IB Math High Level Year 2 Calc Differentiation Practice IB Practice - Calculus - Differentiation (V2 Legacy)

Transcription

1 IB Math High Level Year Calc Differentiation Practice IB Practice - Calculus - Differentiation (V Legac). If =, fin the two values of when = 5. Answer:.. (Total marks). Differentiate = arccos ( ) with respect to, an simplif our answer. Answer:.. (Total marks). Give eact answers in this part of the question. The temperature g (t) at time t of a given point of a heate iron ro is given b ln t g (t) =, t where t > 0. (a) Fin the interval where g (t) > 0. () (b) Fin the interval where g (t) > 0 an the interval where g (t) < 0. (5) (c) Fin the value of t where the graph of g (t) has a point of infleion. () () Let t* be a value of t for which g (t*) = 0 an g (t*) < 0. Fin t*. () (e) Fin the point where the normal to the graph of g (t) at the point (t*, g (t*)) meets the t-ais. () (Total 8 marks). Let f () = ln 5, 0. 5 < <, a, b; (a, b are values of for which f () is not efine). (a) (i) Sketch the graph of f (), inicating on our sketch the number of zeros of f (). Show also the position of an asmptotes. () (ii) Fin all the zeros of f (), (that is, solve f () = 0). () (b) Fin the eact values of a an b. () (c) Fin f (), an inicate clearl where f () is not efine. () Fin the eact value of the -coorinate of the local maimum of f (), for 0 < <.5. (You ma assume that there is no point of infleion.) (e) Write own the efinite integral that represents the area of the region enclose b f () an the -ais. (Do not evaluate the integral.) () 5. Differentiate from first principles f () = cos. (Total 8 marks) 6. For the function f : n, > 0, fin the function f, the erivative of f with respect to. (Total marks) () () C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of

2 IB Math High Level Year Calc Differentiation Practice 7. For the function f : sin + cos, fin the possible values of sin for which f () = 0. Answer:... (Total marks) 8. For what values of m is the line = m + 5 a tangent to the parabola =? Answer:... (Total marks) 9. The tangent to the curve at the point P(, ) meets the -ais at Q an the -ais at R. Fin the ratio PQ : QR. Answer:... (Total marks) 0. (a) Sketch an label the curves = for, an = ln for 0 <. () (b) Fin the -coorinate of P, the point of intersection of the two curves. () (c) If the tangents to the curves at P meet the -ais at Q an R, calculate the area of the triangle PQR. (6) () Prove that the two tangents at the points where = a, a > 0, on each curve are alwas perpenicular. () (Total marks). (a) Let = a bsin, where 0 < a < b. b a sin (i) ( b a )cos Show that =. ( b asin ) () (ii) Fin the maimum an minimum values of. () (iii) Show that the graph of = a bsin, 0 < a < b cannot have a vertical asmptote. b a sin () (b) For the graph of 5sin for 0, 5 sin (i) write own the -intercept; (ii) fin the -intercepts m an n, (where m < n) correct to four significant figures; (iii) sketch the graph. (5) (c) The area enclose b the graph of 5sin an the -ais from = 0 to = n is enote 5 sin b A. Write own, but o not evaluate, an epression for the area A. () (Total 7 marks) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of

3 IB Math High Level Year Calc Differentiation Practice. If f () = ln( ),, fin (a) f (); (b) the value of where the graient of f () is equal to. Answers: (a)... (b)... (Total marks). Fin the -coorinate, between an 0, of the point of infleion on the graph of the function f : e. Give our answer to ecimal places. Answer:.. (Total marks). Fin the graient of the tangent to the curve + = 7 at the point where = an > 0. Answer:... (Total marks) 5. ( sin) The function f is given b f : e, 0. (a) Fin f (). Let n be the value of where the (n + l) th maimum or minimum point occurs, n. (ie 0 is the value of where the first maimum or minimum occurs, is the value of where the secon maimum or minimum occurs, etc). (b) Fin n in terms of n. Answers: (a)... (b)... (Total marks) 6. Let f ( ) ( ),.. (a) Sketch the graph of f (). (An eact scale iagram is not require.) On our graph inicate the approimate position of (i) each zero; (ii) each maimum point; (iii) each minimum point. () (b) (i) Fin f (), clearl stating its omain. (ii) Fin the -coorinates of the maimum an minimum points of f (), for < <. (7) (c) Fin the -coorinate of the point of infleion of f (), where > 0, giving our answer correct to four ecimal places. () (Total marks) 7. The line = 6 9 is a tangent to the curve = + a + b 9 at the point (,7). Fin the values of a an b. (Total marks) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of

