Mark Summer 009 GCE Core Mathematics C (666)
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June 009 666 Core Mathematics C Mark Question Q (a) Iterative formula: n + +, 0. ( ) n + (.)..78....97....6069... An attempt to substitute 0. into the iterative formula. Can be implied by. or.0 Both.(0) and awrt.7 A Both awrt.6 and awrt.60 or.6 A cso () (b) Let f( ) + + 0 Choose suitable interval for, f (.8) 0.00877... e.g. [.8,.9] or tighter f (.9) 0.0086... any one value awrt sf Sign change (and f ( ) is continuous) therefore a root or truncated sf α is such that α (.8,.9) α.9 ( dp) both values correct, sign change and conclusion At a minimum, both values must be correct to sf or truncated sf, candidate states change of sign, hence root. d A () [6] 87_97 GCE Mathematics January 009 7
Q (a) cos θ + sin θ ( cos θ) cos θ sin θ + cos θ cos θ cos θ Dividing cos θ + sin θ by cos θ to give underlined equation. + tan θ sec θ tan θ sec θ (as required) AG Complete proof. No errors seen. A cso () (b) tan θ secθ sec θ, ( eqn *) + + 0 θ < 60 o (sec θ ) + secθ + sec θ Substituting tan θ sec θ into eqn * to get a quadratic in secθ only sec θ + secθ + sec θ sec θ + secθ 0 Forming a three term one sided quadratic epression in sec θ. ( θ )( θ ) sec + sec 0 secθ or secθ Attempt to factorise or solve a quadratic. or cosθ cosθ cos θ ; or cosθ cosθ A; o α 0 or α no solutions o θ 0 0 o A θ 0 o 0 o or θ 60 θ when solving using cos θ... B θ { 0, 0 } o o Note the final A mark has been changed to a B mark. (6) [8] 87_97 GCE Mathematics January 009 8
Q P 80e t (a) t 0 0 P 80e 80() 80 80 B () t 000 t (b) P 000 000 80e e 80 000 t ln 80 Substitutes P 000 and rearranges equation to make e t the subject. t.686... awrt.6 or years A Note t or t awrt.6 t will score A0 () dp t (c) 6e dt (d) 0 6e t ke t and k 80. 6e t A () 0 t ln.6977... 6 { } dp Using 0 and dt an attempt to solve t to find the value of t or. 0 ln 6 P 80e or (.6977... ) P 80e Substitutes their value of t back into the equation for P. d 80(0) P 0 0 or awrt 0 A 6 () [8] 87_97 GCE Mathematics January 009 9
Q (i)(a) y cos Apply product rule: u v cos du dv sin d d Applies vu + uv correctly for their u, u, v, v AND gives an epression of the form dy α cos ± β sin cos sin d Any one term correct Both terms correct and no further simplification to terms in cosα or sin β. A A () (b) y ln( ) + + u du ln( + ) something + + ln( + ) A + ln( ) d + Apply quotient rule: u ln( + ) v + du dv d + d ( + ) ln( + ) + d dy ( + ) vu uv Applying v Correct differentiation with correct bracketing but allow recovery. A () d dy ln( + ) ( + ) {Ignore subsequent working.} 87_97 GCE Mathematics January 009 0
(ii) y +, > At P, y () + 9 At P, y 9 or B ± ( + ) * dy k ( ) + () d ( + ) A aef dy d ( + ) At P, dy d Substituting into an equation ()+ involving d y d ( ) ; Hence m(t) Either T: y ( ) ; or y + c and ( ) c c + ; Either T: y 9 ( ) ; T: y 9 y y m( ) or y y m( their stated ) with their TANGENT gradient and their y ; or uses y m+ c with their TANGENT gradient, their and their y. d*; T: y+ 0 y+ 0 Tangent must be stated in the form a + by + c 0, where a, b and c are integers. A or T: y + (6) T: y + T: y+ 0 [] 87_97 GCE Mathematics January 009
