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⇒= + + Given 0 20C = and 0 15T = . Using the recurrence from the question 1 00 0.8 0.2 5 24CCT= + += Hence 0 52.5 20 32.5C AB AB=++ = ⇒+= − ( ) ( )1 0.6 0.9 52.5 24 0.6 0.9 28.5CA B A B= + += ⇒+= − Using GC 2.5A=− and 30B=− Therefore ( ) ( )52.5 2.5 0.6 30 0.9 nn nC = −−

6 5 Suggested Solution (a) ( ) ( ) 2 22 22 xR y r y r xR − += = ± −− ( ) ( )( ) ( ) ( ) 2222 22 22 2d 22 d 4d Rr Rr Rr Rr Rr Rr V x r xR r xR x x r xR x xr x R x π π π + − + − + − = −− − − −− = −− = −− ⌠⌡ ⌠⌡ ∫ ( ) ( ) , , xRru Rr R r xRru Rr Rr =− = −−= − =+ = +−= And u xR xRu=−⇒=+ ( ) 224d r r V Ru r u uπ − = +−∫ (shown) (b)(i) ( ) ( ) 22Let f u Ru r u= +− 0ur=− ( )f0 r−= 1 2 ru =− 233f 2 2 4 22 r rr r R Rr   −=− =−       2 0u = ( )f0 Rr= 3 2 ru = 3 f 2 22 rr Rr = +   4ur= ( )f0 r = () 42 r rrh −−= = By Simpson’s rule, ( ) 2 3324 0 42 403 22 22 2 4 323 4 23 13 2 , 3 , 1 r rrV R r Rr R r r Rr Rr Rr abc π π π   ≈ ×+− +++ +    = + = + ∴= = = (b)(ii) If r is halved 1 2 r⇒ and R is doubled 2R⇒ , and since 2V Rr∝ ( ) 2 2112 22R r Rr = The answer in (b)(i) would be halved.

7 6 Suggested Solution (a) 12 π1 2 πi ii 00 e ee nk n k n n kk θ θ −−  + = = =∑∑ 2πi i 2πi i2 π i 2πi i2 π 1e e 1e 1ee 1e 0 since e 1 n n n n θ θ  − =   −   −=  − = = But on the other hand, by Euler’s formula, 12 πi 0 1 0 11 00 0e 2 π 2π cos is in 2 π 2π = cos i s in nk n k n k nn kk kk nn kk nn θ θθ θθ −  +  = − = −− = = =   = ++ +      ++ +   ∑ ∑ ∑∑ Equating real parts gives 1 0 2 πcos 0 n k k nθ − =  +=∑ where 1n> . (b) 1nA− 0A 1A P • O • 2A In general, in triangle kPOA , cosine rule gives ( )( ) 222 22 2 π2 cos 2 π 2 cos , 0,1, 2, , 1 kk k kPA OP OA OP OA n kp r pr k nn θ θ =+− + =+− + = −    Adding gives 22 2 01 1 nPA PA PA −+ ++ 

8 1 2 0 n k k PA − = =∑ 1 22 0 2 π2 cos n k kp r pr nθ − =  = +− +  ∑ ( ) 1 22 0 0 2 π2 cos n k kn r p pr nθ − = = +− + ∑     ( ) 22nr p= + (c) When P is the centre O of the circle, 0OP p= = . 2X nr= . When P lies on the circumference of the circle, OP p r= = . 22Y nr= . 2 2 2 2Y nr X nr= = .

9 7 Suggested Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 22 2 2 3 d6 2 ( 1)d 2 2 3 3d d6 ay x a x ya y ax x axx ax x ax ax ax x axa x axa xy x ay = − = − +⋅ − − = −− − =−−−  = −− −−= (b) (c) For region R, 0 xa≤≤ . Considering the upper half of region R where 0y≥ , we have ( ) ( ) 2 2 3 3 x ax x axyy a a −−= ⇒= . ( )( ) ( ) ( ) ( )333d d 2366 33 a x ax ax axy x axxa x x aa aa −− − −= = =   −   ( ) 22 2 d311 d 23 31 12 y ax x ax ax ax −+= +   −= + 2212 6 9 12 ax a ax x ax +− += ( ) 22 2 69 12 3 12 a ax x ax ax ax ++= += Required arc length ( ),0a ( )0,0 2,39 aa  2,39 aa −

10 ( ) ( ) ( ) 2 0 2 0 0 0 11 22 0 13 22 0 d21 d d 32d 12 3 d 3 13 d 3 1 3d 3 1 22 3 1 22 3 1 434 33 a a a a a a y xx ax xax ax x ax ax x ax ax x x a ax x a aa aa aa − = ×+  += × += += = + = +   = + = = ⌠  ⌡ ⌠  ⌡ ⌠⌡ ⌠⌡ ∫ (d) Surface area generated ( ) 2 0 2 0 0 2 0 d21 d d 32d 12 3 d 3 3 d 3 a a a a yπx x x axπx x ax axπx x ax ax xπx ax = +  += += ⋅ += ⌠  ⌡ ⌠  ⌡ ⌠⌡ ⌠ ⌡ 3 2 0 35 22 0 3d 3 26 353 a a π ax x x a πa xx a = + = +  ∫ 55 22 5 222 26 353 28 28 28 3 or 15 453 15 3 π aa a π aa π aπ a = +   = =  Surface Area by the entire loop about y-axis = 2228 3 56 32 45 45a π aπ×=