4 IB Math High Level Year Calc Differentiation Practice 8. Consier the function = tan 8 sin. (a) Fin. (b) Fin the value of cos for which 0. Answers: (a)... (b)... (Total marks) 9. Consier the tangent to the curve = (a) Fin the equation of this tangent at the point where =. (b) Fin the coorinates of the point where this tangent meets the curve again. Answers: (a)... (b)... (Total marks) 0. Let = sin (k) k cos (k), where k is a constant. Show that k sin ( k). (Total marks). Consier the function f ( ), where (a) (b) Show that the erivative +. n f ( ) f ( ). Sketch the function f (), showing clearl the local maimum of the function an its horizontal asmptote. You ma use the fact that n lim 0. (c) Fin the Talor epansion of f () about = e, up to the secon egree term, an show that this polnomial has the same maimum value as f () itself. (5) (Total marks). A curve has equation + =. Fin the equation of the tangent to this curve at the point (, ). Answer:.... The function f is efine b f () = (a) (i) Fin an epression for f (), simplifing our answer. (ii) The tangents to the curve of f () at points A an B are parallel to the -ais. Fin the coorinates of A an of B. (b) (i) Sketch the graph of = f (). (ii) Fin the -coorinates of the three points of infleion on the graph of f. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of () (5) (5)

5 IB Math High Level Year Calc Differentiation Practice (5) (c) Fin the range of (i) f; (ii) the composite function f f. (5) (Total 5 marks). Air is pumpe into a spherical ball which epans at a rate of 8 cm per secon (8 cm s ). Fin the eact rate of increase of the raius of the ball when the raius is cm. Answer: A curve has equation = 8. Fin the equation of the normal to the curve at the point (, ). Answer: The function f is efine b f () =, for > 0. (a) (i) Show that ln f () = (ii) Obtain an epression for f (), simplifing our answer as far as possible. (b) (i) Fin the eact value of satisfing the equation f () = 0 (ii) Show that this value gives a maimum value for f (). (c) Fin the -coorinates of the two points of infleion on the graph of f. 7. Consier the function f (t) = sec t + 5t. (a) Fin f (t). (b) Fin the eact values of (i) f (); (ii) f (); 8. Consier the equation = +. (a) Fin when = an < 0. (b) Fin when = an < 0. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 5 of (5) () () (Total marks) Answers: (a)... (b) (i)... (ii)... Answers: (a)... (b)...

6 IB Math High Level Year Calc Differentiation Practice 9. Let = e sin (). (a) Fin. (b) Fin the smallest positive value of for which = 0. Answers: (a)... (b) An airplane is fling at a constant spee at a constant altitue of km in a straight line that will take it irectl over an observer at groun level. At a given instant the observer notes that the angle is raians an is increasing at 60 raians per secon. Fin the spee, in kilometres per hour, at which the airplane is moving towars the observer. Airplane km Observer Answer:... a. A curve has equation f () =, a 0, b > 0, c > 0. c b e c (a) Show that f () = c ac e e b. c b e () (b) Fin the coorinates of the point on the curve where f () = 0. () (c) Show that this is a point of infleion. () (Total 8 marks). The point P(, p), where p > 0, lies on the curve + = 6. (a) Calculate the value of p. (b) Calculate the graient of the tangent to the curve at P. Answers: (a)... (b).... The function f is efine b f :. Fin the solution of the equation f () =. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 6 of

7 IB Math High Level Year Calc Differentiation Practice. The following iagram shows an isosceles triangle ABC with AB = 0 cm an AC = BC. The verte C is moving in a irection perpenicular to (AB) with spee cm per secon. C A B Calculate the rate of increase of the angle CÂB at the moment the triangle is equilateral. Answer: If = ln ( ), fin. Answer: Fin the equation of the normal to the curve + 9 = 0 at the point (, ). Answer:... π 7. The function f is given b f () = sin 5. (a) Write own f (). π (b) Given that f =, fin f () Fin the graient of the normal to the curve + = at the point (, ) The function f is given b f () = 5, 0. There is a point of infleion on the graph of f at the point P. Fin the coorinates of P. 0. An eperiment is carrie out in which the number n of bacteria in a liqui, is given b the formula n = 650 e kt, where t is the time in minutes after the beginning of the eperiment an k is a constant. The number of bacteria oubles ever 0 minutes. Fin (a) the eact value of k; (b) the rate at which the number of bacteria is increasing when t = 90. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 7 of

