Topic 6 Part 4 [317 marks]

Transcription

1 Topic 6 Part [7 marks] a. ( + tan ) sec tan (+c) M [ marks] [ marks] Some correct answers but too many candidates had a poor approach and did not use the trig identity. b. sin sin (+c) cos M [ marks] Allow integration by parts followed by trig identity. Award M for parts, for trig identity, final answer. [ marks] Total Same as (a).

2 a. f : y y f : y y y y y + M M ( y) y y y y y ( f (y) ) f () ( ) y and might be interchanged earlier. First M is for interchange of variables second M for manipulation Final answer must be a function of Well done. Only a few candidates confused inverse with derivative or reciprocal. b. B A + A( ) + B A A + B equating coefficients and (M) [ marks] A and B Could also be done by division or substitution of values. [ marks] Not enough had the method of polynomial division. c. f() ln + c [ mark] ln accept equivalent e.g. [ mark] Total [7 marks]

3 Reasonable if they had an answer to (b) (follow through was given) usual mistakes with not allowing for the derivative of the bracket.. du e () [7 marks] EITHER integral is e ( e +) + M u + du M Award M only if the integral has completely changed to one in. u du needed for final OR e u integral is du (u ) +6(u )+ M Award M only if the integral has completely changed to one in. u u + du M In both solutions the two method marks are independent. THEN arctan( )(+c) u e + arctan( )(+c) Total [7 marks] () Many good complete answers. Some did not realise it was arctan. Some had poor understanding of the method. a. e + e ( e + e ) M [ marks] [ marks] Well done.

4 b. d let P (n) n y be the statement n + prove for as n M of is which is e + e and is R LHS P () RHS 0 e + e LHS RHS, P () is true n n e n e P (k) P (k + ) assume is true and attempt to prove is true M [7 marks] assuming d k+ y k+ k k e + k e k d d k y d k y k ( ) (M) k k e + k e + k e (k + ) k e + k+ e (as required) Can award the A marks independent of the M marks P () P (k) P (k + ) since is true and is true is true then (by P MI), P (n) is true ( n Z + ) R To gain last R at least four of the above marks must have been gained. [7 marks] The logic of an induction proof was not known well enough. Many candidates used what they had to prove rather than differentiating what they had assumed. They did not have enough eperience in doing Induction proofs. c. e + e e (, ) d y e + e d y, > 0

5 Good, some forgot to test for min/ma, some forgot to give the y value. d. d y e + e e + e M point is e (, ) since the curvature does change (concave down to concave up) it is a point of inflection R Allow derivative is not zero at rd Again quite good, some forgot to check for change in curvature and some forgot the y value. 5a. attempt to differentiate f() + M f () 6 ( ) (Critical values occur at) () 0, f ]0, [ (or 0 < < ) so decreasing on 5b. f () 6 6 setting f () 0 coordinate is (, ) [ marks] Total [7 marks] () M [ marks]

6 6. any attempt at integration by parts du u ln dv v () () [ ln ] M Condone absence of limits at this stage. [ ln ] [ ] 6 Condone absence of limits at this stage. ln ( ) 6 ln 5 6 7a. any attempt to use sine rule M AB sin sin( θ) sin cos θ cos sin θ Condone use of degrees. AB cos θ+ sin θ cos θ+ sin θ AB cos θ+sin θ

7 7b. METHOD setting ) ( sin θ+cos θ) ( cos θ+sin θ) (AB (AB) 0 M M tan θ θ 6 METHOD AB minimum when is maimum M METHOD M () shortest distance from to is perpendicular to R sin sin( θ) AB sin( θ) sin( θ) θ 6 θ θ 6 Total [8 marks] MA B AC AC

8 8. EITHER arctan t OR THEN +t t tan sec + tan + t (M) (M) [8 marks] sin t +t () This is independent of the first two marks + sin +t +( t ) +t M Award M for attempting to obtain integral in terms of and t (+ t )+ t +t arctan( ) [8 marks] + t arctan( tan )(+c) t 9a. g f() [ marks] tan + tan, 0 < [ marks]

9 9b. tan + tan sin +cos sin cos [ marks] sin + cos sin cos M [ marks] 9c. METHOD sin (sin cos )(cos sin ) (sin +cos )(cos +sin ) (sin cos ) ( sin cos cos sin ) ( sin cos + cos + sin ) cos + sin sin cos Substitute into any formula for M 6 sin M() ( ) METHOD + + M 8 8 sec (tan ) (tan ) sec (tan +) sec (tan ) M sec 6 6 (tan ) ( ) ( ) 8 ( ) M Award M for substitution. 6 (+ ) ( ) (+ ) ( ) M

