Algebra y funciones [9 marks] Let f() = 3 ln and g() = ln5 3. a. Epress g() in the form f() + lna, where a Z +. attempt to apply rules of logarithms e.g. ln a b = b lna, lnab = lna + lnb correct application of ln a b = b lna (seen anywhere) e.g. 3 ln = ln 3 correct application of lnab = lna + lnb (seen anywhere) e.g. ln5 3 = ln5 + ln 3 so ln5 3 = ln5 + 3 ln g() = f() + ln5 (accept g() = 3 ln + ln5 ) N The graph of g is a transformation of the graph of f. Give a full geometric description of this transformation. b. transformation with correct name, direction, and value A3 0 e.g. translation by ( ), shift up by ln5, vertical translation of ln5 ln5 In the epansion of (3 ), the term in 5 can be epressed as ( ) (3) p ( ) q. r a. (a) Write down the value of p, of q and of r. (b) Find the coefficient of the term in 5. [5 marks]
(a) p = 5, q = 7, r = 7 (accept r = 5) N3 (b) correct working () ( ) (3) 5 ( ) 7, 79, 43, 7, 4634368 7 coefficient of term in 5 is 4634368 N Note: Do not award the final for an answer that contains. Total [5 marks] b. Write down the value of p, of q and of r. p = 5, q = 7, r = 7 (accept r = 5) N3 c. Find the coefficient of the term in 5. correct working () ( ) (3) 5 ( ) 7, 79, 43, 7, 4634368 7 coefficient of term in 5 is 4634368 N Note: Do not award the final for an answer that contains. Total [5 marks] Let log 3 p = 6 and log 3 q = 7. 3a. (a) Find log 3 p. p q (b) Find log 3 ( ). [7 marks] (c) Find log 3 (9p).
(a) METHOD evidence of correct formula log u n = n logu, log 3 p log 3 ( p ) = METHOD N valid method using p = 3 6 log 3 ( 3 6 ), log3, log 3 3 log 3 ( p ) = N (b) METHOD evidence of correct formula p log( ) = logp logq, 6 7 q p q log 3 ( ) = METHOD N valid method using p = 3 6 and q = 3 7 3 log ( 6 3 ), log3, log 3 3 7 3 p q log 3 ( ) = N (c) METHOD evidence of correct formula log 3 uv = log 3 u + log 3 v, log9 + logp log 3 9 = (may be seen in epression) + logp log 3 (9p) = 8 N METHOD valid method using p = 3 6 log 3 (9 3 6 ), log 3 ( 3 3 6 ) correct working log 3 9 + log 3 3 6, log 3 3 8 log 3 (9p) = 8 N Total [7 marks] 3b. Find log 3p.
METHOD evidence of correct formula log u n = n logu, log 3 p log 3 ( p ) = METHOD N valid method using p = 3 6 log 3 ( 3 6 ), log3, log 3 3 log 3 ( p ) = N p q 3c. Find log 3 ( ). METHOD evidence of correct formula p log( ) = logp logq, 6 7 q p q log 3 ( ) = METHOD N valid method using p = 3 6 and q = 3 7 3 log ( 6 3 ), log3, log 3 3 7 3 p q log 3 ( ) = N 3d. Find log 3 (9p).
