.7n nnX XX+−⇒=− 121.5 0.54nn nXX X −−⇒= − (c) 121.5 0.54nn nXX X −−= − Auxiliary equation: 2 1.5 0.54 0λλ−+= Using GC: 0.6λ = or 0.9λ = Therefore ( ) ( )0.6 0.9 nn nXA B= + for 0n≥ ( ) ( )0.6 0.9 52.5 nn nCA B⇒= + + G