10 9d. Area 6 0 sin +cos sin cos M [ln sin cos ] 6 0 Condone absence of limits and absence of modulus signs at this stage. ln sin cos ln sin 0 cos ln 0 ln( ) ln( ) ln( ) ln( ) ln( + ) + + M M Total 0. ( ) ( e or ) e 0 e (M)()() Award M for integral involving the function given; for correct limits; for and ( ) e Most candidates answered this question correctly. Those candidates who attempted to manipulate the function or attempt an integration wasted time and obtained / marks. The most common errors were an etra factor and a fourth power when attempting to square the function. Many candidates wrote down the correct epression but not all were able to use their calculator correctly.

11 . V 00r () V hr h Allow if value of is substituted later in the question. EITHER dv 00r dr M Award M for an attempt at implicit differentiation. at we have M OR r 0 00 dr dr dv dr r THEN dr dv dv dr 00r M we have M dv dr ( 0.09) (cm s ) This question was well understood and a large percentage appreciated the need for implicit differentiation although some candidates did not recognise the need to treat h as a constant till late in the question. A number of candidates found the answer instead of due to a basic incorrect use of the GDC

12 . f () + e Accept labelled diagram showing the graph y f () above the -ais; y f() do not accept unlabelled graphs nor graph of. EITHER this is always > 0 R so the function is (strictly) increasing and thus OR this is always > 0 (accept 0) R R so there are no turning points and thus R is dependent on the first R. The differentiation was normally completed correctly, but then a large number did not realise what was required to determine the type of the original function. Most candidates scored / and wrote eplanations that showed little or no understanding of the relation between first derivative and the given function. For eample, it was common to see comments about horizontal and vertical line tests but applied to the incorrect function.in term of mathematical language, it was noted that candidates used many terms incorrectly showing no knowledge of the meaning of terms like parabola, even or odd ( or no idea about these concepts).. 0 y () y (0) (M)() [7 marks] The eact answer is y e+ (0) +. e e so gradient of normal is equation of normal is gives y ( ) y c (M) (M)() e The eact answer is y +. e+ Accept y ( 0) [7 marks]

13 Surprisingly many candidates ignored that fact that paper is a calculator paper, attempted an algebraic approach and wasted lots of time. Candidates that used the GDC were in general successful and achieved 7/7. A number of candidates either found the equation of the tangent or used the positive reciprocal for the normal and many did not find the value of corresponding to. y f(0) a. (i) (ii) (iii) at dv a(t) 0 (m s ) t 0 v 00 (m s ) s 0t 5 t (+c) s 000 for t 0 c 000 s 5 t t 0, s 500 (m) M (M) Accept use of definite integrals. Parts (i) and (ii) were well answered by most candidates. In (iii) the constant of integration was often forgotten. Most candidates calculated the displacement and then used different strategies, mostly incorrect, to remove the negative sign from. 500 b. dv [ mark] ( 0 5v) [ mark] Surprisingly part (b) was not well done as the question stated the method. Many candidates simply wrote down dv while others seemed unaware that was the acceleration. dv

14 c. METHOD t 0 5v 5 dv ln( 0 5v)(+c) M Accept equivalent forms using modulus signs. t 0, v 00 0 ln(90) + c 5 c 0 + ln(90) M t 0 + ln 90 ln( 0 5v) Accept equivalent forms using modulus signs v t 0 + ln( ) Accept use of definite integrals. METHOD 0 5v 5 +v t dv dv ln + v (+c) 5 M Accept equivalent forms. t 0, v 00 0 ln 98 + c 5 M ln( 98) If is seen do not award further A marks. c 0 + ln t 0 + ln 98 ln + v Accept equivalent forms v t 0 + ln( ) Accept use of definite integrals. Part (c) was not always well done as it followed from (b) and at times there was very little to allow follow through. Once again some candidates started with what they were trying to prove. Among the candidates that attempted to integrate many did not consider the constant of integration properly.

