Nuevo examen - 20 de Octubre de 2014 [865 marks]

1a. [4 marks]Find .

Markschemecorrect integration A1A1

e.g. , ,

Notes: In the first 2 examples, award A1 for each correct term.

In the third example, award A1 for and A1 for .

substituting limits into their integrated function and subtracting (in any order) (M1)

e.g.

A1 N2

Examiners reportMany candidates answered both parts of this question correctly. In part (b), a large number of successful candidates did not seem tonotice the link between parts (a) and (b), and duplicated the work they had already done in part (a). Also in part (b), a good number ofcandidates squared in their integral, rather than squaring , which of course prevented them from noting thelink between the two parts and obtaining the correct answer.

(x− 4)dx∫ 104

− 4xx2

2[ − 4x]x2

2

10

4

(x−4)2

2

12

(x− 4)2

( − 4(10)) − ( − 4(4)) ,10 − (−8), ( − 0)102

242

212

62

(x− 4)dx = 18∫ 104

(x− 4) x− 4− −−−√

1b. [3 marks]Part of the graph of , for , is shown below. The shaded region R is enclosed by the graph of , the line , and the x-axis.

The region R is rotated about the x-axis. Find the volume of the solid formed.

f(x) = − 4x− −−−√ x ≥ 4 f

x = 10

360∘

Markschemeattempt to substitute either limits or the function into volume formula (M1)

e.g.

Note: Do not penalise for missing or dx.

correct substitution (accept absence of dx and ) (A1)

e.g.

volume = A1 N2

[3 marks]

Examiners reportMany candidates answered both parts of this question correctly. In part (b), a large number of successful candidates did not seem tonotice the link between parts (a) and (b), and duplicated the work they had already done in part (a). Also in part (b), a good number ofcandidates squared in their integral, rather than squaring , which of course prevented them from noting thelink between the two parts and obtaining the correct answer.

π dx, , π∫ 104 f2 ∫ b

a( )x− 4− −−−√ 2 ∫ 10

4 x− 4− −−−√

π

π

π , π (x− 4)dx, (x− 4)dx∫ 104 ( )x− 4− −−−√ 2 ∫ 10

4 ∫ 104

18π

(x− 4) x− 4− −−−√

2a. [2 marks]

Part of the graph of is shown below.

The point P lies on the graph of . At P, x = 1.

Find .

Markscheme A1A1 N2

Note: Award A1 for each correct term.

[2 marks]

Examiners reportA majority of candidates answered part (a) correctly, and a good number earned full marks on both parts of this question. In part (b),some common errors included setting the derivative equal to zero, or substituting 3 for x in their derivative. There were also a fewcandidates who incorrectly tried to work with , rather than , in part (b).

f(x) = a − 6x3 x2

f

(x)f ′

(x) = 3a − 12xf ′ x2

f(x) (x)f ′

2b. [4 marks]The graph of has a gradient of at the point P. Find the value of .f 3 a

Markschemesetting their derivative equal to 3 (seen anywhere) A1

e.g.

attempt to substitute into (M1)

e.g.

correct substitution into (A1)

e.g. ,

A1 N2

[4 marks]

Examiners reportA majority of candidates answered part (a) correctly, and a good number earned full marks on both parts of this question. In part (b),some common errors included setting the derivative equal to zero, or substituting for in their derivative. There were also a fewcandidates who incorrectly tried to work with , rather than , in part (b).

(x) = 3f ′

x = 1 (x)f ′

3a(1 − 12(1))2

(x)f ′

3a− 12 3a = 15

a = 5

3 x

f(x) (x)f ′

3a. [5 marks]

Let , for .

Find .

MarkschemeMETHOD 1

evidence of choosing quotient rule (M1)

e.g.

evidence of correct differentiation (must be seen in quotient rule) (A1)(A1)

e.g. ,

correct substitution into quotient rule A1

e.g. ,

A1 N4

[5 marks]

METHOD 2

evidence of choosing product rule (M1)

e.g. ,

evidence of correct differentiation (must be seen in product rule) (A1)(A1)

e.g. ,

correct working A1

e.g. ,

A1 N4

[5 marks]

f(x) = 6xx+1

x > 0

(x)f ′

v−uu′ v′

v2

(6x) = 6ddx

(x+ 1) = 1ddx

(x+1)6−6x

(x+1)2

6x+6−6x

(x+1)2

(x) =f ′ 6

(x+1)2

6x(x+ 1)−1 u + vv′ u′

(6x) = 6ddx

(x+ 1 = −1(x+ 1 × 1ddx

)−1 )−2

6x× −(x+ 1 + (x+ 1 × 6)−2 )−1 −6x+6(x+1)

(x+1)2

(x) =f ′ 6

(x+1)2

Examiners reportIn part (a), most candidates recognized the need to apply the quotient rule to find the derivative, and many were successful in earningfull marks here.

3b. [4 marks]Let , for .

Show that .

MarkschemeMETHOD 1

evidence of choosing chain rule (M1)

e.g. formula,

correct reciprocal of is (seen anywhere) A1

correct substitution into chain rule A1

e.g. ,

working that clearly leads to the answer A1

e.g. , ,

AG N0

[4 marks]

METHOD 2

attempt to subtract logs (M1)

e.g. ,

correct derivatives (must be seen in correct expression) A1A1

e.g. ,

working that clearly leads to the answer A1

e.g. , ,

AG N0

[4 marks]

Examiners reportIn part (b), many candidates struggled with the chain rule, or did not realize the chain rule was necessary to find the derivative. Again,some candidates attempted to work backward from the given answer, which is not allowed in a "show that" question. A few clevercandidates simplified the situation by applying properties of logarithms before finding their derivative.

g(x) = ln( )6xx+1

x > 0

(x) =g′ 1x(x+1)

× ( )1

( )6xx+1

6xx+1

1

( )6xx+1

x+16x

×1

( )6xx+1

6

(x+1)2( ) ( )6

(x+1)2

x+16x

( ) ( )6(x+1)

16x

( ) ( )1(x+1)2

x+1x

6(x+1)

6x(x+1)2

(x) =g′ 1x(x+1)

lna− lnb ln6x− ln(x+ 1)

−66x

1x+1

−1x

1x+1

x+1−x

x(x+1)6x+6−6x6x(x+1)

6(x+1−x)

6x(x+1)

(x) =g′ 1x(x+1)

3c. [7 marks]Let . The area enclosed by the graph of h , the x-axis and the lines and is . Given that ,

find the value of k .

h(x) = 1x(x+1)

x = 15

x = k ln4 k > 15

Markschemevalid method using integral of h(x) (accept missing/incorrect limits or missing ) (M1)

e.g. ,

recognizing that integral of derivative will give original function (R1)

e.g.

correct substitution and subtraction A1

e.g. ,

setting their expression equal to (M1)

e.g. , ,

correct equation without logs A1

e.g. ,

correct working (A1)

e.g. ,

A1 N4

[7 marks]

Examiners reportFor part (c), many candidates recognized the need to integrate the function, and that their integral would equal . However, manydid not recognize that the integral of h is g . Those candidates who made this link between the parts (b) and (c) often carried oncorrectly to find the value of k , with a few candidates having errors in working with logarithms.

dx

area = h(x)dx∫ k15

∫ ( )1x(x+1)

∫ ( )dx = ln( )1x(x+1)

6xx+1

ln( ) − ln( )6kk+1

6× 15

+115

ln( ) − ln(1)6kk+1

ln4

ln( ) − ln(1) = ln46kk+1

ln( ) = ln46kk+1

h(x)dx = ln4∫ k15

= 46kk+1

6k = 4(k+ 1)

6k = 4k+ 4 2k = 4

k = 2

ln4

s(t) = 2t cost

4a. [4 marks]

A particle’s displacement, in metres, is given by , for , where t is the time in seconds.

On the grid below, sketch the graph of .

s(t) = 2t cost 0 ≤ t ≤ 6

s

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct shape (do not accept line segments).

Only if this A1 is awarded, award the following:

A1 for maximum and minimum within circles,

A1 for x-intercepts between 1 and 2 and between 4 and 5,

A1 for left endpoint at and right endpoint within circle.

[4 marks]

Examiners reportMost candidates sketched an approximately correct shape for the displacement of a particle in the given domain, but many lost marksfor carelessness in graphing the local extrema or the right endpoint.

(0, 0)

4b. [3 marks]Find the maximum velocity of the particle.

Markschemeappropriate approach (M1)

e.g. recognizing that , finding derivative,

valid method to find maximum (M1)

e.g. sketch of , ,

A1 N2

[3 marks]

Examiners reportIn part (b), most candidates knew to differentiate displacement to find velocity, but few knew how to then find the maximum.Occasionally, a candidate would give the time value of the maximum. Others attempted to incorrectly set the first derivative equal tozero and solve analytically rather than take the maximum value from the graph of the velocity function.

v = s′ a = s′′

v (t) = 0v′ t = 5.08698…

v = 10.20025…

v = 10.2 [10.2, 10.3]

5a. [4 marks]

Consider the function .

Sketch the graph of f , for .

Markscheme

A1A1A1A1 N4

Note: The shape must be an approximately correct upwards parabola.

Only if the shape is approximately correct, award the following:

A1 for vertex , A1 for x-intercepts between 0 and 1, and 3 and 4, A1 for correct y-intercept , A1 for correct domain .

Scale not required on the axes, but approximate positions need to be clear.

[4 marks]

Examiners reportA good number of students provided a clear sketch of the quadratic function within the given domain. Some lost marks as they didnot clearly indicate the approximate positions of the most important points of the parabola either by labelling or providing a suitablescale.

f(x) = − 4x+ 1x2

−1 ≤ x ≤ 5

x ≈ 2 (0, 1)[−1, 5]

5b. [1 mark]This function can also be written as .

Write down the value of p .

Markscheme A1 N1

[1 mark]

Examiners reportThere were few difficulties in part (b).

f(x) = (x− p − 3)2

p = 2

( )

5c. [4 marks]The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of .

Show that .

Markschemecorrect vertical reflection, correct vertical translation (A1)(A1)

e.g. , , , ,

transformations in correct order (A1)

e.g. ,

simplification which clearly leads to given answer A1

e.g. ,

AG N0

Note: If working shown, award A1A1A0A0 if transformations correct, but done in reverse order, e.g. .

[4 marks]

Examiners reportIn part (c), candidates often used an insufficient number of steps to show the required result or had difficulty setting out their worklogically.

( )06

g(x) = − + 4x+ 5x2

−f(x) −((x− 2 − 3))2 −y −f(x) + 6 y+ 6

−( − 4x+ 1) + 6x2 −((x− 2 − 3) + 6)2

− + 4x− 1 + 6x2 −( − 4x+ 4 − 3) + 6x2

g(x) = − + 4x+ 5x2

−( − 4x+ 1 + 6)x2

5d. [3 marks]The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of .

The graphs of f and g intersect at two points.

Write down the x-coordinates of these two points.

Markschemevalid approach (M1)

e.g. sketch,

,

(exact), ; A1A1 N3

[3 marks]

Examiners reportPart (d) was generally done well though many candidates gave at least one answer to fewer than three significant figures, potentiallyresulting in more lost marks.

( )06

f = g

−0.449489… 4.449489…

(2 ± )6√ −0.449 [−0.450, − 0.449] 4.45 [4.44, 4.45]

5e. [3 marks]The graph of is obtained by reflecting the graph of in the x-axis, followed by a translation of .

Let R be the region enclosed by the graphs of f and g .

Find the area of R .

g f ( )06

Markschemeattempt to substitute limits or functions into area formula (accept absence of ) (M1)

e.g. , ,

approach involving subtraction of integrals/areas (accept absence of ) (M1)

e.g. ,

A1 N3

[3 marks]

Examiners reportIn part (e), many candidates were unable to connect the points of intersection found in part (d) with the limits of integration. Mistakeswere also made here either using a GDC incorrectly or not subtracting the correct functions. Other candidates tried to divide theregion into four areas and made obvious errors in the process. Very few candidates subtracted from to get a simple functionbefore integrating and there were numerous, fruitless analytical attempts to find the required integral.

dx

((− + 4x+ 5) − ( − 4x+ 1))dx∫ b

ax2 x2 (f− g)∫ −0.449

4.45 ∫ (−2 + 8x+ 4)dxx2

dx

(− + 4x+ 5) − ( − 4x+ 1)∫ b

ax2 ∫ b

ax2 ∫ (f− g)dx

area = 39.19183…

area = 39.2 [39.1, 39.2]

f(x) g(x)

6a. [4 marks]

Consider .

Find .

Markschemeevidence of choosing product rule (M1)

eg

correct derivatives (must be seen in the product rule) , (A1)(A1)

A1 N4

[4 marks]

Examiners reportMany candidates correctly applied the product rule for the derivative, although a common error was to answer .

f(x) = sinxx2

(x)f ′

u + vv′ u′

cosx 2x

(x) = cosx+ 2x sinxf ′ x2

(x) = 2xcosxf ′

6b. [3 marks]Find the gradient of the curve of at .

Markschemesubstituting into their (M1)

eg ,

correct values for both and seen in (A1)

eg

A1 N2

[3 marks]

f x = π

2

π

2(x)f ′

( )f ′ π

2cos( ) + 2 ( ) sin ( )( )π

2

2π

2π

2π

2

sin π

2cos π

2(x)f ′

0 + 2 ( ) × 1π

2

( ) = πf ′ π

2

Examiners reportCandidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was madein the original function. Some candidates did not know the values of sine and cosine at .π

2

7. [6 marks]Let , . The graph of passes through ( , ) .

Find .

Markschemeattempt to integrate which involves (M1)

eg , ,

correct expression (accept absence of )

eg , A2

attempt to substitute (4,0) into their integrated f (M1)

eg ,

(A1)

(accept ) A1 N5

Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve .

[6 marks]

Examiners reportWhile some candidates correctly integrated the function, many missed the division by and answered . Other commonincorrect responses included and . Finding the constant of integration also proved elusive for many. Some eitherdid not remember the or did not try to find its value, while others misunderstood the boundary condition and attempted tocalculate the definite integral from to .

f(x) = ∫ dx122x−5

x > 52

f 4 0

f(x)

ln

ln(2x− 5) 12 ln2x− 5 ln2x

C

12 ln(2x− 5) +C12

6 ln(2x− 5)

0 = 6 ln(2 × 4 − 5) 0 = 6 ln(8 − 5) +C

C = −6 ln3

f(x) = 6 ln(2x− 5) − 6 ln3 (= 6 ln( ))2x−53

6 ln(2x− 5) − ln36

ln

2 12 ln(2x− 5)12x−5xx2

−122(x− 5)−2

+C0 4

8a. [5 marks]

Consider .

Find the set of values of for which is increasing.

Markscheme (seen anywhere) A1A1

Note: Award A1 for and A1 for .

recognizing increasing where (seen anywhere) R1

eg , diagram of signs

attempt to solve (M1)

eg ,

increasing for (accept ) A1 N1

[5 marks]

f(x) = ln( + 1)x4

x f

(x) = × 4f ′ 1+1x4

x3

1+1x4

4x3

f (x) > 0f ′

(x) > 0f ′

(x) > 0f ′

4 = 0x3 > 0x3

f x > 0 x ≥ 0

Examiners reportCandidates who attempted to consider where is increasing generally understood the derivative is needed. However,a number of candidates did not apply the chain rule, which commonly led to answers such as “increasing for all ”.Many set their derivative equal to zero, while neglecting to indicate in their working that for an increasingfunction. Some created a diagram of signs, which provides appropriate evidence as long as it is clear that the signsrepresent .

fx

(x) > 0f ′

f ′

8b. [5 marks]

The second derivative is given by .

The equation has only three solutions, when , .

(i) Find .

(ii) Hence, show that there is no point of inflexion on the graph of at .

Markscheme(i) substituting into (A1)

eg ,

A1 N2

(ii) valid interpretation of point of inflexion (seen anywhere) R1

eg no change of sign in , no change in concavity,

increasing both sides of zero

attempt to find for (M1)

eg , , diagram of signs

correct working leading to positive value A1

eg , discussing signs of numerator and denominator

there is no point of inflexion at AG N0

[5 marks]

Examiners reportFinding proved no challenge, however, using this value to show that no point of inflexion exists proved elusive for many.Some candidates recognized the signs must not change in the second derivative. Few candidates presented evidence in the form of acalculation, which follows from the “hence” command of the question. In this case, a sign diagram without numerical evidence wasnot sufficient.

(x) =f ′′ 4 (3− )x2 x4

( +1)x4 2

(x) = 0f ′′ x = 0 ± 3√4 (±1.316…)

(1)f ′′

f x = 0

x = 1 f ′′

4(3−1)

(1+1)24×2

4

(1) = 2f ′′

(x)f ′′

f ′

(x)f ′′ x < 0

(−1)f ′′ 4 (3− )(−1)2 (−1)4

( +1)(−1)4 2

(−1) = 2f ′′

x = 0

(1)f ′′

8c. [3 marks]There is a point of inflexion on the graph of at .

Sketch the graph of , for .

f x = 3√4 (x = 1.316…)

f x ≥ 0

Markscheme

A1A1A1 N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ( , ) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

Examiners reportFew candidates created a correct graph from the information given or found in the question. This included the point ( , ), the factthat the function is always increasing for , the concavity at and the change in concavity at the given point of inflexion.Many incorrect attempts showed a graph concave down to the right of , changing to concave up.

0 0

0 0x > 0 x = 1

x = 0

9a. [5 marks]

Let . Part of the graph of is shown below.

Show that .

f(x) = 100(1+50 )e−0.2x

f

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

MarkschemeMETHOD 1

setting function ready to apply the chain rule (M1)

eg

evidence of correct differentiation (must be substituted into chain rule) (A1)(A1)

eg ,

correct chain rule derivative A1

eg

correct working clearly leading to the required answer A1

eg

AG N0

METHOD 2

attempt to apply the quotient rule (accept reversed numerator terms) (M1)

eg ,

evidence of correct differentiation inside the quotient rule (A1)(A1)

eg ,

any correct expression for derivative ( may not be explicitly seen) (A1)

eg

correct working clearly leading to the required answer A1

eg ,

AG N0

[5 marks]

Examiners reportIn part (d), the majority of candidates opted to use the quotient rule and did so with some degree of competency, but failed torecognize the command term “show that” and consequently did not show enough to gain full marks. Approaches involving the chainrule were also successful but with the same point regarding sufficiency of work.

