# STEP III - 91*4023334091 Sixth Term Examination Papers...

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Section A: Pure Mathematics 1 Let x 1 , x 2 , . . . , x n and x n +1 be any ﬁxed real numbers. The numbers A and B are deﬁned by A = 1 n n X k =1 x k , B = 1 n n X k =1 ( x k - A ) 2 , and the numbers C and D are deﬁned by C = 1 n + 1 n +1 X k =1 x k , D = 1 n + 1 n +1 X k =1 ( x k - C ) 2 . (i) Express C in terms of A , x n +1 and n . (ii) Show that B = 1 n n X k =1 x 2 k - A 2 . (iii) Express D in terms of B , A , x n +1 and n . Hence show that ( n + 1) D > nB for all values of x n +1 , but that D < B if and only if A - r ( n + 1) B n < x n +1 < A + r ( n + 1) B n . 2 In this question, a is a positive constant. (i) Express cosh a in terms of exponentials. By using partial fractions, prove that Z 1 0 1 x 2 + 2 x cosh a + 1 d x = a 2 sinh a . (ii) Find, expressing your answers in terms of hyperbolic functions, Z 1 1 x 2 + 2 x sinh a - 1 d x and Z 0 1 x 4 + 2 x 2 cosh a + 1 d x . 2 2 9475 Jun10
3 For any given positive integer n , a number a (which may be complex) is said to be a primitive n th root of unity if a n = 1 and there is no integer m such that 0 < m < n and a m = 1. Write down the two primitive 4th roots of unity. Let C n ( x ) be the polynomial such that the roots of the equation C n ( x ) = 0 are the primitive n th roots of unity, the coeﬃcient of the highest power of x is one and the equation has no repeated roots. Show that C 4 ( x ) = x 2 + 1 . (i) Find C 1 ( x ), C 2 ( x ), C 3 ( x ), C 5 ( x ) and C 6 ( x ), giving your answers as unfactorised poly- nomials.

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STEP III - 91*4023334091 Sixth Term Examination Papers...

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