15 d. 5(t 0) ln 98 ( v) +v 98 [ marks] e 5(t 0) v 98e 5(t 0) (M) [ marks] In part (d) many candidates ignored the answer given in (c) and attempted to manipulate different epressions. e. ds at 98 s t + e 5(t 0) 98 5 e 5(t 0) (+k) M 98 t 0, s k k s t e 5(t 0) M Accept use of definite integrals. Part (e) was poorly answered: the constant of integration was often again forgotten and some inappropriate uses of Physics formulas assuming that the acceleration was constant were used. There was unclear thinking with the two sides of an equation being integrated with respect to different variables. f. t 50 for s 0 [ marks] Total [ marks] (M) [ marks] Although part (e) was often incorrect, some follow through marks were gained in part (f).

16 5a. EITHER y ln( a) + b ln(5 + 0) y ln( a) + ln c ln(5 + 0) y ln(c( a)) ln(5 + 0) (M) (M) OR y ln(5 + 0) ln(5( + )) y ln(5) + ln( + ) (M) (M) THEN a, b ln 5 Accept graphical approaches. Accept a, b.6 5b. V e [ln(5 + 0)] 99. e (M) [ marks] [ marks] Total 6a. attempt at implicit differentiation M 5 5y + y 0 [ marks] 5y y 7 for differentiation of, for differentiation of and. 5y + (y 5) 0 5y y 5 [ marks]

17 6b gradient of normal equation of normal substitution of y + 5 (6, ) y + c M Accept y ( 6) 6c. setting y M substituting into original equation () points and () distance 5y y 5 [8 marks] ± (, ) (, ) 8 ( ) Total (M) M [8 marks] 7a. [ marks] s (9t t ) 9 (+c) t t t 0, s c t s s + (9t ) 0 t s

18 7b. correct shape over correct domain maimum at intercept at, intercept at minimum at (, 6.5) t.6 s (5, 9.5) 7c. 9.5 a + b cos 6.5 a + b cos (M) [ marks] Only award M if two simultaneous equations are formed over the correct domain. a 7 b [ marks]

19 7d. at t : 9 + t t ( t) 0 t 9 9 t solving 7 (M) t 5 cos GDC t 6. (M) Accept graphical approaches. Total 8a. cos cos cos ± positive as [ marks] +cos 0 R +cos cos M [ marks] 8b. cosθ sin θ cos sin [ marks] (M) [ marks]

20 8c. 0 cos + sin [ sin cos ] 0 (0) (0 ) 9. (a) [ mark] (b) for point (, 0) for (0, ) for min at in approimately the correct place for (, 0) for shape including continuity at 0 Total

21 0. dv ds s M Award M for s and for the whole epression. a v dv when (M) ds s s s 5 a ( ) () s, a ( 6) (m ) (0.5) 5 s M M is for the substitution of 0.5 into their equation for acceleration. Award MA0 if s 50 is substituted into the correct equation.

22 . (a) METHOD y + +y + 0 M [9 marks] Award M for implicit differentiation, for LHS and for RHS. METHOD y (+ y ) y(+ ) tan tan(arctan ) tan( arctan ) +(tan )(tan(arctan )) + y y (+ ) ( ) (+ ) (+ ) y(+ ) + ( ) (+ ) (M) M (b) y tan tan(arctan ) tan( arctan ) +(tan )(tan(arctan )) (M) (M) The two Ms may be awarded for working in part (a). + y substitution into 6 9 Accept 8 9 etc. Total [9 marks] a. f ln () ln [ marks] M [ marks]

23 b. ln 0 has solution e M y e hence maimum at the point (e, ) e [ marks] [ marks] c. f ( ) ( ln ) () ln M The M should be awarded if the correct working appears in part (b). point of infleion where f () 0 so e, y M C has coordinates e ( e, ) e d. f() 0 f () y + c through (, 0) equation is y () (M)

24 e. METHOD area e ln M [7 marks] order. Award M for integration of difference between line and curve, for correct limits, for correct epressions in either ln (ln ) (+c) (ln) e (+c) ( ) [ ] ( e ) ( ) e e e METHOD area area of triangle e ln M (M) is for correct integral with limits and is dependent on the M. ln (+c) area of triangle [7 marks] (ln ) (M) (e )(e ) M (e )(e ) ( ) e e a. () and ( ) both answers are the same, hence f is continuous (at ) R [ marks] R may be awarded for justification using a graph or referring to limits. Do not award A0R. [ marks]

