Topic 6 Part 2 [456 marks]

Transcription

1 Topic 6 Part [456 marks]. V =.5πr EITHER dv dr dv () () applying chain rule for example dr OR dv dv THEN dr when = πr = 4 dv = = πr = 4 = 4 dr dr dv () πr dr M M 4 r =, = or (cm s ) π 5π Note: Allow h instead of.5 up until the final. [6 marks] [6 marks] There was a large variety of methods used in this question, with most candidates choosing to implicitly differentiate the expression for volume in terms of r.. 8y + 8 lnx 4x + 8y = x dy dy M Note: M for attempt at implicit differentiation. for differentiating 8y lnx, for differentiating the rest. [7 marks] when x =, 8y + 4 y = 7 y 9 4 at = y = (as y > ) dy (, ) = or y = (x ) y = x + 6 [7 marks] (M)

2 The implicit differentiation was generally well done. Some candidates did not realise that they needed to substitute into the original equation to find. Others wasted a lot of time rearranging the derivative to make ythe subject, rather than simply putting in the particular values for dy and. x y a. () sin(π x π ) = = π, π( ) x x =,,,,,,,,, [ marks] [ marks] There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it. b. M [cos(π x n )] n+ = cos(πn) cos(π(n + )) = when n is even and = when n is odd [ marks] [ marks] There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it. c. (M) [ marks] π sin(π ) = = 8. x x [ marks] There were disappointingly few correct answers to part (c) with candidates not realising that it was necessary to combine the previous two parts in order to write down the answer.

3 4a. (i) attempt at chain rule (M) f ln x (x) = x (ii) attempt at chain rule (M) (x) = (iii) g x ln x is positive on g (x) ], [ so is increasing on g(x) AG ], [ [5 marks] [5 marks]

4 4b. (i) rearrange in standard form: () dy x + y =, x > integrating x ln x factor: (ln x) (M) e x ln x = e ln( (ln x) ) () = (lnx) multiply by integrating factor (M) (lnx + y = x M x d attempt (y(lnx) to integrate: ) = x M (or y (lnx) = x ) ) dy ln x (lnx) y = x x + c y = x x+c (ii) (ln attempt x) to use the point to determine c: (e, e ) eg, M or (lne) = e + c or e e e = e+c e (ln e) e = e e + c c = e AG y = x x+e (iii) (ln x) [ marks] graph with correct shape minimum at (accept answers to a minimum of s.f) x =. asymptote shown at x = Note: y-coordinate of minimum not required for ; Equation of asymptote not required for if VA appears on the sketch. Award A for asymptotes if more than one asymptote are shown [ marks]

5 5a. (a) π π (.57), (4.7) hence the coordinates are π π π π (, ), (, ) [ marks] [ marks] 5b. (i) π π ( x (x + cosx) ) Note: Award for, for correct limits and for x (x + cosx. ) π (ii) π A 6 π (= 59.) [5 marks] Notes: Do not award ft from (b)(i). [5 marks]

6 6a. METHOD sketch showing where the lines cross or zeros of (M) y = x(x + ) 6 x () x = and x = () x = the solution is or < x < x > [5 marks] Note: Do not award either final mark if strict inequalities are not given. METHOD separating into two cases and x > (M) x < if then x > always true (M) (x + ) if 6 > then x < (M) (x + ) so the solution 6 < < x < is or < x < x > Note: Do not award either final mark if strict inequalities are not given. METHOD () f(x) = x 7 + x x x x + 9 x + 64x solutions to are (M) x 7 + x x x x + 9 x + 6x = and x =, x = () x = so the solution is or < x < x > Note: Do not award either final mark if strict inequalities are not given. METHOD 4 when f(x) = x x(x + either ) 6 = x or x = () (x + ) 6 = if then (x + ) 6 = so x + = ± or x = (M)()

7 x = the solution is or < x < x > Note: Do not award either final mark if strict inequalities are not given. [5 marks] 6b. METHOD (by substitution) substituting (M) u = x + du = M (u ) u () 6 du = (+c) 8 u8 7 u7 = (x + ) METHOD 8 (x + ) (+c) 8 (by 7 parts) 7 (M)() du dv u = x =, = (x + ) M 6 v = (x + 7 x(x + ) 6 = x (x + ) 7 (x + ) = x(x + ) (x + (+c) METHOD 7 ) 7 (by 56expansion) 8 M f(x) = ( x MA 7 + x x x x + 9 x + 64x) = (+c) 8 x8 7 x7 x 6 x 5 x 4 x x Note: Award M if at least four terms are correct. ) 7 [5 marks] [5 marks] 7a. so x y is an asymptote (M) y = so e x = x = ln is an asymptote (M) x = ln (=.69) [4 marks] [4 marks]