8 IB Math High Level Year Calc Differentiation Practice. Let f () = 5 5,. (a) Fin f (). (b) Solve f () >.. The function f is efine b f () = e p ( + ), here p. (a) (i) Show that f () = e p (p( + ) + ). (ii) Let f (n) () enote the result of ifferentiating f () with respect to, n times. Use mathematical inuction to prove that f (n) () = p n e p (p( + ) + n), n +. (7) (b) When p =, there is a minimum point an a point of infleion on the graph of f. Fin the eact value of the -coorinate of (i) the minimum point; (ii) the point of infleion. () (c) Let p =. Let R be the region enclose b the curve, the -ais an the lines = an =. Fin the area of R. () (Total marks). The iagram shows a trapezium OABC in which OA is parallel to CB. O is the centre of a circle raius r cm. A, B an C are on its circumference. Angle OĈB = θ. O r A C B Let T enote the area of the trapezium OABC. (a) Show that T = r (sin θ + sin θ). For a fie value of r, the value of T varies as the value of θ varies. (b) Show that T takes its maimum value when θ satisfies the equation cos θ + cos θ = 0, an verif that this value of T is a maimum. (c) Given that the perimeter of the trapezium is 75 cm, fin the maimum value of T. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 8 of () (5) (6) (Total 5 marks)

9 IB Math High Level Year Calc Differentiation Practice. Let f be a cubic polnomial function. Given that f (0) =, f (0) =, f () = f () an f ( ) = 6, fin f (). 5. (a) Write own the term in r in the epansion of ( + h) n, where 0 r n, n +. () (b) Hence ifferentiate n, n +, from first principles. (5) (c) Starting from the result n n =, euce the erivative of n, n +. () (Total 0 marks) 6. Let f () = cos ( + ), 0. (a) Fin f (). (b) Fin the eact values of the three roots of f () = Given that + = +, fin. 8. Let f be the function efine for > b f () = ln ( + ). (a) Fin f (). (b) Fin the equation of the normal to the curve = f () at the point where =. Give our answer in the form = a + b where a, b. 9. Let = cos + i sin. (a) Show that θ = i. [You ma assume that for the purposes of ifferentiation an integration, i ma be treate in the same wa as a real constant.] () (b) Hence show, using integration, that = e i. (5) (c) Use this result to euce e Moivre s theorem. () () (i) sin 6θ Given that sin θ = a cos 5 + b cos + c cos, where sin 0, use e Moivre s theorem with n = 6 to fin the values of the constants a, b an c. (ii) sin 6θ Hence euce the value of lim. 0 sin θ (0) (Total 0 marks) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 9 of

10 IB Math High Level Year Calc Differentiation Practice 50. Let f () = ln, 0. (a) Fin f (). (b) Using integration b parts fin 5. Let = arcsin, ], [. Show that = 5. Given that e ln = e for, fin at the point (, ). 5. The following table shows the values of two functions f an g an their first erivatives when = an = 0. f () f () g () g () 0 5 (a) f ( ) Fin the erivative of when = 0. g ( ) (b) Fin the erivative of f (g () + ) when =. 5. The function f is efine b f () = 6 for b where b. (a) Show that f () = ( 6). (ln ). (b) Hence fin the smallest eact value of b for which the inverse function f eists. Justif our answer. 55. Consier the curve with equation + + =. (a) Fin in terms of k, the graient of the curve at the point (, k). (5) (b) Given that the tangent to the curve is parallel to the -ais at this point, fin the value of k. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 0 of. ()

11 IB Math High Level Year Calc Differentiation Practice 56. Anré wants to get from point A locate in the sea to point Y locate on a straight stretch of beach. P is the point on the beach nearest to A such that AP = km an PY = km. He oes this b swimming in a straight line to a point Q locate on the beach an then running to Y. When Anré swims he covers km in 5 5 minutes. When he runs he covers km in 5 minutes. (a) If PQ = km, 0, fin an epression for the time T minutes taken b Anré to reach point Y. (b) Show that T () T (c) (i) Solve 0. (ii) Use the value of foun in part (c) (i) to etermine the time, T minutes, taken for Anré to reach point Y. T 0 5 (iii) Show that an hence show that the time foun in part (c) (ii) is a minimum. () (Total 8 marks) 57. Fin the graient of the tangent to the curve = cos (π) at the point (, ). (Total marks) () C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of

12 IB Math High Level Year - Calc Differentiation Practice - MarkScheme IB Practice - Calculus - Differentiation: (V Legac) MarkScheme. B implicit ifferentiation, ( = ) 6 = 0 (M) 6 (A) When = 5, 50 = = 6 = ± 5 Therefore 6 Note: This can be one eplicitl. (M) (A) (C)(C). Given = arccos ( ) then ( ( ) ) (M) ( ( )) ( ) (M) (A) OR cos = (M) sin = sin ( ) (M) ln t ln t. (a) g (t) =. So g(t) =. (M)(A) t t Hence, g(t) > 0 when > ln t or ln t < or t < e. (M) Since the omain of g (t) is {t:t > 0}, g(t) > 0 when 0 < t < e. (A) (b) ln t t t ( ln t) Since g(t) =, g(t) = t t (M) = t[8 ln t] t (A) Hence g(t) > 0 when 8 ln t < 0 ie t > e 8/. (M) Similarl, g(t) < 0 when 0 < t < e 8/. (A) 5 (c) g(t) = 0 when t = 0 or 8 = ln t. Since, the omain of g is {t:t > 0}, g(t) = 0 when t = e 8/. (M) Since g(t) > 0 when t > e 8/ an g(t) < 0 when t < e 8/, (M) (A) (C) [] [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