100(1 + 50e−0.2x)−1

= −100(1 + 50u′ e−0.2x)−2 = (50 )(−0.2)v′ e−0.2x

(x) = −100(1 + 50 (50 )(−0.2)f ′ e−0.2x)−2 e−0.2x

(x) = 1000 (1 + 50f ′ e−0.2x e−0.2x)−2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

v −uu′ v′

v2

u −vv′ u′

v2

(x) =f ′ (1+50 )(0)−100(50 ×−0.2)e−0.2x e−0.2x

(1+50 )e−0.2x 2

100(−10) −0e−0.2x

(1+50 )e−0.2x 2

0

−100(50 ×−0.2)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 0−100(−10)e−0.2x

(1+50 )e−0.2x 2

−100(−10)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

9b. [4 marks]Find the maximum rate of change of .f

MarkschemeMETHOD 1

sketch of (A1)

eg

recognizing maximum on (M1)

eg dot on max of sketch

finding maximum on graph of A1

eg ( , ) ,

maximum rate of increase is A1 N2

METHOD 2

recognizing (M1)

finding any correct expression for (A1)

eg

finding A1

maximum rate of increase is A1 N2

[4 marks]

Examiners reportPart (e) was poorly done as most were unable to interpret what was required. There were a few responses involving the use of the“trace” feature of the GDC which often led to inaccurate answers and a number of candidates incorrectly reported as theirfinal answer. Some found the maximum value of rather than .

(x)f ′

(x)f ′

(x)f ′

19.6 5 x = 19.560…

5

(x) = 0f ′′

(x) = 0f ′′

(−200 )−(1000 )(2(1+50 )(−10 ))(1+50 )e−0.2x 2 e−0.2x e−0.2x e−0.2x e−0.2x

(1+50 )e−0.2x 4

x = 19.560…

5

x = 19.6f f ′

10. [7 marks]A rocket moving in a straight line has velocity km s and displacement km at time seconds. The velocity is given by . When , .

Find an expression for the displacement of the rocket in terms of .

v –1 s t v

v(t) = 6 + te2t t = 0 s = 10

t

Markschemeevidence of anti-differentiation (M1)

eg

A2A1

Note: Award A2 for , A1 for .

attempt to substitute ( , ) into their integrated expression (even if is missing) (M1)

correct working (A1)

eg ,

A1 N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of .

[7 marks]

Examiners reportA good number of candidates earned full marks on this question, and many others were able to earn at least half of the availablemarks. Most candidates knew to integrate, but there were quite a few who tried to find the derivative instead. Many candidatesintegrated the term incorrectly, but most were able to earn some further method marks for substituting into their integratedfunction. The majority of candidates who substituted ( , ) into their integrated function knew that .

∫ (6 + t)e2t

s = 3 + +Ce2t t2

2

3e2t t2

2

0 10 C

10 = 3 +C C = 7

s = 3 + + 7e2t t2

2

e

6e2t

0 10 = 1e0

11. [7 marks]

The following is the graph of a function , for .

The first part of the graph is a quarter circle of radius with centre at the origin.

(a) Find .

(b) The shaded region is enclosed by the graph of , the -axis, the -axis and the line . The area of this region is .

Find .

f 0 ≤ x ≤ 6

2

f(x)dx∫ 20

f x y x = 6 3π

f(x)dx∫ 62

Markscheme(a) attempt to find quarter circle area (M1)

eg , ,

area of region (A1)

A2 N3

[4 marks]

(b) attempted set up with both regions (M1)

eg , ,

A2 N2

[3 marks]

Total [7 marks]

Examiners reportThere was a minor error on the diagram, where the point on the -axis was labelled (to indicate the length of the radius), rather than

. Examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did notindicate any adverse effect.

While most candidates were able to correctly find the area of the quarter circle in part (a), very few considered that the value of thedefinite integral is negative for the part of the function below the -axis. In part (b), most went on to earn full marks by subtracting thearea of the quarter circle from .

Candidates who did not understand the connection between area and the value of the integral often tried to find a function tointegrate. These candidates were not successful using this method.

(4π)14

πr2

4∫ 2

0 4 − dxx2− −−−−−−√

= π

f(x)dx = −π∫ 20

shaded area − quarter circle 3π−π 3π− f = f∫ 20 ∫ 6

2

f(x)dx = 2π∫ 62

y 2−2

x

3π

12a. [3 marks]

Let , for .

Find .

Markscheme A1A1A1 N3

Note: Award A1 for each term.

[3 marks]

Examiners reportIn part (a), most candidates were able to correctly find the derivative of the function.

f(x) = sinx+ − 2x12x2 0 ≤ x ≤ π

(x)f ′

(x) = cosx+x− 2f ′

12b. [6 marks]

Let be a quadratic function such that . The line is the axis of symmetry of the graph of .

The function can be expressed in the form .

Find the value of for which the tangent to the graph of is parallel to the tangent to the graph of .

g g(0) = 5 x = 2 g

g g(x) = a(x−h + 3)2

x f g

Markscheme

correct derivative of A1A1

eg ,

evidence of equating both derivatives (M1)

eg

correct equation (A1)

eg

working towards a solution (A1)

eg , combining like terms

A1 N0

Note: Do not award final A1 if additional values are given.

[6 marks]

Examiners reportPart (d) required the candidates to find the derivative of , and to equate that to their answer from part (a). Although many candidateswere able to simplify their equation to , many did not know how to solve for at this point. Candidates who had madeerrors in parts (a) and/or (c) were still able to earn follow-through marks in part (d).

g(x) = (x− 2 + 3 = − 2x+ 512

)2 12x2

g

2 × (x− 2)12

x− 2

=f ′ g′

cosx+x− 2 = x− 2

cosx = 0

x = π

2

g

cosx = 0 x

13a. [2 marks]

Consider the functions , and . The following table gives some values associated with these functions.

The following diagram shows parts of the graphs of and .

There is a point of inflexion on the graph of at P, when .

Explain why P is a point of inflexion.

f(x) g(x) h(x)

h h′′

h x = 3

Markscheme (A1)

valid reasoning R1

eg changes sign at , change in concavity of at

so P is a point of inflexion AG N0

[2 marks]

Examiners reportIn part (b), the majority of candidates earned one mark for stating that at point P. As this is not enough to determine a pointof inflexion, very few candidates earned full marks on this question.

(3) = 0h′′

h′′ x = 3 h x = 3

(x) = 0h′′

13b. [7 marks]

Given that ,

find the equation of the normal to the graph of at P.

Markschemerecognizing need to find derivative of (R1)

eg ,

attempt to use the product rule (do not accept ) (M1)

eg ,

correct substitution (A1)

eg

A1

attempt to find the gradient of the normal (M1)

eg ,

attempt to substitute their coordinates and their normal gradient into the equation of a line (M1)

eg , , ,

correct equation in any form A1 N4

eg ,

[7 marks]

Examiners reportPart (d) proved to be quite challenging for even the strongest candidates, as almost none of them used the product rule to find .The most common error was to say . Despite this error, many candidates were able to earn further methodmarks for their work in finding the equation of the normal. There were also a small number of candidates who were able to find theequation for , and from that . These candidates were often successful in earning full marks, although this method wasquite time-consuming.

h(x) = f(x) × g(x)

h

h

h′ (3)h′

= ×h′ f ′ g′

= f + gh′ g′ f ′ (3) = f(3) × (3) + g(3) × (3)h′ g′ f ′

(3) = 3(−3) + (−18) × 1h′

(3) = −27h′

− 1m

− x127

−54 = (3) + b127

0 = (3) + b127

y+ 54 = 27(x− 3) y− 54 = (x+ 3)127

y+ 54 = (x− 3)127

y = x− 54127

19

(3)h′

(3) = (3) × (3)h′ f ′ g′

(x)h′ (x)h′′

14a. [5 marks]

Let and , where , and . Let be the region enclosed by the -axis, the graph of , and the graphof .

Let .

Find the area of .

f(x) = ex

4 g(x) = mx m ≥ 0 −5 ≤ x ≤ 5 R y f

g

m = 1

R

Markschemeattempt to find intersection of the graphs of and (M1)

eg

A1

valid attempt to find area of (M1)

eg , ,

area A2 N3

[5 marks]

Examiners reportThere was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed whatwas intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notifythe IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

In part (a)(ii), most candidates found the intersection correctly. Those who used their GDC to evaluate the integral numerically wereusually successful, unlike those who attempted to solve with antiderivatives. A common error was to find the area of the regionenclosed by and (although it involved a point of intersection outside of the given domain), rather than the area of the regionenclosed by and and the -axis.

f g

= xex

4

x = 1.42961…

R

∫ (x− )dxex

4 (g − f)∫ 10 ∫ (f− g)

= 0.697

f g

f g y

14b. [8 marks]Consider all values of such that the graphs of and intersect. Find the value of that gives the greatest value for the areaof .

Markschemerecognize that area of is a maximum at point of tangency (R1)

eg

equating functions (M1)

eg ,

(A1)

equating gradients (A1)

eg ,

attempt to solve system of two equations for (M1)

eg

(A1)

attempt to find (M1)

eg ,

(exact), A1 N3

[8 marks]

Examiners reportThere was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed whatwas intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notifythe IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

While some candidates were able to show some good reasoning in part (b), fewer were able to find the value of which maximizedthe area of the region. In addition to the answer obtained from the restricted domain, full marks were awarded for the answer obtainedby using the point of tangency.

m f g m

R

R

m = (x)f ′

f(x) = g(x) = mxex

4

(x) =f ′ 14

ex

4

(x) = (x)f ′ g′ = m14

ex

4

x

×x =14

ex

4 ex

4

x = 4

m

(4)f ′ 14

e44

m = e14

0.680

m

15. [4 marks]

The velocity of a particle in ms is given by , for .

Write down the positive -intercept.

Markschemerecognizing distance is area under velocity curve (M1)

eg , shading on diagram, attempt to integrate

valid approach to find the total area (M1)

eg , , ,

correct working with integration and limits (accept or missing ) (A1)

eg , ,

distance (m) A1 N3

[4 marks]

Examiners reportIn (b)(ii), most appreciated that the definite integral would give the distance travelled but few could write a valid expression andnormally just integrated from to without considering the part of the graph below the -axis. Again, analytic approaches toevaluating their integral predominated over simpler GDC approaches and some candidates had their calculator set in degree moderather than radian mode.

−1 v = − 1esin t 0 ≤ t ≤ 5

t

s = ∫ v

area A + area B ∫ vdt− ∫ vdt vdt+ vdt∫ 3.140 ∫ 5

3.14 ∫ |v|

dx dt

vdt+ vdt∫ 3.140 ∫ 3.14

5 3.067… + 0.878… − 1∫ 50 ∣∣esin t ∣∣

= 3.95

t = 0 t = 5 t

16a. [4 marks]

The following diagram shows the graph of a quadratic function f , for .

The graph passes through the point P(0, 13) , and its vertex is the point V(2, 1) .

The function can be written in the form .

(i) Write down the value of h and of k .

(ii) Show that .

0 ≤ x ≤ 4

f(x) = a(x−h + k)2

a = 3

Markscheme(i) , A1A1 N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f M1

e.g.

working towards solution A1

e.g.

AG N0

[4 marks]

Examiners reportIn part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able tosuccessfully show that . Unfortunately, a few candidates did not understand the "show that" command, and simply verifiedthat would work, rather than showing how to find .

h = 2 k = 1

13 = a(0 − 2 + 1)2

13 = 4a+ 1

a = 3

a = 3a = 3 a = 3

16b. [3 marks]Find , giving your answer in the form .

Markschemeattempting to expand their binomial (M1)

e.g. ,

correct working (A1)

e.g.

(accept , , ) A1 N2

[3 marks]

Examiners reportIn part (b), most candidates were able to find in the required form. For a few candidates, algebraic errors kept them from findingthe correct function, even though they started with correct values for a, h and k.

f(x) A +Bx+Cx2

f(x) = 3( − 2 × 2x+ 4) + 1x2 (x− 2 = − 4x+ 4)2 x2

f(x) = 3 − 12x+ 12 + 1x2

f(x) = 3 − 12x+ 13x2 A = 3 B = −12 C = 13

f(x)

16c. [8 marks]Calculate the area enclosed by the graph of f , the x-axis, and the lines and .x = 2 x = 4

MarkschemeMETHOD 1

integral expression (A1)

e.g. ,

A1A1A1

Note: Award A1 for , A1 for , A1 for .

correct substitution of correct limits into their expression A1A1

e.g. ,

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working (A1)

e.g.

A1 N3

[8 marks]

METHOD 2

integral expression (A1)

e.g. ,

A2A1

Note: Award A2 for , A1 for .

correct substitution of correct limits into their expression A1A1

e.g. , ,

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working (A1)

e.g.

A1 N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4 (R1)

e.g. sketch,

integral expression (A1)

e.g. ,

A1A1A1

Note: Award A1 for , A1 for , A1 for .

correct substitution of correct limits into their expression A1(A1)

e.g. ,

Note: Award A1 for substituting 2, (A1) for substituting 0.

A1 N3

[8 marks]

(3 − 12x+ 13)∫ 42 x2 ∫ fdx

Area = [ − 6 + 13xx3 x2 ]42

x3 −6x2 13x

( − 6 × + 13 × 4) − ( − 6 × + 13 × 2)43 42 23 22 64 − 96 + 52 − (8 − 24 + 26)

64 − 96 + 52 − 8 + 24 − 26,20 − 10

Area = 10

(3 + 1)∫ 42 (x− 2)2 ∫ fdx

Area = [(x− 2 +x)3 ]42

(x− 2)3 x

(4 − 2 + 4 − [(2 − 2 + 2])3 )3 + 4 − ( + 2)23 03 + 4 − 223

8 + 4 − 2

Area = 10

f = f∫ 42 ∫ 2

0

(3 − 12x+ 13)∫ 20 x2 ∫ fdx

Area = [ − 6 + 13xx3 x2 ]20

x3 −6x2 13x

( − 6 × + 13 × 2) − ( − 6 × + 13 × 0)23 22 03 02 8 − 24 + 26

Area = 10

Examiners reportIn part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic andarithmetic errors prevented many from finding the correct area.

17a. [6 marks]

Let for , , . The graph of is given below.

The graph of has a local minimum at A( , ) and a local maximum at B.

Use the quotient rule to show that .

Markschemecorrect derivatives applied in quotient rule (A1)A1A1

,

Note: Award (A1) for 1, A1 for and A1 for , only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule A1

e.g. ,

correct working (A1)

e.g.

expression clearly leading to the answer A1

e.g.

AG N0

[6 marks]

Examiners reportWhile most candidates answered part (a) correctly, there were some who did not show quite enough work for a "show that" question.A very small number of candidates did not follow the instruction to use the quotient rule.

f(x) = x

−2 +5x−2x2−2 ≤ x ≤ 4 x ≠ 1

2x ≠ 2 f

f 1 1

(x) =f ′ 2 −2x2

(−2 +5x−2)x2 2

1 −4x+ 5

−4x 5

1×(−2 +5x−2)−x(−4x+5)x2

(−2 +5x−2)x2 2

−2 +5x−2−x(−4x+5)x2

(−2 +5x−2)x2 2

−2 +5x−2−(−4 +5x)x2 x2

(−2 +5x−2)x2 2

−2 +5x−2+4 −5xx2 x2

(−2 +5x−2)x2 2

(x) =f ′ 2 −2x2

(−2 +5x−2)x2 2

17b. [3 marks]Given that the line does not meet the graph of f , find the possible values of k .

Markschemerecognizing values between max and min (R1)

A2 N3

[3 marks]

Examiners reportIn part (c), a significant number of candidates seemed to think that the line was a vertical line, and attempted to find the verticalasymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates whorecognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between thegiven local minimum and the local maximum they had found in part (b).

y = k

< k < 119

y = k

18a. [2 marks]

Let , for .

Find .

Markscheme A1A1 N2

[2 marks]

Examiners reportMany students failed in applying the chain rule to find the correct derivative, and some inappropriately used the product rule.However, many of those obtained full follow through marks in part (b) for the sketch of the function they found in part (a).

f(x) = cos( )ex −2 ≤ x ≤ 2

(x)f ′

(x) = − sin( )f ′ ex ex

18b. [4 marks]On the grid below, sketch the graph of .(x)f ′

Markscheme

A1A1A1A1 N4

Note: Award A1 for shape that must have the correct domain (from to ) and correct range (from to ), A1 for minimum incircle, A1 for maximum in circle and A1 for intercepts in circles.

[4 marks]

Examiners reportMany students failed in applying the chain rule to find the correct derivative, and some inappropriately used the product rule.However, many of those obtained full follow through marks in part (b) for the sketch of the function they found in part (a).

Most candidates sketched an approximately correct shape in the given domain, though there were some that did not realize they hadto set their GDC to radians, producing a meaningless sketch.

It is very important to stress to students that although they are asked to produce a sketch, it is still necessary to show its key featuressuch as domain and range, stationary points and intercepts.

−2 +2 −6 4

19a. [3 marks]

A particle moves in a straight line with velocity , for , where v is in centimetres per second and t is in seconds.

Find the acceleration of the particle after 2.7 seconds.

Markschemerecognizing that acceleration is the derivative of velocity (seen anywhere) (R1)

e.g.

correctly substituting 2.7 into their expression for a (not into v) (A1)

e.g.

(exact), A1 N3

[3 marks]

v = 12t− 2 − 1t3 t ≥ 0

a = , ,12 − 6sd2

dt2v′ t2

(2.7)s′′

acceleration = −31.74 −31.7

Examiners reportThis question was well answered by many candidates, although there were some who did not recognize the relationship betweenvelocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most ofthe marks. Very few appear to have used their GDC for the integration.

19b. [3 marks]Find the displacement of the particle after 1.3 seconds.

Markschemerecognizing that displacement is the integral of velocity R1

e.g.

correctly substituting 1.3 (A1)

e.g.

(exact), (cm) A1 N2

[3 marks]

Examiners reportThis question was well answered by many candidates, although there were some who did not recognize the relationship betweenvelocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most ofthe marks. Very few appear to have used their GDC for the integration.

s = ∫ v

vdt∫ 1.30

displacement = 7.41195 7.41

20a. [2 marks]

Let , where a , b and c are real numbers. The graph of f passes through the point (2, 9) .

Show that .

Markschemeattempt to substitute coordinates in f (M1)

e.g.

correct substitution A1

e.g.

AG N0

[2 marks]

Examiners reportPart (a) was generally well done, with a few candidates failing to show a detailed substitution. Some substituted 2 in place of x, butdidn't make it clear that they had substituted in y as well.

f(x) = a + b + cx3 x2

8a+ 4b+ c = 9

f(2) = 9

a× + b× + c = 923 22

8a+ 4b+ c = 9

20b. [7 marks]The graph of f has a local minimum at .

Find two other equations in a , b and c , giving your answers in a similar form to part (a).

(1, 4)

Markschemerecognizing that is on the graph of f (M1)

e.g.

correct equation A1

e.g.

recognizing that at minimum (seen anywhere) (M1)

e.g.

(seen anywhere) A1A1

correct substitution into derivative (A1)

e.g.

correct simplified equation A1

e.g.

[7 marks]

Examiners reportA great majority could find the two equations in part (b). However there were a significant number of candidates who failed toidentify that the gradient of the tangent is zero at a minimum point, thus getting the incorrect equation .

(1, 4)

f(1) = 4

a+ b+ c = 4

= 0f ′

(1) = 0f ′

(x) = 3a + 2bxf ′ x2

3a× + 2b× 1 = 012

3a+ 2b = 0

3a+ 2b = 4

20c. [4 marks]Find the value of a , of b and of c .