8 7b. (i) M (x) = f ( e x ) e x ( e x +) e x ( e x ) [8 marks] = ex 4 e x e x ( e x ) (ii) f when (x) = M e x 4 e x e x = e x ( e x 4e x ) = e x =, e x =.6, e x = 4.4 (or e x = ± 5 ) Note: Award for zero, for other two solutions. Accept any answers which show a zero, a negative and a positive. as exactly one solution > e x R Note: Do not award marks for purely graphical solution. (iii) (.44, 8.47) [8 marks] 7c. () f () = 4 so gradient of normal is 4 (M) () f() = so equation of is L y = x [4 marks] 4 [4 marks]

9 7d. M (x) = so f 4 (M) x =.46 () f(.46) = 8.47 equation of is L y 8.47 = (x.46) (or 4 ) y = x + 8. [5 marks] 4 [5 marks] 8a. (a) M f ( x +) x(x+) ( x +) x x+ ( x +) [(x) marks] = (= ) [ marks] 8b. [ mark] x x+ = ( x +) x = ± [ mark]

10 8c. (x) = f ( x ) ( x +) (x)( x +)( x x+) ( x 4 +) Note: Award for or equivalent. ( x )( + ) x Note: Award for or equivalent. (x)( x + )( x x + ) [ marks] ( x )( x +) 4x( x x+) = = ( x +) x +6x 6x ( x +) ( x +x x ) [ (= marks] ) ( x +) 8d. recognition that is a factor (R) (x ) M (x )( x + bx + c) = ( x + x x ) x + 4x + = x = ± [4 marks] Note: Allow long division / synthetic division. [4 marks] 8e. M x+ M x + x+ x = + x + x + x + = ln( x + ) + arctan(x) M = [ ln( x + ) + arctan(x)] = ln + arctan ln arctan( ) π = ln [6 marks] 4 [6 marks]

11 9a. [ marks] Note:. y = ± π for correct shape, for asymptotic behaviour at [ marks] 9b. h g(x) = arctan( ) domain of x is equal to the domain of h g g : x, x [ marks] [ marks]

12 9c. (i) f(x) = arctan(x) + arctan( ) M x f (x) = + () +x + x x f (x) = x + +x +x = +x = (ii) METHOD f is a constant when x > M f() = π + AG 4 = π π 4 METHOD + x x R [7 marks] from diagram θ = arctan x α = arctanx R θ + α = π hence AG f(x) = π METHOD M tan(f(x)) = tan(arctan(x) + arctan( )) x = x+ x x( x ) denominator =, so R π f(x) = (for x > ) [7 marks] 9d. (i) Nigel is correct. METHOD is an odd function and arctan(x) is an odd function composition x of two odd functions is an odd function and sum of two odd functions is an odd function METHOD f( x) = arctan( x) + arctan( ) = arctan(x) arctan( ) = f(x) therefore f is an odd function. Rx x (ii) f(x) = π [ marks] R [ marks]

13 a. x A =.87 x B [ marks] = 6.78 [ marks] b. (M)() K sin x x.877k e x = 6.76 [ marks] Note: Award (M) for definite integral and () for a correct definite integral. [ marks]

14 . METHOD volume of a cone is V = π h given r M h = r, V = () dv = πh when dh M dv h = 4, = π 4.5 (using = ) dv dv dh dh dv = 8π (= 5.) (cm mi ) METHOD n volume of a cone is V = π h given r M h = r, V = π h dv = π h when M dv h = 4, = π 4.5 dv = 8π (= 5.) (cm mi ) METHOD n V = π h M r dr = π(rh + r dh ) dv Note: π h dh Award M for attempted implicit differentiation and for each correct term on the RHS. [5 marks] when M dv h = 4, r = 4, = π( ) dv = 8π (= 5.) (cm mi ) [5 marks] n

15 a. METHOD expanding the brackets first: M x 4 + x y + y 4 = 4xy M 4 x + 4x y + 4x dy y + 4 y dy = 4 y + 8xy Note: Award M for an attempt at implicit differentiation. Award for each side correct. dy [5 marks] or equivalent dy x + = METHOD x xy+ x y y y y M ( x + y dy )(x + y ) = 4 y + 8xy Note: Award M for an attempt at implicit differentiation. Award for each side correct. dy ( x M + y dy )(x + y ) = y dy + xy x or + x y + x + = + xy equivalent dy y y dy y dy x = x y + y [5 marks] yx xy+ y dy b. METHOD at (, ), is undefined dy y = METHOD gradient of normal M = = dy at (, ) gradient = y = [ marks] M (yx xy+ y ) ( x x y + y ) [ marks]