METHOD evidence of correct formula log 3 uv = log 3 u + log 3 v, log9 + logp log 3 9 = (may be seen in epression) + logp log 3 (9p) = 8 N METHOD valid method using p = 3 6 log 3 (9 3 6 ), log 3 ( 3 3 6 ) correct working log 3 9 + log 3 3 6, log 3 3 8 log 3 (9p) = 8 N Total [7 marks] 4. The constant term in the epansion of ( + ), where a R is 80. Find a. a a 6 [7 marks] evidence of binomial epansion selecting correct term, ( ) evidence of identifying constant term in epansion for power 6 th r = 3, 4 term evidence of correct term (may be seen in equation) 0 a6, ( 6 ) a 3 3 ( ) a a 3 a ( 3 ) 6 a ( ) 0 6 + ( ) ( ) 5 a ( ) + a A () attempt to set up their equation 6 ( ) ( ) 3 a ( 3 ) = 80, a 3 = 80 correct equation in one variable a () 0 a 3 = 80, a 3 = 64 a = 4 N4 [7 marks] a 3 Write down the value of (i) log ; 5a. 3 7 [ mark] (i) log 3 7 = 3 N [ mark]
5b. (ii) log ; 8 8 [ mark] (ii) log 8 = N [ mark] 8 5c. (iii) log. 6 4 [ mark] (iii) log 6 4 = N [ mark] 5d. Hence, solve log 3 7 + log 8 log 4 =. 8 6 log 4 correct equation with their three values correct working involving powers = 8 3 = log 4, 3 + ( ) = log 4 = 4 3, 4 3 log = 4 4 N () () Consider the epansion of ( + 3) 0. 6a. Write down the number of terms in this epansion. [ mark] terms N [ mark] 6b. Find the term containing 3. evidence of binomial epansion ( n ) a n r b r, attempt to epand r evidence of choosing correct term () 8 th 0 term, r = 7, ( ), () 3 (3) 7 7 correct working () 0 ( ) () 3 (3) 7 0, ( )() 3 (3) 7, 7 3 6440 3 (accept 6000 3 ) N3 8
7. Consider the epansion of (3 + ). The constant term is 6 8. Find k. valid approach ( 8 ), r (3 ) 8 r r ( ) (3 ) 8 + ( 8 ) ( ) + ( ) +, Pascal s triangle to line (3 ) 7 k 8 (3 ) 6 ( k ) 9 th attempt to find value of r which gives term in 0 eponent in binomial must give, ( ) 8 r ( ) = correct working () (8 r) r =, 8 3r = 0, r + ( 8 + r) = evidence of correct term () 8 8 ( ), ( ), r = 6, r = 6 (3 ) ( k ) equating their term and 68 to solve for k 8 ( ) = 68, = 6 (3 ) ( k 6 ) k = ± N k k 8 k 6 68 8(9) M k r 0 [7 marks] Note: If no working shown, award N0 for k =. Total [7 marks] Let f() = p 3 + p + q. Find f (). 8a. f () = 3p + p + q A N Note: Award if only error. Given that f () 0, show that. 8b. p 3pq [5 marks] evidence of discriminant (must be seen eplicitly, not in quadratic formula) correct substitution into discriminant (may be seen in inequality) f () 0 then f has two equal roots or no roots (R) recognizing discriminant less or equal than zero R correct working that clearly leads to the required answer p b [5 marks] 4ac (p) 4 3p q, 4p pq Δ 0, 4p pq 0 p 3pq 0, 4p pq 3pq AG N0
π 4 Let f() = cos( ) + sin( ), for 4 4. π 4 9a. Sketch the graph of f. N3 Note: Award for approimately correct sinusoidal shape. Only if this is awarded, award the following: for correct domain, for approimately correct range. 9b. Find the values of where the function is decreasing. [5 marks] recognizes decreasing to the left of minimum or right of maimum, (R) -values of minimum and maimum (may be seen on sketch in part (a)) two correct intervals N5 [5 marks] f () < 0 = 3, (,.4) 4 < < 3, 4; < 3, ()() 9c. The function f can also be written in the form f() = asin( ( + c)), where a R, and 0 c. Find the value of a; π 4 recognizes that a is found from amplitude of wave (R) y-value of minimum or maimum ( 3,.4), (,.4) a =.44 a =, (eact),.4, N3 () 9d. The function f can also be written in the form f() = asin( ( + c)), where a R, and 0 c. Find the value of c. π 4
4 METHOD recognize that shift for sine is found at -intercept attempt to find -intercept π 4 cos( ) + sin( ) = 0, = 3 + 4k, k Z = () c = N4 π 4 (R) METHOD attempt to use a coordinate to make an equation π 4 sin( c) =, sin( (3 c)) = 0 attempt to solve resulting equation sketch, = 3 + 4k, k Z = () c = N4 π 4 (R) Let f() = 3, where q. q 0a. Write down the equations of the vertical and horizontal asymptotes of the graph of f. = q, y = 3 (must be equations) N 0b. The vertical and horizontal asymptotes to the graph of f intersect at the point Q(, 3). Find the value of q. recognizing connection between point of intersection and asymptote = q = N (R) 0c. The vertical and horizontal asymptotes to the graph of f intersect at the point Q(, 3). The point P(, y) lies on the graph of f. Show that PQ = ( ) 3 + ( ).