25 b. reflection in the y-ais +, f( ) { ( + ), < (M) Award M for evidence of reflecting a graph in y-ais. translation ( ) 0 g() {, 0, < 0 (M) Award (M) for attempting to substitute ( ) for, or translating a graph along positive -ais. Award for the correct domains (this mark can be awarded independent of the M). Award for the correct epressions.. asecθ dθ new limits: a θ using asecθ tanθ a θ a sec θ tan θ dθ a sec θ a sec θ a cos θ a cos a () and () dθ θ (cosθ + ) [ sin θ + θ] a M M or equivalent ( + ) or equivalent ( + 6) a [7 marks] [7 marks]

26 5a. (a) f ( +) (+) ( +) () ( ) [ marks] + ( +) M [ marks] 5b. + 0 ( +) ± [ mark] [ mark] 5c. f ( ) ( +) ()( +)( +) () ( +) [ marks] Award for ( )( + ) or equivalent. Award for ()( + )( + ) or equivalent. ( )( +) ( +) ( +) +6 6 ( +) [ marks] ( + ) ( ) ( +)

27 5d. recognition that ( ) is a factor (R) ( )( + b + c) ( + ) ± M Allow long division / synthetic division. 5e. 0 + M ln( + ) + arctan() [ ln( 0 + ) + arctan()] ln M ln + arctan0 ln arctan( ) M 6. use of the quotient rule or the product rule M C (+ t (t) ) t t 6 t ( t ) or (+ t ) (+ t ) +t (+ ) t Award for a correct numerator and for a correct denominator in the quotient rule, and for each correct term in the product rule. attempting to solve t ± (minutes) C (t) 0 for t (M) C ( ) (mg l ) or equivalent. This question was generally well done. A significant number of candidates did not calculate the maimum value of. C

28 7. du (u )du Award the for any correct relationship between and. + (u ) du u (M) du Award the M for an attempt at substitution resulting in an integral only involving. u + du u () u u + ln u(+c) + ln( + )(+C) u Award the for a correct epression in, but not necessarily fully epanded/simplified. Many candidates worked through this question successfully. A significant minority either made algebraic mistakes with the substitution or tried to work with an integral involving both and. u 8a. p () f ()g() + g ()f() (M) [ marks] Award M if the derivative is in terms of or. + [ marks] This was a problem question for many candidates. Some quite strong candidates, on the evidence of their performance on other questions, did not realise that composite functions and functions of a function were the same thing, and therefore that the chain rule applied. 8b. h () g (f()) f () (M)() h () g () f () 6 Total

29 This was a problem question for many candidates. Some quite strong candidates, on the evidence of their performance on other questions, did not realise that composite functions and functions of a function were the same thing, and therefore that the chain rule applied. 9a. (i) e y+ M The M is for switching variables and can be awarded at any stage. Further marks do not rely on this mark being awarded. taking the natural logarithm of both sides and attempting to transpose ( f ()) (ln ) R + > 0 (ii) or equivalent, for eample. M Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper. 9b. ln (ln ) ln ln (or equivalent) M ln (or equivalent) e coordinates of P are ( e, ) Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper. 9c. coordinates of are ( ) seen anywhere [ marks] M at Q, y Q, 0 [ marks]

30 Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper. 9d. let the required area be A A ln e e M The M is for a difference of integrals. Condone absence of limits here. attempting to use integration by parts to find [ ] [ ln e ] e ln (M) ln Award for and for. The second M and second are independent of the first M and the first. e e e e ( ) A productive question for many candidates, but some didn t realise that a difference of areas/integrals was required.

31 9e. (i) METHOD consider for eample h() ln h() 0 and h () () as h () 0 for, then h() 0 for R as h () 0 for 0 <, then h() 0 for 0 < so g(), R + R METHOD g () g () < 0 R + (concave down) for R the graph of is below its tangent R so y g() (y at ) g(), R + The reasoning may be supported by drawn graphical arguments. METHOD

32 clear correct graphs of y and ln for > 0 ln statement to the effect that the graph of is below the graph of its tangent at R (ii) replacing by to obtain ln( ) ( ) M ln ( ) ln ( ) so g(), R + () Total [ marks] (i) Many candidates adopted a graphical approach, but sometimes with unconvincing reasoning. (ii) Poorly answered. Many candidates applied the suggested substitution only to one side of the inequality, and then had to fudge the answer.

