Nuevo examen - 20 de Octubre de 2014
[865 marks]
1a. Find ∫ 10 (x − 4)dx .
4
[4 marks]
Markscheme
correct integration
e.g.
x2
2
− 4x, [
x2
2
A1A1
10
− 4x] ,
4
(x−4)2
2
Notes: In the first 2 examples, award A1 for each correct term.
In the third example, award A1 for 12 and A1 for (x − 4)2 .
substituting limits into their integrated function and subtracting (in any order)
(M1)
e.g. ( 102 − 4(10)) − ( 42 − 4(4)) ,10 − (−8), 12 (62 − 0)
2
∫410 (x − 4)dx = 18
2
A1
N2
Examiners report
Many candidates answered both parts of this question correctly. In part (b), a large number of successful candidates did not seem to
notice 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 of
−−−−
candidates squared (x − 4) in their integral, rather than squaring √x − 4 , which of course prevented them from noting the
link between the two parts and obtaining the correct answer.
−−−−
1b. Part of the graph of f(x) = √x − 4 , for x ≥ 4 , is shown below. The shaded region R is enclosed by the graph of f , the line
x = 10 , and the x-axis.
The region R is rotated 360∘ about the x-axis. Find the volume of the solid formed.
[3 marks]
Markscheme
attempt to substitute either limits or the function into volume formula
e.g.
10
b
π ∫4 f 2 dx, ∫a
(M1)
−−−− 2
10 −−−−
(√x − 4) , π ∫4 √x − 4
Note: Do not penalise for missing π or dx.
correct substitution (accept absence of dx and π) (A1)
−−−− 2
e.g. π∫410 (√x − 4) , π ∫410 (x − 4)dx, ∫410 (x − 4)dx
volume = 18π
A1
N2
[3 marks]
Examiners report
Many candidates answered both parts of this question correctly. In part (b), a large number of successful candidates did not seem to
notice 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 of
−−−−
candidates squared (x − 4) in their integral, rather than squaring √x − 4 , which of course prevented them from noting the
link between the two parts and obtaining the correct answer.
Part of the graph of f(x) = ax3 − 6x2 is shown below.
The point P lies on the graph of f . At P, x = 1.
2a. Find f ′ (x) .
[2 marks]
Markscheme
f ′ (x) = 3ax2 − 12x
A1A1
N2
Note: Award A1 for each correct term.
[2 marks]
Examiners report
A 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 few
candidates who incorrectly tried to work with f(x) , rather than f ′ (x) , in part (b).
2b. The graph of f has a gradient of 3 at the point P. Find the value of a .
[4 marks]
Markscheme
setting their derivative equal to 3 (seen anywhere)
A1
′
e.g. f (x) = 3
attempt to substitute x = 1 into f ′ (x)
(M1)
e.g. 3a(1)2 − 12(1)
correct substitution into f ′ (x)
(A1)
e.g. 3a − 12 , 3a = 15
a=5
A1
N2
[4 marks]
Examiners report
A 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 few
candidates who incorrectly tried to work with f(x) , rather than f ′ (x) , in part (b).
6x
Let f(x) = x+1
, for x > 0 .
3a. Find f ′ (x) .
[5 marks]
Markscheme
METHOD 1
evidence of choosing quotient rule
e.g.
(M1)
u′v−uv′
v2
evidence of correct differentiation (must be seen in quotient rule)
e.g.
d
dx
d
dx
(6x) = 6 ,
(x + 1) = 1
A1
correct substitution into quotient rule
e.g.
(x+1)6−6x
(x+1)2
f ′ (x) =
(A1)(A1)
6x+6−6x
,
(x+1)2
6
A1
(x+1)2
N4
[5 marks]
METHOD 2
evidence of choosing product rule
−1
e.g. 6x(x + 1)
,
uv′
(M1)
+ vu′
evidence of correct differentiation (must be seen in product rule)
e.g.
d
dx
(6x) = 6 ,
correct working
d
dx
(x + 1)−1 = −1(x + 1)−2 × 1
A1
e.g. 6x × −(x + 1)−2 + (x + 1)−1 × 6 ,
f ′ (x) =
6
(x+1)2
[5 marks]
A1
N4
−6x+6(x+1)
(x+1)2
(A1)(A1)
Examiners report
In part (a), most candidates recognized the need to apply the quotient rule to find the derivative, and many were successful in earning
full marks here.
3b. Let g(x) = ln (
6x
)
x+1
, for x > 0 .
1
x(x+1)
Show that g′ (x) =
[4 marks]
.
Markscheme
METHOD 1
(M1)
evidence of choosing chain rule
e.g. formula,
(
1
6x
)
x+1
6x
× ( x+1
)
correct reciprocal of
1
(
6x
)
x+1
is
x+1
6x
(seen anywhere)
correct substitution into chain rule
e.g.
(
1
×
6x
)
x+1
6
(x+1)2
,(
6
(x+1)2
A1
) ( x+1
)
6x
working that clearly leads to the answer
e.g. (
6
)
(x+1)
g′ (x) =
1
( 6x
),(
1
x(x+1)
1
)
(x+1)2
AG
A1
( x+1
),
x
A1
6(x+1)
6x(x+1)2
N0
[4 marks]
METHOD 2
attempt to subtract logs
(M1)
e.g. ln a − ln b , ln 6x − ln(x + 1)
correct derivatives (must be seen in correct expression)
e.g.
6
6x
1
− x+1
,
1
x
1
− x+1
working that clearly leads to the answer
e.g.
x+1−x
x(x+1)
,
A1A1
6x+6−6x
6x(x+1)
1
g′ (x) = x(x+1)
,
AG
A1
6(x+1−x)
6x(x+1)
N0
[4 marks]
Examiners report
In 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 clever
candidates simplified the situation by applying properties of logarithms before finding their derivative.
3c. Let h(x) =
1
x(x+1)
. The area enclosed by the graph of h , the x-axis and the lines x =
find the value of k .
1
5
and x = k is ln 4 . Given that k > 15 ,
[7 marks]
Markscheme
valid method using integral of h(x) (accept missing/incorrect limits or missing dx )
e.g. area = ∫ 1 h(x)dx , ∫ (
k
5
(M1)
1
)
x(x+1)
recognizing that integral of derivative will give original function
(R1)
6x
1
e.g. ∫ ( x(x+1)
)dx = ln ( x+1
)
correct substitution and subtraction
6k
e.g. ln ( k+1
) − ln ( 1
6× 1
5
5
+1
A1
6k
) , ln ( k+1
) − ln(1)
setting their expression equal to ln 4
(M1)
6k
6k
e.g. ln ( k+1
) − ln(1) = ln 4 , ln ( k+1
) = ln 4 , ∫ 1 h(x)dx = ln 4
k
5
correct equation without logs
6k
e.g. k+1
A1
= 4 , 6k = 4(k + 1)
correct working
(A1)
e.g. 6k = 4k + 4 , 2k = 4
k=2
A1
N4
[7 marks]
Examiners report
For part (c), many candidates recognized the need to integrate the function, and that their integral would equal ln 4 . However, many
did not recognize that the integral of h is g . Those candidates who made this link between the parts (b) and (c) often carried on
correctly to find the value of k , with a few candidates having errors in working with logarithms.
s(t) = 2t cos t
A particle’s displacement, in metres, is given by s(t) = 2t cos t , for 0 ≤ t ≤ 6 , where t is the time in seconds.