13 IB Math High Level Year - Calc Differentiation Practice - MarkScheme 8 / 8 / e, e is the point of infleion. The require value of t is e 8/. (A) Note: Awar (A) for evaluating t as e 8/. () g(t) = 0 when ln t = or t = e. (M) Also g(e ) = e [8 ln e ] 6 5 e e < 0 (M) Hence t* = e (A) (e) At (t*, g (t*)) the tangent is horizontal. (M) So the normal at the point (t*, g (t*)) is the line t = t*. (M) Thus, it meets the t ais at the point t = t* = e an hence the point is (e, 0). (A). (a) (i) = ln 5.5 [8] asmptote asmptote (G) Note: Awar (G) for correct shape, incluing three zeros, an (G) for both asmptotes (ii) f () = 0 for = 0.599,.5,.5 (G)(G)(G ) 5 (b) f () is unefine for ( 5 ) = 0 (M) ( ) = 0 Therefore, = 0 or = / (A) (c) f () = or 5 (M)(A) f () is unefine at = 0 an = / (A) () For the -coorinate of the local maimum of f (), where 0 < <.5 put f () = 0 (R) 5 6 = 0 (M) (e).5 6 = 5 (A) The require area is.5 A = f ( ) (A) 5. Using first principles Note: Awar (A) for each correct limit. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

14 IB Math High Level Year - Calc Differentiation Practice - MarkScheme f ( h) f ( ) f () = lim h0 h cos( h) cos( ) = lim h0 h (M) cos( )cos( h) sin( )sin( h) cos( ) = lim h0 h (M)(A) cos( h) sin( h) = cos() lim sin( ) lim h0 h h0 h (M)(A) sin( h) cos( h) But lim = an lim = 0 h0 h h0 h (C)(C) Therefore, f () = sin (A) OR f ( h) f ( ) f () = lim h0 h cos( h) cos( ) = lim h0 h (M) sin( h)sin h = lim (using an metho) h0 h (M)(A) sin h = lim sin h h0 h (M) sin h But = lim = an = lim sin h = sin h0 h0 h (C) Therefore, f () = sin (A) 8 6. f () = ln f () = ln + (M)(M) = ln + (A) (C) f : ln + 7. f () = sin + cos f () = cos sin (M) = sin sin = ( + sin )( sin ) (M) = 0 when sin = or (A) (C) [8] [] [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

15 IB Math High Level Year - Calc Differentiation Practice - MarkScheme 8. Metho : = m = = m when = (M) m m Thus,, lies on = m + 5. (R) m m Then, + 5, so m = m =. (C) Metho : For intersection: m + 5 = or + m + = 0. For tangenc: iscriminant = 0 Thus, m = 0 m = (C) (A) (M) (M) 9. = so =. At P(, ),. (M) The tangent is =, giving Q =, 0 an R = 0,. (A) Therefore, PQ : QR = : or : = :. (A) (C) 0. (a) (A) [] [] = R Q O = ln Note: Awar (C) for =, (C) for = (b) + ln =0 when = P (C) Therefore, the -coorinate of P is (G) (c) The tangent at P to = has equation = , (G) an the tangent at P to = ln has equation = (G) Thus, the area of triangle PQR = ( )(0.58). (M) = 0.0 ( sf) (A) OR = = (M) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5 ln.

16 IB Math High Level Year - Calc Differentiation Practice - MarkScheme Therefore, the tangent at (p, p ) has equation p = p. (C) = ln (M) Therefore, the tangent at (p, p ) has equation + p = p + p. (C) Thus, Q = (0, p ) an R = (0, p + ). Thus, the area of the triangle PQR = (p + )p (M) = 0.0 ( sf) (A) 6 () = when = a, = a (C) = ln when = a, a (a > 0) (C) Now, (a) = for all a > 0. a (M) Therefore, the tangents to the curve at = a on each curve are alwas perpenicular. (R)(AG). (a) (i) a bsin =, 0 < a < b b a sin ( b a sin )( b cos ) ( a bsin )( a cos ) ( b asin ) (M)(C) b cos absin cos a cos absin cos = (M)(C) ( b a sin ) b a cos = (AG) b a sin (ii) = 0 cos = 0 since b a 0. π This gives = (+k, k ) (M)(C) π a b When =, = b a =, π a b an when =, = b a =. Therefore, maimum = an minimum =. (A) (iii) A vertical asmptote at the point eists if an onl if b + a sin = 0. (R) b Then, since 0 < a < b, sin =, which is impossible. a (R) Therefore, no vertical asmptote eists. (AG) (b) (i) -intercept = 0.8 (A) (ii) For -intercepts, sin = =.069, (A) (iii) [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 5 of 5