Markschemevalid method for solving system of equations (M1)

e.g. inverse of a matrix, substitution

, , A1A1A1 N4

[4 marks]

Examiners reportA considerable number of candidates only had 2 equations, so that they either had a hard time trying to come up with a third equation(incorrectly combining some of the information given in the question) to solve part (c) or they completely failed to solve it.

Despite obtaining three correct equations many used long elimination methods that caused algebraic errors. Pages of calculationsleading nowhere were seen.

Those who used matrix methods were almost completely successful.

a = 2 b = −3 c = 5

21a. [1 mark]

Let .

Write down .

Markscheme A1 N1

[1 mark]

f(x) = e6x

(x)f ′

(x) = 6f ′ e6x

Examiners reportOn the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponentialfunction and successfully finding the equation of the tangent line.

21b. [4 marks]The tangent to the graph of f at the point has gradient m .

(i) Show that .

(ii) Find b .

Markscheme(i) evidence of valid approach (M1)

e.g. ,

correct manipulation A1

e.g. ,

AG N0

(ii) evidence of finding (M1)

e.g.

A1 N2

[4 marks]

Examiners reportOn the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponentialfunction and successfully finding the equation of the tangent line. Some candidates lost a mark in (b)(i) for not showing sufficientworking leading to the given answer.

P(0, b)

m = 6

(0)f ′ 6e6×0

6e0 6 × 1

m = 6

f(0)

y = e6(0)

b = 1

21c. [1 mark]Hence, write down the equation of this tangent.

Markscheme A1 N1

[1 mark]

Examiners reportOn the whole, candidates handled this question quite well.

y = 6x+ 1

22a. [4 marks]

In this question, you are given that , and .

The displacement of an object from a fixed point, O is given by for .

In this interval, there are only two values of t for which the object is not moving. One value is .

Find the other value.

cos =π

312

sin =π

33√

2

s(t) = t− sin 2t 0 ≤ t ≤ π

t = π

6

Markschemeevidence of valid approach (M1)

e.g. setting

correct working A1

e.g. ,

, , (A1)

A1 N3

[4 marks]

Examiners reportIn part (b), most candidates understood that they needed to set their derivative equal to zero, but fewer were able to take the next stepto solve the resulting double angle equation. Again, some candidates over-complicated the equation by using the double angleidentity. Few ended up with the correct answer .

(t) = 0s′

2 cos2t = 1 cos2t = 12

2t = π

35π3

…

t = 5π6

5π6

22b. [3 marks]Show that between these two values of t .

Markschemeevidence of valid approach (M1)

e.g. choosing a value in the interval

correct substitution A1

e.g.

A1

AG N0

[3 marks]

Examiners reportIn part (c), many candidates knew they needed to test a value between and their value from part (b), but fewer were able tosuccessfully complete that calculation. Some candidates simply tested their boundary values while others unsuccessfully attempted tomake use of the second derivative.

(t) > 0s′

< t <π

65π6

( ) = 1 − 2 cosπs′ π

2

( ) = 3s′ π

2

(t) > 0s′

π/6

22c. [5 marks]Find the distance travelled between these two values of t .

Markschemeevidence of approach using s or integral of (M1)

e.g. ; , ;

substituting values and subtracting (M1)

e.g. ,

correct substitution A1

e.g. ,

distance is A1A1 N3

Note: Award A1 for , A1 for .

[5 marks]

Examiners reportAlthough many candidates did not attempt part (d), those who did often demonstrated a good understanding of how to use thedisplacement function s or the integral of their derivative from part (a). Candidates who had made an error in part (b) often could notfinish, as could not be evaluated at their value without a calculator. Of those who had successfully found the other boundaryof , a common error was giving the incorrect sign of the value of . Again, this part was a good discriminator betweenthe grade 6 and 7 candidates.

s′

∫ (t)dts′ s( )5π6

s( )π6

[t− sin 2t]5π6π

6

s( ) − s( )5π6

π

6( − ) − ( − (− ))π

63√

25π6

3√2

− sin − [ − sin ]5π6

5π3

π

6π

3( − (− )) − ( − )5π

63√

2π

63√

2

+2π3

3√

2π3

3√

sin(2t)5π/6 sin(5π/3)

23a. [2 marks]

The graph of , for , is shown below.

The graph has -intercepts at , , and .

Find k .

Markschemeevidence of valid approach (M1)

e.g. ,

A1 N2

[2 marks]

y = (x− 1)sinx 0 ≤ x ≤ 5π2

x 0 1 π k

y = 0 sinx = 0

2π = 6.283185…

k = 6.28

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.π

23b. [3 marks]The shaded region is rotated about the x-axis. Let V be the volume of the solid formed.

Write down an expression for V .

Markschemeattempt to substitute either limits or the function into formula (M1)

(accept absence of )

e.g. , ,

correct expression A2 N3

e.g. ,

[3 marks]

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.

360∘

dx

V = π dx∫ k

π(f(x))2 π∫ ((x− 1)sinx)2 π dx∫ 6.28…

πy2

π xdx∫ 6.28π

(x− 1)2sin2 π dx∫ 2ππ

((x− 1)sinx)2

π

23c. [2 marks]The shaded region is rotated about the x-axis. Let V be the volume of the solid formed.

Find V .

Markscheme

A2 N2

[2 marks]

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx, using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.

360∘

V = 69.60192562…

V = 69.6

π

24. [6 marks]Let . Given that , find .(x) = 3 + 2f ′ x2 f(2) = 5 f(x)

Markschemeevidence of anti-differentiation (M1)

e.g. ,

(seen anywhere, including the answer) A1A1

attempt to substitute (2, 5) (M1)

e.g. ,

finding the value of c (A1)

e.g. ,

A1 N5

[6 marks]

Examiners reportThis question, which required candidates to integrate a simple polynomial and then substitute an initial condition to solve for "c", wasvery well done. Nearly all candidates who attempted this question were able to earn full marks. The very few mistakes that were seeninvolved arithmetic errors when solving for "c", or failing to write the final answer as the equation of the function.

∫ (x)f ′ ∫ (3 + 2)dxx2

f(x) = + 2x+ cx3

f(2) = (2 + 2(2))3 5 = 8 + 4 + c

5 = 12 + c c = −7

f(x) = + 2x− 7x3

25a. [3 marks]

The following diagram shows the graph of , for .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i) a ;

(ii) c ;

(iii) d .

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

f(x) = asin(b(x− c)) + d 2 ≤ x ≤ 10

a = 8

c = 2

d = 4

Examiners reportPart (a) of this question proved challenging for most candidates.

25b. [2 marks]Show that .

MarkschemeMETHOD 1

recognizing that period (A1)

correct working A1

e.g. ,

AG N0

METHOD 2

attempt to substitute M1

e.g.

correct working A1

e.g.

AG N0

[2 marks]

Examiners reportAlthough a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize thatthis period could be found using the given coordinates of the maximum and minimum points.

b = π

4

= 8

8 = 2πb

b = 2π8

b = π

4

12 = 8 sin(b(4 − 2)) + 4

sin 2b = 1

b = π

4

25c. [3 marks]Find .

Markschemeevidence of attempt to differentiate or choosing chain rule (M1)

e.g. ,

(accept ) A2 N3

[3 marks]

Examiners reportIn part (c), not many candidates found the correct derivative using the chain rule.

(x)f ′

cos (x− 2)π

4× 8π

4

(x) = 2πcos( (x− 2))f ′ π

42πcos (x− 2)π

4

25d. [6 marks]At a point R, the gradient is . Find the x-coordinate of R.−2π

Markschemerecognizing that gradient is (M1)

e.g.

correct equation A1

e.g. ,

correct working (A1)

e.g.

using (seen anywhere) (A1)

e.g.

simplifying (A1)

e.g.

A1 N4

[6 marks]

Examiners reportFor part (d), a good number of candidates correctly set their expression equal to , but errors in their previous values kept mostfrom correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

(x)f ′

(x) = mf ′

−2π = 2πcos( (x− 2))π

4−1 = cos( (x− 2))π

4

(−1) = (x− 2)cos−1 π

4

(−1) = πcos−1

π = (x− 2)π

4

4 = (x− 2)

x = 6

−2π

26a. [4 marks]

Let . The line L is the tangent to the curve of f at (4, 6) .

Find the equation of L .

Markschemefinding A1

attempt to find (M1)

correct value A1

correct equation in any form A1 N2

e.g. ,

[4 marks]

Examiners reportWhile most candidates answered part (a) correctly, finding the equation of the tangent, there were some who did not consider thevalue of their derivative when .

f(x) = + 214x2

(x) = xf ′ 12

(4)f ′

(4) = 2f ′

y− 6 = 2(x− 4) y = 2x− 2

x = 4

( ) = 90

26b. [6 marks]

Let , for . The following diagram shows the graph of g .

Find the area of the region enclosed by the curve of g , the x-axis, and the lines and . Give your answer in theform , where .

Markscheme

correct integral A1A1

e.g.

substituting limits and subtracting (M1)

e.g. ,

correct working (A1)

e.g.

correct application of (A1)

e.g.

A1 N4

[6 marks]

Examiners reportIn part (b), most candidates knew that they needed to integrate to find the area, but errors in integration, and misapplication of therules of logarithms kept many from finding the correct area.

g(x) = 903x+4

2 ≤ x ≤ 12

x = 2 x = 12a lnb a,b ∈ Z

area = dx∫ 122

903x+4

30 ln(3x+ 4)

30 ln(3 × 12 + 4) − 30 ln(3 × 2 + 4) 30 ln40 − 30 ln10

30(ln40 − ln10)

lnb− lna

30 ln 4010

area = 30 ln4

26c. [3 marks]The graph of g is reflected in the x-axis to give the graph of h . The area of the region enclosed by the lines L , , and the x-axis is 120 .

Find the area enclosed by the lines L , , and the graph of h .

Markschemevalid approach (M1)

e.g. sketch, area h = area g , 120 + their answer from (b)

A2 N3

[3 marks]

x = 2 x = 12120 cm2

x = 2 x = 12

area = 120 + 30 ln4

Examiners reportIn part (c), it was clear that a significant number of candidates understood the idea of the reflected function, and some recognized thatthe integral was the negative of the integral from part (b), but only a few recognized the relationship between the areas. Many thoughtthe area between h and the x-axis was 120.

27a. [4 marks]

Let , where . The function v is obtained when the graph of f is transformed by

a stretch by a scale factor of parallel to the y-axis,

followed by a translation by the vector .

Find , giving your answer in the form .

Markschemeapplies vertical stretch parallel to the y-axis factor of (M1)

e.g. multiply by , ,

applies horizontal shift 2 units to the right (M1)

e.g. ,

applies a vertical shift 4 units down (M1)

e.g. subtracting 4, ,

A1 N4

[4 marks]

Examiners reportWhile a number of candidates had an understanding of each transformation, most had difficulty applying them in the correct order,and few obtained the completely correct answer in part (a). Many earned method marks for discerning three distinct transformations.Few candidates knew to integrate to find the distance travelled. Many instead substituted time values into the velocity function or itsderivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather thanusing the GDC, which often proved unsuccessful.

f(t) = 2 + 7t2 t > 0

13

( )2−4

v(t) a(t− b + c)2

13

13

f(t)13

× 213

f(t− 2) t− 2

f(t) − 4 − 473

v(t) = (t− 2 −23

)2 53

27b. [3 marks]A particle moves along a straight line so that its velocity in ms , at time t seconds, is given by v . Find the distance theparticle travels between and .

−1

t = 5.0 t = 6.8

Markschemerecognizing that distance travelled is area under the curve M1

e.g. , sketch

distance = 15.576 (accept 15.6) A2 N2

[3 marks]

Examiners reportWhile a number of candidates had an understanding of each transformation, most had difficulty applying them in the correct order,and few obtained the completely correct answer in part (a). Many earned method marks for discerning three distinct transformations.Few candidates knew to integrate to find the distance travelled. Many instead substituted time values into the velocity function or itsderivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather thanusing the GDC, which often proved unsuccessful.

∫ v, (t− 2 − t29

)3 53

28a. [3 marks]

Let , for .

Sketch the graph of f .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape with inflexion/change of curvature, A1 for maximum skewed to the left, A1 forasymptotic behaviour to the right.

[3 marks]

Examiners reportMany candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find themaximum value.

f(x) = 20xe0.3x

0 ≤ x ≤ 20

28b. [3 marks](i) Write down the x-coordinate of the maximum point on the graph of f .

(ii) Write down the interval where f is increasing.

Markscheme(i) A1 N1

(ii) correct interval, with right end point A1A1 N2

e.g. ,

Note: Accept any inequalities in the right direction.

[3 marks]

Examiners reportMany candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find themaximum value.

x = 3.33

3 13

0 < x ≤ 3.33 0 ≤ x < 3 13

28c. [5 marks]Show that .

Markschemevalid approach (M1)

e.g. quotient rule, product rule

2 correct derivatives (must be seen in product or quotient rule) (A1)(A1)

e.g. , or

correct substitution into product or quotient rule A1

e.g. ,

correct working A1

e.g. , ,

AG N0

[5 marks]

Examiners reportMost had a valid approach in part (c) using either the quotient or product rule, but many had difficulty applying the chain rule with afunction involving e and simplifying.

(x) =f ′ 20−6xe0.3x

20 0.3e0.3x −0.3e−0.3x

20 −20x(0.3)e0.3x e0.3x

( )e0.3x 220 + 20x(−0.3)e−0.3x e−0.3x

20 −6xe0.3x e0.3x

e0.6x

(20−20x(0.3))e0.3x

( )e0.3x 2(20 + 20x(−0.3))e−0.3x

(x) =f ′ 20−6xe0.3x

28d. [4 marks]Find the interval where the rate of change of f is increasing.

Markschemeconsideration of or (M1)

valid reasoning R1

e.g. sketch of , is positive, , reference to minimum of

correct value (A1)

correct interval, with both endpoints A1 N3

e.g. ,

[4 marks]

Examiners reportPart (d) was difficult for most candidates. Although many associated rate of change with derivative, only the best-prepared studentshad valid reasoning and could find the correct interval with both endpoints.

f ′ f ′′

f ′ f ′′ = 0f ′′ f ′

6.6666666… (6 )23

6.67 < x ≤ 20 6 ≤ x < 2023

29a. [4 marks]

Let , for .

Use the quotient rule to show that .

Markscheme , (seen anywhere) A1A1

attempt to substitute into the quotient rule (do not accept product rule) M1

e.g.

correct manipulation that clearly leads to result A1

e.g. , , ,

AG N0

[4 marks]

Examiners reportMany candidates clearly knew their quotient rule, although a common error was to simplify as and then "cancel" theexponents.

g(x) = ln xx2

x > 0

(x) =g′ 1−2 ln xx3

lnx =ddx

1x

= 2xddxx2

( )−2x ln xx2 1x

x4

x−2x ln xx4

x(1−2 ln x)

x4

x

x4

2x ln xx4

(x) =g′ 1−2 ln xx3

2x lnx 2 lnx2

29b. [3 marks]The graph of g has a maximum point at A. Find the x-coordinate of A.

Markschemeevidence of setting the derivative equal to zero (M1)

e.g. ,

A1

A1 N2

[3 marks]

(x) = 0g′ 1 − 2 lnx = 0

lnx = 12

x = e12

Examiners reportFor (b), those who knew to set the derivative to zero typically went on find the correct x-coordinate, which must be in terms of e, asthis is the calculator-free paper. Occasionally, students would take and attempt to solve from .= 01−2 ln x

x31 − 2 lnx = x3

30a. [1 mark]

The velocity v ms of a particle at time t seconds, is given by , for .

Write down the velocity of the particle when .

Markscheme A1 N1

[1 mark]

Examiners reportMany candidates gave a correct initial velocity, although a substantial number of candidates answered that .

−1 v = 2t+ cos2t 0 ≤ t ≤ 2

t = 0

v = 1

0 + cos0 = 0

30b. [8 marks]When , the acceleration is zero.

(i) Show that .

(ii) Find the exact velocity when .

Markscheme(i) A1

A1A1

Note: Award A1 for coefficient 2 and A1 for .

evidence of considering acceleration = 0 (M1)

e.g. ,

correct manipulation A1

e.g. ,

(accept ) A1

AG N0

(ii) attempt to substitute into v (M1)

e.g.

A1 N2

[8 marks]

Examiners reportFor (b), students commonly applied the chain rule correctly to achieve the derivative, and many recognized that the acceleration mustbe zero. Occasionally a student would use a double-angle identity on the velocity function before differentiating. This is not incorrect,but it usually caused problems when trying to show . At times students would reach the equation and then substitutethe , which does not satisfy the “show that” instruction.

t = k

k = π

4

t = π

4

(2t) = 2ddt

(cos2t) = −2 sin 2tddt

−sin 2t

= 0dvdt

2 − 2 sin 2t = 0

sin 2k = 1 sin 2t = 1

2k = π

22t = π

2

k = π

4

t = π

4

2 ( ) + cos( )π

42π4

v = π

2

k = π

4sin 2k = 1

π

4

30c. [4 marks]When , and when , .

Sketch a graph of v against t .

t < π

4> 0dv

dtt > π

4> 0dv

dt

Markscheme

A1A1A2 N4

Notes: Award A1 for y-intercept at , A1 for curve having zero gradient at , A2 for shape that is concave down to the leftof and concave up to the right of . If a correct curve is drawn without indicating , do not award the second A1 for the zerogradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.

[4 marks]

Examiners reportThe challenge in this question is sketching the graph using the information achieved and provided. This requires students to makegraphical interpretations, and as typical in section B, to link the early parts of the question with later parts. Part (a) provides the y-intercept, and part (b) gives a point with a horizontal tangent. Plotting these points first was a helpful strategy. Few understood eitherthe notation or the concept that the function had to be increasing on either side of the , with most thinking that the point was either amax or min. It was the astute student who recognized that the derivatives being positive on either side of creates a point of inflexion.

Additionally, important points should be labelled in a sketch. Indicating the on the x-axis is a requirement of a clear graph.Although students were not penalized for not labelling the on the y-axis, there should be a recognition that the point is higher thanthe y-intercept.

(0, 1) t = π

4π

4π

4t = π

4

π

4π

4

π

4π

2

30d. [3 marks]Let d be the distance travelled by the particle for .

(i) Write down an expression for d .

(ii) Represent d on your sketch.

0 ≤ t ≤ 1

(2t+ cos2t)dt∫ 10 [ + ]t2 sin 2t

2

1

01 + sin 2

2vdt∫ 1

0

t = 1 t = π

4

Examiners reportWhile some candidates recognized that the distance is the area under the velocity graph, surprisingly few included neither the limits ofintegration in their expression, nor the “dt”. Most unnecessarily attempted to integrate the function, often giving an answer with “+C”,and only earned marks if the limits were included with their result. Few recognized that a shaded area is an adequate representation ofdistance on the sketch, with most fruitlessly attempting to graph a new curve.