4a. On the grid below, sketch the graph of s .
[4 marks]
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 (0, 0) and right endpoint within circle.
[4 marks]
Examiners report
Most candidates sketched an approximately correct shape for the displacement of a particle in the given domain, but many lost marks
for carelessness in graphing the local extrema or the right endpoint.
4b. Find the maximum velocity of the particle.
[3 marks]
Markscheme
appropriate approach
(M1)
e.g. recognizing that v = s ′ , finding derivative, a = s ′′
valid method to find maximum
e.g. sketch of v ,
v′ (t)
(M1)
= 0 , t = 5.08698 …
v = 10.20025 …
v = 10.2 [10.2, 10.3] A1
N2
[3 marks]
Examiners report
In 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 to
zero and solve analytically rather than take the maximum value from the graph of the velocity function.
Consider the function f(x) = x2 − 4x + 1 .
5a. Sketch the graph of f , for −1 ≤ x ≤ 5 .
[4 marks]
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 x ≈ 2 , A1 for x-intercepts between 0 and 1, and 3 and 4, A1 for correct y-intercept (0, 1), A1 for correct domain
[−1, 5].
Scale not required on the axes, but approximate positions need to be clear.
[4 marks]
Examiners report
A good number of students provided a clear sketch of the quadratic function within the given domain. Some lost marks as they did
not clearly indicate the approximate positions of the most important points of the parabola either by labelling or providing a suitable
scale.
5b. This function can also be written as f(x) = (x − p)2 − 3 .
[1 mark]
Write down the value of p .
Markscheme
p=2
A1
N1
[1 mark]
Examiners report
There were few difficulties in part (b).
( )
5c.
0
The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of ( ) .
6
[4 marks]
Show that g(x) = −x2 + 4x + 5 .
Markscheme
(A1)(A1)
correct vertical reflection, correct vertical translation
2
e.g. −f(x) , −((x − 2) − 3) , −y , −f(x) + 6 , y + 6
transformations in correct order
(A1)
e.g. −(x2 − 4x + 1) + 6 , −((x − 2)2 − 3) + 6
simplification which clearly leads to given answer
e.g.
−x2
+ 4x − 1 + 6 ,
g(x) = −x2 + 4x + 5
−(x2
AG
A1
− 4x + 4 − 3) + 6
N0
Note: If working shown, award A1A1A0A0 if transformations correct, but done in reverse order, e.g. −(x2 − 4x + 1 + 6).
[4 marks]
Examiners report
In part (c), candidates often used an insufficient number of steps to show the required result or had difficulty setting out their work
logically.
5d.
0
The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of ( ) .
6
[3 marks]
The graphs of f and g intersect at two points.
Write down the x-coordinates of these two points.
Markscheme
valid approach
(M1)
e.g. sketch, f = g
−0.449489 … , 4.449489 …
(2 ± √6) (exact), −0.449 [−0.450, − 0.449] ; 4.45 [4.44, 4.45]
A1A1
N3
[3 marks]
Examiners report
Part (d) was generally done well though many candidates gave at least one answer to fewer than three significant figures, potentially
resulting in more lost marks.
5e.
0
The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of ( ) .
6
Let R be the region enclosed by the graphs of f and g .
Find the area of R .
[3 marks]
Markscheme
attempt to substitute limits or functions into area formula (accept absence of dx )
e.g.
∫ab
((−x2 + 4x + 5) − (x2 − 4x + 1))dx ,
−0.449
∫4.45
(f − g)
, ∫ (−2x2 + 8x + 4)dx
approach involving subtraction of integrals/areas (accept absence of dx )
e.g.
∫ab
(−x2 + 4x + 5) − ∫ab
(M1)
(M1)
(x2 − 4x + 1) , ∫ (f − g)dx
area = 39.19183 …
area = 39.2 [39.1, 39.2]
A1
N3
[3 marks]
Examiners report
In part (e), many candidates were unable to connect the points of intersection found in part (d) with the limits of integration. Mistakes
were also made here either using a GDC incorrectly or not subtracting the correct functions. Other candidates tried to divide the
region into four areas and made obvious errors in the process. Very few candidates subtracted f(x) from g(x) to get a simple function
before integrating and there were numerous, fruitless analytical attempts to find the required integral.
Consider f(x) = x2 sin x .
6a. Find f ′ (x) .
[4 marks]
Markscheme
evidence of choosing product rule
(M1)
eg uv′ + vu′
correct derivatives (must be seen in the product rule) cos x , 2x
f ′ (x) = x2 cos x + 2x sin x
(A1)(A1)
A1 N4
[4 marks]
Examiners report
Many candidates correctly applied the product rule for the derivative, although a common error was to answer f ′ (x) = 2x cos x .
6b. Find the gradient of the curve of f at x =
π
2
.
[3 marks]
Markscheme
substituting
π
2
into their f ′ (x)
(M1)
2
eg f ′ ( π2 ) , ( π2 ) cos ( π2 ) + 2 ( π2 ) sin ( π2 )
correct values for both sin π2 and cos π2 seen in f ′ (x)
eg 0 + 2 ( π2 ) × 1
f ′ ( π2 ) = π
[3 marks]
A1 N2
(A1)
Examiners report
Candidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was made
in the original function. Some candidates did not know the values of sine and cosine at π2 .
7.
Let f(x) = ∫
12
dx
2x−5
, x > 52 . The graph of f passes through (4, 0) .
[6 marks]
Find f(x) .
Markscheme
attempt to integrate which involves ln
(M1)
eg ln(2x − 5) , 12 ln 2x − 5 , ln 2x
correct expression (accept absence of C)
eg 12 ln(2x − 5) 12 + C , 6 ln(2x − 5)
A2
attempt to substitute (4,0) into their integrated f
(M1)
eg 0 = 6 ln(2 × 4 − 5) , 0 = 6 ln(8 − 5) + C
C = −6 ln 3
(A1)
f(x) = 6 ln(2x − 5) − 6 ln 3 (= 6 ln ( 2x−5
)) (accept 6 ln(2x − 5) − ln 36 )
3
A1
N5
Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve ln.
[6 marks]
Examiners report
While some candidates correctly integrated the function, many missed the division by 2 and answered 12 ln (2x − 5) . Other common
incorrect responses included x212x
and −122(x − 5)−2 . Finding the constant of integration also proved elusive for many. Some either
−5x
did not remember the +C or did not try to find its value, while others misunderstood the boundary condition and attempted to
calculate the definite integral from 0 to 4.
Consider f(x) = ln(x4 + 1) .
8a. Find the set of values of x for which f is increasing.
[5 marks]
Markscheme
f ′ (x) =
1
x4+1
× 4x3 (seen anywhere)
Note: Award A1 for
1
x4+1
A1A1
and A1 for 4x3 .
recognizing f increasing where f ′ (x) > 0 (seen anywhere)
′
eg f (x) > 0 , diagram of signs
attempt to solve f ′ (x) > 0
(M1)
eg 4x3 = 0 , x3 > 0
f increasing for x > 0 (accept x ≥ 0 )
[5 marks]
A1
N1
R1
Examiners report
Candidates who attempted to consider where f 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 x”.
