NYJC TJC VJC FM 2024 Prelim P2 Solution

1 2024 TJC H2 FM Prelim Paper 2 [100 marks] Section A: Pure Mathematics [50 marks] 1 (a) Solve the equation 3 32 32 3iz =− , giving your answers 1z , 2z and 3z in exact form ie,r  where 0r  ,   −   and 1 2 3arg arg argz z z . [3] (b) The roots 1z , 2z and 3z in part (a) are represented by points 1P , 2P and 3P . Find the exact perimeter of triangle 1 2 3P P P . [2] Solution (a) 32 32 3i 32 1 3i− = − ( ) 2232 1 3 64= + = ( ) ( )arg 32 32 3i arg 1 3i− =− + 3 =− 3 32 32 3iz =− i23 364e k z  −+= ( )i 1 63 364e k z  −+ = ( )i 1 694e k z  −+ = where 0, 1k = 75i i i9 9 9 1 2 3 4e , 4e , 4ez z z   −−  = = = (b) Since the points are equally space apart along the circle, triangle 1 2 3P P P is an equilateral triangle. Length 1 3 1 3PP z z=− Method 1 Using cosine rule

2 2 2 2 1 3 1 1 1 1 22 cos 3z z z z z z − = + − ( ) 22 14 4 2 16 2 = + − −  ( ) 234= 13 43zz − = Therefore, perimeter of the triangle is 3 4 3 12 3= Method 2 5ii99 1 3 1 3 4e 4ePP z z − = − = − 6ii994e 1 e − =−   2 2 2i i i6 6 64 e e e   −=−   4 2isin 3 =−  43= Therefore, perimeter of the triangle is 3 4 3 12 3=

3 2 The Fibonacci numbers nF are defined by the conditions 0 0F = , 1 1F = and 11n n nF F F+−=+ for all 1n . (a) Compute 23,FF , 4F and 5F . [1] (b) Compute 2 11n n nF F F+− − for 1,2,3,4n= . [2] (c) Conjecture and prove by induction, for all 1n , an expression for 2 11n n nF F F+− − . [6] Solution (a) 24 531, 2, 3, 5F F F F= = = = (b) When 1n= ( )2 2 0 1 1 0 1 1F F F− = − =− When 2n= ( )2 3 1 2 2 1 1 1F F F− = − = When 3n= ( )43 2 2 2 3 1 2 1F F F− = − =− When 4n= ( )54 2 3 2 5 2 3 1F F F− = − = (c) Conjecture: ( ) 2 11 1 n n n nF F F+− − = − Let nP be the statement that ( ) 2 11 1 n n n nF F F+− − = − for n + . When 1n= , it is proved in part (a) Assume kP is true for some k + , ie. ( ) 2 11 1 k kk kF F F+− − = − . When 1nk=+ , ( )( ) 22 21 1 1 1 1kk k k k k k kF F F F F F F F++ + + − +− = + − − 22 1 1 1 1 1 1k k k k k k k kF F F F F F F F+ + + − − += + − − − 1 1 1 1k k k k k kF F F F F F+ + − −= − − ( ) 22 1 1 1 1 1 k k k k k k k k kF F F F F F F F+ + − −− −= − + − − ( ) 12 11 1 k k k k kkF F F F F + +−= − − + − ( ) ( ) 1 11 1 k k k k kF F F F + +−= − − + − ( ) 1 1 k+ =− Since 11 kkkF F F+−=+ Alternative solution: When 1nk=+ ,

4 ( )1 22 2 1 1k k k kk kkF F F F F F F++ + +− = + − ( ) 2 1 1 1k k k k k kF F F F F F+ + −= + − + 2 1 1 1 1k k k k k k kF F F F F F F+ + + −= + − − 2 11k k kF F F +−=− ( ) 2 11k k kF F F+−=− − ( )1 k =− − ( ) 1 1 k+ =− Thus 1kP + is true. Since 1P is true, and kP is true implies 1kP + is true. By Mathematical Induction, nP is true for all positive integer n. Extension of question: Explain why 2 consecutive Fibonacci numbers have no common factor greater than 1. [Solution]: ( ) 2 11 1 n n n nF F F+− − = − Consider the common factor between 1nF − and nF . It divides the LHS of the expression above, this also implies it divides ( )1 n − . This means the common factor must only be 1. Thus 2 consecutive Fibonacci numbers have no common factor greater than 1.