31. [6 marks]Let and , for .

Find the area of the region enclosed by the graphs of f and g .

Markschemeevidence of finding intersection points (M1)

e.g. , , sketch showing intersection

, (may be seen as limits in the integral) A1A1

evidence of approach involving integration and subtraction (in any order) (M1)

e.g. , ,

A2 N3

[6 marks]

Examiners reportThis question was poorly done by a great many candidates. Most seemed not to understand what was meant by the phrase "regionenclosed by" as several candidates assumed that the limits of the integral were those given in the domain. Few realized what area wasrequired, or that intersection points were needed. Candidates who used their GDCs to first draw a suitable sketch could normallyrecognize the required region and could find the intersection points correctly. However, it was disappointing to see the number ofcandidates who could not then use their GDC to find the required area or who attempted unsuccessful analytical approaches.

f(x) = cos( )x2 g(x) = ex −1.5 ≤ x ≤ 0.5

f(x) = g(x) cos =x2 ex

x = −1.11 x = 0

cos −∫ 0−1.11 x2 ex ∫ (cos − )dxx2 ex ∫ g − f

area = 0.282

32a. [2 marks]

The following diagram shows a waterwheel with a bucket. The wheel rotates at a constant rate in an anticlockwise (counter-clockwise)direction.

The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above the water level. After t seconds, the height of the bucketabove the water level is given by .

Show that .

h = asin bt+ 2

a = 4

MarkschemeMETHOD 1

evidence of recognizing the amplitude is the radius (M1)

e.g. amplitude is half the diameter

A1

AG N0

METHOD 2

evidence of recognizing the maximum height (M1)

e.g. ,

correct reasoning

e.g. and has amplitude of 1 A1

AG N0

[2 marks]

Examiners reportParts (a) and (b) were generally well done.

a = 82

a = 4

h = 6 asin bt+ 2 = 6

asin bt = 4 sin bt

a = 4

32b. [2 marks]The wheel turns at a rate of one rotation every 30 seconds.

Show that .

MarkschemeMETHOD 1

period = 30 (A1)

A1

AG N0

METHOD 2

correct equation (A1)

e.g. ,

A1

AG N0

[2 marks]

Examiners reportParts (a) and (b) were generally well done, however there were several instances of candidates working backwards from the givenanswer in part (b).

b = π

15

b = 2π30

b = π

15

2 = 4 sin 30b+ 2 sin 30b = 0

30b = 2π

b = π

15

32c. [6 marks]In the first rotation, there are two values of t when the bucket is descending at a rate of .

Find these values of t .

0.5 ms−1

Markschemerecognizing (seen anywhere) R1

attempting to solve (M1)

e.g. sketch of , finding

correct work involving A2

e.g. sketch of showing intersection,

, A1A1 N3

[6 marks]

Examiners reportParts (c) and (d) proved to be quite challenging for a large proportion of candidates. Many did not attempt these parts. The mostcommon error was a misinterpretation of the word "descending" where numerous candidates took to be 0.5 instead of but

incorrect derivatives for h were also widespread. The process required to solve for t from the equation

overwhelmed those who attempted algebraic methods. Few could obtain both correct solutions, more had one correct while othersincluded unreasonable values including .

(t) = −0.5h′

h′ h′

h′

h′ −0.5 = cos( t)4π15

π

15

t = 10.6 t = 19.4

(t)h′ −0.5

−0.5 = cos( t)4π15

π

15

t < 0

32d. [4 marks]In the first rotation, there are two values of t when the bucket is descending at a rate of .

Determine whether the bucket is underwater at the second value of t .

MarkschemeMETHOD 1

valid reasoning for their conclusion (seen anywhere) R1

e.g. so underwater; so not underwater

evidence of substituting into h (M1)

e.g. ,

correct calculation A1

e.g.

correct statement A1 N0

e.g. the bucket is underwater, yes

METHOD 2

valid reasoning for their conclusion (seen anywhere) R1

e.g. so underwater; so not underwater

evidence of valid approach (M1)

e.g. solving , graph showing region below x-axis

correct roots A1

e.g. ,

correct statement A1 N0

e.g. the bucket is underwater, yes

[4 marks]

Examiners reportIn part (d), not many understood that the condition for underwater was and had trouble interpreting the meaning of "secondvalue". Many candidates, however, did recover to gain some marks in follow through.

0.5 ms−1

h(t) < 0 h(t) > 0

h(19.4) 4 sin + 219.4π15

h(19.4) = −1.19

h(t) < 0 h(t) > 0

h(t) = 0

17.5 27.5

h(t) < 0

33a. [2 marks]

The following diagram shows the graph of .

The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.

Identify the two points of inflexion.

MarkschemeB, D A1A1 N2

[2 marks]

Examiners reportMost candidates were able to recognize the points of inflexion in part (a).

f(x) = e−x2

33b. [5 marks](i) Find .

(ii) Show that .

Markscheme(i) A1A1 N2

Note: Award A1 for and A1 for .

(ii) finding the derivative of , i.e. (A1)

evidence of choosing the product rule (M1)

e.g.

A1

AG N0

[5 marks]

Examiners reportMost candidates were able to recognize the points of inflexion in part (a) and had little difficulty with the first and second derivativesin part (b). A few did not recognize the application of the product rule in part (b).

(x)f ′

(x) = (4 − 2)f ′′ x2 e−x2

(x) = −2xf ′ e−x2

e−x2 −2x

−2x −2

−2e−x2 −2x× −2xe−x2

−2 + 4e−x2x2e−x2

(x) = (4 − 2)f ′′ x2 e−x2

33c. [4 marks]Find the x-coordinate of each point of inflexion.

Markschemevalid reasoning R1

e.g.

attempting to solve the equation (M1)

e.g. , sketch of

, A1A1 N3

[4 marks]

Examiners reportObtaining the x-coordinates of the inflexion points in (c) usually did not cause many problems.

(x) = 0f ′′

(4 − 2) = 0x2 (x)f ′′

p = 0.707 (= )12√

q = −0.707 (= − )12√

33d. [4 marks]Use the second derivative to show that one of these points is a point of inflexion.

Markschemeevidence of using second derivative to test values on either side of POI M1

e.g. finding values, reference to graph of , sign table

correct working A1A1

e.g. finding any two correct values either side of POI,

checking sign of on either side of POI

reference to sign change of R1 N0

[4 marks]

Examiners reportOnly the better-prepared candidates understood how to set up a second derivative test in part (d). Many of those did not show, orclearly indicate, the values of x used to test for a point of inflexion, but merely gave an indication of the sign. Some candidates simply

resorted to showing that , completely missing the point of the question. The necessary condition for a point of

inflexion, i.e. and the change of sign for , seemed not to be known by the vast majority of candidates.

f ′′

f ′′

(x)f ′′

(± ) = 0f ′′ 12√

(x) = 0f ′′ (x)f ′′

34. [6 marks]Let . Find .h(x) = 6xcos x (0)h′

MarkschemeMETHOD 1 (quotient)

derivative of numerator is 6 (A1)

derivative of denominator is (A1)

attempt to substitute into quotient rule (M1)

correct substitution A1

e.g.

substituting (A1)

e.g.

A1 N2

METHOD 2 (product)

derivative of 6x is 6 (A1)

derivative of is (A1)

attempt to substitute into product rule (M1)

correct substitution A1

e.g.

substituting (A1)

e.g.

A1 N2

[6 marks]

Examiners reportThe majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question.However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakesin simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A fewcandidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotientrule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.

−sinx

(cos x)(6)−(6x)(− sinx)

(cos x)2

x = 0

(cos 0)(6)−(6×0)(− sin 0)

(cos 0)2

(0) = 6h′

h(x) = 6x× (cosx)−1

(cosx)−1 (−(cosx (−sinx)))−2

(6x)(−(cosx (−sinx)) + (6)(cosx)−2 )−1

x = 0

(6 × 0)(−(cos0 (−sin 0)) + (6)(cos0)−2 )−1

(0) = 6h′

f(x) = 2 2

35a. [5 marks]

The following diagram shows part of the graph of the function .

The line T is the tangent to the graph of f at .

Show that the equation of T is .

Markscheme (A1)

A1

evidence of finding the gradient of f at M1

e.g. substituting into

finding gradient of f at A1

e.g.

evidence of finding equation of the line M1

e.g. ,

AG N0

[5 marks]

Examiners reportThe majority of candidates seemed to know what was meant by the tangent to the graph in part (a), but there were many who did notfully show their work, which is of course necessary on a "show that" question. While many candidates knew they needed to find thederivative of f , some failed to substitute the given value of x in order to find the gradient of the tangent.

f(x) = 2x2

x = 1

y = 4x− 2

f(1) = 2

(x) = 4xf ′

x = 1

x = 1 (x)f ′

x = 1

(1) = 4f ′

y− 2 = 4(x− 1) 2 = 4(1) + b

y = 4x− 2

35b. [2 marks]Find the x-intercept of T .

Markschemeappropriate approach (M1)

e.g.

A1 N2

[2 marks]

Examiners reportPart (b), finding the x-intercept, was answered correctly by nearly every candidate.

4x− 2 = 0

x = 12

35c. [9 marks]The shaded region R is enclosed by the graph of f , the line T , and the x-axis.

(i) Write down an expression for the area of R .

(ii) Find the area of R .

Markscheme(i) bottom limit (seen anywhere) (A1)

approach involving subtraction of integrals/areas (M1)

e.g. ,

correct expression A2 N4

e.g. , ,

(ii) METHOD 1 (using only integrals)

correct integration (A1)(A1)(A1)

,

substitution of limits (M1)

e.g.

area = A1 N4

METHOD 2 (using integral and triangle)

area of triangle = (A1)

correct integration (A1)

substitution of limits (M1)

e.g. ,

correct simplification (A1)

e.g.

area = A1 N4

[9 marks]

Examiners reportIn part (c), most candidates struggled with writing an expression for the area of R . Many tried to use the difference of the twofunctions over the entire interval 0–1, not noticing that the area from 0–0.5 only required the use of function f . Many of thesecandidates were able to earn follow-through marks in the second part of (c) for their correct integration. There were a few candidateswho successfully found the area under the line as the area of a triangle.

x = 0

∫ f(x) − area of triangle ∫ f− ∫ l

2 dx− (4x− 2)dx∫ 10 x2 ∫ 1

0.5 f(x)dx−∫ 10

12

2 dx+ (f(x) − (4x− 2))dx∫ 0.50 x2 ∫ 1

0.5

∫ 2 dx =x2 2x3

3∫ (4x− 2)dx =2 − 2xx2

+ − 2 + 2 − ( − + 1)112

23

112

12

16

12

∫ 2 dx =x2 2x3

3

(1 − (023

)3 23

)3 − 023

−23

12

16

36a. [2 marks]

The following diagram shows part of the graph of a quadratic function f .

The x-intercepts are at and , and the y-intercept is at .

Write down in the form .

Markscheme A1A1 N2

[2 marks]

Examiners reportParts (a) and (c) of this question were very well done by most candidates.

(−4, 0) (6, 0) (0, 240)

f(x) f(x) = −10(x− p)(x− q)

f(x) = −10(x+ 4)(x− 6)

36b. [4 marks]Find another expression for in the form .

MarkschemeMETHOD 1

attempting to find the x-coordinate of maximum point (M1)

e.g. averaging the x-intercepts, sketch, , axis of symmetry

attempting to find the y-coordinate of maximum point (M1)

e.g.

A1A1 N4

METHOD 2

attempt to expand (M1)

e.g.

attempt to complete the square (M1)

e.g.

A1A1 N4

[4 marks]

f(x) f(x) = −10(x−h + k)2

= 0y′

k = −10(1 + 4)(1 − 6)

f(x) = −10(x− 1 + 250)2

f(x)

−10( − 2x− 24)x2

−10((x− 1 − 1 − 24))2

f(x) = −10(x− 1 + 250)2

Examiners reportIn part (b), many candidates attempted to use the method of completing the square, but were unsuccessful dealing with the coefficientof . Candidates who recognized that the x-coordinate of the vertex was 1, then substituted this value into the function from part(a), were generally able to earn full marks here.

−10

36c. [2 marks]Show that can also be written in the form .

Markschemeattempt to simplify (M1)

e.g. distributive property,

correct simplification A1

e.g. ,

AG N0

[2 marks]

Examiners reportParts (a) and (c) of this question were very well done by most candidates.

f(x) f(x) = 240 + 20x− 10x2

−10(x− 1)(x− 1) + 250

−10( − 6x+ 4x− 24)x2 −10( − 2x+ 1) + 250x2

f(x) = 240 + 20x− 10x2

36d. [7 marks]A particle moves along a straight line so that its velocity, , at time t seconds is given by , for .

(i) Find the value of t when the speed of the particle is greatest.

(ii) Find the acceleration of the particle when its speed is zero.

Markscheme(i) valid approach (M1)

e.g. vertex of parabola,

A1 N2

(ii) recognizing (M1)

A1A1

speed is zero (A1)

( ) A1 N3

[7 marks]

Examiners reportIn part (d), it was clear that many candidates were not familiar with the relationship between velocity and acceleration, and did notunderstand how those concepts were related to the graph which was given. A large number of candidates used time in part b(ii),rather than . To find the acceleration, some candidates tried to integrate the velocity function, rather than taking the derivative ofvelocity. Still others found the derivative in part b(i), but did not realize they needed to use it in part b(ii), as well.

v ms−1 v = 240 + 20t− 10t2

0 ≤ t ≤ 6

(t) = 0v′

t = 1

a(t) = (t)v′

a(t) = 20 − 20t

⇒ t = 6

a(6) = −100 ms−2

t = 1t = 6

37. [8 marks]A gradient function is given by . When , . Find the value of y when .= 10 − 5dydx

e2x x = 0 y = 8 x = 1

MarkschemeMETHOD 1

evidence of anti-differentiation (M1)

e.g.

A2A1

Note: Award A2 for , A1 for . If “C” is omitted, award no further marks.

substituting (M1)

e.g.

(A1)

substituting (M1)

A1 N4

METHOD 2

evidence of definite integral function expression (M2)

e.g. ,

initial condition in definite integral function expression (A2)

e.g. ,

correct definite integral expression for y when (A2)

e.g.

A1 N4

[8 marks]

Examiners reportAlthough a pleasing number of candidates recognized the requirement of integration, many did not correctly apply the reverse of thechain rule to integration. While some candidates did not write the constant of integration, many did, earning additional follow-throughmarks even with an incorrect integral. Weaker candidates sometimes substituted into or attempted some work with a tangentline equation, earning no marks.

∫ (10 − 5)dxe2x

y = 5 − 5x+Ce2x

5e2x −5x

(0, 8)

8 = 5 +C

C = 3 (y = 5 − 5x+ 3)e2x

x = 1

y = 34.9 (5 − 2)e2

(t)dt =f(x) − f(a)∫ x

af ′ (10 − 5)∫ x

0 e2x

(10 − 5)dt = y− 8∫ x

0 e2t (10 − 5)dx+ 8∫ x

0 e2x

x = 1

(10 − 5)dx+ 8∫ 10 e2x

y = 34.9 (5 − 2)e2

x = 1 dydx

38a. [4 marks]

Let .

Find .

Markschemeevidence of choosing the product rule (M1)

e.g.

correct derivatives , 2 (A1)(A1)

A1 N4

[4 marks]

Examiners reportMost candidates answered part (a) correctly, using the product rule to find the derivative, and earned full marks here. There weresome who did not know to use the product rule, and of course did not find the correct derivative.

g(x) = 2x sinx

(x)g′

u + vv′ u′

cosx

(x) = 2xcosx+ 2 sinxg′

38b. [3 marks]Find the gradient of the graph of g at .

Markschemeattempt to substitute into gradient function (M1)

e.g.

correct substitution (A1)

e.g.

A1 N2

[3 marks]

Examiners reportIn part (b), many candidates substituted correctly into their derivatives, but then used incorrect values for and , leading tothe wrong gradient in their final answers.

x = π

(π)g′

2πcosπ+ 2 sinπ

gradient = −2π

sinx cosx

39. [6 marks]The graph of the function passes through the point . The gradient function of f is given as

. Find .

Markschemeevidence of integration

e.g. (M1)

A1A1

substituting initial condition into their expression (even if C is missing) M1

e.g.

(A1)

A1 N5

[6 marks]

Examiners reportWhile most candidates realized they needed to integrate in this question, many did so unsuccessfully. Many did not account for thecoefficient of x, and failed to multiply by . Some of the candidates who substituted the initial condition into their integral were notable to solve for "c", either because of arithmetic errors or because they did not know the correct value for .

y = f(x) ( ,4)32

(x) = sin(2x− 3)f ′

f(x)

f(x) = ∫ sin(2x− 3)dx

= − cos(2x− 3) +C12

4 = − cos0 +C12

C = 4.5

f(x) = − cos(2x− 3) + 4.512

12

cos0

f(x) = 3

40a. [7 marks]

Let . The following diagram shows part of the graph of f .

The point , where , lies on the graph of f . The tangent at P crosses the x-axis at the point . This tangent intersects

the graph of f at the point R(−2, −8) .

(i) Show that the gradient of [PQ] is .

(ii) Find .

(iii) Hence show that .

f(x) = x3

P(a,f(a)) a > 0 Q( ,0)23

a3

a− 23

(a)f ′

a = 1

Markscheme(i) substitute into gradient (M1)

e.g.

substituting

e.g. A1

gradient AG N0

(ii) correct answer A1 N1

e.g. , ,

(iii) METHOD 1

evidence of approach (M1)

e.g. ,

simplify A1

e.g.

rearrange A1

e.g.

evidence of solving A1

e.g.

AG N0

METHOD 2

gradient RQ A1

simplify A1

e.g.

evidence of approach (M1)

e.g. , ,

simplify A1

e.g. ,

AG N0

[7 marks]

Examiners reportPart (a) seemed to be well-understood by many candidates, and most were able to earn at least partial marks here. Part (ai) was a"show that" question, and unfortunately there were some candidates who did not show how they arrived at the given expression.

= −y1 y2−x1 x2

f(a)−0

a− 23

f(a) = a3

−0a3

a− 23

a3

a− 23

3a2 (a) = 3f ′ (a) =f ′ a3

a− 23

(a) = gradientf ′ 3 =a2 a3

a− 23

3 (a− ) =a2 23

a3

3 − 2 =a3 a2 a3

2 − 2 = 2 (a− 1) = 0a3 a2 a2

a = 1

= −8

−2− 23

,3−8

− 83

(a) = gradientf ′ 3 =a2 −8

−2− 23

= 3a3

a− 23

3 = 3a2 = 1a2

a = 1

40b. [9 marks]

The equation of the tangent at P is . Let T be the region enclosed by the graph of f , the tangent [PR] and the line , between and where . This is shown in the diagram below.

Given that the area of T is , show that k satisfies the equation .