′
Many set their derivative equal to zero, while neglecting to indicate in their working that f (x) > 0 for an increasing
function. Some created a diagram of signs, which provides appropriate evidence as long as it is clear that the signs
′
represent f .
The second derivative is given by f ′′ (x) =
4x2(3− x4)
(x4+1)2
.
4
The equation f ′′ (x) = 0 has only three solutions, when x = 0 , ±√
3 (±1.316 …) .
8b. (i)
(ii)
Find f ′′ (1) .
[5 marks]
Hence, show that there is no point of inflexion on the graph of f at x = 0 .
Markscheme
(i)
eg
substituting x = 1 into f ′′
4(3−1)
(1+1)2
,
4×2
4
f ′′ (1) = 2
(ii)
(A1)
A1
N2
valid interpretation of point of inflexion (seen anywhere)
R1
′′
eg no change of sign in f (x) , no change in concavity,
f ′ increasing both sides of zero
attempt to find f ′′ (x) for x < 0
eg f ′′ (−1) ,
2
4
4(−1) (3− (−1) )
2
((−1)4+1)
(M1)
, diagram of signs
correct working leading to positive value
A1
eg f ′′ (−1) = 2 , discussing signs of numerator and denominator
there is no point of inflexion at x = 0
AG
N0
[5 marks]
Examiners report
Finding f ′′ (1) 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 a
calculation, which follows from the “hence” command of the question. In this case, a sign diagram without numerical evidence was
not sufficient.
8c. There is a point of inflexion on the graph of f at x = √3 (x = 1.316 …) .
4
Sketch the graph of f , for x ≥ 0 .
[3 marks]
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 (0, 0) , A1 for increasing throughout.
Sketch need not be drawn to scale. Only essential features need to be clear.
[3 marks]
Examiners report
Few candidates created a correct graph from the information given or found in the question. This included the point (0, 0), the fact
that the function is always increasing for x > 0 , the concavity at x = 1 and the change in concavity at the given point of inflexion.
Many incorrect attempts showed a graph concave down to the right of x = 0 , changing to concave up.
Let f(x) =
100
(1+50e−0.2x )
. Part of the graph of f is shown below.
9a. Show that f ′ (x) =
1000e−0.2x
(1+50e−0.2x )2
.
[5 marks]
Markscheme
METHOD 1
setting function ready to apply the chain rule
eg
(M1)
100(1 + 50e−0.2x )−1
evidence of correct differentiation (must be substituted into chain rule)
eg u′ =
−100(1 + 50e−0.2x )−2
correct chain rule derivative
′
eg f (x) =
, v′ = (50e−0.2x )(−0.2)
A1
−100(1 + 50e−0.2x )−2 (50e−0.2x )(−0.2)
A1
correct working clearly leading to the required answer
′
eg f (x) =
f ′ (x) =
(A1)(A1)
1000e−0.2x (1 + 50e−0.2x )−2
1000e−0.2x
(1+50e−0.2x )2
AG
N0
METHOD 2
attempt to apply the quotient rule (accept reversed numerator terms)
eg
vu′−uv′
v2
,
(A1)(A1)
evidence of correct differentiation inside the quotient rule
eg f ′ (x) =
(1+50e
−0.2x
)(0)−100(50e
−0.2x
(1+50e−0.2x )2
×−0.2)
,
100(−10)e
−0.2x
−0
(1+50e−0.2x )2
any correct expression for derivative (0 may not be explicitly seen)
eg
(M1)
uv′−vu′
v2
(A1)
−100(50e−0.2x ×−0.2)
(1+50e−0.2x )2
correct working clearly leading to the required answer
eg f ′ (x) =
f ′ (x) =
0−100(−10)e−0.2x
(1+50e−0.2x )2
1000e−0.2x
(1+50e−0.2x )2
AG
,
A1
−100(−10)e−0.2x
(1+50e−0.2x )2
N0
[5 marks]
Examiners report
In part (d), the majority of candidates opted to use the quotient rule and did so with some degree of competency, but failed to
recognize the command term “show that” and consequently did not show enough to gain full marks. Approaches involving the chain
rule were also successful but with the same point regarding sufficiency of work.
9b. Find the maximum rate of change of f .
[4 marks]
Markscheme
METHOD 1
sketch of f ′ (x)
(A1)
eg
recognizing maximum on f ′ (x)
(M1)
eg dot on max of sketch
finding maximum on graph of f ′ (x)
A1
eg (19.6, 5) , x = 19.560 …
maximum rate of increase is 5
A1 N2
METHOD 2
recognizing f ′′ (x) = 0
(M1)
finding any correct expression for f ′′ (x) = 0
eg
(1+50e
−0.2x 2
) (−200e
−0.2x
finding x = 19.560 …
)−(1000e
−0.2x
)(2(1+50e
−0.2x
(1+50e−0.2x )4
(A1)
)(−10e−0.2x ))
A1
maximum rate of increase is 5
A1
N2
[4 marks]
Examiners report
Part (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 x = 19.6 as their
final answer. Some found the maximum value of f rather than f ′ .
10. A rocket moving in a straight line has velocity v km s–1 and displacement s km at time t seconds. The velocity v is given by
v(t) = 6e2t + t . When t = 0 , s = 10 .
Find an expression for the displacement of the rocket in terms of t .
[7 marks]
Markscheme
evidence of anti-differentiation
eg ∫
(6e2t
(M1)
+ t)
2
s = 3e2t + t2 + C
A2A1
Note: Award A2 for 3e2t , A1 for
t2
2
.
attempt to substitute (0, 10) into their integrated expression (even if C is missing)
correct working
(M1)
(A1)
eg 10 = 3 + C , C = 7
2
s = 3e2t + t2 + 7
A1
N6
Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of e.
[7 marks]
Examiners report
A good number of candidates earned full marks on this question, and many others were able to earn at least half of the available
marks. Most candidates knew to integrate, but there were quite a few who tried to find the derivative instead. Many candidates
integrated the term 6e2t incorrectly, but most were able to earn some further method marks for substituting into their integrated
function. The majority of candidates who substituted (0, 10) into their integrated function knew that e0 = 1 .
The following is the graph of a function f , for 0 ≤ x ≤ 6 .
The first part of the graph is a quarter circle of radius 2 with centre at the origin.
11. (a)
(b)
Find ∫02 f(x)dx .
The shaded region is enclosed by the graph of f , the x-axis, the y-axis and the line x = 6 . The area of this region is 3π .
Find ∫26 f(x)dx .
[7 marks]
Markscheme
(a)
eg
attempt to find quarter circle area
1
(4π)
4
,
πr2
4
,
area of region = π
2
∫0 f(x)dx
= −π
(M1)
−−−−−−−
∫02 √4 − x2 dx
(A1)
A2
N3
[4 marks]
(b)
attempted set up with both regions
(M1)
2
6
eg shaded area − quarter circle , 3π − π , 3π − ∫0 f = ∫2 f
6
∫2 f(x)dx = 2π
A2
N2
[3 marks]
Total [7 marks]
Examiners report
There was a minor error on the diagram, where the point on the y-axis was labelled 2 (to indicate the length of the radius), rather than
−2. Examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not
indicate 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 the
definite integral is negative for the part of the function below the x-axis. In part (b), most went on to earn full marks by subtracting the
area of the quarter circle from 3π.
