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2003_Paper I

# 2003_Paper I - STEP I 2003 2 Section A Pure Mathematics n 1...

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STEP I, 2003 2 Section A: Pure Mathematics 1 It is given that n r = - 1 r 2 can be written in the form pn 3 + qn 2 + rn + s , where p , q , r and s are numbers. By setting n = - 1, 0, 1 and 2, obtain four equations that must be satisﬁed by p , q , r and s and hence show that n r =0 r 2 = 1 6 n ( n + 1)(2 n + 1) . Given that n r = - 2 r 3 can be written in the form an 4 + bn 3 + cn 2 + dn + e , show similarly that n r =0 r 3 = 1 4 n 2 ( n + 1) 2 . 2 The ﬁrst question on an examination paper is: Solve for x the equation 1 x = 1 a + 1 b . where (in the question) a and b are given non-zero real numbers. One candidate writes x = a + b as the solution. Show that there are no values of a and b for which this will give the correct answer. The next question on the examination paper is: Solve for x the equation 1 x = 1 a + 1 b + 1 c . where (in the question) a , b and c are given non-zero numbers. The candidate uses the same technique, giving the answer as x = a + b + c . Show that the candidate’s answer will be correct if and only if a , b and c satisfy at least one of the equations a + b = 0 , b + c = 0 or c + a = 0 . (i) 3 Show that 2 sin( 1 2 θ ) = sin θ if and only if sin( 1 2 θ ) = 0 . (ii) Solve the equation 2 tan( 1 2 θ ) = tan θ . (iii) Show that 2 cos( 1 2 θ ) = cos θ if and only if θ = (4 n + 2) π ± 2 φ where φ is deﬁned by cos φ = 1 2 ( 3 - 1) , 0 6 φ 6 π/ 2, and n is any integer.

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STEP I, 2003 3 4 Solve the inequality sin θ + 1 cos θ 6 1 where 0 6 θ < 2 π and cos θ 6 = 0 . (i)
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2003_Paper I - STEP I 2003 2 Section A Pure Mathematics n 1...

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