Q (a) y Curve retains shape when > ln k B ( 0, k ) Curve reflects through the -ais when < ln k B O ( ) ln k, 0 ( 0, k ) and ( ln k, 0) marked in the correct positions. B (b) ( k, 0) y ( 0, lnk ) Correct shape of curve. The curve should be contained in quadrants, and (Ignore asymptote) B () O ( k, 0) and ( 0, lnk) B (c) Range of f: f( ) > k or y> k or ( k, ) (d) y e k y+ k e ( ) ( ) Either f( ) > k or y> k or ( k, ) or f > kor Range > k. Attempt to make (or swapped y) the subject ln y+ k ln y+ k Makes e the subject and takes ln of both sides B () () Hence f ( ) ln( + k) ln( k) + or ln ( + k ) A cao () (e) f ( ) : Domain: > k or ( k, ) Either > k or ( k, ) or Domain > kor ft one sided inequality their part (c) RANGE answer B () [0] 87_97 GCE Mathematics January 009
Q6 (a) ( ) A B cos A+ A cosa cos Acos A sin Asin A Applies A B to cos( A+ B) to give the underlined equation or cos A cos A sin A cos A cos A sin A and gives cos A + sin A cos A sin A sin A sin A (as required) Complete proof, with a link between LHS and RHS. No errors seen. A AG () (b) C C sin sin cos Eliminating y correctly. cos sin cos ( ) sin cos cos sin cos cos Using result in part (a) to substitute for sin as ± ± cos or ksin as cos k ± ± to produce an equation in only double angles. sin + cos Rearranges to give correct result A AG () (c) sin + cos R cos( α ) sin + cos Rcoscosα + Rsinsinα Equate sin : Rsinα Equate cos : Rcosα R + ; R B tanα α 6.8698976... o tanα ± or tanα ± or sinα ± or cosα ± their R their R awrt 6.87 A Hence, sin + cos cos( 6.87) () 87_97 GCE Mathematics January 009
(d) sin + cos cos( 6.87) ( ) cos( their α ) cos 6.87 ± their R ( ) 6.87 66.8... ( ) 6.87 60 66.8... o o awrt 66 A Hence,.69... o, 6.09... o One of either awrt.6 or awrt.7 or awrt 6. or awrt 6. A Both awrt.6 AND awrt 6. A If there are any EXTRA solutions inside the range 0 < 80 then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < 80. () [] 87_97 GCE Mathematics January 009
Q7 8 f( ) + ( + ) ( )( + ) R,,. (a) f( ) An attempt to combine to one ( )( + ) ( ) + 8 fraction ( )( + ) Correct result of combining all three fractions A + 8 + + 8 ( )( + ) + [( + )( ) ] Simplifies to give the correct numerator. Ignore omission of denominator A ( + )( ) [( + )( ) ] An attempt to factorise the numerator. d ( ) ( ) Correct result A cso AG () (b) g( ) e e R, ln. Apply quotient rule: u e v e du dv e e d d g( ) e (e ) e (e ) (e ) vu uv Applying v Correct differentiation A e e e + e (e ) e (e ) Correct result A AG cso () 87_97 GCE Mathematics January 009
(c) e g( ) (e ) e (e ) e e e e + Puts their differentiated numerator equal to their denominator. e e + 0 e e + A (e )(e ) 0 e or e Attempt to factorise or solve quadratic in e ln or 0 both 0, ln A () [] 87_97 GCE Mathematics January 009 6
Q8 (a) sin sin cos sin cos B aef () (b) cosec 8cos 0, 0 < < π 8cos 0 sin Using cosec sin sin 8cos 8sincos ( ) sin cos sin sin sin k, where < k < and k 0 sin A { } { } Radians 0.68...,.8889... Degrees.77..., 6... { } { } Radians 0.6...,... Degrees 7.87..., 8.76... Either arwt 7. or 8.76 or 0. or. or. or awrt 0.0π or awrt 0.6 π. Both 0. and. Solutions for the final two A marks must be given in only. If there are any EXTRA solutions inside the range 0 < < π then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < < π. A A cao () [6] 87_97 GCE Mathematics January 009 7
87_97 GCE Mathematics January 009 8