17 IB Math High Level Year - Calc Differentiation Practice - MarkScheme 0 m n (C) 5 (c).069 Area = 5sin 5. 5sin 0 5 sin.0695 sin (M)(C) OR sin Area = 0 5 sin (M)(C). (a) If f () = ln( ), Then f () = (A) (b) Put =.8 (using a graphic ispla calculator or the quaratic formula). If f : () e then f () = e + e f () = e + e + e = e ( + + ) For a point of infleion solve f () = 0 f () = 0 at = (using a graphic ispla calculator or the quaratic formula) (Since f () 0 at this value, then it is a point of infleion.) Note: Some caniates ma fin the value of from f() b fining the minimum turning point using a graphic ispla calculator. + = 7 When =, = (since > 0) (M) ( + = 7) 6 +8 = 0 (A) The graient where = an = is (A)(C) OR + = 7 = 7, since > 0 (M) (A) (7 ), (A) (A) (A) (A) [7] [] [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 6 of 5

18 IB Math High Level Year - Calc Differentiation Practice - MarkScheme =, when = (A)(C) 5. (a) f () = cos ()e (+sin) (A) (C) (b) For maimum or minimum points, f () = 0 cos π = 0 (M) k = n then n = (A)(C) 6. (a) f () = ( ).5 [] [] ma ma zero zero 0.5 min 0.5 min zero (b) (i) Let f () =.5 (A) Notes: Awar (A) for the shape, incluing the two cusps (sharp points) at = ±. (i) Awar (A) for the zeros at = ± an = 0. (ii) Awar (A) for the maimum at = an the minimum at =. (iii) Awar (A) for the maimum at appro. = 0.65, an the minimum at appro. = 0.65 There are no asmptotes. The caniates are not require to raw a scale. Then f () = ( ) ( ) (M)(A) f () = ( f () = ( f () = ( (or equivalent) (or equivalent) The omain is.., ± (accept. < <., ±) (ii) For the maimum or minimum points let f () = 0 ie (7 ) = 0 or use the graph. C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 7 of 5 ) ( ) 7 ) 7 ) (A) (M)

19 IB Math High Level Year - Calc Differentiation Practice - MarkScheme Therefore, the -coorinate of the maimum point is = (or 0.655) an (A) 7 the -coorinate of the minimum point is = 7 (or 0.655). (A) 7 Notes: Caniates ma o this using a GDC, in that case awar (M)(G). (c) The -coorinate of the point of infleion is = ±.9 (G) OR (7 9) f () =, ± 9 ( ) (M) For the points of infleion let f () = 0 an use the graph, ie = 9 7 =.9. (A) Note: Caniates ma o this b plotting f () an fining the - coorinate of the minimum point. There are other possible methos. 7. For the curve, = 7 when = a + b =, an (M) = 6 + a + b = 6 when = a + b = 0. (M) Solving gives a = an b = 8. (A)(C) 8. (a) = sec 8 cos (A) (C) 8cos (b) (M) cos = 0 cos = (A)(C) 9. METHOD (a) The equation of the tangent is = 8. (G)(C) (b) The point where the tangent meets the curve again is (, 0). (G)(C) METHOD (a) = an = = at =. (M) Therefore, the tangent equation is = 8. (A)(C) (b) This tangent meets the curve when 8 = which gives = 0 ( + ) ( + ) = 0. The require point of intersection is (, 0). (A)(C) 0. = sin (k) k cos (k) = k cos (k) k{cos (k) + [ k sin (k)]} (M)(C) = k cos (k) k cos (k) + k sin (k) (C) = k sin (k) (AG) [] [] [] [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 8 of 5

20 IB Math High Level Year - Calc Differentiation Practice - MarkScheme. (a) The erivative can be foun b logarithmic ifferentiation. Let = f (). (b) = ln ln ln ln (M) (M)(M) ln = ln that is, f () = f () (AG) This function is efine for positive an real numbers onl. To fin the eact value of the local maimum: = 0 ln = = e (M) = e e To fin the horizontal asmptote: (A) ln lim( ) lim ln lim = 0 lim = (M)(A) [] ( ) (e,e e = (A) 5 (c) B Talor s theorem we have f (e) P () = f (e) + f (e)( e) + ( e) (A) ln ln f () = f () f ( ) (M) Also, f (e) = 0, an f (e) = 0 + f (e) e e e e (M)(A) e e hence P () = e e e e ( e) which is a parabola with verte at = e an P (e) = e e = f (e) (R)(AG) = 0 (M)(A) [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 9 of 5