Markschemeapproach to find area of T involving subtraction and integrals (M1)

e.g. , ,

correct integration with correct signs A1A1A1

e.g. ,

correct limits and k (seen anywhere) A1

e.g. ,

attempt to substitute k and (M1)

correct substitution into their integral if 2 or more terms A1

e.g.

setting their integral expression equal to (seen anywhere) (M1)

simplifying A1

e.g.

AG N0

[9 marks]

Examiners reportIn part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtractingthe two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errorswhich prevented candidates from correctly showing the desired final result.

y = 3x− 2 x = k

x = −2 x = k −2 < k < 1

2k+ 4 − 6 + 8 = 0k4 k2

∫ f− (3x− 2)dx (3x− 2) −∫ k

−2 ∫ k

−2x3 ∫ ( − 3x+ 2)x3

− + 2x14x4 3

2x2 − 2x−3

2x2 1

4x4

−2

( − 3x+ 2)dx∫ k

−2 x3 [ − + 2x]14x4 3

2x2

k

−2

−2

( − + 2k) − (4 − 6 − 4)14k4 3

2k2

2k+ 4

− + 2 = 014k4 3

2k2

− 6 + 8 = 0k4 k2

f(x) = 16 − 4 2− −−−−−−√

41. [6 marks]

The graph of , for , is shown below.

The region enclosed by the curve of f and the x-axis is rotated about the x-axis.

Find the volume of the solid formed.

Markschemeattempt to set up integral expression M1

e.g. , ,

, (seen anywhere) A1A1

evidence of substituting limits into the integrand (M1)

e.g. ,

volume A2 N3

[6 marks]

Examiners reportMany candidates correctly integrated using , although some neglected to square the function and mired themselves in awkwardintegration attempts. Upon substituting the limits, many were unable to carry the calculation to completion. Occasionally the wasneglected in a final answer. Weaker candidates considered the solid formed to be a sphere and did not use integration.

f(x) = 16 − 4x2− −−−−−−√ −2 ≤ x ≤ 2

360∘

π∫ dx16 − 4x2− −−−−−−√ 2 2π (16 − 4 )∫ 20 x2 ∫ dx16 − 4x2− −−−−−−√ 2

∫ 16dx = 16x ∫ 4 dx =x2 4x3

3

(32 − ) − (−32 + )323

323

64 − 643

= 128π3

f(x)π

f(x) = − − 3x1 3 2

42. [6 marks]

Let . Part of the graph of f is shown below.

There is a maximum point at A and a minimum point at B(3, − 9) .

Write down the coordinates of

(i) the image of B after reflection in the y-axis;

(ii) the image of B after translation by the vector ;

(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor .

Markscheme(i) A1 N1

(ii) A1A1 N2

(iii) reflection gives (A1)

stretch gives A1A1 N3

[6 marks]

Examiners reportCandidates were generally successful in finding images after single transformations in part (b). Common incorrect answers for (biii)

included , (6, 9) and (6, 18) , demonstrating difficulty with images from horizontal stretches.

f(x) = − − 3x12x3 x2

( )−25

12

(−3, − 9)

(1, − 4)

(3, 9)

( , 9)32

( , )32

92

43a. [5 marks]

Let , for .

Use the quotient rule to show that .

f(x) = cos xsinx

sinx ≠ 0

(x) =f ′ −1xsin2

Markscheme , (seen anywhere) (A1)(A1)

evidence of using the quotient rule M1

correct substitution A1

e.g. ,

A1

AG N0

[5 marks]

Examiners reportMany candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identityachieves the numerator of the answer given. Whether changing to , or applying the quotient rule a second time, mostcandidates neglected the chain rule in finding the second derivative.

sinx = cosxddx

cosx = −sinxddx

sinx(− sinx)−cos x(cos x)

xsin2

− x− xsin2 cos2

xsin2

(x) =f ′ −( x+ x)sin2 cos2

xsin2

(x) =f ′ −1xsin2

−(sinx)−2

43b. [3 marks]Find .

MarkschemeMETHOD 1

appropriate approach (M1)

e.g.

A1A1 N3

Note: Award A1 for , A1 for .

METHOD 2

derivative of (seen anywhere) A1

evidence of choosing quotient rule (M1)

e.g. , ,

A1 N3

[3 marks]

Examiners reportWhether changing to , or applying the quotient rule a second time, most candidates neglected the chain rule in finding thesecond derivative.

(x)f ′′

(x) = −(sinxf ′ )−2

(x) = 2( x)(cosx)f ′′ sin−3 (= )2 cos xxsin3

2 xsin−3 cosx

x = 2 sinxcosxsin2

u = −1 v = xsin2 =f ′′ x×0−(−1)2 sinx cos xsin2

( x)sin2 2

(x) =f ′′ 2 sinx cos x

( x)sin2 2(= )2 cos x

xsin3

−(sinx)−2

( ) ( )

43c. [3 marks]

In the following table, and . The table also gives approximate values of and near .

Find the value of p and of q.

Markschemeevidence of substituting M1

e.g. ,

, A1A1 N1N1

[3 marks]

Examiners reportThose who knew the trigonometric ratios at typically found the values of p and of q, sometimes in follow-through from an incorrect

.

( ) = pf ′ π

2( ) = qf ′′ π

2(x)f ′ (x)f ′′ x = π

2

π

2

−1sin2 π

2

2 cos π2

sin3 π2

p = −1 q = 0

π

2(x)f ′′

44. [6 marks]Let . The point lies on the curve of f . At P, the normal to the curve is parallel to . Find the valueof k.

Markschemegradient of tangent (seen anywhere) (A1)

(seen anywhere) A1

recognizing the gradient of the tangent is the derivative (M1)

setting the derivative equal to 8 (A1)

e.g. ,

substituting (seen anywhere) (M1)

A1 N4

[6 marks]

Examiners reportCandidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient ofthe normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks.Those who were unclear often either gained a few marks for finding the derivative and substituting , or no marks for workingthat did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting thederivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among thecandidates who demonstrated greater understanding, more used the gradient of the normal (the equation ) than thegradient of the tangent ( ) ; this led to more algebraic errors in obtaining the final answer of . A number of unsuccessfulcandidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.

f(x) = kx4 P(1, k) y = − x18

= 8

(x) = 4kf ′ x3

4k = 8x3 k = 2x3

x = 1

k = 2

x = 1

− k = −14

18

4k = 8 k = 2

45a. [2 marks]

A function f is defined for . The graph of f is given below.

The graph has a local maximum when , and local minima when , .

Write down the x-intercepts of the graph of the derivative function, .

Markschemex-intercepts at , 0, 2 A2 N2

[2 marks]

Examiners reportCandidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left thisquestion blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often hadtrouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Othercandidates listed individual x-values here rather than a range of values.

−4 ≤ x ≤ 3

x = 0 x = −3 x = 2

f ′

−3

45b. [2 marks]Write down all values of x for which is positive.

Markscheme , A1A1 N2

[2 marks]

Examiners reportCandidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left thisquestion blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often hadtrouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Othercandidates listed individual x-values here rather than a range of values.

(x)f ′

−3 < x < 0 2 < x < 3

45c. [2 marks]At point D on the graph of f , the x-coordinate is . Explain why at D.

Markschemecorrect reasoning R2

e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing

therefore the second derivative is negative AG

[2 marks]

−0.5 (x) < 0f ′′

Examiners reportMany candidates had difficulty explaining why the second derivative is negative in part (c). A number claimed that since the point Dwas “close” to a maximum value, the second derivative must be negative; this incorrect appeal to the second derivative test indicates alack of understanding of how the test works and the relative concept of closeness. Some candidates claimed D was a point ofinflexion, again demonstrating poor understanding of the second derivative. Among candidates who answered part (c) correctly, somestated that f was concave down while others gave well-formed arguments for why the first derivative was decreasing. A fewcandidates provided nicely sketched graphs of and and used them in their explanations.f ′ f ′′

46a. [3 marks]

Consider the function f with second derivative . The graph of f has a minimum point at A(2, 4) and a maximum point at

.

Use the second derivative to justify that B is a maximum.

Markschemesubstituting into the second derivative M1

e.g.

A1

since the second derivative is negative, B is a maximum R1 N0

[3 marks]

Examiners reportMany candidates were successful with this question. In part (a), some candidates found and were unclear how to conclude,

but most demonstrated a good understanding of the second derivative test.

(x) = 3x− 1f ′′

B(− , )43

35827

3 × (− ) − 143

(− ) = −5f ′′ 43

(− )f ′′ 43

46b. [4 marks]Given that , show that .

Markschemesetting equal to zero (M1)

evidence of substituting (or ) (M1)

e.g.

correct substitution A1

e.g. ,

correct simplification

e.g. , , A1

AG N0

[4 marks]

Examiners reportA large percentage of candidates were successful in showing that but there were still some who worked backwards from theanswer. Others did not use the given information and worked from the second derivative, integrated, and then realized that p was theconstant of integration. Candidates who evaluated the derivative at but set the result equal to 4 clearly did not understand theconcept being assessed. Few candidates used the point B with fractional coordinates.

(x) = −x+ pf ′ 32x2 p = −4

(x)f ′

x = 2 x = − 43

(2)f ′

(2 − 2 + p32

)2 − (− ) + p32

(− )43

243

6 − 2 + p = 0 + + p = 083

43

4 + p = 0

p = −4

p = −4

x = 2

f(x)

46c. [7 marks]Find .

Markschemeevidence of integration (M1)

A1A1A1

substituting (2, 4) or into their expression (M1)

correct equation A1

e.g. , ,

A1 N4

[7 marks]

Examiners reportCandidates often did well on the first part of (c), knowing to integrate and successfully finding some or all terms. Some had troublewith the fractions or made careless errors with the signs; others did not use the value of and so could not find the third termwhen integrating. It was very common for candidates to either forget the constant of integration or to leave it in without finding itsvalue.

f(x)

f(x) = − − 4x+ c12x3 1

2x2

(− , )43

35827

× − × − 4 × 2 + c = 412

23 12

22 × 8 − × 4 − 4 × 2 + c = 412

12

4 − 2 − 8 + c = 4

f(x) = − − 4x+ 1012x3 1

2x2

p = −4

47a. [5 marks]

Let . Part of the graph of f is shown below.

The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for

(i) ;

(ii) .

Markscheme(i) A1

, A1A1 N2

(ii) A1

A1 N1

[5 marks]

f(x) = 6 + 6 sinx

0 ≤ x < 2π

6 + 6 sinx = 6

6 + 6 sinx = 0

sinx = 0

x = 0 x = π

sinx = −1

x = 3π2

Examiners reportMany candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not showsufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidatesworked in degrees; the majority worked in radians.

47b. [1 mark]Write down the exact value of the x-intercept of f , for .

Markscheme A1 N1

[1 mark]

Examiners reportWhile some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), mostused their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

0 ≤ x < 2π

3π2

47c. [6 marks]The area of the shaded region is k . Find the value of k , giving your answer in terms of .

Markschemeevidence of using anti-differentiation (M1)

e.g.

correct integral (seen anywhere) A1A1

correct substitution (A1)

e.g. ,

A1A1 N3

[6 marks]

Examiners reportMost candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end.Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; withoutthis working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with thecommon trigonometric values.

π

(6 + 6 sinx)dx∫3π2

0

6x− 6 cosx

6 ( ) − 6 cos( ) − (−6 cos0)3π2

3π2

9π− 0 + 6

k = 9π+ 6

47d. [2 marks]Let . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Markschemetranslation of A1A1 N2

[2 marks]

g(x) = 6 + 6 sin (x− )π2

( )π

2

0

Examiners reportWhile there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d).This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by .π

2

47e. [3 marks]Let . The graph of f is transformed to the graph of g.

Given that and , write down the two values of p.

Markschemerecognizing that the area under g is the same as the shaded region in f (M1)

, A1A1 N3

[3 marks]

Examiners reportMost candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majoritytried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in thewording of the question did not appear to have been noticed by candidates in this part either.

g(x) = 6 + 6 sin (x− )π2

g(x)dx = k∫ p+ 3π2

p0 ≤ p < 2π

p = π

2p = 0

48a. [3 marks]

The velocity v ms of an object after t seconds is given by , for .

On the grid below, sketch the graph of v , clearly indicating the maximum point.

−1 v(t) = 15 − 3tt√ 0 ≤ t ≤ 25

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape, A1 for right endpoint at and A1 for maximum point in circle.

[3 marks]

Examiners reportThe graph in part (a) was well done. It was pleasing to see many candidates considering the domain as they sketched their graph.

(25, 0)

48b. [4 marks](i) Write down an expression for d .

(ii) Hence, write down the value of d .

Markscheme(i) recognizing that d is the area under the curve (M1)

e.g.

correct expression in terms of t, with correct limits A2 N3

e.g. ,

(ii) (m) (accept 149 to 3 sf) A1 N1

[4 marks]

Examiners reportPart (b) (i) asked for an expression which bewildered a great many candidates. However, few had difficulty obtaining the correctanswer in (b) (ii).

∫ v(t)

d = (15 − 3t)dt∫ 90 t√ d = vdt∫ 9

0

d = 148.5

49a. [3 marks]

Let .

There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.

(x) = −24 + 9 + 3x+ 1f ′ x3 x2

Markschemevalid approach R1

e.g. , the max and min of gives the points of inflexion on f

(accept ( ) and ( ) A1A1 N1N1

[3 marks]

Examiners reportThere were mixed results in part (a). Students were required to understand the relationships between a function and its derivative andoften obtained the correct solutions with incorrect or missing reasoning.

(x) = 0f ′′ f ′

−0.114, 0.364 −0.114, 0.811 0.364, 2.13)

49b. [2 marks]Let . Explain why the graph of g has no points of inflexion.

MarkschemeMETHOD 1

graph of g is a quadratic function R1 N1

a quadratic function does not have any points of inflexion R1 N1

METHOD 2

graph of g is concave down over entire domain R1 N1

therefore no change in concavity R1 N1

METHOD 3

R1 N1

therefore no points of inflexion as R1 N1

[2 marks]

Examiners reportIn part (b), the question was worth two marks and candidates were required to make two valid points in their explanation. There weremany approaches to take here and candidates often confused their reasoning or just kept writing hoping that somewhere along theway they would say something correct to pick up the points. Many confused and .

g(x) = (x)f ′′

(x) = −144g′′

(x) ≠ 0g′′

f ′ g′

f(x) = x ln(4 − )2

50a. [3 marks]

Let , for . The graph of f is shown below.

The graph of f crosses the x-axis at , and .

Find the value of a and of b .

Markschemeevidence of valid approach (M1)

e.g. , graph

, A1A1 N3

[3 marks]

Examiners reportThis question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. Forexample, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.

f(x) = x ln(4 − )x2 −2 < x < 2

x = a x = 0 x = b

f(x) = 0

a = −1.73 b = 1.73 (a = − , b = )3√ 3√

50b. [2 marks]The graph of f has a maximum value when .

Find the value of c .

Markschemeattempt to find max (M1)

e.g. setting , graph

(accept (1.15, 1.13)) A1 N2

[2 marks]

Examiners reportThis question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. Forexample, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.

x = c

(x) = 0f ′

c = 1.15

50c. [3 marks]The region under the graph of f from to is rotated about the x-axis. Find the volume of the solid formed.x = 0 x = c 360∘

Markschemeattempt to substitute either limits or the function into formula M1

e.g. , ,

A2 N2

[3 marks]

Examiners reportMost candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find thecorrect answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) werealways futile and no marks were gained.

V = π dx∫ c

0 [f(x)]2 π∫ [x ln(4 − )]x2 2π dx∫ 1.149…

0 y2

V = 2.16

50d. [4 marks]Let R be the region enclosed by the curve, the x-axis and the line , between and .

Find the area of R .

Markschemevalid approach recognizing 2 regions (M1)

e.g. finding 2 areas

correct working (A1)

e.g. ,

area (accept 2.06) A2 N3

[4 marks]

Examiners reportMost candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find thecorrect answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) werealways futile and no marks were gained.

x = c x = a x = c

f(x)dx+ f(x)dx∫ −1.73…0 ∫ 1.149…

0 − f(x)dx+ f(x)dx∫ 0−1.73… ∫ 1.149…

0

= 2.07

51a. [5 marks]

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is , where .

Show that the area of the window is given by .

2 m θ 0 < θ < π

2

y = 4 sin θ + 2 sin 2θ

Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g. ,

attempt to replace by M1

e.g.

AG N0

[5 marks]

Examiners reportAs the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) wasgenerally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in aclear and organized manner. Some tried a "working backwards" approach, earning no marks.

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

51b. [4 marks]Zoe wants a window to have an area of . Find the two possible values of .

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

, A1A1 N3

[4 marks]

Examiners reportIn part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and insteadattempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a secondsolution outside the domain.

5 m2 θ

y = 5 4 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856 ( )49.0∘ θ = 1.25 ( )71.4∘

51c. [7 marks]John wants two windows which have the same area A but different values of .

Find all possible values for A .

θ

Markschemerecognition that lower area value occurs at (M1)

finding value of area at (M1)

e.g. , draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

Examiners reportA pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or themaximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident fromcandidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken intoconsideration for future paper writing.

θ = π

2

θ = π

2

4 sin ( ) + 2 sin (2 × )π

2π

2

A = 4

A = 5.19615…

4 < A < 5.20 4 < A < 5.19

52a. [2 marks]

Consider , , where p is a constant.

Find .

Markscheme A1A1 N2

Note: Award A1 for , A1 for .

[2 marks]

Examiners reportCandidates did well on (a).

f(x) = +x2 p

xx ≠ 0

(x)f ′

(x) = 2x−f ′ p

x2

2x − p

x2

52b. [4 marks]There is a minimum value of when . Find the value of .

Markschemeevidence of equating derivative to 0 (seen anywhere) (M1)

evidence of finding (seen anywhere) (M1)

correct equation A1

e.g. ,

A1 N3

[4 marks]

f(x) x = −2 p

(−2)f ′

−4 − = 0p

4−16 − p = 0

p = −16

Examiners reportFor (b), a great number of candidates substituted into the function instead of into the derivative.

The derivate of was calculated without difficulties, but there were numerous problems regarding the derivative of . There wereseveral candidates who considered both p and x as variables; some tried to use the quotient rule and had difficulties, others usednegative exponents and were not successful.

x2 p

x

53a. [7 marks]

Let , for . The graph of f is given below.

The y-intercept is at the point A.

(i) Find the coordinates of A.

(ii) Show that at A.

Markscheme(i) coordinates of A are A1A1 N2

(ii) derivative of (seen anywhere) (A1)

evidence of correct approach (M1)

e.g. quotient rule, chain rule

finding A2

e.g. ,

substituting into (do not accept solving ) M1

at A AG N0

[7 marks]

Examiners reportAlmost all candidates earned the first two marks in part (a) (i), although fewer were able to apply the quotient rule correctly.

f(x) = 3 + 20−4x2

x ≠ ±2

(x) = 0f ′

(0, − 2)

− 4 = 2xx2

(x)f ′

(x) = 20 × (−1) × ( − 4 × (2x)f ′ x2 )−2 ( −4)(0)−(20)(2x)x2

( −4)x2 2

x = 0 (x)f ′ (x) = 0f ′

(x) = 0f ′

40(3 +4)2

53b. [6 marks]The second derivative . Use this to

(i) justify that the graph of f has a local maximum at A;

(ii) explain why the graph of f does not have a point of inflexion.

