# ARC 2000S - SUGGESTED SOLUTIONS MATHEMATICS 3 UNIT 2000...

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Unformatted text preview: SUGGESTED SOLUTIONS, MATHEMATICS 3 UNIT - 2000 Question 1 (a) u = e x – 1 : when x = 0, u = e – 1 = 1 – 1 = 0 du = e x dx : when x = ln2, u = e ln2 – 1 = 2 – 1 = 1 ∴ dx e e x x ∫- 2 ln 1 ( 29 3 2 3 2 1 2 3 1 2 3 2 3 1 2 1 2 ln 1 3 2 3 2 1 =- = = = =- = ∫ ∫ u u du u dx e e x x (b) LHS x x cos 3 cos = RHS sin 4 1 sin 2 sin 2 1 cos sin . cos sin 2 cos ) sin 2 1 ( cos sin 2 sin cos 2 cos cos ) 2 cos( 2 2 2 2 =- =-- =-- =- = + = x x x x x x x x x x x x x x x x x (2 marks) (c) ( 29 ) integer an ( ) ( ) 1 ( sin sin 2 3 sin 3 2 cosec 2 cosec 3 3 3 n n n π θ θ θ θ θ π π +-- = ∴- = ∴- = ∴- = ⇒- = (2 marks) (d) (i) ) 3 )( 1 2 ( 3 5 2 2 ≥ +- ≥- + x x x x Test x = 0 : invalid ∴ Solution is 2 1 or 3 ≥- ≤ x x (ii) 1 3 5 2 2 ≥-- + x x x On multiplying both sides by ( x – 1 ) 2 : ) 1 )( 3 )( 1 2 ( ) 1 )( 3 5 2 ( 2 ≥- +- ≥-- + x x x x x x (2 marks) Test x = 0 : valid ∴ Solution is ) 1 since ( 1 3 2 1 ≠ ≤ ≤- x x or x (e) (i) Number of ways = 90720 ! 2 ! 2 ! 9 = (ii) Number of ways with the letters “I” next to each other: (3 marks) I I 20160 ! 2 ! 7 8 = × = ∴ The probability of the letters “I” being together 9 2 90720 20160 ) ( ) ( = = = S n E n-3 0 ½-3 0 ½ 1 Question Two (a) Put n = 1, 2 n + n 3 = 2 + 1 = 3 which is divisible by 3 ∴ The statement is true for n = 1 Now assume that it is true for n = k ( k a positive integer) ie. 2 k + k 3 = 3 M ( M an integer)………………………..(1) Let n = k + 1 ∴ 2( k + 1 ) + ( k + 1 ) 3 = 2 k + 2 + k 3 + 3 k 2 + 3 k + 1 = 2 k + k 3 + 3 k 2 + 3 k + 3 But from (1) 2 k + k 3 = 3 M ∴ Expression = 3 M + 3 k 2 + 3 k + 3 = 3( M + k 2 + k + 1 ) which is divisible by 3. ∴ It is true for n = k + 1 if it is true for n = k . But it is true for n = 1. ∴ True for n = 1 + 1 = 2, and 2 + 1 = 3 and so on. ∴ By Mathematical Induction, it is true for all positive integers n . (3 marks) (b) Let n n ap ap a p a S + + + + = ... ……………………………..(1) then 1 ... + + + + + + = × n n n ap ap ap a S p …………………….(2) (1) – (2) gives: 2 2 2 2 1 ) 1 ( ) 1 ( p p ap a p p ap a S p ap a p S ap p a S p S n n n n n n n n-- =-- = ∴- =-- = ×- + + + + (3 marks) (c) (i) ) 1 )( 2 ( ) 2 ( ) 2 ( 2 2 2- + = +- + =-- + x b x b x b x x b x bx x 4 4 4 4 8 8 4 ) 2 )( 1 ( 2 ) 2 )( 1 2 ( ) 2 ( ) 2 ( 4 ) 1 ( 2 ) 1 2 ( ) ( 2 2 3 3 2 3 2 3 =- +- + +- =--- +-...
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ARC 2000S - SUGGESTED SOLUTIONS MATHEMATICS 3 UNIT 2000...

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