21 IB Math High Level Year - Calc Differentiation Practice - MarkScheme = (A) At (,), = (A) Equation of tangent is = l( ) or + = (A) (C6). (a) (i) f ' () = (M)(A) (b) = (A) (ii) f () = 0 => = ± A, B(, ) or A (, ) B, (A)(A) 5 (i) (c) (ii) Note: Awar (G) for general shape an (G) for inication of scale. The points of infleion can be foun b locating the ma/min on the graph of f '. This gives =.5, 0.7,.88. OR f () = (G) (G) (M) f () = 0 = 0 (A) =.5, 0.7,.88 (G) 5 The graph of = f () helps: / + C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 0 of 5

22 IB Math High Level Year - Calc Differentiation Practice - MarkScheme V. = 8 (cm s ), V = r t V => = r (M)(A) r V V r r V V = => = (M) t r t t t r r When r =, = 8 ( ) (M)(A) t = (cm s ) (o not accept 0.59) (A) (C6) π 5. METHOD + = 0 (M)(A) At (, ), + 6 = 0 (M) = (A) Graient of normal = (A) Equation of normal is = ( ) (A) (C6) METHOD = (A) 5 = (M)(A) = when = (A) (i) Range of f is,. (A)(A) (ii) We require the image set of,. f =, f () = = 9 9 (M) 7 Range of g is,. (A)(A) 5 Note: Since the question i not specif eact ranges accept open intervals or numerical approimations (eg [0., 0.58]). [5] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

23 IB Math High Level Year - Calc Differentiation Practice - MarkScheme Graient of normal = (A) Equation of normal is = ( ) (A) (C6) ln 6. (a) (i) f () = (M)(A) ln = (AG) ln ln ln (ii) f "() = (M)(A) ln ln = (A) 5 Note: Awar the secon (A) for some form of simplification, ln ln eg accept. (b) (i) ln = 0 giving = ln (M)(A) Note: Awar (M)(A0) for =.89. (ii) With this value of, 8 f () = venumber < 0 (M)(A) Therefore, a maimum. (AG) (c) Points of infleion satisf f (0) = 0, ie (ln ) ln + = 0 (M) = ln (A) = (= 0.85,.9) (A) ln OR = 0.85,.9 8 ln ln (M)(G)(G ) 7. (a) METHOD f (t) = sec t + 5t f (t) = (cos t) +5t f (t) = 6(cos t) ( sin t) + 5 (M)(A) 6sin t = cos t + 5 (C) METHOD f (t) = sec t(sec t tan t) + 5 (M)(A) = 6sec t tan t + 5(= 6 tan t + 6 tan t + 5) (C) [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

24 IB Math High Level Year - Calc Differentiation Practice - MarkScheme (b) f (π) = (cosπ) + 5π (M) = + 5π (A) (C) 6sin π f (π) = (cosπ) +5 (M) = 5 (A) (C) 8. = + (a) = = 0 (M) = or = < 0 = (A) (C) (b) + = + (M)(M)(A ) (, ) (A) 5 (C) 9. = e sin (π) (a) = e sin (π) + πe cos (π) (M)(A)(A ) (C) (b) 0 = e ( sin (π) + π cos (π)) π tan (π) = (M) π = (M) = 0.76 (0.7 to sf) (A) (C) 0. tan θ = sec θ (M) t t π when θ =, = an sec θ = (A)(A) sec (M) t t () t 60 C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

25 IB Math High Level Year - Calc Differentiation Practice - MarkScheme km s t 5 = 0 km h (A) t The airplane is moving towars him at 0 km h (A) (C6) Note: Awar (C5) if the answer is given as 0 km h. a. f () =, a 0, b > 0, c > 0 c b e c c ( b e )(0) ( a)( ce ) (a) f () = (M) c ( b e ) f () = ace = (A) c ( b e ) = ( b e c c bac e c ) ( ac e c ac c (e ( b e ) ( ace ( b e c ) c c ) c c ) ac )( b e )( ce C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5 (M) ac (e ) bac e = (A) c ( b e ) c c ac e (e b) = (AG) c ( b e ) (b) f () = 0 e c = b c = ln b a a = ln b lnb c b e b a b = a So coorinate = b c b (A)(A) (c) Now e c > b on one sie of = ln b an e c < b on the other sie. c (R) f () changes sign at this point. (R) It is a point of infleion. (AG). (a) p + p = 6 p = (M)(A) (C) 8 Note: Do not penalize if p = also appears. (b) = 0 (A)(A) Note: Awar (A) for + an (A) for 6 = 0. at P(, ), = 0 = 8 (A) graient = (= 0.57) 7 (A)(C) (e c ) c c ) [8]