Markscheme(i) reference to (seen anywhere) (R1)

reference to is negative (seen anywhere) R1

evidence of substituting into M1

finding A1

then the graph must have a local maximum AG

(ii) reference to at point of inflexion (R1)

recognizing that the second derivative is never 0 A1 N2

e.g. , , , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

Examiners reportMany candidates were able to state how the second derivative can be used to identify maximum and inflection points, but fewer wereactually able to demonstrate this with the given function. For example, in (b)(ii) candidates often simply said "the second derivativecannot equal 0" but did not justify or explain why this was true.

(x) =f ′′ 40(3 +4)x2

( −4)x2 3

(x) = 0f ′

(0)f ′′

x = 0 (x)f ′′

(0) =f ′′ 40×4

(−4)3(= − )5

2

(x) = 0f ′′

40(3 + 4) ≠ 0x2 3 + 4 ≠ 0x2 ≠ −x2 43

53c. [1 mark]Describe the behaviour of the graph of for large .

Markschemecorrect (informal) statement, including reference to approaching A1 N1

e.g. getting closer to the line , horizontal asymptote at

[1 mark]

Examiners reportNot too many candidates could do part (c) correctly.

f |x|

y = 3

y = 3 y = 3

54a. [4 marks]

Let . Line L is the normal to the graph of f at the point (4, 2) .

Show that the equation of L is .

f(x) = x√

y = −4x+ 18

Markschemefinding derivative (A1)

e.g.

correct value of derivative or its negative reciprocal (seen anywhere) A1

e.g. ,

gradient of normal = (seen anywhere) A1

e.g. ,

substituting into equation of line (for normal) M1

e.g.

AG N0

[4 marks]

Examiners reportParts (a) and (b) were well done by most candidates.

(x) = ,f ′ 12x

12

12 x√

12 4√

14

1gradient of tangent

− = −41(4)f′

−2 x√

y− 2 = −4(x− 4)

y = −4x+ 18

54b. [3 marks]

In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .

Find an expression for the area of R .

Markschemesplitting into two appropriate parts (areas and/or integrals) (M1)

correct expression for area of R A2 N3

e.g. area of R = , (triangle)

Note: Award A1 if dx is missing.

[3 marks]

Examiners reportWhile quite a few candidates understood that both functions must be used to find the area in part (c), very few were actually able towrite a correct expression for this area and this was due to candidates not knowing that they needed to integrate from to and thenfrom to .

dx+ (−4x+ 18)dx∫ 40 x√ ∫ 4.5

4 dx+ × 0.5 × 2∫ 40 x√ 1

2

0 44 4.5

54c. [8 marks]The region R is rotated about the x-axis. Find the volume of the solid formed, giving your answer in terms of .

Markschemecorrect expression for the volume from to (A1)

e.g. , ,

A1

(A1)

A1

finding the volume from to

EITHER

recognizing a cone (M1)

e.g.

(A1)

A1

total volume is A1 N4

OR

(M1)

A1

A1

total volume is A1 N4

[8 marks]

Examiners reportOn part (d), some candidates were able to earn follow through marks by setting up a volume expression, but most of theseexpressions were incorrect. If they did not get the expression for the area correct, there was little chance for them to get part (d)correct.

For those candidates who used their expression in part (c) for (d), there was a surprising amount of them who incorrectly applieddistributive law of the exponent with respect to the addition or subtraction.

360∘ π

x = 0 x = 4

V = π[f ]dx∫ 40 (x)2 dx∫ 4

0 π x√ 2 πxdx∫ 40

V = [ π ]12x2

4

0

V = π( × 16 − × 0)12

12

V = 8π

x = 4 x = 4.5

V = π h13r2

V = π(2 ×13

)2 12

= 2π3

8π+ π23

(= π)263

V = π dx∫ 4.54 (−4x+ 18)2

= π(16 − 144x+ 324)dx∫ 4.54 x2

= π[ − 72 + 324x]163x3 x2

4.5

4

= 2π3

8π+ π23

(= π)263

55a. [2 marks]

Let and .

Find .

f(x) = cos2x g(x) = ln(3x− 5)

(x)f ′

(x) = −sin 2x× 2(= −2 sin 2x)f ′

−sin 2x

Examiners reportAlmost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of thechain rule, forgetting to account for the coefficient of in their derivatives. A few did not know how to use the product rule, eventhough it is in the information booklet.

x

55b. [2 marks]Find .

Markscheme A1A1 N2

Note: Award A1 for 3, A1 for .

[2 marks]

Examiners reportAlmost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of thechain rule, forgetting to account for the coefficient of in their derivatives. A few did not know how to use the product rule, eventhough it is in the information booklet.

(x)g′

(x) = 3 ×g′ 13x−5

(= )33x−5

13x−5

x

55c. [2 marks]Let . Find .

Markschemeevidence of using product rule (M1)

A1 N2

[2 marks]

Examiners reportAlmost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of thechain rule, forgetting to account for the coefficient of x in their derivatives. A few did not know how to use the product rule, eventhough it is in the information booklet.

h(x) = f(x) × g(x) (x)h′

(x) = (cos2x)( ) + ln(3x− 5)(−2 sin 2x)h′ 33x−5

56. [7 marks]Consider the curve with equation , where p and q are constants. The point lies on the curve. Thetangent to the curve at A has gradient . Find the value of p and of q .

Markschemesubstituting , into (M1)

A1

finding derivative (M1)

A1

correct substitution, A1

, A1A1 N2N2

[7 marks]

f(x) = p + qxx2 A(1, 3)8

x = 1 y = 3 f(x)

3 = p+ q

(x) = 2px+ qf ′

2p+ q = 8

p = 5 q = −2

Examiners reportA good number of candidates were able to obtain an equation by substituting the point into the function’s equation. Not as manyknew how to find the other equation by using the derivative. Some candidates thought they needed to find the equation of the tangentline rather than recognising that the information about the tangent provided the gradient of the function at the point. While they wereusually able to find this equation correctly, it was irrelevant to the question asked.

1, 3

57. [7 marks]

A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m , as shown below.

The fencing used for side AB costs per metre. The fencing for the other three sides costs per metre. The farmer createsan enclosure so that the cost is a minimum. Find this minimum cost.

2

$11 $3

MarkschemeMETHOD 1

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using $3 for three sides and $11 for one side (M1)

e.g.

correct expression for cost A2

e.g. , ,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,

minimum cost is 420 (dollars) A1 N4

OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.

minimum cost is 420 (dollars) A1 N4

METHOD 2

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using for three sides and for one side (M1)

e.g.

correct expression for cost A2

e.g. , ,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,

minimum cost is 420 (dollars) A1 N4

OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.

minimum cost is 420 (dollars) A1 N4

[7 marks]

AB = x AD = 525x

3(AD + BC + CD) + 11AB

× 3 + × 3 + 11x+ 3x525x

525x

× 3 + × 3 + 11AB + 3AB525AB

525AB

+ 14x3150x

x = 15

(x) = + + 14C′ −1575x2

−1575x2

+ 14 = 0−3150x2

AD = x AB = 525x

$3 $11

3(AD + BC + CD) + 11AB

3 (x+x+ ) + × 11525x

525x

3 (AD + AD + ) + × 11525AD

525AD

6x+ 7350x

x = 35

(x) = 6 −C′ 7350x2

6 − = 07350x2

Examiners reportAlthough this question was a rather straight-forward optimisation question, the lack of structure caused many candidates difficulty.Some were able to calculate cost values but were unable to create an algebraic cost function. Those who were able to create a costfunction in two variables often could not use the area relationship to obtain a function in a single variable and so could make nofurther progress. Of those few who created a correct cost function, most set the derivative to zero to find that the minimum costoccurred at , leading to . Although this is a correct approach earning full marks, candidates seem not to recognise that theresult can be obtained from the GDC, without the use of calculus.

x = 15 $420

58a. [3 marks]

Let , for .

Find .

Markschemeevidence of choosing the product rule (M1)

e.g.

A1A1 N3

[3 marks]

Examiners reportThis problem was well done by most candidates. There were some candidates that struggled to apply the product rule in part (a) andoften wrote nonsense like .

f(x) = xcosx 0 ≤ x ≤ 6

(x)f ′

x× (−sinx) + 1 × cosx

(x) = cosx−x sinxf ′

−x sinx = −sinx2

58b. [4 marks]On the grid below, sketch the graph of .y = (x)f ′

Markscheme

A1A1A1A1 N4

Note: Award A1 for correct domain, with endpoints in circles, A1 for approximately correct shape, A1 for local minimumin circle, A1 for local maximum in circle.

[4 marks]

Examiners reportIn part (b), few candidates were able to sketch the function within the required domain and a large number of candidates did not havetheir calculator in the correct mode.

0 ≤ x ≤ 6

59. [7 marks]The acceleration, , of a particle at time t seconds is given by

The particle is at rest when .

Find the velocity of the particle when .

Markschemeevidence of integrating the acceleration function (M1)

e.g.

correct expression A1A1

evidence of substituting (1, 0) (M1)

e.g.

(A1)

(A1)

(accept the exact answer ) A1 N3

[7 marks]

a ms−2

a = + 3 sin 2t, for t ≥ 1.1t

t = 1

t = 5

∫ ( + 3 sin 2t)dt1t

ln t− cos2t+ c32

0 = ln1 − cos2 + c32

c = −0.624 (= cos2 − ln1 or cos2)32

32

v = ln t− cos2t− 0.62432

(= ln t− cos2t+ cos2 or lnt− cos2t+ cos2 − ln1)32

32

32

32

v(5) = 2.24 ln5 − 1.5cos10 + 1.5cos2

Examiners reportThis problem was not well done. A large number of students failed to recognize that they needed to integrate the accelerationfunction. Even among those who integrated the function, there were many who integrated incorrectly. A great number of candidateswere not able to handle the given initial condition to find the integration constant but incorrectly substituted directly into theirexpression.

t = 5

60a. [2 marks]

Let . Part of the graph of f is shown below.

The y-intercept is at (0, 13) .

Show that .

Markschemesubstituting (0, 13) into function M1

e.g.

A1

AG N0

[2 marks]

Examiners reportThis question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by mostcentres. Parts (a) and (b) proved very accessible to many candidates.

f(x) = A + 3ekx

A = 10

13 = A + 3e0

13 = A+ 3

A = 10

60b. [3 marks]Given that (correct to 3 significant figures), find the value of k.

Markschemesubstituting into A1

e.g. ,

evidence of solving equation (M1)

e.g. sketch, using

(accept ) A1 N2

[3 marks]

f(15) = 3.49

f(15) = 3.49

3.49 = 10 + 3e15k 0.049 = e15k

ln

k = −0.201 ln 0.04915

Examiners reportThis question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by mostcentres. Parts (a) and (b) proved very accessible to many candidates.

60c. [5 marks](i) Using your value of k , find .

(ii) Hence, explain why f is a decreasing function.

(iii) Write down the equation of the horizontal asymptote of the graph f .

Markscheme(i)

A1A1A1 N3

Note: Award A1 for , A1 for , A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative R1 N1

e.g. , derivative always negative

(iii) A1 N1

[5 marks]

Examiners reportThe chain rule in part (c) was also carried out well. Few however, recognized the command term “hence” and that guarantees a decreasing function. A common answer for the equation of the asymptote was to give or .

(x)f ′

f(x) = 10 + 3e−0.201x

f(x) = 10 × −0.201e−0.201x (= −2.01 )e−0.201x

10e−0.201x × − 0.201

(x) < 0f ′

y = 3

(x) < 0f ′

y = 0 x = 3

60d. [6 marks]Let .

Find the area enclosed by the graphs of f and g .

Markschemefinding limits , (seen anywhere) A1A1

evidence of integrating and subtracting functions (M1)

correct expression A1

e.g. ,

area A2 N4

[6 marks]

Examiners reportIn part (d), it was again surprising and somewhat disappointing to see how few candidates were able to use their GDC effectively tofind the area between curves, often not finding correct limits, and often trying to evaluate the definite integral without the GDC,which led nowhere.

g(x) = − + 12x− 24x2

3.8953… 8.6940…

g(x) − f(x)dx∫ 8.693.90 [(− + 12x− 24) − (10 + 3)]dx∫ 8.69

3.90 x2 e−0.201x

= 19.5

f(x) = sin 2x+ 10x

61a. [1 mark]

Let , for . Part of the graph of f is given below.

There is an x-intercept at the point A, a local maximum point at M, where and a local minimum point at N, where .

Write down the x-coordinate of A.

Markscheme A1 N1

[1 mark]

Examiners reportParts (a) and (b) were generally well answered, the main problem being the accuracy.

f(x) = sin 2x+ 10ex 0 ≤ x ≤ 4

x = p x = q

2.31

61b. [2 marks]Find the value of

(i) p ;

(ii) q .

Markscheme(i) 1.02 A1 N1

(ii) 2.59 A1 N1

[2 marks]

Examiners reportParts (a) and (b) were generally well answered, the main problem being the accuracy.

61c. [3 marks]Find . Explain why this is not the area of the shaded region.

Markscheme A1 N1

split into two regions, make the area below the x-axis positive R1R1 N2

[3 marks]

f(x)dx∫ q

p

f(x)dx = 9.96∫ q

p

Examiners reportMany students lacked the calculator skills to successfully complete (6)(c) in that they could not find the value of the definite integral.Some tried to find it by hand. When trying to explain why the integral was not the area, most knew the region under the x-axis wasthe cause of the integral not giving the total area, but the explanations were not sufficiently clear. It was often stated that the areabelow the axis was negative rather than the integral was negative.

62a. [5 marks]

Consider , for . The graph of f is given below.

Let P and Q be points on the curve of f where the tangent to the graph of f is parallel to the x-axis.

(i) Find the x-coordinate of P and of Q.

(ii) Consider . Write down all values of k for which there are exactly two solutions.

Markscheme(i) A1A1 N2

(ii) recognizing that it occurs at P and Q (M1)

e.g. ,

, A1A1 N3

[5 marks]

Examiners reportMany candidates correctly found the x-coordinates of P and Q in (a)(i) with their GDC. In (a)(ii) some candidates incorrectlyinterpreted the words “exactly two solutions” as an indication that the discriminant of a quadratic was required. Many failed to realisethat the values of k they were looking for in this question were the y-coordinates of the points found in (a)(i).

f(x) = x ln(4 − )x2 −2 < x < 2

f(x) = k

−1.15, 1.15

x = −1.15 x = 1.15

k = −1.13 k = 1.13

62b. [4 marks]Let , for .

Show that .

g(x) = ln(4 − )x3 x2 −2 < x < 2

(x) = + 3 ln(4 − )g′ −2x4

4−x2x2 x2

Markschemeevidence of choosing the product rule (M1)

e.g.

derivative of is (A1)

derivative of is (A1)

correct substitution A1

e.g.

AG N0

[4 marks]

Examiners reportMany candidates were unclear in their application of the product formula in the verifying the given derivative of g. Showing that thederivative was the given expression often received full marks though it was not easy to tell in some cases if that demonstration camethrough understanding of the product and chain rules or from reasoning backwards from the given result.

u + vv′ u′

x3 3x2

ln(4 − )x2 −2x4−x2

× + ln(4 − ) × 3x3 −2x4−x2

x2 x2

(x) = + 3 ln(4 − )g′ −2x4

4−x2x2 x2

62c. [2 marks]Let , for .

Sketch the graph of .

Markscheme

A1A1 N2

[2 marks]

Examiners reportSome candidates drew their graphs of the derivative in (c) on their examination papers despite clear instructions to do their work onseparate sheets. Most who tried to plot the graph in (c) did so successfully.

g(x) = ln(4 − )x3 x2 −2 < x < 2

g′

62d. [3 marks]Let , for .

Consider . Write down all values of w for which there are exactly two solutions.

Markscheme , A1A2 N2

[3 marks]

g(x) = ln(4 − )x3 x2 −2 < x < 2

(x) = wg′

w = 2.69 w < 0

Examiners reportCorrect solutions to 10(d) were not often seen.

63a. [3 marks]

The diagram shows part of the graph of . The x-intercepts are at points A and C. There is a minimum at B, and a maximum at D.

(i) Write down the value of at C.

(ii) Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.

Markscheme(i) A1 N1

(ii) METHOD 1

to the left of C, to the right of C R1R1 N2

METHOD 2

R2 N2

[3 marks]

Examiners reportThe variation in successful and unsuccessful responses to this question was remarkable. Many candidates did not even attempt it.Candidates could often determine from the graph, the minimum and maximum values of the original function, but few could correctlyuse the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking atthe graph of and not the graph of .

y = (x)f ′

(x)f ′

(x) = 0f ′

(x) < 0f ′ (x) > 0f ′

(x) > 0f ′′

f ′ f

63b. [1 mark]Which of the points A, B, D corresponds to a maximum on the graph of f ?

MarkschemeA A1 N1

[1 mark]

Examiners reportThe variation in successful and unsuccessful responses to this question was remarkable. Many candidates did not even attempt it.Candidates could often determine from the graph, the minimum and maximum values of the original function, but few could correctlyuse the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking atthe graph of and not the graph of .f ′ f

63c. [3 marks]Show that B corresponds to a point of inflexion on the graph of f .

MarkschemeMETHOD 1

R2

discussion of sign change of R1

e.g. to the left of B and to the right of B; changes sign either side of B

B is a point of inflexion AG N0

METHOD 2

B is a minimum on the graph of the derivative R2

discussion of sign change of R1

e.g. to the left of B and to the right of B; changes sign either side of B

B is a point of inflexion AG N0

[3 marks]

Examiners reportIn part (c), many candidates once more failed to respect the command term "show" and often provided an incomplete answer.Candidates should be encouraged to refer to the number of marks available for a particular part when deciding how much informationshould be given.

(x) = 0f ′′

(x)f ′′

(x) < 0f ′′ (x) > 0f ′′ (x)f ′′

f ′

(x)f ′′

(x) < 0f ′′ (x) > 0f ′′ (x)f ′′

64a. [2 marks]

The acceleration, , of a particle at time t seconds is given by .

Find the acceleration of the particle at .

Markschemesubstituting (M1)

e.g.

A1 N2

[2 marks]

Examiners reportParts (a) and (b) of this question were generally well done.

a ms−2 a = 2t+ cost

t = 0

t = 0

a(0) = 0 + cos0

a(0) = 1

64b. [5 marks]Find the velocity, v, at time t, given that the initial velocity of the particle is .ms−1

Markschemeevidence of integrating the acceleration function (M1)

e.g.

correct expression A1A1

Note: If " " is omitted, award no further marks.

evidence of substituting (2,0) into indefinite integral (M1)

e.g. ,

A1 N3

[5 marks]

Examiners reportParts (a) and (b) of this question were generally well done.

