2007 SRJC Prelim Paper 2 Solutions

2007 SRJC Prelim Paper 2 Solutions - 1 2007 JC2 H2 MATH...

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1 [Turn Over 2007 JC2 H2 MATH PRELIM PAPER 2 1 The curve C is defined parametrically by sec xa and tan ya , where a is a fixed constant. It is known that the point P with coordinates ( sec , tan ) aa  lies on C . Show that the equation of the tangent at P is given by sin cos x y a  and the equation normal at P is given by sin 2 tan x y a  . The tangent and the normal at P cut the x -axis at T and N respectively. Prove that OT ON = 2 a 2 , where O is the origin. < Solutions > C: sec , tan x a y a  d sec tan d x a and 2 d sec d y a 2 1 d sec sec 1 cos cos sin d sec tan tan sin cos ec    At P( P( sec , tan ) , Equation of tangent:   1 tan sec sin y a x a   2 2 sin sin tan sec 1 sin sin sec sin tan sin cos cos 1 sin cos cos cos sin cos (shown) y a x a x y a a a a a x y a     Equation of normal:   tan sin sec y a x a   sin tan sin 2 tan (shown) x y a   At y= 0, cos (from eqn of tgt) i.e. T (acos ,0)  22 sin 2 tan (fr eqn of normal) i.e. N ( ,0) cos cos x a x 2 2 OT×ON cos 2 (shown) cos a [5] [3]

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2 2 (a) Find the modulus and argument of the complex number z , where 2 4 (1 ) 3 ) i z i , giving your answers in exact form. Hence, using de Moivre’s theorem, evaluate 6 z . < Solutions > 2 2 44 1 ( 2) 1 8 ( 4) 13 i z i (can use g.c.) 2 4 43 5 6 ) arg( ) arg 2arg(1 ) 4arg(1 3 ) 3 ) 2( ) 4( ) i z i i i     (can use g.c) Thus   55 66 1 cos sin 8 zi    6 6 6 1 cos sin 8   6 11 cos5 sin5 88   2 (b) (i) Show that for any complex number i e , cot 2 2 2 1 i i e i e     . (ii) Write down the solution to the equation 5 10 z  leaving your answers in the form i re where r > 0 and   . Hence, find all possible values of the complex number w that satisfies 5 1 1 w w , leaving your answers in the form x yi where , xy . [2] [2] [2] [3] [1]
3 [Turn Over < Solutions > (i) ( 1) 1 1 ( 1) ( 1 ( ) 1 i i i i i i i i i e e e e e e e e e  1 (cos sin ) 2 2cos 1 sin 2 2(1 cos ) i i    22 2 2 2sin cos 1 2 2 1 (1 2sin ) i   11 cot 2 2 2 i    (shown) (ii) 2 5 5 (2 ) 5 1 0 1 , 0, 1, 2 k i ik z z e z e k          i.e 24 55 1, , i z e e   For 5 1 1 w w , Let 1 w z w 1 z w z Thus 2 5 2 5 cot , 1 2 2 5 1 k i k i z e k wi z e 0, 1, 2 k   0.5 0.688 , 0.5 0.162 w i i

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4 3 A differential equation is given by d ( ) ( ) d n y P x y Q x y x  .
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2007 SRJC Prelim Paper 2 Solutions - 1 2007 JC2 H2 MATH...

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