16 . (a) (M) C = AX 5k + XB k Note: Award (M) for attempting to express the cost in terms of AX, XB and k. [5 marks] = 5k 45 + x + ( x)k AG = 5k 5 + x + ( x)k [ marks] (b) dc Note: (i) M 5 x = k[ ] = k( ) 5+x 5x 5+x Award M for an attempt to differentiate and for the correct derivative. dc (ii) M = attempting to solve () 5 5+x = 75 6 x = 9.9 METHOD (m) (= (m)) for example, at M dc x = 9 =.895k < at dc x = 9 =.56k > Note: Award M for attempting to find the gradient either side of and for two correct values. x = 9.9 thus gives a minimum x = 9.9 METHOD AG d C = at (M) d x = 9.9 C =.45k > Note: 5k ( x + 5) Award M for attempting to find the second derivative and for the correct value. Note: If is obtained and its value at d C is not calculated, award (M) for correct reasoning eg, both numerator and denominator are positive at x = 9.9. x = 9.9 thus gives a minimum AG x = 9.9 METHOD versus x. dc Sketching the graph of either C versus x or M Clearly indicating that gives the minimum on their graph. x = 9.9

17 x = 9.9 [7 marks] (c) C min = 5k Note: Accept k. Accept 4k. [ mark] (d) M 45 arctan( ) = K K =.57K = [ marks] K (e) (i) when θ = (ii), x = 6 (m) ( (m)) 45 M.78K K = 4.7 (% ) [ marks] % Total [5 marks] 4a. (a) [ marks] for correct shape and correct domain [ (.4, marks].884) (, ) 6

18 4b. EITHER [4 marks] u = t du = t OR t = u = THEN du u M t du = +t M 4 +u u = or equivalent arctan( )(+c) t [4 = marks] arctan( )(+c) 4 4c. (M) 6 t M +t 4 t = [ arctan( 6 )] = (arctan( )) (= (arctan( ))) (m) 4 Note: Accept or equivalent. arctan(6 ) [ marks] 4 [ marks] 4d. dv ds () = s( s) a = v dv (M) ds a = arcsin( s) a = arcsin(.) a =.56 (m s ) [ marks] s( s)..9 [ marks]

19 5. MM x y + x dy y + x y dy dy + 9 = Note: First M for attempt at implicit differentiation, second M for use of product rule. [7 marks] dy x y +x ( () = ) y x y 9 x + x y = x ( + ) = y x = Note:. y = ± Do not award if extra solutions given eg substituting into original equation x = y 9y = y(y + )(y ) = y =, y = ± coordinates (,),(,),(, ) [7 marks] (M) The majority of candidates were able to apply implicit differentiation and the product rule correctly to obtain. The better then recognised that x was ( + y the only ) = possible solution. Such candidates usually went on to obtain full marks. A number decided that x = though then made no further progress. The solution set y = ± and x = was also occasionally seen. A small minority found the correct x and y values for the three co-ordinates but then surprisingly y = ±i expressed them as and (, ), (, ). (, ) 6a. (i) M f (x) = e x xe x (ii) f (x) = x = coordinates (, e ) [ marks] [ marks] Part a) proved to be an easy start for the vast majority of candidates.

20 6b. f (x) = e x e x + x e x (= e x ( x)) substituting into x = M f (x) hence maximum RAG f () (= e ) < [ marks] [ marks] Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so. Part c) again proved to be an easily earned marks. 6c. M f (x) = ( x = ) coordinates (, e ) [ marks] [ marks] Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so. Part c) again proved to be an easily earned marks.

21 6d. (i) g(x) = x x (ii) coordinates e of maximum (, e ) (iii) equating and attempting to solve f(x) = g(x) x e x = () x(e x e x ) = x = e x = e x x e x or [5 marks] e x = x = ln (ln4) Note: Award first () only if factorisation seen or if two correct solutions are seen. Many candidates lost their way in part d). A variety of possibilities for were suggested, commonly g(x), xe x xe or similar variations. Despite section ii) being worth only one mark, (and state being present in the question), many laborious attempts at further differentiation were seen. Part diii was usually answered well by those who gave the correct function for. g(x)

22 6e. [4 marks] A4 Note: Award for shape of, including domain extending beyond f. x = Ignore any graph shown for. x < Award for A and B correctly identified. Award for shape of, including domain extending beyond g. x = Ignore any graph shown for. Allow follow through from x <. f Award for C, D and E correctly identified (D and E are interchangeable). [4 marks] Part e) was also answered well by those who had earned full marks up to that point. 6f. M A = x x e = [ x e x ] e x Note: Condone absence of limits or incorrect limits. [ marks] = e [ e x ] [ = marks] e (e ) = e While the integration by parts technique was clearly understood, it was somewhat surprising how many careless slips were seen in this part of the question. Only a minority gained full marks for part f).