∫ (2t+ cost)dt

+ sin t+ ct2

+c

2 = 0 + sin 0 + c c = 2

v(t) = + sin t+ 2t2

64c. [7 marks]Find , giving your answer in the form .

Markscheme A1A1A1

Note: Award A1 for each correct term.

evidence of using (M1)

correct substitution A1

e.g. ,

(accept , ) A1A1 N3

[7 marks]

Examiners reportProblems arose in part (c) with many candidates not substituting correctly, leading to only a partially correct final answer.There were also a notable few who were not aware that in both parts (a) and (c).

vdt∫ 30 p− qcos3

∫ ( + sin t+ 2)dt = − cost+ 2tt2 t3

3

v(3) − v(0)

(9 − cos3 + 6) − (0 − cos0 + 0) (15 − cos3) − (−1)

16 − cos3 p = 16 q = −1

s(3) − s(0)cos0 = 1

64d. [2 marks]What information does the answer to part (c) give about the motion of the particle?

Markschemereference to motion, reference to first 3 seconds R1R1 N2

e.g. displacement in 3 seconds, distance travelled in 3 seconds

[2 marks]

Examiners reportThere were a variety of interesting answers about the motion of the particle, few being able to give both parts of the answer correctly.

f(x) = 5 cos x g(x) = −0.5 + 5x− 82

65. [5 marks]

Let and for .

Let R be the region enclosed by the graphs of f and g . Find the area of R.

MarkschemeMETHOD 1

intersect when and (may be seen as limits of integration) A1A1

evidence of approach (M1)

e.g. , ,

A2 N3

METHOD 2

intersect when and (seen anywhere) A1A1

evidence of approach using a sketch of g and f , or . (M1)

e.g. area = ,

A2 N3

[5 marks]

Examiners reportPart (d) proved elusive to many candidates. Some used creative approaches that split the area into parts above and below the x-axis;while this leads to a correct result, few were able to achieve it. Many candidates were unable to use their GDCs effectively to findpoints of intersection and the subsequent area.

f(x) = 5 cos xπ4

g(x) = −0.5 + 5x− 8x2 0 ≤ x ≤ 9

x = 2 x = 6.79

∫ g − f ∫ f(x)dx− ∫ g(x)dx ((−0.5 + 5x− 8) − (5 cos x))∫ 6.792 x2 π

4

area = 27.6

x = 2 x = 6.79

g − f

A+B−C 12.7324 + 16.0938 − 1.18129…

area = 27.6

66. [6 marks]Let . Find the gradient of the normal to the curve of f at .f(x) = cosxex x = π

Markschemeevidence of choosing the product rule (M1)

A1A1

substituting (M1)

e.g. , ,

taking negative reciprocal (M1)

e.g.

gradient is A1 N3

[6 marks]

Examiners reportCandidates familiar with the product rule easily found the correct derivative function. Many substituted to find the tangent gradient,but surprisingly few candidates correctly considered that the gradient of the normal is the negative reciprocal of this answer.

(x) = × (−sinx) + cosx×f ′ ex ex (= cosx− sinx)ex ex

π

(π) = cosπ− sinπf ′ eπ eπ (−1 − 0)eπ −eπ

− 1(π)f′

1eπ

π

67a. [4 marks]

The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object as functions of time, t.

Complete the following table by noting which graph A, B or C corresponds to each function.

Markscheme

A2A2 N4

[4 marks]

Examiners reportMany candidates answered this question completely and correctly, showing a good understanding of the graphical relationshipbetween displacement, velocity and acceleration. When done incorrectly, many answered with the displacement as graph B andacceleration as graph C.

67b. [2 marks]Write down the value of t when the velocity is greatest.

Markscheme A2 N2

[2 marks]

Examiners reportMany candidates found the value of t which gave a maximum in the remaining graph, and were awarded follow through marks.

t = 3

68. [7 marks]The graph of between and is rotated about the x-axis. The volume of the solid formed is . Findthe value of a.

Markschemeattempt to substitute into formula (M1)

integral expression A1

e.g. ,

correct integration (A1)

e.g.

correct substitution (A1)

equating their expression to M1

e.g.

A2 N2

[7 marks]

Examiners reportDespite the “reverse” nature of this question, many candidates performed well with the integration. Some forgot to square thefunction, while others did not discard the negative value of a. Some attempted to equate to the formula for volume of a sphere,which suggests this topic was not fully covered in some centres.

y = x√ x = 0 x = a 360∘ 32π

V = ∫ π dxy2

π ( dx∫ a

0 x√ )2 π∫ x

∫ xdx = 12x2

V = π[ ]12a2

32π

π[ ] = 32π12a2

= 64a2

a = 8

32π

69. [8 marks]

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is radians, where .

(i) Find .

(ii) Hence, find the exact value of which maximizes the area of the rectangle.

(iii) Use the second derivative to justify that this value of does give a maximum.

Markscheme(i) A2 N2

(ii) for setting derivative equal to 0 (M1)

e.g. ,

(A1)

A1 N2

(iii) valid reason (seen anywhere) R1

e.g. at , ; maximum when

finding second derivative A1

evidence of substituting M1

e.g. , ,

produces the maximum area AG N0

[8 marks]

Examiners reportAs the area function was given in part (b), many candidates correctly found the derivative in (c) and knew to set this derivative to zerofor a maximum value. Many gave answers in degrees, however, despite the given domain in radians.

Although some candidates found the second derivative function correctly, few stated that the second derivative must be negative at amaximum value. Simply calculating a negative value is not sufficient for a justification.

θ 0 ≤ θ ≤ π

2

dAdθ

θ

θ

= 36 cos2θdAdθ

36 cos2θ = 0 = 0dAdθ

2θ = π

2

θ = π

4

π

4< 0Ad2

dθ2(x) < 0f ′′

= −72 sin 2θAd2

dθ2

π

4

−72 sin (2 × )π4

−72 sin ( )π2

−72

θ = π

4

f(x) = ax

70a. [7 marks]

Let , , .The graph of f is shown below.

The region between and is shaded.

Given that , find the coordinates of all points of inflexion.

Markschemeevidence of appropriate approach (M1)

e.g.

to set the numerator equal to 0 (A1)

e.g. ;

(0, 0) , , (accept , etc) A1A1A1A1A1 N5

[7 marks]

Examiners reportAlthough many recognized the requirement to set the second derivative to zero in (b), a majority neglected to give their answers asordered pairs, only writing the x-coordinates. Some did not consider the negative root.

f(x) = ax

+1x2−8 ≤ x ≤ 8 a ∈ R

x = 3 x = 7

(x) =f ′′ 2ax( −3)x2

( +1)x2 3

(x) = 0f ′′

2ax( − 3) = 0x2 ( − 3) = 0x2

( , )3√ a 3√4

(− ,− )3√ a 3√4

x = 0 y = 0

70b. [7 marks]It is given that .

(i) Find the area of the shaded region, giving your answer in the form .

(ii) Find the value of .

∫ f(x)dx = ln( + 1) +Ca

2x2

p lnq

2f(x− 1)dx∫ 84

Markscheme(i) correct expression A2

e.g. , ,

area = A1A1 N2

(ii) METHOD 1

recognizing the shift that does not change the area (M1)

e.g. ,

recognizing that the factor of 2 doubles the area (M1)

e.g.

(i.e. their answer to (c)(i)) A1 N3

METHOD 2

changing variable

let , so

(M1)

substituting correct limits

e.g. , , (M1)

A1 N3

[7 marks]

Examiners reportFor those who found a correct expression in (c)(i), many finished by calculating . Few recognized that thetranslation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).

[ ln( + 1)]a

2x2

7

3ln50 − ln10a

2a

2(ln50 − ln10)a

2

ln5a

2

f(x− 1)dx = f(x)dx∫ 84 ∫ 7

3 ln5a

2

2f(x− 1)dx =2 f(x− 1)dx∫ 84 ∫ 8

4 (= 2 f(x)dx)∫ 73

2f(x− 1)dx = a ln5∫ 84 2 ×

w = x− 1 = 1dwdx

2 ∫ f(w)dw = ln( + 1) + c2a2

w2

[a ln[ + 1]](x− 1)28

4[a ln( + 1)]w2 7

3a ln50 − a ln10

2f(x− 1)dx = a ln5∫ 84

ln50 − ln10 = ln40

71a. [4 marks]

Let .

Use the formula to show that the derivative of is .

Markschemeevidence of substituting (M1)

correct substitution A1

e.g.

simplifying A1

e.g.

factoring out h A1

e.g.

AG N0

[4 marks]

f(x) = − 4x+ 1x3

(x) =f ′ limh→0f(x+h)−f(x)

hf(x) 3 − 4x2

x+h

(x) =f ′ limh→0−4(x+h)+1−( −4x+1)(x+h)3 x3

h

( +3 h+3x + −4x−4h+1− +4x−1)x3 x2 h2 h3 x3

h

h(3 +3xh+ −4)x2 h2

h

(x) = 3 − 4f ′ x2

Examiners reportIn part (b), it was clear that many candidates had difficulty with differentiation from first principles. Those that successfully set theanswer up, often got lost in the simplification.

71b. [4 marks]The tangent to the curve of f at the point is parallel to the tangent at a point Q. Find the coordinates of Q.

Markscheme (A1)

setting up an appropriate equation M1

e.g.

at Q, (Q is ) A1 A1

[4 marks]

Examiners reportPart (c) was poorly done with many candidates assuming that the tangents were horizontal and then incorrectly estimating themaximum of f as the required point. Many candidates unnecessarily found the equation of the tangent and could not make any furtherprogress.

P(1, − 2)

(1) = −1f ′

3 − 4 = −1x2

x = −1,y = 4 (−1, 4)

72a. [2 marks]

A function f has its first derivative given by .

Find the second derivative.

MarkschemeMETHOD 1

A2 N2

METHOD 2

attempt to expand (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMany candidates completed parts (a) and (b) successfully.

(x) = (x− 3f ′ )3

(x) = 3(x− 3f ′′ )2

(x− 3)3

(x) = − 9 + 27x− 27f ′ x3 x2

(x) = 3 − 18x+ 27f ′′ x2

72b. [1 mark]Find and .

Markscheme , A1 N1

[1 mark]

Examiners reportMany candidates completed parts (a) and (b) successfully.

(3)f ′ (3)f ′′

(3) = 0f ′ (3) = 0f ′′

72c. [2 marks]The point P on the graph of f has x-coordinate . Explain why P is not a point of inflexion.

MarkschemeMETHOD 1

does not change sign at P R1

evidence for this R1 N0

METHOD 2

changes sign at P so P is a maximum/minimum (i.e. not inflexion) R1

evidence for this R1 N0

METHOD 3

finding and sketching this function R1

indicating minimum at R1 N0

[2 marks]

Examiners reportA rare few earned any marks in part (c) - most justifying the point of inflexion with the zero answers in part (b), not thinking thatthere is more to consider.

3

f ′′

f ′

f(x) = (x− 3 + c14

)4

x = 3

73a. [2 marks]

Let and .

Write down

(i) ;

(ii) .

Markscheme(i) A1 N1

(ii) A1 N1

[4 marks]

Examiners reportA good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixedsuccess.

f(x) = e−3x g(x) = sin (x− )π3

(x)f ′

(x)g′

−3e−3x

cos(x− )π3

73b. [4 marks]

Let . Find the exact value of .h(x) = sin (x− )e−3x π

3( )h′ π

3

Markschemeevidence of choosing product rule (M1)

e.g.

correct expression A1

e.g.

complete correct substitution of (A1)

e.g. � � � � � � � � �

A1 N3

[4 marks]

Examiners reportOften the substitution of was incomplete or not done at all.

u + vv′ u′

−3 sin (x− ) + cos(x− )e−3x π

3e−3x π

3

x = π

3

−3 sin ( − ) + cos( − )e−3 π

3π

3π

3e−3 π

3π

3π

3

( ) =h′ π

3e−π

π

3

74a. [6 marks]

In this question s represents displacement in metres and t represents time in seconds.

The velocity v m s of a moving body is given by where a is a non-zero constant.

(i) If when , find an expression for s in terms of a and t.

(ii) If when , write down an expression for s in terms of a and t.

MarkschemeNote: In this question, do not penalize absence of units.

(i) (M1)

(A1)(A1)

substituting when ( ) (M1)

A1 N5

(ii) A1 N1

[6 marks]

Examiners reportPart (a) proved accessible for most.

–1 v = 40 − at

s = 100 t = 0

s = 0 t = 0

s = ∫ (40 − at)dt

s = 40t− a + c12t2

s = 100 t = 0 c = 100

s = 40t− a + 10012t2

s = 40t− a12t2

74b. [6 marks]

Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by , where at P. The station is 500 m from P.

A train M slows down so that it comes to a stop at the station.

(i) Find the time it takes train M to come to a stop, giving your answer in terms of a.

(ii) Hence show that .

v = 40 − at

t = 0

a = 85

Markscheme(i) stops at station, so (M1)

(seconds) A1 N2

(ii) evidence of choosing formula for s from (a) (ii) (M1)

substituting (M1)

e.g.

setting up equation M1

e.g. , ,

evidence of simplification to an expression which obviously leads to A1

e.g. , ,

AG N0

[6 marks]

Examiners reportPart (b), simple as it is, proved elusive as many candidates did not make the connection that when the train stops. Instead, manyattempted to find the value of t using .

v = 0

t = 40a

t = 40a

40 × − a×40a

12

402

a2

500 = s 500 = 40 × − a×40a

12

402

a2500 = −1600

a

800a

a = 85

500a = 800 5 = 8a

1000a = 3200 − 1600

a = 85

v = 0a = 8

5

74c. [5 marks]For a different train N, the value of a is 4.

Show that this train will stop before it reaches the station.

MarkschemeMETHOD 1

, stops when

(A1)

A1

substituting into expression for s M1

A1

since (allow FT on their s, if ) R1

train stops before the station AG N0

METHOD 2

from (b) A2

substituting into expression for s

e.g. M1

A1

since R1

train stops before the station AG N0

METHOD 3

a is deceleration A2

A1

so stops in shorter time (A1)

so less distance travelled R1

so stops before station AG N0

[5 marks]

Examiners reportFew were successful in part (c).

v = 40 − 4t v = 0

40 − 4t = 0

t = 10

s = 40 × 10 − × 4 ×12

102

s = 200

200 < 500 s < 500

t = = 10404

s = 40 × 10 − × 4 ×12

102

s = 200

200 < 500

4 > 85

75a. [2 marks]

Consider the curve . Let P be the point on the curve where .

Write down the gradient of the curve at P.

Markschemegradient is A2 N2

[2 marks]

Examiners reportAlthough the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding thederivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earnedfollow through marks in part (b).

y = ln(3x− 1) x = 2

0.6

75b. [5 marks]The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.

Markschemeat R, (seen anywhere) A1

at , (A1)

gradient of normal (A1)

evidence of finding correct equation of normal A1

e.g. ,

(accept 2.96) A1

coordinates of R are (2.97,0) N3

[5 marks]

Examiners reportAlthough the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding thederivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earnedfollow through marks in part (b).

y = 0

x = 2 y = ln5 (= 1.609…)

= −1.6666…

y = ln5 = − (x− 2)53

y = −1.67x+ c

x = 2.97

76a. [3 marks]

Let , for . The following diagram shows the graph of f .

Let R be the region enclosed by the x-axis and the curve of f .

Find the area of R.

Markschemefinding the limits , (A1)

integral expression A1

e.g.

area = 52.1 A1 N2

[3 marks]

Examiners reportMany candidates set up a completely correct equation for the area enclosed by the x-axis and the curve. Also, many of them tried ananalytic approach which sometimes returned incorrect answers. Using the wrong limits and was a common error.

f(x) = x(x− 5)2 0 ≤ x ≤ 6

x = 0 x = 5

f(x)dx∫ 50

0 6

76b. [4 marks]Find the volume of the solid formed when R is rotated through about the x-axis.360∘

Markschemeevidence of using formula (M1)

correct expression A1

e.g. volume

volume = 2340 A2 N2

[4 marks]

Examiners reportThe formula for the volume of revolution given in the data booklet was seen many times in part (b). Some candidates wrote theintegrand incorrectly, either missing the or not squaring. A good number of students could write a completely correct integralexpression for the volume of revolution but fewer could evaluate it correctly as many started an analytical approach instead of usingtheir GDC.

Many candidates did not use a GDC at all in this question. Pages of calculations were produced in an effort to find the area and thevolume of revolution. This probably caused a shortage of time for later questions.

v = ∫ π dxy2

= π dx∫ 50 x

2(x− 5)4

π

77. [2 marks]

Let , for .

Write down one value of x such that .

Markschemeevidence of correct approach (M1)

e.g. max/min, sketch of indicating roots

one 3 s.f. value which rounds to one of , , , A1 N2

[2 marks]

Examiners reportThe most common approach in part (d) was to differentiate and set . Fewer students found the values of x given by themaximum or minimum values on their graphs.

f(x) = 3 sinx+ 4 cosx −2π ≤ x ≤ 2π

(x) = 0f ′

(x)f ′

−5.6 −2.5 0.64 3.8

(x) = 0f ′

78. [4 marks]

Let , .

The graph of f is revolved about the x-axis from to . Find the volume of the solid formed.

f(x) = xcos(x− sinx) 0 ≤ x ≤ 3

360∘ x = 0 x = a

Markschemeevidence of using (M1)

fully correct integral expression A2

e.g. , A1 N2

[4 marks]

Examiners reportA good number of candidates could set up the correct integral expression for volume, but surprisingly few were able to use theirGDC to find the correct value. Some attempted to analytically integrate the square of this unusual function, expending valuable timein this effort. A small but significant number of candidates wrote a final answer as , which accrued the accuracy penalty.

V = π∫ dx[f(x)]2

V = π dx∫ 2.310 [xcos(x− sinx)]2 V = π dx∫ 2.31

0 [f(x)]2

V = 5.90

1.88π

79a. [3 marks]

Let , .

Show that .

Markschemecorrectly finding the derivative of , i.e. A1

correctly finding the derivative of , i.e. A1

evidence of using the product rule, seen anywhere M1

e.g.

AG N0

[3 marks]

Examiners reportA good number of candidates demonstrated the ability to apply the product and chain rules to obtain the given derivative.

f(x) = cosxe2x −1 ≤ x ≤ 2

(x) = (2 cosx− sinx)f ′ e2x

e2x 2e2x

cosx −sinx

(x) = 2 cosx− sinxf ′ e2x e2x

(x) = 2 (2 cosx− sinx)f ′ e2x

79b. [5 marks]Let the line L be the normal to the curve of f at .

Find the equation of L .

Markschemeevidence of finding , seen anywhere A1

attempt to find the gradient of f (M1)

e.g. substituting into

value of the gradient of f A1

e.g. , equation of tangent is

gradient of normal (A1)

A1 N3

[5 marks]

x = 0

f(0) = 1

x = 0 (x)f ′

(0) = 2f ′ y = 2x+ 1

= − 12

y− 1 = − x(y = − x+ 1)12

12

Examiners reportWhere candidates recognized that the gradient of the tangent is the derivative, many went on to correctly find the equation of thenormal.