Candidates who did not understand the connection between area and the value of the integral often tried to find a function to
integrate. These candidates were not successful using this method.
Let f(x) = sin x + 12 x2 − 2x , for 0 ≤ x ≤ π .
12a. Find f ′ (x) .
[3 marks]
Markscheme
f ′ (x) = cos x + x − 2
A1A1A1
N3
Note: Award A1 for each term.
[3 marks]
Examiners report
In part (a), most candidates were able to correctly find the derivative of the function.
Let g be a quadratic function such that g(0) = 5 . The line x = 2 is the axis of symmetry of the graph of g .
The function g can be expressed in the form g(x) = a(x − h)2 + 3 .
12b. Find the value of x for which the tangent to the graph of f is parallel to the tangent to the graph of g .
[6 marks]
Markscheme
g(x) = 12 (x − 2)2 + 3 = 12 x2 − 2x + 5
correct derivative of g
A1A1
eg 2 × 12 (x − 2) , x − 2
evidence of equating both derivatives
(M1)
eg f ′ = g′
correct equation
(A1)
eg cos x + x − 2 = x − 2
working towards a solution
(A1)
eg cos x = 0 , combining like terms
x = π2
A1
N0
Note: Do not award final A1 if additional values are given.
[6 marks]
Examiners report
Part (d) required the candidates to find the derivative of g, and to equate that to their answer from part (a). Although many candidates
were able to simplify their equation to cos x = 0, many did not know how to solve for x at this point. Candidates who had made
errors in parts (a) and/or (c) were still able to earn follow-through marks in part (d).
Consider the functions f(x) , g(x) and h(x) . The following table gives some values associated with these functions.
The following diagram shows parts of the graphs of h and h′′ .
There is a point of inflexion on the graph of h at P, when x = 3 .
13a. Explain why P is a point of inflexion.
[2 marks]
Markscheme
h′′ (3) = 0
(A1)
R1
valid reasoning
′′
eg h changes sign at x = 3 , change in concavity of h at x = 3
AG
so P is a point of inflexion
N0
[2 marks]
Examiners report
In part (b), the majority of candidates earned one mark for stating that h′′ (x) = 0 at point P. As this is not enough to determine a point
of inflexion, very few candidates earned full marks on this question.
Given that h(x) = f(x) × g(x) ,
13b. find the equation of the normal to the graph of h at P.
[7 marks]
Markscheme
recognizing need to find derivative of h
′
(R1)
′
eg h , h (3)
attempt to use the product rule (do not accept h′ = f ′ × g′ )
′
eg h =
fg ′
′
′
+ gf , h (3) =
correct substitution
(M1)
f(3) × g′ (3) + g(3) × f ′ (3)
(A1)
eg h′ (3) = 3(−3) + (−18) × 1
h′ (3) = −27
A1
attempt to find the gradient of the normal
(M1)
eg − m1 , − 271 x
attempt to substitute their coordinates and their normal gradient into the equation of a line
eg −54 =
1
(3) + b
27
,0=
1
(3) + b
27
correct equation in any form
eg y + 54 =
1
(x − 3)
27
,y=
A1
, y + 54 = 27(x − 3) , y − 54 =
(M1)
1
(x + 3)
27
N4
1
x − 54 19
27
[7 marks]
Examiners report
Part (d) proved to be quite challenging for even the strongest candidates, as almost none of them used the product rule to find h′ (3).
The most common error was to say h′ (3) = f ′ (3) × g′ (3). Despite this error, many candidates were able to earn further method
marks for their work in finding the equation of the normal. There were also a small number of candidates who were able to find the
equation for h′ (x) , and from that h′′ (x). These candidates were often successful in earning full marks, although this method was
quite time-consuming.
x
Let f(x) = e 4 and g(x) = mx , where m ≥ 0 , and −5 ≤ x ≤ 5 . Let R be the region enclosed by the y-axis, the graph of f , and the graph
of g .
Let m = 1.
14a. Find the area of R .
[5 marks]
Markscheme
attempt to find intersection of the graphs of f and g
(M1)
x
4
eg e = x
x = 1.42961 …
A1
valid attempt to find area of R
(M1)
eg ∫ (x − e )dx , ∫01 (g − f) , ∫ (f − g)
x
4
area = 0.697
A2
N3
[5 marks]
Examiners report
There was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed what
was intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notify
the 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 were
usually successful, unlike those who attempted to solve with antiderivatives. A common error was to find the area of the region
enclosed by f and g (although it involved a point of intersection outside of the given domain), rather than the area of the region
enclosed by f and g and the y-axis.
14b. Consider all values of m such that the graphs of f and g intersect. Find the value of m that gives the greatest value for the area [8 marks]
of R .
Markscheme
recognize that area of R is a maximum at point of tangency
(R1)
eg m = f ′ (x)
(M1)
equating functions
x
4
eg f(x) = g(x) , e = mx
x
f ′ (x) = 14 e 4
(A1)
equating gradients
(A1)
x
eg f ′ (x) = g′ (x) , 14 e 4 = m
attempt to solve system of two equations for x
eg
1
e
4
x=4
x
4
×x = e
(M1)
x
4
(A1)
attempt to find m
eg f ′ (4) , 14 e
(M1)
4
4
m = 14 e (exact), 0.680
A1
N3
[8 marks]
Examiners report
There was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed what
was intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notify
the 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 m which maximized
the area of the region. In addition to the answer obtained from the restricted domain, full marks were awarded for the answer obtained
by using the point of tangency.
The velocity of a particle in ms−1 is given by v = esin t − 1 , for 0 ≤ t ≤ 5 .
15. Write down the positive t-intercept.
[4 marks]
Markscheme
recognizing distance is area under velocity curve
(M1)
eg s = ∫ v , shading on diagram, attempt to integrate
valid approach to find the total area
(M1)
3.14
eg area A + area B , ∫ vdt − ∫ vdt , ∫0
5
vdt + ∫3.14 vdt , ∫ |v|
correct working with integration and limits (accept dx or missing dt )
eg
∫03.14
vdt + ∫53.14
distance = 3.95 (m)
vdt , 3.067 … + 0.878 … ,
A1
∫05 ∣esin t
(A1)
− 1∣
N3
[4 marks]
Examiners report
In (b)(ii), most appreciated that the definite integral would give the distance travelled but few could write a valid expression and
normally just integrated from t = 0 to t = 5 without considering the part of the graph below the t-axis. Again, analytic approaches to
evaluating their integral predominated over simpler GDC approaches and some candidates had their calculator set in degree mode
rather than radian mode.
The following diagram shows the graph of a quadratic function f , for 0 ≤ x ≤ 4 .
The graph passes through the point P(0, 13) , and its vertex is the point V(2, 1) .
16a. The function can be written in the form f(x) = a(x − h)2 + k .
(i)
Write down the value of h and of k .
(ii) Show that a = 3 .
