2000_Paper II

# 2000_Paper II - STEP II, 2000 2 Section A: Pure Mathematics...

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Unformatted text preview: STEP II, 2000 2 Section A: Pure Mathematics 1 A number of the form 1 /N , where N is an integer greater than 1, is called a unit fraction . Noting that 1 2 = 1 3 + 1 6 and 1 3 = 1 4 + 1 12 , guess a general result of the form 1 N = 1 a + 1 b ( * ) and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions. By writing ( * ) in the form ( a- N )( b- N ) = N 2 and by considering the factors of N 2 , show that if N is prime, then there is only one way of expressing 1 /N as the sum of two distinct unit fractions. Prove similarly that any fraction of the form 2 /N , where N is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions. 2 Prove that if ( x- a ) 2 is a factor of the polynomial p( x ), then p ( a ) = 0. Prove a corresponding result if ( x- a ) 4 is a factor of p( x ) . Given that the polynomial x 6 + 4 x 5- 5 x 4- 40 x 3- 40 x 2 + 32 x + k has a factor of the form ( x- a ) 4 , find k . 3 The lengths of the sides BC , CA , AB of the triangle ABC are denoted by a , b , c , respectively. Given that b = 8 + 1 , c = 3 + 2 , A = / 3 + 3 , where 1 , 2 , and 3 are small, show that a 7 + , where = (13 1- 2 2 + 24 3 3 ) / 14. Given now that | 1 | 6 2 10- 3 , | 2 | 6 4 9 10- 2 , | 3 | 6 3 10- 3 , find the range of possible values of . STEP II, 2000 3 4 Prove that (cos + i sin )(cos + i sin ) = cos( + ) + i sin( + ) and that, for every positive integer n , (cos + i sin ) n = cos n + i sin n. By considering (5- i ) 2 (1 + i ), or otherwise, prove that arctan(7 / 17) + 2 arctan(1 / 5) = / 4 . Prove also that 3 arctan(1 / 4) + arctan(1 / 20) + arctan(1 / 1985) = / 4 . [Note that arctan is another notation for tan- 1 .] 5 It is required to approximate a given function f( x ), over the interval 0 6 x 6 1, by the linear function x , where is chosen to minimise Z 1 f( x )- x 2 d x....
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## 2000_Paper II - STEP II, 2000 2 Section A: Pure Mathematics...

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