23 7a. M z n + z n = cosnθ + i sin nθ + cos( nθ) + i sin( nθ) = cosnθ + cosnθ + i sin nθ i sin nθ AG = cosnθ [ marks] [ marks] Part a) has appeared several times before, though with it again being a show that question, some candidates still need to be more aware of the need to show every step in their working, including the result that. sin( nθ) = sin(nθ) 7b. (b) (z + ) z 4 Note: Accept. (z + ) z 4 [ mark] = z z ( ) + 6 z ( ) + 4z( ) + z = 6cos 4 θ z z z 4 [ mark] Part b) was usually answered correctly. 7c. METHOD M (z + z 4 ) = ( z 4 + ) + 4 ( z + ) + 6 z 4 z ( cosθ ) 4 = cos4θ + 8 cosθ + 6 Note: Award for RHS, for LHS, independent of the M. [4 marks] θ = cos4θ + cosθ + cos 4 8 (or p =, q =, r = ) METHOD 8 8 M cos 4 cos θ+ θ = ( ) = ( θ + cosθ + ) 4 cos cos 4θ+ = ( + cosθ + ) 4 θ = cos4θ + cosθ + cos 4 8 (or p =, q =, r = ) [4 marks] Part c) was again often answered correctly, though some candidates often less successfully utilised a trig-only approach rather than taking note of part b).

24 7d. M (z + ) z 6 = z z 5 ( ) + 5 z ( ) + ( ) + 5 ( ) + 6z( ) + z 4 z z z z z 4 z 5 z 6 (z + ) = ( z 6 + ) + 6 ( z 4 + ) + 5 ( z + ) + z 6 z 4 z ( cosθ ) 6 = cos6θ + cos4θ + cosθ + z 6 Note: Award for RHS, for LHS, independent of the M. [ marks] AG cos 6 θ = 5 cos6θ + cos4θ + cosθ + Note: Accept a purely trigonometric solution as for (c). [ marks] Part d) was a good source of marks for those who kept with the spirit of using complex numbers for this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions using this approach were extremely rare. 7e. [ marks] π cos θdθ = ( cos6θ + cos4θ + cosθ + ) dθ M 6 π 6 π = [ sin 6θ + sin 4θ + sin θ + θ] = 5π [ marks] 5 5 Part e) was well answered, though numerical slips were often common. A small number integrated as sin nθ. n cosnθ A large number of candidates did not realise the help that part e) inevitably provided for part f). Some correctly expressed the volume as and thus gained the first marks but were able to progress no further. Only a small number of able candidates were able to obtain the π cos correct 4 x π cos x answer of 6. π 7f. M V = π π sin xcos 4 x M = π π cos 4 x π π cos 6 x π cos 4 π x = 6 π V = 5π = Note: 6 π Follow through from an incorrect r in (c) provided the final answer is positive. [4 marks]

25 7g. (i) constant term = ( k (k)! ) (k)! = k = (accept C ) k!k! (k!) k k (ii) k π cos k (k)!π θdθ = (k!) π [ marks] π cos k (k)!π θdθ = [ marks] k ( )π k or k+ k+ (k!) Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest scoring candidates gain all marks. 8. EITHER du = sec u (M) sec udu 4tan u 4+4tan u sec udu 4tan u sec u du sec (= or = udu ) OR 4sin u tan u+ 4tan u 4sec u [7 marks] u = arctan x du = (M) x +4 4tan u+4du 4tan u sec udu THEN 4tan u sec udu tan u = 4 cos u = cosec ucotudu (= du) 4 4 sin u = cosec u(+c) (= (+C)) use of 4 either 4 sin u or an appropriate trigonometric identity u = x either or sin u = x (or equivalent) x +4 cosec u = AG x = +4 (+C) [7 marks] 4x x +4 x M Most candidates found this a challenging question. A large majority of candidates were able to change variable from x to u but were not able to make any further progress.

26 9a. (i) either counterexample or sketch or recognising that intersects the graph of y = k (k > ) twice M y = f(x) function is not (does not obey horizontal line test) so does not exist f (ii) AG () f (x) = ( ) () ex e x f (ln) = 4 (=.) M m = 4 5 f(ln) = (=.67) EITHER y M 5 = x ln 4 4y = x + ln OR 5 M = ln + c 4 5 c = + ln 4 y = x + + ln 4 y = 9x ln THEN AG 9x + y 9 ln = (iii) The tangent at has equation (a, f(a)). (M) y f(a) = f (or equivalent) (a)(x a) () f (or f(a) (a) = equivalent) a e attempting a e a e = a + e a to solve a for a a = ±. [4 marks] 5 4 (M) R [4 marks] In part (a) (i), successful candidates typically sketched the graph of, applied the horizontal line test to the graph and concluded that the function was not y = f(x) (it did not obey the horizontal line test). In part (a) (ii), a large number of candidates were able to show that the equation of the normal at point P was. A few candidates used the gradient of the tangent rather than using it to find the gradient of the normal. 9x + y 9 ln = Part (a) (iii) challenged most candidates. Most successful candidates graphed and y = f(x) on the same set of axes and found the x-coordinates of the intersection points. y = x f (x)