79c. [6 marks]The graph of f and the line L intersect at the point (0, 1) and at a second point P.

(i) Find the x-coordinate of P.

(ii) Find the area of the region enclosed by the graph of f and the line L .

Markscheme(i) evidence of equating correct functions M1

e.g. , sketch showing intersection of graphs

A1 N1

(ii) evidence of approach involving subtraction of integrals/areas (M1)

e.g. ,

fully correct integral expression A2

e.g. ,

A1 N2

[6 marks]

Examiners reportFew candidates showed the setup of the equation in part (c) before writing their answer from the GDC. Although a good number ofcandidates correctly expressed the integral to find the area between the curves, surprisingly few found a correct answer. Although thisis a GDC paper, some candidates attempted to integrate this function analytically.

cosx = − x+ 1e2x 12

x = 1.56

∫ [f(x) − g(x)]dx ∫ f(x)dx− area under trapezium

[ cosx− (− x+ 1)]dx∫ 1.560 e2x 1

2cosxdx− 0.951…∫ 1.56

0 e2x

area = 3.28

80a. [2 marks]Find .

Markscheme (accept ) A1A1 N2

[2 marks]

Examiners reportMany candidates were unable to correctly integrate but did recognize that the integral involved the natural log function; they mostoften missed the factor or replaced it with 2.

∫ dx12x+3

∫ dx = ln(2x+ 3) +C12x+3

12

ln |(2x+ 3)| +C12

12

80b. [4 marks]Given that , find the value of P.dx = ln∫ 30

12x+3

P−−√

Markscheme

evidence of substitution of limits (M1)

e.g.

evidence of correctly using (seen anywhere) (A1)

e.g.

evidence of correctly using (seen anywhere) (A1)

e.g.

(accept ) A1 N2

[4 marks]

Examiners reportPart (b) proved difficult as many were unable to use the basic rules of logarithms.

dx =∫ 30

12x+3

[ ln(2x+ 3)]12

3

0

ln9 − ln312

12

lna− lnb = ln a

b

ln312

a lnb = lnba

ln 93

−−√P = 3 ln 3√

81. [7 marks]A particle moves along a straight line so that its velocity, at time t seconds is given by . When , thedisplacement, s, of the particle is 7 metres. Find an expression for s in terms of t.

Markschemeevidence of anti-differentiation (M1)

e.g.

A2A1

substituting , (M1)

A1

A1 N3

[7 marks]

Examiners reportThere were a number of completely correct solutions to this question. However, there were many who did not know the relationshipbetween velocity and position. Many students differentiated rather than integrated and those who did integrate often had difficultywith the term involving e. Many who integrated correctly neglected the C or made .

v ms−1 v = 6 + 4e3t t = 0

s = ∫ (6 + 4)dxe3x

s = 2 + 4t+Ce3t

t = 0

7 = 2 +C

C = 5

s = 2 + 4t+ 5e3t

C = 7

( ) = + 2 − 51 3 2

82a. [3 marks]

Consider . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.

Find .

Markscheme A1A1A1 N3

[3 marks]

Examiners reportThis question was very well done with most candidates showing their work in an orderly manner. There were a number ofcandidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expressionrather than an equation.

f(x) = + 2 − 5x13x3 x2

(x)f ′

(x) = + 4x− 5f ′ x2

82b. [4 marks]Find the x-coordinate of M.

Markschemeevidence of attempting to solve (M1)

evidence of correct working A1

e.g. , , sketch

, (A1)

so A1 N2

[4 marks]

Examiners reportThis question was very well done with most candidates showing their work in an orderly manner. There were a number ofcandidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expressionrather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was amaximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which wasunnecessary and used up valuable time on the exam.

(x) = 0f ′

(x+ 5)(x− 1) −4± 16+20√2

x = −5 x = 1

x = −5

82c. [3 marks]Find the x-coordinate of N.

MarkschemeMETHOD 1

(may be seen later) A1

evidence of setting second derivative = 0 (M1)

e.g.

A1 N2

METHOD 2

evidence of use of symmetry (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculation A1

e.g.

A1 N2

[3 marks]

Examiners reportThis question was very well done with most candidates showing their work in an orderly manner. There were a number ofcandidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expressionrather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was amaximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which wasunnecessary and used up valuable time on the exam.

(x) = 2x+ 4f ′′

2x+ 4 = 0

x = −2

−5+12

x = −2

82d. [4 marks]The line L is the tangent to the curve of f at . Find the equation of L in the form .

Markschemeattempting to find the value of the derivative when (M1)

A1

valid approach to finding the equation of a line M1

e.g. ,

A1 N2

[4 marks]

Examiners reportThis question was very well done with most candidates showing their work in an orderly manner. There were a number ofcandidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expressionrather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was amaximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which wasunnecessary and used up valuable time on the exam.

(3, 12) y = ax+ b

x = 3

(3) = 16f ′

y− 12 = 16(x− 3) 12 = 16 × 3 + b

y = 16x− 36

y = (x)′

83a. [2 marks]

The diagram below shows part of the graph of the gradient function, .

On the grid below, sketch a graph of , clearly indicating the x-intercept.

Markscheme

A1A1 N2

Note: Award A1 for negative gradient throughout, A1 for x-intercept of q. It need not be linear.

[2 marks]

Examiners reportSeveral candidates had a correct sketch in part (a).

y = (x)f ′

y = (x)f ′′

83b. [2 marks]Complete the table, for the graph of .y = f(x)

Markscheme

A1A1 N1N1

Examiners reportThe majority of the errors occurred in parts (b) and (c). In part (b), some seemed to just guess while others left it blank.

83c. [2 marks]Justify your answer to part (b) (ii).

MarkschemeMETHOD 1

Second derivative is zero, second derivative changes sign. R1R1 N2

METHOD 2

There is a maximum on the graph of the first derivative. R2 N2

Examiners reportIn part (c), justification lacked completeness. For example, many stated that the second derivative must equal zero but said nothing ofits change in sign.

84a. [6 marks]

The following diagram shows the graphs of and , for .

Let A be the area of the region enclosed by the curves of f and g.

(i) Find an expression for A.

(ii) Calculate the value of A.

f(x) = ln(3x− 2) + 1 g(x) = −4 cos(0.5x) + 2 1 ≤ x ≤ 10

Markscheme(i) intersection points , (may be seen as the limits) (A1)(A1)

approach involving subtraction and integrals (M1)

fully correct expression A2

e.g. ,

(ii) A1 N1

[6 marks]

Examiners reportMany candidates did not make good use of the GDC in this problem. Most had the correct expression but incorrect limits. Some triedto integrate to find the area without using their GDC. This became extremely complicated and time consuming.

x = 3.77 x = 8.30

((−4 cos(0.5x) + 2) − (ln(3x− 2) + 1))dx∫ 8.303.77 g(x)dx− f(x)dx∫ 8.30

3.77 ∫ 8.303.77

A = 6.46

84b. [4 marks](i) Find .

(ii) Find .

Markscheme(i) A1A1 N2

Note: Award A1 for numerator (3), A1 for denominator ( ) , but penalize 1 mark for additional terms.

(ii) A1A1 N2

Note: Award A1 for 2, A1 for , but penalize 1 mark for additional terms.

[4 marks]

Examiners reportIn part (b), the chain rule was not used by some.

(x)f ′

(x)g′

(x) =f ′ 33x−2

3x− 2

(x) = 2 sin(0.5x)g′

sin(0.5x)

85a. [2 marks]

Let .

Show that .

Markschemeevidence of factorising 3/division by 3 A1

e.g. , , (do not accept 4 as this is show that)

evidence of stating that reversing the limits changes the sign A1

e.g.

AG N0

[2 marks]

Examiners reportThis question was very poorly done. Very few candidates provided proper justification for part (a), a common error being to write

. What was being looked for was that and .

3f(x)dx = 12∫ 51

f(x)dx = −4∫ 15

3f(x)dx = 3 f(x)dx∫ 51 ∫ 5

1123

∫ 51

3f(x)dx3

f(x)dx = − f(x)dx∫ 15 ∫ 5

1

f(x)dx = − 4∫ 15

f(x)dx =f(5) − f(1)∫ 51 3f(x)dx =3 f(x)dx∫ 5

1 ∫ 51 f(x)dx = − f(x)dx∫ 1

5 ∫ 51

85b. [5 marks]Find the value of .

Markschemeevidence of correctly combining the integrals (seen anywhere) (A1)

e.g.

evidence of correctly splitting the integrals (seen anywhere) (A1)

e.g.

(seen anywhere) A1

A1

A1 N3

[5 marks]

Examiners reportPart (b) had similar problems with neither the combining of limits nor the splitting of integrals being done very often. A common errorwas to treat as 1 in order to make and then write .

(x+ f(x))dx+ (x+ f(x))dx∫ 21 ∫ 5

2

I = (x+ f(x))dx+ (x+ f(x))dx = (x+ f(x))dx∫ 21 ∫ 5

2 ∫ 51

I = xdx+ f(x)dx∫ 51 ∫ 5

1

∫ xdx = x2

2

xdx = = −∫ 51 [ ]x2

2

5

1

252

12

(= ,12)242

I = 16

f(x) f(x)dx = 4∫ 51 ∫ 5

1 (x+ f(x))dx = [x+ 1]51

86a. [2 marks]

Let .

Find , giving your answer in the form where .

Markscheme A2 N2

[2 marks]

Examiners reportThis question was not done well by most candidates.

f : x ↦ xsin3

(x)f ′ a x xsinp cosq a, p, q ∈ Z

(x) = 3 xcosxf ′ sin2

86b. [7 marks]Let for . Find the volume generated when the curve of g is revolved through about thex-axis.

g(x) = sinx(cosx3√ )12 0 ≤ x ≤ π

22π

V = π dx∫ b

ay2

V = π( sinx x dx∫π

20 3√ cos

12 )2

= π 3 xcosxdx∫π

20 sin2

V = π[ x]sin3π

2

0(= π( ( ) − 0))sin3 π

2sin3

sin = 1π

2sin 0 = 0

π (1 − 0)

V = π

Examiners reportThis question was not done well by most candidates. No more than one-third of them could correctly give the range of and few could provide adequate justification for there being exactly one solution to in the interval . Finding thederivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of theformula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the

necessary function and limits into and fewer still could square correctly. From those who did squarecorrectly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.

f(x) = xsin3

f(x) = 1 [0, 2π]

π dx∫ b

ay2 sinx x3√ cos

12

87a. [8 marks]

Let S be the total area of the two segments shaded in the diagram below.

Find the value of when S is a local minimum, justifying that it is a minimum.

MarkschemeMETHOD 1

attempt to differentiate (M1)

e.g.

setting derivative equal to 0 (M1)

correct equation A1

e.g. , ,

A1 N3

EITHER

evidence of using second derivative (M1)

A1

A1

it is a minimum because R1 N0

OR

evidence of using first derivative (M1)

for (may use diagram) A1

for (may use diagram) A1

it is a minimum since the derivative goes from negative to positive R1 N0

METHOD 2

is minimum when is a maximum R3

is a maximum when (A2)

A3 N3

[8 marks]

θ

= −4 cosθdSdθ

−4 cosθ = 0 cosθ = 0 4 cosθ = 0

θ = π

2

(θ) = 4 sin θS′′

( ) = 4S′′ π

2

( ) > 0S′′ π

2

θ < , (θ) < 0π

2S′

θ > , (θ) > 0π

2S′

2π− 4 sin θ 4 sin θ

4 sin θ sin θ = 1

θ = π

2

Examiners reportOnly a small number of candidates recognized the fact S would be minimum when sin was maximum, leading to a simple non-calculus solution. Those who chose the calculus route often had difficulty finding the derivative of S, failing in a significant number ofcases to recognize that the derivative of a constant is 0, and also going through painstaking application of the product rule to find thesimple derivative. When it came to justify a minimum, there was evidence in some cases of using some form of valid test, butexplanation of the test being used was generally poor.

87b. [2 marks]Find a value of for which S has its greatest value.

MarkschemeS is greatest when is smallest (or equivalent) (R1)

(or ) A1 N2

[2 marks]

Examiners reportCandidates who answered part (d) correctly generally did well in part (e) as well, though answers outside the domain of werefrequently seen.

θ

4 sin θ

θ = 0 π

θ

88a. [3 marks]

Let .

Show that .

Markschemeevidence of using the product rule M1

A1A1

Note: Award A1 for , A1 for .

AG N0

[3 marks]

Examiners reportMany candidates clearly applied the product rule to correctly show the given derivative. Some candidates missed the multiplicativenature of the function and attempted to apply a chain rule instead.

f(x) = (1 − )ex x2

(x) = (1 − 2x− )f ′ ex x2

(x) = (1 − ) + (−2x)f ′ ex x2 ex

(1 − )ex x2 (−2x)ex

(x) = (1 − 2x− )f ′ ex x2

y = f(x)

88b. [4 marks]

Part of the graph of , for , is shown below. The x-coordinates of the local minimum and maximum points are r and srespectively.

Write down the value of r and of s.

Markschemeat the local maximum or minimum point

(M1)

(M1)

A1A1 N2N2

[4 marks]

Examiners reportAlthough part (c) was a “write down” question where no working is required, a good number of candidates used an algebraic methodof solving for r and s which sometimes returned incorrect answers. Those who used their GDC usually found correct values, althoughnot always to three significant figures.

y = f(x) −6 ≤ x ≤ 2

(x) = 0f ′ ( (1 − 2x− ) = 0)ex x2

⇒ 1 − 2x− = 0x2

r = −2.41 s = 0.414

88c. [4 marks]Let L be the normal to the curve of f at . Show that L has equation .

Markscheme A1

gradient of the normal A1

evidence of substituting into an equation for a straight line (M1)

correct substitution A1

e.g. , ,

AG N0

[4 marks]

Examiners reportIn part (d), many candidates showed some skill showing the equation of a normal, although some tried to work with the gradient ofthe tangent.

P(0, 1) x+ y = 1

(0) = 1f ′

= −1

y− 1 = −1(x− 0) y− 1 = −x y = −x+ 1

x+ y = 1

y = f(x)

88d. [5 marks]Let R be the region enclosed by the curve and the line L.

(i) Find an expression for the area of R.

(ii) Calculate the area of R.

Markscheme(i) intersection points at and (may be seen as the limits) (A1)

approach involving subtraction and integrals (M1)

fully correct expression A2 N4

e.g. ,

(ii) area A1 N1

[5 marks]

Examiners reportSurprisingly few candidates set up a completely correct expression for the area between curves that considered both integration andthe correct subtraction of functions. Using limits of and 2 was a common error, as was integrating on alone. Wherecandidates did write a correct expression, many attempted to perform analytic techniques to calculate the area instead of using theirGDC.

y = f(x)

x = 0 x = 1

( (1 − ) − (1 −x))dx∫ 10 ex x2 f(x)dx− (1 −x)dx∫ 1

0 ∫ 10

R = 0.5

−6 f(x)

89a. [1 mark]

A toy car travels with velocity v ms for six seconds. This is shown in the graph below.

Write down the car’s velocity at .

Markscheme A1 N1

[1 mark]

Examiners report[N/A]

−1

t = 3

4 (m )s−1

89b. [2 marks]Find the car’s acceleration at .t = 1.5

Markschemerecognizing that acceleration is the gradient M1

e.g.

A1 N1

[2 marks]

Examiners report[N/A]

a(1.5) = 4−02−0

a = 2 (m )s−2

89c. [3 marks]Find the total distance travelled.

Markschemerecognizing area under curve M1

e.g. trapezium, triangles, integration

correct substitution A1

e.g. ,

distance 18 (m) A1 N2

[3 marks]

Examiners report[N/A]

(3 + 6)412

|v(t)|dt∫ 60

90. [3 marks]Find .

Markschemeattempt to use substitution or inspection M1

e.g. so

correct working A1

e.g.

A1 N3

[3 marks]

Examiners report[N/A]

∫ dxex

1+ex

u = 1 + ex =dudx

ex

∫ = lnuduu

ln(1 + ) +Cex

91a. [4 marks]

Given that , answer the following.

Find the first four derivatives of .

f(x) = 1x

f(x)

Markscheme (or ) A1 N1

(or ) A1 N1

(or ) A1 N1

(or ) A1 N1

[4 marks]

Examiners report[N/A]

(x) = −f ′ x−2 − 1x2

(x) = 2f ′′ x−3 2x3

(x) = −6f ′′′ x−4 − 6x4

(x) = 24f(4) x−5 24x5

91b. [3 marks]Write an expression for in terms of x and n .

Markscheme or A1A1A1 N3

[3 marks]

Examiners report[N/A]

(x)f(n)

(x) =f(n) n!(−1)n

xn+1(−1 n!( ))n x−(n+1)

92a. [2 marks]

Let , . The following diagram shows the graph of .

The -intercept is at ( , ) , there is a minimum point at A ( , ) and a maximum point at B.

Find .

Markscheme A1A1 N2

[2 marks]

Examiners report[N/A]

f(x) = cosx+ sinx3√ 0 ≤ x ≤ 2π f

y 0 1 p q

(x)f ′

(x) = −sinx+ cosxf ′ 3√

92b. [10 marks]Hence

(i) show that ;

(ii) verify that A is a minimum point.

Markscheme(i) at A, R1

correct working A1

e.g.

A1

, A1

attempt to substitute their x into M1

e.g.

correct substitution A1

e.g.

correct working that clearly leads to A1

e.g.

AG N0

(ii) correct calculations to find either side of A1A1

e.g. , changes sign from negative to positive R1

so A is a minimum AG N0

[10 marks]

Examiners report[N/A]

q = −2

(x) = 0f ′

sinx = cosx3√

tanx = 3√

x = π

34π3

f(x)

cos( ) + sin ( )4π3

3√ 4π3

− + (− )12

3√ 3√2

−2

− −12

32

q = −2

(x)f ′ x = 4π3

(π) = 0 −f ′ 3√ (2π) = 0 +f ′ 3√

(x)f ′

93. [1 mark]

Let , for .

Write down the gradient of the graph of f at .

Markschemegradient is 1.28 A1 N1

[1 mark]

Examiners report[N/A]

f(x) = 4x− − 3ex−2 0 ≤ x ≤ 5

x = 3

94. [5 marks]

Let , .

Let R be the region in the first quadrant enclosed by the graph of h , the x-axis and the line .

(i) Find the area of R.

(ii) Write down an expression for the volume obtained when R is revolved through about the x-axis.

h(x) = 2x−1x+1

x ≠ −1

x = 3

360∘

Printed for sek-catalunya

© International Baccalaureate Organization 2014 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme(i) A2 N2

(ii) attempt to substitute into volume formula (do not accept ) M1

volume A2 N3

[5 marks]

Examiners report[N/A]

area = 2.06

π dx∫ b

ay2

= π dx∫ 312

( )2x−1x+1

2

Typesetting math: 100%