[4 marks]
Markscheme
(i) h = 2 , k = 1
A1A1
N2
(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f
M1
2
e.g. 13 = a(0 − 2) + 1
working towards solution
A1
e.g. 13 = 4a + 1
a=3
AG
N0
[4 marks]
Examiners report
In part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able to
successfully show that a = 3 . Unfortunately, a few candidates did not understand the "show that" command, and simply verified
that a = 3 would work, rather than showing how to find a = 3 .
16b. Find f(x) , giving your answer in the form Ax2 + Bx + C .
[3 marks]
Markscheme
attempting to expand their binomial
e.g. f(x) =
3(x2
correct working
e.g. f(x) =
3x2
(M1)
− 2 × 2x + 4) + 1 , (x − 2)2 = x2 − 4x + 4
(A1)
− 12x + 12 + 1
f(x) = 3x2 − 12x + 13 (accept A = 3 , B = −12 , C = 13 )
A1
N2
[3 marks]
Examiners report
In part (b), most candidates were able to find f(x) in the required form. For a few candidates, algebraic errors kept them from finding
the correct function, even though they started with correct values for a, h and k.
16c. Calculate the area enclosed by the graph of f , the x-axis, and the lines x = 2 and x = 4 .
[8 marks]
Markscheme
METHOD 1
(A1)
integral expression
e.g.
4
∫2 (3x2
− 12x + 13) , ∫ fdx
Area = [x3 − 6x2 + 13x]42
A1A1A1
Note: Award A1 for x3 , A1 for −6x2 , A1 for 13x .
correct substitution of correct limits into their expression
A1A1
e.g. (43 − 6 × 42 + 13 × 4) − (23 − 6 × 22 + 13 × 2) , 64 − 96 + 52 − (8 − 24 + 26)
Note: Award A1 for substituting 4, A1 for substituting 2.
(A1)
correct working
e.g. 64 − 96 + 52 − 8 + 24 − 26,20 − 10
Area = 10
A1
N3
[8 marks]
METHOD 2
(A1)
integral expression
e.g.
4
∫2 (3(x − 2)2
+ 1) , ∫ fdx
3
Area = [(x − 2) + x]42
A2A1
Note: Award A2 for (x − 2)3 , A1 for x .
correct substitution of correct limits into their expression
A1A1
e.g. (4 − 2)3 + 4 − [(2 − 2)3 + 2] , 23 + 4 − (03 + 2) , 23 + 4 − 2
Note: Award A1 for substituting 4, A1 for substituting 2.
correct working
(A1)
e.g. 8 + 4 − 2
Area = 10
A1
N3
[8 marks]
METHOD 3
recognizing area from 0 to 2 is same as area from 2 to 4
4
(R1)
2
e.g. sketch, ∫2 f = ∫0 f
(A1)
integral expression
2
e.g. ∫0 (3x2 − 12x + 13) , ∫ fdx
Area = [x3 − 6x2 + 13x]20
Note: Award A1 for
x3
A1A1A1
, A1 for −6x2 , A1 for 13x .
correct substitution of correct limits into their expression
3
2
3
2
A1(A1)
e.g. (2 − 6 × 2 + 13 × 2) − (0 − 6 × 0 + 13 × 0) , 8 − 24 + 26
Note: Award A1 for substituting 2, (A1) for substituting 0.
Area = 10
[8 marks]
A1
N3
Examiners report
In part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic and
arithmetic errors prevented many from finding the correct area.
Let f(x) =
x
−2x2+5x−2
for −2 ≤ x ≤ 4 , x ≠ 12 , x ≠ 2 . The graph of f is given below.
The graph of f has a local minimum at A(1, 1) and a local maximum at B.
17a. Use the quotient rule to show that f ′ (x) =
2x2−2
(−2x2+5x−2)2
.
[6 marks]
Markscheme
(A1)A1A1
correct derivatives applied in quotient rule
1, −4x + 5
Note: Award (A1) for 1, A1 for −4x and A1 for 5, only if it is clear candidates are using the quotient rule.
correct substitution into quotient rule
e.g.
1×(−2x2+5x−2)−x(−4x+5)
(−2x2+5x−2)2
correct working
e.g.
,
A1
−2x2+5x−2−x(−4x+5)
(−2x2+5x−2)2
(A1)
2
−2x +5x−2−(−4x2+5x)
(−2x2+5x−2)2
expression clearly leading to the answer
e.g.
A1
−2x2+5x−2+4x2−5x
(−2x2+5x−2)2
f ′ (x) =
2x2−2
(−2x2+5x−2)2
AG
N0
[6 marks]
Examiners report
While 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.
17b. Given that the line y = k does not meet the graph of f , find the possible values of k .
[3 marks]
Markscheme
recognizing values between max and min
1
9
<k<1
A2
(R1)
N3
[3 marks]
Examiners report
In part (c), a significant number of candidates seemed to think that the line y = k was a vertical line, and attempted to find the vertical
asymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates who
recognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between the
given local minimum and the local maximum they had found in part (b).
Let f(x) = cos(ex ) , for −2 ≤ x ≤ 2 .
18a. Find f ′ (x) .
[2 marks]
Markscheme
f ′ (x) = −ex sin(ex )
A1A1
N2
[2 marks]
Examiners report
Many 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).
18b. On the grid below, sketch the graph of f ′ (x) .
[4 marks]
Markscheme
A1A1A1A1
N4
Note: Award A1 for shape that must have the correct domain (from −2 to +2 ) and correct range (from −6 to 4 ), A1 for minimum in
circle, A1 for maximum in circle and A1 for intercepts in circles.
[4 marks]
Examiners report
Many 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 had
to 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 features
such as domain and range, stationary points and intercepts.
A particle moves in a straight line with velocity v = 12t − 2t3 − 1 , for t ≥ 0 , where v is in centimetres per second and t is in seconds.
19a. Find the acceleration of the particle after 2.7 seconds.
[3 marks]
Markscheme
recognizing that acceleration is the derivative of velocity (seen anywhere)
e.g. a =
2
ds
, v′ ,12 − 6t2
dt2
correctly substituting 2.7 into their expression for a (not into v)
e.g.
s ′′ (2.7)
acceleration = −31.74 (exact), −31.7
[3 marks]
A1
N3
(A1)
(R1)
Examiners report
This question was well answered by many candidates, although there were some who did not recognize the relationship between
velocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most of
the marks. Very few appear to have used their GDC for the integration.
19b. Find the displacement of the particle after 1.3 seconds.
[3 marks]
Markscheme
recognizing that displacement is the integral of velocity
R1
e.g. s = ∫ v
(A1)
correctly substituting 1.3
e.g.
1.3
∫0 vdt
displacement = 7.41195 (exact), 7.41 (cm)
A1
N2
[3 marks]
Examiners report
This question was well answered by many candidates, although there were some who did not recognize the relationship between
velocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most of
the marks. Very few appear to have used their GDC for the integration.
Let f(x) = ax3 + bx2 + c , where a , b and c are real numbers. The graph of f passes through the point (2, 9) .
20a. Show that 8a + 4b + c = 9 .
[2 marks]
Markscheme
attempt to substitute coordinates in f
(M1)
e.g. f(2) = 9
correct substitution
3
A1
2
e.g. a × 2 + b × 2 + c = 9
8a + 4b + c = 9
AG
N0
[2 marks]
Examiners report
Part (a) was generally well done, with a few candidates failing to show a detailed substitution. Some substituted 2 in place of x, but
didn't make it clear that they had substituted in y as well.