27 9b. (i) y = e M x + e x e x y e x + = Note: Award M for either attempting to rearrange or interchanging x and y. [8 marks] e x y± 4y = 4 e x = y ± y x = ln(y ± y ) f (x) = ln(x + x ) Note: Award for correct notation and for stating the positive branch. (ii) (M)() V = π (ln(y + y )) dy Note:. V = π 5 d c = 7. (unit s ) [8 marks] Award M for attempting to use x dy Part (b) (i) challenged most candidates. While a large number of candidates seemed to understand how to find an inverse function, poor algebra skills (e.g. erroneously taking the natural logarithm of both sides) meant that very few candidates were able to form a quadratic in either. or e x e y a. METHOD M f (x) = q x = () = q 6 = [4 marks] f q = 6 f() = p = 5 p = 4 METHOD M f(x) = (x ) + 5 = x + 6x 4 q = 6, p = 4 [4 marks] M In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values.

28 b. M g(x) = 4 + 6(x ) (x ) (= + x x ) Note: Accept any alternative form which is correct. Award MA for a substitution of (x + ). [ marks] [ marks] In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values. a. dy = x x x( ) = x x =, ± at dy = (, ), (, ), (, ) Note: Award A for all three x-values correct with errors/omissions in y-values. [4 marks] [4 marks] The whole of this question seemed to prove accessible to a high proportion of candidates. (a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at. (b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c). The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. b. [4 marks] = y = (x ) (y = x + ) = (x ) x = (,)

29 The whole of this question seemed to prove accessible to a high proportion of candidates. (a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at. (b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c). The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. c. gradient of normal () = equation of normal is (M) 8 y = (x ) (y = x + ) at x =, y = 8 Note: In the following, allow FT on incorrect coordinates of T and N. [7 marks] lengths of, PN = 9 PT = 5 area of triangle 9 M PTN = (or equivalent e.g. 9 = ) [78 marks] 5 9 The whole of this question seemed to prove accessible to a high proportion of candidates. (a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at. (b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c). The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.

30 a. (i) [9 marks] A Note: Award for correct, for correct sin x. sin x Note: Award A for two correct shapes with and/or missing. π Note: Condone graph outside the domain. (ii), sin x = sin x M x π sin xcosx sin x = sin x( cosx ) = NN x =, π (iii) area M = π Note: (sin x sin x) Award M for an integral that contains limits, not necessarily correct, with and sin x subtracted in either order. sin x

31 π = [ cosx + cosx] (M) π π = ( cos + cos ) ( cos + cos) 4 = = 4 [9 marks] A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by and so omit the x = value. Many recognised the value from the graph and corrected this in their final solution. sin x The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well. b. M x = 8 sin θ cosθdθ Note: π 6 4sin θ 4 x Award M for substitution 4 4sin θ and reasonable attempt at finding expression for in terms of, first for correct limits, second for correct substitution for. dθ π 6 8sin θdθ π M cosθdθ π 6 = [4θ sin θ] (M) π π = ( sin ) π = [8 marks] [8 marks] A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by and so omit the x = value. Many recognised the value from the graph and corrected this in their final solution. sin x The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

32 c. (i) [8 marks] M from the diagram above the shaded area R = a f(x) = ab b (y)dy AG f = ab b (x) (ii) f x f(x) = arcsin f (x) = 4 sin x 4 M π x π 6 arcsin( ) = 4 sin x 4 Note: Award for the limit seen anywhere, for all else correct. π 6 π = [ 4 cosx] π 6 π = 4 + Note: Award no marks for methods using integration by parts. [8 marks]

33 A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by and so omit the x = value. Many recognised the value from the graph and corrected this in their final solution. sin x The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well. a. (M)() a = e.t at, t = a =.5 (m s ) (accept e ) [ marks] [ marks] Part (a) was generally correctly answered. A few candidates suffered the Arithmetic Penalty for giving their answer to more than sf. A smaller number were unable to differentiate the exponential function correctly. Part (b) was less well answered, many candidates not thinking clearly about the position and direction associated with the initial conditions. b. METHOD (M) d = 5( e.t ) = 8.8 so distance above ground = 7 (m) ( sf) (accept 76 (m)) METHOD M s = 5( e.t ) = 5t + 5 e.t (+c) Taking s = when t = gives c = 5 M So when t =, s = 8... so distance above ground = 7 (m) ( sf) (accept 76 (m)) [ marks] [ marks]

34 Part (a) was generally correctly answered. A few candidates suffered the Arithmetic Penalty for giving their answer to more than sf. A smaller number were unable to differentiate the exponential function correctly. Part (b) was less well answered, many candidates not thinking clearly about the position and direction associated with the initial conditions.