26 IB Math High Level Year - Calc Differentiation Practice - MarkScheme. f () = f () = ln (M) f () = (ln ) (A) (ln ) = = (ln) (A) ln = ln (M) (ln) ln (ln) = (A) ln = 0.60 (A)(C6). Let h = height of triangle an θ = CÂB. (A) Then, h = 5 tan h 5sec t t (A) (M)(A) π Put θ =. = 5 t (A) 8 ra per sec Accept per seconor 5.7 per secon t 0 π (A)(A) (C6) Note: Awar (A) for the correct value, an (A) for the correct units. 5. = ln ( ) ( ) ( ) () = or ( ) ( ) = 0 Differentiating w.r.t Note: Awar (A)for, an (A) for EITHER (M)(A) (A) (M)(A) (A)(C6) (A)(A) (A) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 5 of 5

27 IB Math High Level Year - Calc Differentiation Practice - MarkScheme at point (, ) graient = 0.8. Graient of normal =.5 OR 9 Graient of normal 9 at point (, ), graient is.5 THEN Equation of normal is given b =.5( ) or = (A) (A) 7. (a) Using the chain rule f () = cos 5 5 (M) = 0 cos 5 A (b) f () = f ( ) (A) (A) π = cos5 5 + c A Substituting to fin c, f π = π π cos5 + c = 5 M c = + cos = + = (A) f () = cos A N (A)(C6) 8. Attempting to ifferentiate implicitl (M) + = = 0 A Substituting = an = (M) = 0 A 5 = = 5 A Graient of normal is 5. A N 9. METHOD f ( ) f ( ) f ( ) 0 (M)(A) (A) (M) 0.80 an.08 (accept.07 ) (A)(A) The point of infleion is ( 0.80,.08) or, 5 METHOD (C5)(C) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 6 of 5

28 IB Math High Level Year - Calc Differentiation Practice - MarkScheme f ( ) (M)(A) f( ) has a maimum when 0.80 (M)(A) (C5).08 (accept.07 ) (A) (C) 0 0. (a) e k (M)(A) ln k (A) (C) 0 n kt (b) 650ke (M)(A) t n when t 90, to sf (A) (C) t. METHOD (a) (b) METHOD (a) f ( ) f( ) ( ) (b) f( ) ( )( 5) ( 5 5) 5 ( ) 5 ( ) ( ) ( ) ( ) (M)(A) (A) (M)(A) (A) (M) (A) (A) p p. (a) (i) f ( ) pe ( ) e (A) p e p ( ) (ii) The result is true for n since LHS e p p ( ) (AG) an RHS e p p p p( ) e p( ). (M) ( k) k p Assume true for n k : f ( ) p e p( ) k ( k ) ( k) k p k p f ( ) f ( ) p pe p( ) k p e p k p p e p( ) k Therefore, true for nk true for nk an the (M) (M)(A) (A) (C) (C) (A) (M)(A) (C) (C) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 7 of 5

29 IB Math High Level Year - Calc Differentiation Practice - MarkScheme. proposition is prove b inuction. (R) 7 (b) (i) f ( ) e ( ) 0 (M) (c) (ii) area OR area f ( ) 8.08 f ( ) e ( ) f ( ) e ( ) EITHER 8.08 f ( ) f ( ) (A) (M) (M) (A) (M) N (A)N N (A) N [] O r A C N B (a) h rsin CB CN rcos h Using T ( rcb) (M) (A)(A) r T (sin sin cos ) r (sin sin ) (A) (AG) N0 (b) T r (cos cos ) 0 (for ma) cos (cos ) cos cos 0 (M) (M)(AG) cos 0.59 ( 0.959) (A) T r ( sin sin ) (M) T r 0 there is a maimum (when ) (R) 5 (c) In triangle AOB: AB r sin (M)(A) Perimeter OABC r r cos r sin 75 (M) C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 8 of 5

30 IB Math High Level Year - Calc Differentiation Practice - MarkScheme When 0.959, r 8.5 cm (A) r 8.5 Area OABC (sin sin ) (sin sin.87) (M) 96 cm (A) N 6. f () = a + b + c + (M) f () = a + b + c f () = 6a + b (M) f (0) = = (A) f () = f () a + b + c + = a + b + c = a + b f (0) = = c (A) f () = 6 = 6a + b b, a 5 5 f Accept a, b, c, (A)(A) (C6) 5. (a) r th n r nr n! r nr term = (A) h h n r r! n r! (b) n n n h lim h 0 h (M) n n n n n n n h h... h = lim h 0 h (A) n n nn n n n n h h... h = lim h 0 h (A) n nn n n = limn h... h h 0 (A) Note: Accept first, secon an last terms in the lines above. = n n (A) (c) n n = n n n n 0 (M) n n n n n 0 (A) n n n 0 (A) [5] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 9 of 5