20b. The graph of f has a local minimum at (1, 4) .
Find two other equations in a , b and c , giving your answers in a similar form to part (a).
[7 marks]
Markscheme
recognizing that (1, 4) is on the graph of f
(M1)
e.g. f(1) = 4
A1
correct equation
e.g. a + b + c = 4
recognizing that f ′ = 0 at minimum (seen anywhere)
(M1)
′
e.g. f (1) = 0
f ′ (x) = 3ax2 + 2bx (seen anywhere)
correct substitution into derivative
A1A1
(A1)
e.g. 3a × 12 + 2b × 1 = 0
correct simplified equation
A1
e.g. 3a + 2b = 0
[7 marks]
Examiners report
A great majority could find the two equations in part (b). However there were a significant number of candidates who failed to
identify that the gradient of the tangent is zero at a minimum point, thus getting the incorrect equation 3a + 2b = 4 .
20c. Find the value of a , of b and of c .
[4 marks]
Markscheme
valid method for solving system of equations
(M1)
e.g. inverse of a matrix, substitution
a = 2 , b = −3 , c = 5
A1A1A1
N4
[4 marks]
Examiners report
A 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 calculations
leading nowhere were seen.
Those who used matrix methods were almost completely successful.
Let f(x) = e6x .
21a. Write down f ′ (x) .
[1 mark]
Markscheme
f ′ (x) = 6e6x
[1 mark]
A1
N1
Examiners report
On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential
function and successfully finding the equation of the tangent line.
21b. The tangent to the graph of f at the point P(0, b) has gradient m .
(i)
[4 marks]
Show that m = 6 .
(ii) Find b .
Markscheme
(i) evidence of valid approach
(M1)
′
e.g. f (0) , 6e6×0
correct manipulation
e.g.
6e0
A1
, 6×1
m=6
AG
N0
(ii) evidence of finding f(0)
(M1)
e.g. y = e6(0)
b=1
A1
N2
[4 marks]
Examiners report
On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential
function and successfully finding the equation of the tangent line. Some candidates lost a mark in (b)(i) for not showing sufficient
working leading to the given answer.
21c. Hence, write down the equation of this tangent.
[1 mark]
Markscheme
y = 6x + 1
A1
N1
[1 mark]
Examiners report
On the whole, candidates handled this question quite well.
In this question, you are given that cos π3 = 12 , and sin π3 =
√3
2
.
The displacement of an object from a fixed point, O is given by s(t) = t − sin 2t for 0 ≤ t ≤ π .
22a. In this interval, there are only two values of t for which the object is not moving. One value is t =
Find the other value.
π
6
.
[4 marks]
Markscheme
(M1)
evidence of valid approach
e.g. setting
s ′ (t)
=0
correct working
A1
e.g. 2 cos 2t = 1 , cos 2t =
2t = π3 ,
t=
5π
3
5π
6
,…
A1
1
2
(A1)
N3
[4 marks]
Examiners report
In part (b), most candidates understood that they needed to set their derivative equal to zero, but fewer were able to take the next step
to solve the resulting double angle equation. Again, some candidates over-complicated the equation by using the double angle
identity. Few ended up with the correct answer 5π
.
6
22b. Show that s ′ (t) > 0 between these two values of t .
[3 marks]
Markscheme
evidence of valid approach
(M1)
e.g. choosing a value in the interval π6 < t <
correct substitution
5π
6
A1
e.g. s ′ ( π2 ) = 1 − 2 cos π
s ′ ( π2 ) = 3
s ′ (t) > 0
A1
AG
N0
[3 marks]
Examiners report
In part (c), many candidates knew they needed to test a value between π/6 and their value from part (b), but fewer were able to
successfully complete that calculation. Some candidates simply tested their boundary values while others unsuccessfully attempted to
make use of the second derivative.
22c. Find the distance travelled between these two values of t .
[5 marks]
Markscheme
evidence of approach using s or integral of s ′
e.g. ∫ s ′ (t)dt ; s ( 5π
) , s ( π6 ) ; [t − sin 2t]
6
substituting values and subtracting
e.g. s ( 5π
) − s ( π6 ) , ( π6 −
6
correct substitution
e.g.
5π
6
√3
)
2
(M1)
− ( 5π
− (−
6
2π
3
+ √3
Note: Award A1 for
√3
))
2
A1
− sin 5π
− [ π6 − sin π3 ] , ( 5π
− (−
3
6
distance is
(M1)
5π
6
π
6
A1A1
2π
3
√3
))
2
− ( π6 −
√3
)
2
N3
, A1 for √3 .
[5 marks]
Examiners report
Although many candidates did not attempt part (d), those who did often demonstrated a good understanding of how to use the
displacement function s or the integral of their derivative from part (a). Candidates who had made an error in part (b) often could not
finish, as sin(2t) could not be evaluated at their value without a calculator. Of those who had successfully found the other boundary
of 5π/6 , a common error was giving the incorrect sign of the value of sin(5π/3) . Again, this part was a good discriminator between
the grade 6 and 7 candidates.
The graph of y = (x − 1) sin x , for 0 ≤ x ≤
5π
2
, is shown below.
The graph has x-intercepts at 0, 1, π and k .
23a. Find k .
[2 marks]
Markscheme
evidence of valid approach
e.g. y = 0 , sin x = 0
2π = 6.283185 …
k = 6.28
[2 marks]
A1
N2
(M1)
Examiners report
Candidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of course
included the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their function
and/or the omission of π . There were still many who were unable to use their calculator successfully to find the required volume.
23b. The shaded region is rotated 360∘ about the x-axis. Let V be the volume of the solid formed.
[3 marks]
Write down an expression for V .
Markscheme
attempt to substitute either limits or the function into formula
(M1)
(accept absence of dx )
6.28…
e.g. V = π ∫π (f(x))2 dx , π ∫ ((x − 1) sin x)2 , π ∫π
k
A2
correct expression
e.g.
π ∫π6.28
y2 dx
N3
(x − 1) sin2 xdx , π ∫π2π ((x − 1) sin x)2 dx
2
[3 marks]
Examiners report
Candidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of course
included the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their function
and/or the omission of π . There were still many who were unable to use their calculator successfully to find the required volume.
23c. The shaded region is rotated 360∘ about the x-axis. Let V be the volume of the solid formed.
[2 marks]
Find V .
Markscheme
V = 69.60192562 …
V = 69.6
A2
N2
[2 marks]
Examiners report
Candidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of course
included the omission of dx, using the given domain as the upper and lower bounds of integration, forgetting to square their function
and/or the omission of π . There were still many who were unable to use their calculator successfully to find the required volume.