35 4. let x = distance from observer to rocket let h = the height of the rocket above the ground METHOD dh = when h = 8 M x = h + 6 = ( h + 6 ) h = dh h when h = +6 8 M dh = dh h = h +6 = 4 (m s ) [6 marks] METHOD M h + 6 = x h = x dh dh h x = 8 4 = (= ) 5 dh = M = dh dh 4 = 5 = 4 (m s ) [6 marks] METHOD M x = 6 + h dh x = h when h = 8, x = M 8 = = 4 (m s ) [6 marks] METHOD 4 dh Distance between the observer and the rocket M = ( ) = Component of the velocity in the line of sight [6 marks] = sin θ (where angle of elevation) θ = sin θ = 8 M

36 sin θ = 8 component = 4 (m s ) [6 marks] Questions of this type are often open to various approaches, but most full solutions require the application of related rates of change. Although most candidates realised this, their success rate was low. This was particularly apparent in approaches involving trigonometric functions. Some candidates assumed constant speed this gained some small credit. Candidates should be encouraged to state what their symbols stand for. 5. [8 marks] x + y = a M + = x y dy dy = x y = x y Note: Accept from making y the subject of the equation, and all correct subsequent working a = dy x therefore the gradient at the point P is given by dy q = p equation of tangent is M q y q = (x p) p q (y = x + q + ) p q p x-intercept: y =, q p n = + p = + p y-intercept: q q p x =, m = q p + q M n + m = q p + p + q p + q = q p + p + q = ( p + q) AG = a [8 marks] Many candidates were able to perform the implicit differentiation. Few gained any further marks.

37 6a. prove that [8 marks] + ( ) + ( ) + 4 ( ) +...+n ( ) = 4 for n = n n+ n + LHS =, RHS = 4 = 4 = so true for n = R assume true for n = k M so + ( ) + ( ) + 4 ( ) +...+k ( ) = 4 now for n = k + k k+ k LHS: + ( ) + ( ) + 4 ( ) +...+k ( k ) + (k + ) ( k ) M k+ = 4 + (k + ) ( k ) k (or equivalent) (k+) k+ = 4 + k k (accept = 4 (k+)+ (k+) ) 4 k+ k Therefore if it is true for n = k it is true for n = k +. It has been shown to be true for n = so it is true for all. R n ( Z + ) Note: To obtain the final R mark, a reasonable attempt at induction must have been made. [8 marks] Part A: Given that this question is at the easier end of the proof by induction spectrum, it was disappointing that so many candidates failed to score full marks. The n = case was generally well done. The whole point of the method is that it involves logic, so let n = k or put n = k, instead of assume... to be true for n = k, gains no marks. The algebraic steps need to be more convincing than some candidates were able to show. It is astonishing that the R mark for the final statement was so often not awarded.

38 6b. (a) METHOD M e x sin x = cosx e x + e x cosx = cosx e x + e x sin x 4e x sin x M 5 e x sin x = cosx e x + e x sin x AG e x sin x = ( sin x cosx) + C 5 ex METHOD M sin xe x sin xe = x e cosx x sin xe = x e cosx x e sin x x 4 4 M 5 e x e sin x = x sin x cos xe x 4 4 AG e x sin x = ( sin x cosx) + C 5 ex [6 marks] [7 marks] (b) M dy = e x sin x y arcsin y = ( sin x cosx)(+c) 5 ex when M x =, y = C = 5 y = sin( ( sin x cosx) + ) 5 ex 5 [5 marks] (c) (i)

39 P is (.6, ) Note: Award for.6 seen anywhere, for complete sketch. Note: Allow FT on their answer from (b) (ii) M V =.6... πy A =.5 Note: Allow FT on their answers from (b) and (c)(i). [6 marks] Part B: Part (a) was often well done, although some faltered after the first integration. Part (b) was also generally well done, although there were some errors with the constant of integration. In (c) the graph was often attempted, but errors in (b) usually led to manifestly incorrect plots. Many attempted the volume of integration and some obtained the correct value.

40 7a. M I = π sin x e x Note: Award M for [6 marks] I = sin x π e x Attempt at integration by parts, even if inappropriate modulus signs are present. M or = [ e x π cosx] π cosx e x = [ e x π sin x] π cosx e x or = [ e x π cosx] [ e x π sin x] π sin x e x = [ e x sin x + e x π cosx] π sin x e x or = [ e x π cosx] [ e x π sin x] I M [ e x sin x + e x π cosx] I Note: Do not penalise absence of limits at this stage I = e π + I AG I = ( + e π ) Note: If modulus signs are used around cos x, award no accuracy marks but do not penalise modulus signs around sin x. [6 marks] Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the I significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.

