Mark (Results) June 03 GCE Further Pure Mathematics FP (6668/0) Original Paper
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June 03 Further Pure Mathematics FP 6668 Mark Question (a) y''' + xy'' + y' sin x so y''' xy'' y' sin x M, A, A (b) y ''' + 3 0 so y ''' 3 B d y (c) Substitute to give d Use Taylor Expansion to give x B 3 x y + 3 x+ x... M A () (7). (a) B (V shape) B (Parabola) B (positions correct) (b) Put 4 x x 3 or 4 x x+ 3 M Solve x + x 7 0, to give + 4+ 8 x + M A Solve x x 0, to give 4+ 4 x M A So < x < B (6) (9) 4
Question 3. (a) kcos kx M A f( x) + sinkx () (b) ( sin ) sin cos ( cos ) + kx k kx k kx k kx f ( x) (+ sin kx) M k sin kx k (sin kx+ cos kx) so f ( x) (+ sin kx) f() x k ( sin kx) (+ sin kx) + k * + sinkx and use sin kx + cos kx M Acao 3 k cos kx (c) f B ( x) (+ sin kx) Finds f(0), f (0), f (0) and f (0) M 3 k k 3 Uses MacLaurin Expansion to obtain f(x) 0 + kx x + x... 6 M A (4) Alternative method for (c) Uses ln ( + y) Use sinkx Obtains 3 y y 3 y + + with y sinkx B 3 3 k x kx in ln expansion M 6.. 3 k k 3 M A 0 + kx x + x... 6 (4) 5
Question 4. + x cotx B IF d x x Obtains ln x + ln sin x M A {So IF xsinx} d sin x (IF y) IF dx x Aft M cosx x M A yxsinx sin xdx dx sin x ( + c ) 4 So y cosec c x cos x+ M A x xsinx (9) 6
Question 5 (a) + r ( r+ ) ( r+ ) M A A (b) LHS + 3 + + 3 4 + + 3 4 5 + ( n ) ( n ) + n M + + ( n ) n ( n+ ) + + n ( n+ ) ( n+ ) A + n + n + n n n n + 3 + 4+ + ( n+ )( n+ ) M dep n + 3n ( n+ )( n+ ) nn ( + 3) ( n+ )( n+ ) A (4) 7
Question 6 Modulus 3 B Argument 5π arctan( ) 3 6 MA 5π 5π π π 5 5 z 3 (cos( ) + isin( )) (cos + isin ) 6 6 6 6 M A OR π nπ π nπ (cos + + isin + ) 6 5 6 5 M 7π 7π (cos + isin ) 30 30, 9π 9π (cos + isin ) 30 30, A 7π 7π (cos + isin ) 30 30, 9π 9π (cos + isin ) 30 30 A (8) 7(a) Differentiate twice and obtaining dy x x d y x x x λxe + λe and 4λxe + λe + λe dx dx M A 3 Substitute to give λ M A (4) (b) Complementary function is y A B x x e + e M A So general solution is y A B x x e + e + 3 e x x A (7) 8
Question 8. (a) B () (b) w(z-i) z+7i so zw ( ) 7i + iw and z M 7i + iw A w So 7i+ iw w M Using w u + iv, ( ) ( 7) ( ) v + u+ u + v M So 3u + 3v + 30u+ 48 0, which is a circle equation A As ( u+ 5) + v 3 So centre is 5 and radius is 3 M A (5) () (8) 9
Question Q9 (a) sinθ sinθ 0.5 π 5π θ or 6 6, M A, π 5π Points are (, 6 ) and (, ) A 6 (b) Uses Area θ M ( sin θ ) d (4 8sin θ + ( cos θ))dθ M [(6 θ 8cos θ sin θ ))] + A Uses limits π and 6 π to give 7 3 π or 4 7 3 π M A Finds area of sector of circle π 3 B So required area is 8 π 7 3 3 M A (8) 0
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