correct substitution into distance formula ( ) + (y 3) attempt to substitute y = 3 ( ) 3 + ( 3) correct simplification of ( 3) () correct epression clearly leading to the required answer 3 3+3, ( ) 3 3+3 + ( ) PQ = ( ) 3 + ( ) AG N0 3 3( ) 3 0d. The vertical and horizontal asymptotes to the graph of f intersect at the point Q(, 3). Hence find the coordinates of the points on the graph of f that are closest to (, 3). [6 marks] recognizing that closest is when PQ is a minimum (R) sketch of PQ, (PQ) () = 0 = 0.7305 =.7305 (seen anywhere) attempt to find y-coordinates f( 0.7305) ( 0.7305,.67949),(.7305, 4.7305) ( 0.73,.7),(.73, 4.73) N4 [6 marks] Let f() = 5. Part of the graph of fis shown in the following diagram. The graph crosses the -ais at the points A and B. a. Find the -coordinate of A and of B.
recognizing f() = 0 f = 0, = 5 = ±.3606 = ± 5 (eact), = ±.4 N3 b. The rion enclosed by the graph of f and the -ais is revolved 360 about the -ais. Find the volume of the solid formed. attempt to substitute either limits or the function into formula involving f π (5 ) d, π.4 ( 0 + 5), π.4 4 87.38 volume = 87 A N3 0 f Let f() = 3 and g() = 5, for 0. 3 a. Find f (). interchanging and y = 3y f + 3 () = (accept y =, ) + 3 + 3 N b. Show that (g )() =. f 5 + attempt to form composite (in any order) correct substitution + g ( ), 3 5 + 3( ) 3 5 + 3 3 (g f 5 )() = + AG N0 Let h() = 5, for 0. The graph of h has a horizontal asymptote at y = 0. + c. Find the y-intercept of the graph of h.
valid approach 5 h(0), 5 0+ y = (accept (0,.5)) N d. Hence, sketch the graph of h. A N3 Notes: Award for approimately correct shape (reciprocal, decreasing, concave up). Only if this is awarded, award A for all the following approimately correct features: y-intercept at (0,.5), asymptotic to - ais, correct domain 0. If only two of these features are correct, award. e. For the graph of h, write down the -intercept; [ mark] 5 = (accept (.5, 0)) [ mark] N f. For the graph of h, write down the equation of the vertical asymptote. [ mark] = 0 (must be an equation) N [ mark] g. Given that h (a) = 3, find the value of a.
METHOD attempt to substitute 3 into h (seen anywhere) correct equation a = METHOD N () attempt to find inverse (may be seen in (d)) h(3), a = correct equation, a = 5 3+ 5 3+ 5 y+, h(3) = a =, =, + h 5 5 N = 3 5 () Let f() = 5, for 5. Find f. 3a. () METHOD attempt to set up equation = y 5, = 5 correct working () 4 = y 5, = + 5 f () = 9 N METHOD interchanging and y (seen anywhere) = y 5 correct working () = y 5, y = + 5 f () = 9 N Let g be a function such that g eists for all real numbers. Given that, find. 3b. g(30) = 3 (f g )(3) recognizing g (3) = 30 f(30) correct working () (f g )(3) = 30 5, 5 (f g )(3) = 5 N Note: Award A0 for multiple values, ±5.
Consider f() = ln( 4 + ). Find the value of f(0). 4a. substitute 0 into f ln(0 + ), ln f(0) = 0 N Find the set of values of for which f is increasing. 4b. [5 marks] () = 4 f + 4 3 (seen anywhere) Note: Award for and for 4 3. 4 + f recognizing f increasing where f () > 0 (seen anywhere) R () > 0, diagram of signs attempt to solve f () > 0 4 3 = 0, 3 > 0 f increasing for > 0 (accept 0 ) N [5 marks] f 4 (3 4 ) ( 4 +) The second derivative is given by () =. The equation f () = 0 has only three solutions, when = 0, ± 4 3 (±.36 ). (i) Find f. 4c. () (ii) Hence, show that there is no point of infleion on the graph of f at = 0. [5 marks]
(i) substituting = into f () 4(3 ) 4 (+) 4, f () = N (ii) valid interpretation of point of infleion (seen anywhere) R no change of sign in f f increasing both sides of zero attempt to find f () for < 0 f 4 ( ) (3 ( ) 4 ) (( ) 4 +) (), no change in concavity, ( ),, diagram of signs correct working leading to positive value f ( ) =, discussing signs of numerator and denominator there is no point of infleion at = 0 AG N0 [5 marks] There is a point of infleion on the graph of f at = 4 3 ( =.36 ). 4d. Sketch the graph of f, for 0. N3 Notes: Award for shape concave up left of POI and concave down right of POI. Only if this is awarded, then award the following: for curve through ( 0, 0), for increasing throughout. Sketch need not be drawn to scale. Only essential features need to be clear.