41 7b. [4 marks] I n = e sin x nπ x Attempt to use the substitution (n+)π M y = x nπ (putting, y = x nπ and dy = ) [nπ, (n + )π] [, π] so I n = π sin(y + nπ) dy e (y+nπ) = e nπ π sin(y + nπ) dy e y = e nπ π sin ydy e y AG = e nπ I [4 marks] Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the I significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 7c. M e sin x = x n= () = e nπ I n= the term is an infinite geometric series with common ratio (M) e π therefore In () e x I sin x = +e = π e (= π + ) ( e ) [5 marks] π ( e π ) e π [5 marks]

42 Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the I significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 8a. (i) the period is [ marks] (ii) (M) ds v = = πcos(πt) + πcos(πt) (M) dv a = = π sin(πt) 4π sin(πt) (iii) v = π(cos(πt) + cos(πt)) = EITHER M cos(πt) + cos (πt) = () ( cos(πt) )(cos(πt) + ) = cos(πt) = or cos(πt) = t =, t = 5 7 t =, t =, t =, t = OR M πt πt cos( )cos( ) = πt πt cos( ) = or cos( ) = t =, 5 7 t =,,, [ marks] In (a), only a few candidates gave the correct period but the expressions for velocity and acceleration were correctly obtained by most candidates. In (a)(iii), many candidates manipulated the equation v = correctly to give the two possible values for but then failed to find all the possible values of t. cos(πt)

43 8b. [8 marks] P(n) : f (n) (x) = ( ) n (Aa n sin(ax) + Bb n sin(bx)) M P() : f (x) = (Aacos(ax) + Bb cos(bx)) = Aa sin(ax) Bb sin(bx) = (Aa sin(ax) + Bb sin(bx)) true P() assume that is true M P(k) : f (k) (x) = ( ) k (Aa k sin(ax) + Bb k sin(bx)) consider P(k + ) M f (k+) (x) = ( ) k (Aa k+ cos(ax) + Bb k+ cos(bx)) f (k+) (x) = ( ) k ( Aa k+ sin(ax) Bb k+ sin(bx)) = ( ) k+ (Aa k+ sin(ax) + Bb k+ sin(bx)) true implies P(k) true, P(k + ) true so P() true P(n) R n Z + Note: Award the final R only if the previous three M marks have been awarded. [8 marks] Solutions to (b) were disappointing in general with few candidates giving a correct solution.

44 9a. [ marks] Note: Award for shape, for x-intercept is.8, accept sin( ) or sin() for y-intercept is.4. [ marks] Many candidates attempted this question successfully. In (a), however, a large number of candidates did not use the zoom feature of the GDC to draw an accurate sketch of the given function. In (b), some candidates used the domain as the limits of the integral. Other candidates did not take the absolute value of the integral.

45 9b. (M) A =.8 x + sin(x ).86 sq units [ marks] [ marks] Many candidates attempted this question successfully. In (a), however, a large number of candidates did not use the zoom feature of the GDC to draw an accurate sketch of the given function. In (b), some candidates used the domain as the limits of the integral. Other candidates did not take the absolute value of the integral.. [6 marks] π V = h M r dv π dr = [rh + r ] at the given instant dh M dv π = [(4)()( ) + 4 ()] π = = 5. 5 hence, the volume is decreasing (at approximately 5 per century) mm [6 marks] R Few candidates applied the method of implicit differentiation and related rates correctly. Some candidates incorrectly interpreted this question as one of constant linear rates.

46 a. METHOD [5 marks] +e y = ln( x ) M e = x = dy now M = +e x () e x = e y () dy = (+ e x ) e y + e x = e y e y + e y e x +e x Note: Only one of the two above marks may be implied. AG dy e = y = Note: Candidates may find as a function of x and then work backwards from the given answer. Award full marks if done correctly. dy METHOD +e y = ln( x ) M e y = x = ln(e y ) M = dy = AG dy e = y [5 marks] +e x e x = e y dy e y e y e y e y Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, e y although partial credit was given for this. In this case, they often found problems in simplifying the algebra.

47 b. [ marks] METHOD when x =, y = ln = when dy x =, = = d y M = when e y dy d x =, y = = M d y e = y ( dy ) when e y 4 d y d x =, y = = y = f() + ()x + + (M) y = x two of the above terms are zero AG f f () x f () x!! 4 8 x x METHOD when x =, y = ln = when dy x =, = = M d y e = y dy e = y e ( y e ) = y + 4 when e y d x =, y = + = 4 M d y e = ( y e y ) when dy 4 d x =, y = ( ) = y = f() + ()x + + (M) y = x two of the above terms are zero AG f f () x f () x!! 8 x x [ marks]

48 Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, e y although partial credit was given for this. In this case, they often found problems in simplifying the algebra.. METHOD area = arctanx attempting to integrate by parts M = [x arctanx] x +x = [x arctanx] [ ln( + x )] Note: Award even if limits are absent. [6 marks] π = ln4 π (= ln) METHOD π M π area = tan ydy M π = + [ln cosy ] π π = + ln π (= ln) [6 marks] Many candidates were able to write down the correct expression for the required area, although in some cases with incorrect integration limits. However, very few managed to achieve any further marks due to a number of misconceptions, in particular. Candidates who realised they should use integration by parts were in general very successful in answering this question. It was pleasing arctanx to = see cotx a few = cos alternative x correct approaches to this question. sin x