31 IB Math High Level Year - Calc Differentiation Practice - MarkScheme 6. (a) f () = cos ( + ) ( sin ( + )) AAA f () = cos ( + ) (sin ( + )) Note: Awar A for cos ( + ), A for sin ( + ) an A for. (b) f () = 0 cos ( + ) = 0 or sin ( + ) = 0, or 8 8 AAA Note: Do not penalize the inclusion of aitional answers. 7. (ln ) + = + n n n n n Note: Awar A for (ln ) +, A for an A for. (A) (M)AAA [0] ((ln ) + ) = (ln ) + M ln = ln A N0 8. (a) f MA N (b) Hence when =, graient of tangent 7 (A) 7 graient of normal is 7 ln 7 7 ln 7 (A) M A N (accept = ) 9. (a) sin θ icos θ A EITHER i sin θ icos θ θ A = i (cos + i sin ) A = i AG N0 C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 0 of 5

32 IB Math High Level Year - Calc Differentiation Practice - MarkScheme OR i = i(cos + i sin) (= i cos + i sin) A = i cos sin A = θ AG N0 (b) i θ MA ln = i + c A Substituting (0, ) 0 = 0 + c c = 0 A ln = i A = e i AG N0 (c) cos n + i sin n = e in M = (e i ) n A = (cos + i sin ) n AG N0 Note: Accept this proof in reverse. () (i) cos 6 + i sin 6 = (cos + i sin) 6 M Epaning rhs using the binomial theorem MA = cos cos 5 i sin + 5 cos (i sin) + 0 cos (i sin) + 5 cos (i sin) + 6 cos (i sin) 5 + (i sin) 6 Equating imaginar parts (M) sin 6 = 6 cos 5 sin 0 cos sin + 6 cos sin 5 A sin 6θ 6 cos 5 0 cos ( cos ) + 6 cos ( cos ) sin θ A = cos 5 cos + 6 cos (a =, b =, c = 6) A N0 (ii) sin 6θ 5 lim lim cos θ cos θ 6cos θ θ 0 sin θ θ 0 M = + 6 = 6 A N0 50. (a) f ln (M) = ln A N (b) Using integration b parts METHOD ln ln ln AA 5. = arcsin = ln ln (A) = (ln ) ( ln ) + C A (= (ln ) ln + + C) METHOD ln ln ln ln = (ln ) ln ( ln ) + C A (= (ln ) ln + + C) Note: Do not penalize the absence of + C. AAA [0] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

33 IB Math High Level Year - Calc Differentiation Practice - MarkScheme C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5 MA MA = A = = A = AG Note: The final AA are for equivalent algebraic manipulations leaing to AG. 5. e ln = Differentiating implicitl (M) AA Notes: Awar A for, A for an 0. N.B. Incorrect manipulation of e can lea to the correct final result. EITHER Collecting terms (M) (A) A OR Substituting = = (M) arcsin 0 ln e e ln e ln e e ln e e e

34 IB Math High Level Year - Calc Differentiation Practice - MarkScheme e ln 0 e e f g f g 5. (a) erivative = M g when = 0 erivative = = A N0 (b) erivative = f (g () + )(g () + ) M when = erivative = f ( + ) ( + ) (A) = ()() = A N0 5. (a) Use of quotient (or prouct) rule (M) 6 f (A) A (A) = AG N0 6 (b) Solving f () = 0 for (M) 6 A f has to be for f to eist an so the least value of b is the larger of the two -coorinates (accept a labelle sketch) R Hence b 6 A N 55. (a) Attempting implicit ifferentiation M 0 A EITHER Substituting =, = k eg k k 0 M Attempting to make the subject M OR Attempting to make the subject eg = M THEN k k 5 Substituting =, = k into (b) Solving = 0 for k gives k = A A M A N C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

35 IB Math High Level Year - Calc Differentiation Practice - MarkScheme 56. (a) AQ = (km) (A) QY = ( ) (km) T 5 5 AQ5QY = (mins) A (A) (M) (b) Attempting to use the chain rule on 5 5 (M) T (c) (i) 5 or equivalent A A A AG N0 Squaring both sies an rearranging to obtain 5 = + M = A N Note: Do not awar the final A for stating a negative solution in final answer. (ii) T M = 0 (mins) A N Note: Allow FT on incorrect value. (iii) METHOD Attempting to use the quotient rule M u, v u, an v (A) / T 5 5 A Attempt to simplif (M) 5 5 or equivalent / A 0 5 / 0 5 When =, > 0 an hence T = 0 / is a minimum Note: Allow FT on incorrect value, 0. METHOD Attempting to use the prouct rule AG R N0 M C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page of 5

36 IB Math High Level Year - Calc Differentiation Practice - MarkScheme u v / u, v, an T / 5 5 / / / Attempt to simplif / / 0 5 When =, > 0 an hence T = 0 is a / minimum Note: Allow FT on incorrect value, 0. (A) A (M) A AG R N0 57. METHOD sin AAA At (, ), 0 MA A METHOD sin AAA A sin At (, ), 0 5 / sin MA [8] [] C:\Users\Bob\Documents\Dropbo\Desert\HL\7Calculus\HL.CalcDiffPractice.oc on 08/6/06 at 6:0 PM Page 5 of 5