The velocity of a particle in ms is given by v = e sin t, for 0 t 5. On the grid below, sketch the graph of v. 5a. N3 Note: Award for approimately correct shape crossing -ais with 3 < < 3.5. Only if this is awarded, award the following: for maimum in circle, for endpoints in circle. Find the total distance travelled by the particle in the first five seconds. 5b. [ mark] t = π (eact), 3.4 N [ mark] 5c. Write down the positive t -intercept.
recognizing distance is area under velocity curve s = v, shading on diagram, attempt to intrate valid approach to find the total area area A + area B, vdt vdt, vdt+ vdt, v 3.4 correct working with intration and limits (accept d or missing dt ) () vdt+ vdt, 3.067 + 0.878, 5 0 e sin t 3.4 3.4 0 distance = 3.95 (m) N3 3.4 5 0 5 Let f and g be functions such that g() = f( + ) + 5. (a) The graph of f is mapped to the graph of g under the following transformations: 6a. [6 marks] vertical stretch by a factor of k, followed by a translation ( p ). q Write down the value of (i) k ; (ii) p ; (iii) q. (b) Let h() = g(3). The point A( 6, 5) on the graph of g is mapped to the point A on the graph of h. Find A. (a) (i) k = N (ii) p = N (iii) q = 5 N (b) recognizing one transformation horizontal stretch by A is (, 5) N3 3, reflection in -ais Total [6 marks] The graph of f is mapped to the graph of g under the following transformations: 6b. vertical stretch by a factor of k, followed by a translation ( p ). q Write down the value of (i) k ; (ii) p ; (iii) q.
(i) k = N (ii) p = N (iii) q = 5 N 6c. Let h() = g(3). The point A( 6, 5) on the graph of g is mapped to the point A on the graph of h. Find A. recognizing one transformation horizontal stretch by A is (, 5) N3 3, reflection in -ais Total [6 marks] 00 (+50 e 0. ) Let f() =. Part of the graph of f is shown below. Write down f(0). 7a. [ mark] f(0) = 00 5 [ mark] (eact),.96 N 7b. Solve f() = 95. setting up equation 95 = 00 +50e 0., sketch of graph with horizontal line at y = 95 = 34.3 N
Find the range of f. 7c. upper bound of y is 00 () lower bound of y is 0 () range is 0 < y < 00 N3 000e 0. 7d. Show that f () =. (+50 e 0. ) [5 marks] METHOD setting function ready to apply the chain rule evidence of correct differentiation (must be substituted into chain rule) u = 00( + 50e 0. ), v = (50 e 0. )( 0.) correct chain rule derivative correct working clearly leading to the required answer 00( + 50e 0. ) f () = 00( + 50 e 0. ) (50 e 0. )( 0.) f () = 000 e 0. ( + 50e 0. ) f () = METHOD 000e 0. (+50 e 0. ) AG N0 attempt to apply the quotient rule (accept reversed numerator terms) ()() vu uv v, uv vu v evidence of correct differentiation inside the quotient rule ()() f (+50 e 0. )(0) 00(50 e 0. 0.) 00( 0) e 0. 0 (+50 e 0. ) (+50 e 0. ) () =, any correct epression for derivative ( 0 may not be eplicitly seen) () 00(50 e 0. 0.) (+50 e 0. ) correct working clearly leading to the required answer 0 00( 0)e 0. f () =, (+50 e 0. ) 00( 0)e 0. (+50 e 0. ) f () = [5 marks] 000e 0. (+50 e 0. ) AG N0 7e. Find the maimum rate of change of f.
METHOD sketch of f () () recognizing maimum on dot on ma of sketch finding maimum on graph of () ( 9.6, 5), = 9.560 maimum rate of increase is 5 N METHOD recognizing f () = 0 finding any correct epression for f () = 0 f f () () (+50 e 0. ) ( 00 e 0. ) (000 e 0. )((+50 e 0. )( 0 e 0. )) finding = 9.560 maimum rate of increase is 5 N (+50 e 0. ) 4 Let f() = sin +, for. 0 π 8a. Find f (). f () = cos + Note: Award for each term. N3 Let g be a quadratic function such that g(0) = 5. The line = is the ais of symmetry of the graph of g. Find g(4). 8b.