49 . attempt at implicit differentiation M e let (x+y) dy dy ( + ) = sin(xy)(x + y), x = M y = dy e ( + ) = dy = let, x = π y = π dy dy so e ( + ) = sin( π)(x + y) = dy = since both points lie on the line this is a common tangent R y = x Note: must be seen for the final R. It is not sufficient to note that the gradients are equal. y = x [7 marks] [7 marks] Implicit differentiation was attempted by many candidates, some of whom obtained the correct value for the gradient of the tangent. However, very few noticed the need to go further and prove that both points were on the same line. 4a. (i) M f (x) = x x ln x x = ln x x so when, f i.e. (x) = lnx = x = e [5 marks] (ii) when f and (x) > x when < e f (x) R< x > e hence local maximum AG Note: Accept argument using correct second derivative. (iii) y e [5 marks]

50 Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly. 4b. f M (x) = x x ( ln x)x x 4 x x+x ln x = x 4 + ln x = x Note: May be seen in part (a). [5 marks] f (M) (x) = + lnx = x = e since when f and (x) < x when < e f (x) R > x > e then point of inflexion [5 ( e marks], ) e Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.

51 4c. [ marks] Note: Award for the maximum and intercept, for a vertical asymptote and for shape (including turning concave up). [ marks] Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.

52 4d. (i) [6 marks] Note: Award for each correct branch. (ii) all real values (iii) (M)() Note: Award (M)() for sketching the graph of h, ignoring any graph of g. (accept ) e < x < x < [6 marks]

53 Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly. 5a. ( ) f (x) (M) = x 6x 9 = (x + )(x ) = ; x = x = (max) (, 5); (min) (, 7) Note: The coordinates need not be explicitly stated but the values need to be seen. [4 marks] N y = 8x + 7 [4 marks] There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points. 5b. inflexion (, ) f (x) = 6x 6 = which lies on RAG y = 8x + 7 [ marks] [ marks] There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points. There were many correct demonstrations of the show that in (b). 6a. equation of line in graph 5 a = t (a = t + 5) [ mark] [ mark]

54 This question was well answered by a large number of candidates and indicated a good understanding of calculus, kinematics and use of the graphing calculator. Some candidates worked in and xrather than, y aand vbut mostly obtained correct solutions. Although the majority of candidate used integration throughout the question some correct solutions t were obtained by considering the areas in the diagram. 6b. (M) dv 5 = t + 5 () 5 v = + 5t + c 4 when t, t ms = v = 5 5 v = + 5t from graph t or by finding time when a = maximum ms = 95 [4 marks] [4 marks] This question was well answered by a large number of candidates and indicated a good understanding of calculus, kinematics and use of the graphing calculator. Some candidates worked in and xrather than, y aand vbut mostly obtained correct solutions. Although the majority of candidate used integration throughout the question some correct solutions t were obtained by considering the areas in the diagram.

55 6c. EITHER [ marks] graph drawn and intersection with ms (M)() v = 95 \(t = = 4.8\) OR ; 5 95 = + 5t + 5 t = t t = = 4.8(8 ) [ marks] This question was well answered by a large number of candidates and indicated a good understanding of calculus, kinematics and use of the graphing calculator. Some candidates worked in and xrather than, y aand vbut mostly obtained correct solutions. Although the majority of candidate used integration throughout the question some correct solutions t were obtained by considering the areas in the diagram. 7a. volume (M) = π h dy x M π h ydy y π [ h πh ] = [ marks] [ marks]

56 This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer. 7b. volume (M) = π h dy x M π h ydy y π [ h πh ] = [ marks] [ marks] This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer. 7c. surface area dv = surface area (M) = πx = πh M πh V = V h π dv V = π π AG dv = πv Note: Assuming that without justification gains no marks. dh = [6 marks] [6 marks] This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer.

57 7d. surface area dv = surface area (M) = πx = πh M πh V = V h π dv V = π π AG dv = πv Note: Assuming that without justification gains no marks. dh = [6 marks] [6 marks] This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer. 7e. ( Vcm = ) 5π = 57 dv = πv attempting to separate variables M [7 marks] EITHER dv = π V V = πt + c c = 5π M V = hours 5π t = = π OR M dv = π T 5π V Note: Award M for attempt to use definite integrals, for correct limits and for correct integrands. [ V ] = π T 5π hours 5π T = = [7 marks] π This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer.

58 7f. ( Vcm = ) 5π = 57 dv = πv attempting to separate variables M [7 marks] EITHER dv = π V V = πt + c c = 5π M V = hours 5π t = = π OR M dv = π T 5π V Note: Award M for attempt to use definite integrals, for correct limits and for correct integrands. [ V ] = π T 5π hours 5π T = = [7 marks] π This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer. International Baccalaureate Organization 7 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Sierra High School