5 3 At the start of the month in January 2023, Sandy started her career as a lifestyle vlog content creator with 60 subscribers for her channel. By the end of each month, she lost 5% of her existing subscribers but gained 100 new subscribers each month through her publicity efforts. Let nu be the number of subscribers at the end of the nth month taking January 2023 as the first month. (a) Write down a recurrence relation to model the number of subscribers at the end of the nth month. [2] (b) Find the general formula of nu in terms of n. Hence show that the number of subscribers at the end of December 2023 is 952, correct to the nearest integer. [4] (c) Based on the model in (a), comment on the long-term prospect of Sandy’s career. [2] At the start of January 2024, Sandy decided to collaborate with another well -known content creator to gain more exposure for her own channel. By the end of each month, she gained k% of the number of existing subscribers but lost 100 existing subscribers. Let nw be the number of subscribers at the end of the nth month taking January 2024 as the first month. (d) By considering the general formula of nw , find an ineuuality in terms of k that will lead to an increase in the number of subscribers in the long run. Hence find the least integer value of k. [4] Solution (a) 10.95 100nnuu −=+ , 1n 0 60u = (b) ( )20.95 0.95 100 100nnuu −= + + ( )2 20.95 0.95 100 100nnuu −= + + ( ) ( )3 20.95 0.95 100 0.95 100 100nnuu −= + + + ( ) ( ) 2 3 30.95 0.95 100 0.95 100 100nnuu −= + + + ( ) ( ) ( ) 12 0.95 0.95 100 0.95 100 100 n n nn nnuu − −−= + + + + ( )0 210.95 100 1 0.95 0.95 0.95nn nuu −= + + + + + (essential working) ( ) 1 0.950.95 60 100 1 0.95 n n n u  −=+  − ( ) ( )0.95 60 2000 1 0.95nn nu = + − ( )2000 1940 0.95 n nu =−

6 Alternative method: 10.95 100nnuu −=+ Let ( )10.95nnuu  −− = − 0.95 100 2000  − =  = 1 2000 0.952000 n n u u − − =− ( )02000 2000 0.95 n nuu− = − ( )60 2000 0.95 2000n nu = − + ( )2000 1940 0.95 n nu =− At the end of December 2023, n = 12 ( ) 12 12 2000 1940 0.95 951.7014 952u = − = = (nearest integer) (c) Using the model ( )2000 1940 0.95 n nu =− as n→ , 0.95 0n → 2000nu → . From the GC, we observe that the number of subscribers increases slowly and stabilise to 2000 in the long run. Which means she will never have more than 2000 subscribers. (d) At the start of January 2024, she should have 952 subscriber from the previous model. 11 100100 nn kww − = + − , 1n 0 952u = Using the pattern from (b) 2 0 1 1 100 1 1 1 1100 100 100 100 n n n k k k kwu −       = + − + + + + + + +                 ( ) 11 1001 952 100100 11 100 n n n k kw k  −+  = + −   −+  ( ) 100001 952 1 1100 100 n nn kkw k    = + + − +          10000 100001 952100 n n kw kk    = + − +       In the long run 1 100 n k+ → therefore to have a positive increase

7 10000952 0 k− 10000 10.504952k = Alternative method: 11 100100 nn kww − = + − Consider: ( )1 , where 1 100 nn kw r w r −− = − = + 100 10000100 1r rk  − =−  = = − 1 n n w rw  − − =− ( )( )0 n nw w r− = − ( )10000 10000952 n nwr kk − = −  10000 10000952 1 100 n n kw kk   = − + +     In the long run, 1 100 n k+ → , therefore to have a positive increase 10000952 0 k− 10000 10.504952k = Therefore, least integer value of k is 11.

8 4 A car travels over a rough surface. The vertical motion of the front suspension is modelled by the differential euuation 2 2 d 25 30cos5 d y yt t += where y is the vertical displacement of the top of the suspension and t is time. (a) Find the general solution of the above di