24. Let f ′ (x) = 3x2 + 2 . Given that f(2) = 5 , find f(x) .
[6 marks]
Markscheme
evidence of anti-differentiation
′
e.g. ​ ∫ f (x) , ∫
(3x2
(M1)
+ 2)dx
f(x) = x3 + 2x + c (seen anywhere, including the answer)
attempt to substitute (2, 5)
A1A1
(M1)
3
e.g. f(2) = (2) + 2(2) , 5 = 8 + 4 + c
finding the value of c
(A1)
e.g. 5 = 12 + c , c = −7
f(x) = x3 + 2x − 7
A1
N5
[6 marks]
Examiners report
This question, which required candidates to integrate a simple polynomial and then substitute an initial condition to solve for "c", was
very well done. Nearly all candidates who attempted this question were able to earn full marks. The very few mistakes that were seen
involved arithmetic errors when solving for "c", or failing to write the final answer as the equation of the function.
The following diagram shows the graph of f(x) = a sin(b(x − c)) + d , for 2 ≤ x ≤ 10 .
There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .
25a. Use the graph to write down the value of
(i)
a;
(ii) c ;
(iii) d .
Markscheme
(i) a = 8
A1
N1
(ii) c = 2
A1
N1
(iii) d = 4
[3 marks]
A1
N1
[3 marks]
Examiners report
Part (a) of this question proved challenging for most candidates.
25b. Show that b =
π
4
.
[2 marks]
Markscheme
METHOD 1
recognizing that period = 8
A1
correct working
e.g. 8 =
2π
b
b = π4
AG
,b=
(A1)
2π
8
N0
METHOD 2
M1
attempt to substitute
e.g. 12 = 8 sin(b(4 − 2)) + 4
correct working
A1
e.g. sin 2b = 1
b = π4
AG
N0
[2 marks]
Examiners report
Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that
this period could be found using the given coordinates of the maximum and minimum points.
25c. Find f ′ (x) .
[3 marks]
Markscheme
evidence of attempt to differentiate or choosing chain rule
e.g.
cos π4
(x − 2) ,
π
4
(M1)
×8
f ′ (x) = 2π cos ( π4 (x − 2)) (accept 2π cos π4 (x − 2) )
A2
N3
[3 marks]
Examiners report
In part (c), not many candidates found the correct derivative using the chain rule.
25d. At a point R, the gradient is −2π . Find the x-coordinate of R.
[6 marks]
Markscheme
recognizing that gradient is f ′ (x)
(M1)
′
e.g. f (x) = m
A1
correct equation
e.g. −2π = 2π cos ( π4 (x − 2)) , −1 = cos ( π4 (x − 2))
correct working
e.g.
cos−1 (−1)
(A1)
= (x − 2)
π
4
using cos−1 (−1) = π (seen anywhere)
(A1)
e.g. π = (x − 2)
π
4
simplifying
(A1)
e.g. 4 = (x − 2)
x=6
A1
N4
[6 marks]
Examiners report
For part (d), a good number of candidates correctly set their expression equal to −2π , but errors in their previous values kept most
from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.
Let f(x) = 14 x2 + 2 . The line L is the tangent to the curve of f at (4, 6) .
26a. Find the equation of L .
[4 marks]
Markscheme
finding f ′ (x) = 12 x
A1
attempt to find f ′ (4)
′
correct value f (4) = 2
(M1)
A1
correct equation in any form
A1
N2
e.g. y − 6 = 2(x − 4) , y = 2x − 2
[4 marks]
Examiners report
While most candidates answered part (a) correctly, finding the equation of the tangent, there were some who did not consider the
value of their derivative when x = 4 .
( )=
90
Let g(x) =
90
3x+4
, for 2 ≤ x ≤ 12 . The following diagram shows the graph of g .
26b. Find the area of the region enclosed by the curve of g , the x-axis, and the lines x = 2 and x = 12 . Give your answer in the
form a ln b , where a,b ∈ Z .
[6 marks]
Markscheme
area = ∫212
90
dx
3x+4
A1A1
correct integral
e.g. 30 ln(3x + 4)
substituting limits and subtracting
(M1)
e.g. 30 ln(3 × 12 + 4) − 30 ln(3 × 2 + 4) , 30 ln 40 − 30 ln 10
correct working
(A1)
e.g. 30(ln 40 − ln 10)
correct application of ln b − ln a
e.g. 30 ln
(A1)
40
10
area = 30 ln 4
A1
N4
[6 marks]
Examiners report
In part (b), most candidates knew that they needed to integrate to find the area, but errors in integration, and misapplication of the
rules of logarithms kept many from finding the correct area.
26c. 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 , x = 2 , x = 12 [3 marks]
and the x-axis is 120 120 cm2 .
Find the area enclosed by the lines L , x = 2 , x = 12 and the graph of h .
Markscheme
valid approach
(M1)
e.g. sketch, area h = area g , 120 + their answer from (b)
area = 120 + 30 ln 4
[3 marks]
A2
N3
Examiners report
In part (c), it was clear that a significant number of candidates understood the idea of the reflected function, and some recognized that
the integral was the negative of the integral from part (b), but only a few recognized the relationship between the areas. Many thought
the area between h and the x-axis was 120.
Let f(t) = 2t2 + 7 , where t > 0 . The function v is obtained when the graph of f is transformed by
a stretch by a scale factor of 13 parallel to the y-axis,
followed by a translation by the vector (
2
).
−4
27a. Find v(t) , giving your answer in the form a(t − b)2 + c .
[4 marks]
Markscheme
applies vertical stretch parallel to the y-axis factor of
e.g. multiply by
1
3
1
3
(M1)
, 13 f(t) , 13 × 2
applies horizontal shift 2 units to the right
(M1)
e.g. f(t − 2) , t − 2
applies a vertical shift 4 units down
(M1)
e.g. subtracting 4, f(t) − 4 , 73 − 4
v(t) = 23 (t − 2)2 − 53
A1
N4
[4 marks]
Examiners report
While 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 its
derivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather than
using the GDC, which often proved unsuccessful.
27b. A particle moves along a straight line so that its velocity in ms−1 , at time t seconds, is given by v . Find the distance the
particle travels between t = 5.0 and t = 6.8 .
[3 marks]
Markscheme
recognizing that distance travelled is area under the curve
e.g. ∫ v,
2
(t − 2)3
9
−
5
t
3
M1
, sketch
distance = 15.576 (accept 15.6)
A2
N2
[3 marks]
Examiners report
While 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 its
derivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather than
using the GDC, which often proved unsuccessful.
Let f(x) =
20x
e0.3x
, for 0 ≤ x ≤ 20 .
28a. Sketch the graph of f .
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for approximately correct shape with inflexion/change of curvature, A1 for maximum skewed to the left, A1 for
asymptotic behaviour to the right.
[3 marks]
Examiners report
Many candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find the
maximum value.
28b. (i)
Write down the x-coordinate of the maximum point on the graph of f .
[3 marks]
(ii) Write down the interval where f is increasing.
Markscheme
(i) x = 3.33
A1
N1
(ii) correct interval, with right end point 3 13
A1A1
N2
e.g. 0 < x ≤ 3.33 , 0 ≤ x < 3 13
Note: Accept any inequalities in the right direction.
[3 marks]
Examiners report
Many candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find the
maximum value.
28c. Show that f ′ (x) =
20−6x
e0.3x
.
[5 marks]
Markscheme
(M1)
valid approach
e.g. quotient rule, product rule
2 correct derivatives (must be seen in product or quotient rule)
(A1)(A1)
e.g. 20 , 0.3e0.3x or −0.3e−0.3x
correct substitution into product or quotient rule
e.g.
