1999_Paper II - STEP II, 1999 Section A: 1 2 Pure...

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STEP II, 1999 2 Section A: Pure Mathematics 1 Let x = 10 100 , y = 10 x , z = 10 y , and let a 1 = x ! , a 2 = x y , a 3 = y x , a 4 = z x , a 5 = e xyz , a 6 = z 1 /y , a 7 = y z/x . (i) Use Stirling’s approximation n ! 2 π n n + 1 2 e - n , which is valid for large n , to show that log 10 (log 10 a 1 ) 102. (ii) Arrange the seven numbers a 1 , . . . , a 7 in ascending order of magnitude, justifying your result. 2 Consider the quadratic equation nx 2 + 2 x ( pn 2 + q ) + rn + s = 0 , ( * ) where p > 0, p ± = r and n = 1, 2, 3, . . . . (i) For the case where p = 3, q = 50, r = 2, s = 15, find the set of values of n for which equation ( * ) has no real roots. (ii) Prove that if p < r and 4 q ( p - r ) > s 2 , then ( * ) has no real roots for any value of n . (iii) If n = 1, p - r = 1 and q = s 2 / 8, show that ( * ) has real roots if, and only if, s 6 4 - 2 2 or s > 4 + 2 2 . 3 Let S n ( x ) = e x 3 d n d x n e - x 3 · . Show that S 2 ( x ) = 9 x 4 - 6 x and find S 3 ( x ). Prove by induction on n that S n ( x ) is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of x . Show also that if d S n d x = 0 for some value a of x , then S n ( a ) S n +1 ( a ) 6 0.
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STEP II, 1999 3 4 By considering the expansions in powers of x of both sides of the identity (1 + x ) n (1 + x ) n (1 + x ) 2 n , show that n X s =0 ± n s 2 = ± 2 n n , where ± n s = n ! s ! ( n - s )! . By considering similar identities, or otherwise, show also that: (i) if n is an even integer, then n X s =0 ( - 1) s ± n s 2 = ( - 1) n/ 2 ± n n/ 2 ; (ii) n X t =1 2 t ± n t 2 = n ± 2 n n . 5 Show that if α is a solution of the equation 5cos x + 12sin x = 7 , then either cos α = 35 - 12 120 169 or cos α has one other value which you should find.
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1999_Paper II - STEP II, 1999 Section A: 1 2 Pure...

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