recognizing g(0) = 5 gives the point ( 0, 5) (R) recognize symmetry verte, sketch g(4) = 5 N3 The function g can be epressed in the form g() = a( h ) + 3. (i) 8c. Write down the value of h. (ii) Find the value of a. (i) h = N (ii) substituting into g() = a( ) + 3 (not the verte) 5 = a(0 ) + 3, 5 = a(4 ) + 3 working towards solution () 5 = 4a + 3, 4a = a = N 8d. Find the value of for which the tangent to the graph of f is parallel to the tangent to the graph of g. [6 marks]
g() = ( ) + 3 = + 5 correct derivative of g ( ), evidence of equating both derivatives correct equation () working towards a solution = π f = g cos + = () cos = 0, combining like terms N0 Note: Do not award final if additional values are given. [6 marks] (ln ) Let f() =, for > 0. 9a. Show that f ln () =. METHOD correct use of chain rule ln, ln ln Note: Award for, for. f () = METHOD AG N0 correct substitution into quotient rule, with derivatives seen correct working f () = ln ln 0 (ln ) 4 4 ln 4 ln AG N0 9b. There is a minimum on the graph of f. Find the -coordinate of this minimum. setting derivative = 0 f ln () = 0, = 0 correct working () ln = 0, = e 0 = N
Let g() =. The following diagram shows parts of the graphs of f and g. The graph of f has an -intercept at = p. 9c. Write down the value of p. intercept when f p = N () = 0 9d. The graph of g intersects the graph of f when = q. Find the value of q. equating functions correct working f ln = g, = ln = () q = e (accept = e) N 9e. The graph of g intersects the graph of f when = q. Let R be the rion enclosed by the graph of f, the graph of g and the line = p. Show that the area of is. R [5 marks]
evidence of intrating and subtracting functions (in any order, seen anywhere) correct intration ln A substituting limits into their intrated function and subtracting (in any order) e q ln f ( ) d, g (ln e) (ln ) (lne ln) ( ) (ln ) Note: Do not award M if the intrated function has only one term. correct working area = ( 0) ( 0), AG N0 Notes: Candidates may work with two separate intrals, and only combine them at the end. Award marks in line with the markscheme. [5 marks] Let f() = ( )( 4). 0a. Find the -intercepts of the graph of f. valid approach f() = 0, sketch of parabola showing two -intercepts =, = 4 (accept (, 0), (4, 0)) N3 0b. The rion enclosed by the graph of f and the -ais is rotated 360 about the -ais. Find the volume of the solid formed. attempt to substitute either limits or the function into formula involving d, π (f()) (( )( 4)) volume = 8.π (eact), 5.4 A N3 4 f
A particle moves along a straight line such that its velocity, v ms, is given by v(t) = 0te.7t, for t 0. a. On the grid below, sketch the graph of v, for 0 t 4. A N3 Notes: Award for approimately correct domain 0 t 4. The shape must be approimately correct, with maimum skewed left. Only if the shape is approimately correct, award A for all the following approimately correct features, in circle of tolerance where drawn (accept seeing correct coordinates for the maimum, even if point outside circle): Maimum point, passes through origin, asymptotic to t-ais (but must not touch the ais). If only two of these features are correct, award. b. Find the distance travelled by the particle in the first three seconds. valid approach (including 0 and 3) 3 0te dt, 3 f(), area from 0 to 3 (may be shaded in diagram) 0 0 distance = 3.33 (m) N c. Find the velocity of the particle when its acceleration is zero.
recognizing acceleration is derivative of velocity (R) a = dv dv, attempt to find, reference to maimum on the graph of v dt dt valid approach to find v when a = 0 (may be seen on graph) dv = 0, 0e.7t 7t e.7t = 0, t = 0.588 dt velocity =.6 (m s ) N3 Note: Award RMA0 for (0.588, 6) if velocity is not identified as final answer Let f() = 3 4. The following diagram shows part of the curve of f. The curve crosses the -ais at the point P. a. Write down the -coordinate of P. [ mark] = (accept (, 0) ) N [ mark] Write down the gradient of the curve at P. b. evidence of finding gradient of f at = e.g. f () the gradient is 0 N c. Find the equation of the normal to the curve at P, giving your equation in the form y = a + b.
evidence of native reciprocal of gradient e.g., f () 0 evidence of correct substitution into equation of a line e.g. y 0 = ( ), 0 = 0.() + b y = + 0 0 0 (accept a = 0., b = 0. ) N () International Baccalaureate Organization 06 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Colio Aleman de Barranquilla