33 0. METHOD attempt to set up (diagram, vectors) (M) correct distances 5t, y 0t () () the distance between the two cyclists at time is t s (5t) + (0t) 5t (km) ds 5 (km ) h hence the rate is independent of time METHOD + y s attempting to differentiate implicitly (M) + y s ds () the distance between the two cyclists at time is t (5t) + (0t) 5t (km) () (5t)(5 + (0t)(0) (5t) ds M ds Award M for substitution of correct values into their equation involving. ds 5 (km ) h hence the rate is independent of time METHOD s + y () ds +y + y (M)() Award M for attempting to differentiate the epression for. s ds (5t)(5)+(0t)(0) (5t) + (0t) M ds Award M for substitution of correct values into their. ds 5 (km ) h hence the rate is independent of time ds Reasonably well done. Most successful candidates determined that s 5t 5 from 5t and y 0t. A number of candidates did not use calculus while a few candidates correctly used implicit differentiation.

34 a. t 0 t 6 (s) (M) [ marks] Award A0 if either or or both are stated with. [ marks] t 0.6 t. t 6 5 (t ) 0 t Part (a) was not done as well as epected. A large number of candidates attempted to solve for. Some candidates attempted to find when the particle s acceleration was zero. b. let be the distance travelled before coming to rest d d 5 (t ) + 6 t 0 (M)() Award M for two correct integrals even if the integration limits are incorrect. The second integral can be specified as the area of a triangle. d 7 attempting to solve T M T.9 (s) ( 5.7) (m) Total [7 marks] T 6 t () ( ) 7 (or equivalent) for Most candidates had difficulty with part (b) with a variety of errors committed. A significant proportion of candidates did not understand what was required. Many candidates worked with indefinite integrals rather than with definite integrals. Only a small percentage of candidates started by correctly finding the distance travelled by the particle before coming to rest. The occasional candidate made adroit use of a GDC and found the correct value of t by finding where the graph of 5 (t ) + t crossed the horizontal ais. 0 a. use of A qr sin θ to obtain A ( + )(5 ) sin 0 M ( + )(5 0 + ) A ( ) [ marks] [ marks] This question was generally well done. Parts (a) and (b) were straightforward and well answered.

35 b. (i) da ( 6 + 5) ( )( 5) [ marks] (ii) METHOD EITHER da (( ) 6 ( ) + 5) 0 M OR da THEN ( ( ) ) (( ) 5) 0 da so 0 when METHOD da solving 0 for M < < 5 M da so 0 when METHOD da a correct graph of versus M da the graph clearly showing that 0 when da so 0 when [ marks] This question was generally well done. Parts (a) and (b) were straightforward and well answered. c. [7 marks] d A ( 8) d A,.5 (< 0) P QR A ma 7 (.7) (c m ) 7 PQ (cm) PR ( ) (cm) Q R 7 ( ) + ( ) ( ) ( ) cos QR 9.8 (cm) 7

36 This question was generally well done. Parts (c) (i) and (ii) were also well answered with most candidates correctly applying the second derivative test and displaying sound reasoning skills. Part (c) (iii) required the use of the cosine rule and was reasonably well done. The most common error committed by candidates in attempting to find the value of QR was to use PR (cm) rather than PR ( ) (cm). The occasional candidate used cos 0. a. attempting to use attempting to epress in terms of y ie (y + 6) (M) for y h, V [ marks] V h 0 h V ( + 6h) b a y + 6 (M) [ marks] This question was done reasonably well by a large proportion of candidates. Many candidates however were unable to show the required result in part (a). A number of candidates seemingly did not realize how the container was formed while other candidates attempted to fudge the result. b. EITHER [ marks] the depth stabilizes when attempting to solve dv 50 h (h+6) 0 ie h (h+6) for h (M) R OR the depth stabilizes when dh (h+6) 50 h (h+6) 0 ie (8.5 ) 0 R attempting to solve (h+6) 50 h (h+6) (8.5 ) 0 for h (M) THEN h 5.06 (cm) [ marks] Total dv dh In part (c), a pleasing number of candidates realized that the water depth stabilized when either 0 or 0, sketched an appropriate graph and found the correct value of h. Some candidates misinterpreted the situation and attempted to find the coordinates of the local minimum of their graph.

37 . f(0) 0 f () e cos e sin + f (0) 0 f () e (M) sin f (0) 0 f () () e sin + e cos f () (0) M [7 marks] the first non-zero term is! ( ) Award no marks for using known series. [7 marks] Most students had a good understanding of the techniques involved with this question. A surprising number forgot to show f(0) 0. Some candidates did not simplify the second derivative which created etra work and increased the chance of errors being made. International Baccalaureate Organization 07 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Sierra High School