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=+ ,

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