20e
0.3x
−20x(0.3)e
(e0.3x )2
20e
0.3x
f ′ (x) =
−6xe
e0.6x
0.3x
20−6x
e0.3x
A1
, 20e−0.3x + 20x(−0.3)e−0.3x
A1
correct working
e.g.
0.3x
,
e0.3x (20−20x(0.3))
AG
(e0.3x )2
, e−0.3x (20 + 20x(−0.3))
N0
[5 marks]
Examiners report
Most had a valid approach in part (c) using either the quotient or product rule, but many had difficulty applying the chain rule with a
function involving e and simplifying.
28d. Find the interval where the rate of change of f is increasing.
[4 marks]
Markscheme
consideration of f ′ or f ′′
(M1)
R1
valid reasoning
′
e.g. sketch of f , f ′′ is positive, f ′′ = 0 , reference to minimum of f ′
correct value 6.6666666 … (6 23 )
correct interval, with both endpoints
(A1)
A1
N3
2
3
e.g. 6.67 < x ≤ 20 , 6 ≤ x < 20
[4 marks]
Examiners report
Part (d) was difficult for most candidates. Although many associated rate of change with derivative, only the best-prepared students
had valid reasoning and could find the correct interval with both endpoints.
Let g(x) =
ln x
x2
, for x > 0 .
29a. Use the quotient rule to show that g ′ (x) =
1−2 ln x
x3
.
[4 marks]
Markscheme
d
dx
d
dx
ln x = x1 ,
x2 = 2x (seen anywhere)
A1A1
attempt to substitute into the quotient rule (do not accept product rule)
e.g.
x2( 1x )−2x ln x
x4
correct manipulation that clearly leads to result
e.g.
M1
x−2x ln x
x4
,
x(1−2 ln x)
g′ (x) = 1−2 3ln x
x
x4
AG
,
x
x4
,
A1
2x ln x
x4
N0
[4 marks]
Examiners report
Many candidates clearly knew their quotient rule, although a common error was to simplify 2x ln x as 2 ln x2 and then "cancel" the
exponents.
29b. The graph of g has a maximum point at A. Find the x-coordinate of A.
Markscheme
evidence of setting the derivative equal to zero
e.g.
g′ (x)
ln x =
x=e
1
2
1
2
[3 marks]
= 0 , 1 − 2 ln x = 0
A1
A1
N2
(M1)
[3 marks]
Examiners report
For (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, as
this is the calculator-free paper. Occasionally, students would take 1−2 3ln x = 0 and attempt to solve from 1 − 2 ln x = x3 .
x
The velocity v ms−1 of a particle at time t seconds, is given by v = 2t + cos 2t , for 0 ≤ t ≤ 2 .
30a. Write down the velocity of the particle when t = 0 .
[1 mark]
Markscheme
v=1
A1
N1
[1 mark]
Examiners report
Many candidates gave a correct initial velocity, although a substantial number of candidates answered that 0 + cos 0 = 0 .
30b. When t = k , the acceleration is zero.
Show that k =
(i)
π
4
[8 marks]
.
(ii) Find the exact velocity when t = π4 .
Markscheme
(i) dtd (2t) = 2
d
dt
A1
(cos 2t) = −2 sin 2t
A1A1
Note: Award A1 for coefficient 2 and A1 for − sin 2t .
evidence of considering acceleration = 0
e.g.
dv
dt
(M1)
= 0 , 2 − 2 sin 2t = 0
correct manipulation
A1
e.g. sin 2k = 1 , sin 2t = 1
2k = π2 (accept 2t = π2 )
k = π4
AG
A1
N0
(ii) attempt to substitute t = π4 into v
(M1)
e.g. 2 ( π4 ) + cos ( 2π
)
4
v = π2
A1
N2
[8 marks]
Examiners report
For (b), students commonly applied the chain rule correctly to achieve the derivative, and many recognized that the acceleration must
be 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 k = π4 . At times students would reach the equation sin 2k = 1 and then substitute
the π4 , which does not satisfy the “show that” instruction.
30c. When t <
π
4
,
dv
dt
> 0 and when t > π4 ,
Sketch a graph of v against t .
dv
dt
>0 .
[4 marks]
Markscheme
A1A1A2
N4
Notes: Award A1 for y-intercept at (0, 1) , A1 for curve having zero gradient at t = π4 , A2 for shape that is concave down to the left
of π4 and concave up to the right of π4 . If a correct curve is drawn without indicating t = π4 , do not award the second A1 for the zero
gradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.
[4 marks]
Examiners report
The challenge in this question is sketching the graph using the information achieved and provided. This requires students to make
graphical interpretations, and as typical in section B, to link the early parts of the question with later parts. Part (a) provides the yintercept, and part (b) gives a point with a horizontal tangent. Plotting these points first was a helpful strategy. Few understood either
the notation or the concept that the function had to be increasing on either side of the π4 , with most thinking that the point was either a
max or min. It was the astute student who recognized that the derivatives being positive on either side of π4 creates a point of inflexion.
Additionally, important points should be labelled in a sketch. Indicating the π4 on the x-axis is a requirement of a clear graph.
Although students were not penalized for not labelling the π2 on the y-axis, there should be a recognition that the point is higher than
the y-intercept.
30d. Let d be the distance travelled by the particle for 0 ≤ t ≤ 1 .
(i)
Write down an expression for d .
(ii) Represent d on your sketch.
∫0 (2t + cos 2t)dt [ t2 + sin22t ]
1
t=1
1
0
1
1 + sin2 2 ∫0 vdt
t = π4
[3 marks]
Examiners report
While some candidates recognized that the distance is the area under the velocity graph, surprisingly few included neither the limits of
integration 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 of
distance on the sketch, with most fruitlessly attempting to graph a new curve.
31. Let f(x) = cos(x2 ) and g(x) = ex , for −1.5 ≤ x ≤ 0.5 .
[6 marks]
Find the area of the region enclosed by the graphs of f and g .
Markscheme
evidence of finding intersection points
(M1)
e.g. f(x) = g(x) , cos x2 = ex , sketch showing intersection
x = −1.11 , x = 0 (may be seen as limits in the integral)
A1A1
evidence of approach involving integration and subtraction (in any order)
(M1)
0
e.g. ∫−1.11
cos x2 − ex , ∫ (cos x2 − ex )dx , ∫ g − f
area = 0.282
A2
N3
[6 marks]
Examiners report
This question was poorly done by a great many candidates. Most seemed not to understand what was meant by the phrase "region
enclosed by" as several candidates assumed that the limits of the integral were those given in the domain. Few realized what area was
required, or that intersection points were needed. Candidates who used their GDCs to first draw a suitable sketch could normally
recognize the required region and could find the intersection points correctly. However, it was disappointing to see the number of
candidates who could not then use their GDC to find the required area or who attempted unsuccessful analytical approaches.
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 bucket
above the water level is given by h = a sin bt + 2 .
32a. Show that a = 4 .
[2 marks]
Markscheme
METHOD 1
evidence of recognizing the amplitude is the radius
(M1)
e.g. amplitude is half the diameter
a=
8
2
a=4
A1
AG
N0
METHOD 2
evidence of recognizing the maximum height
(M1)
e.g. h = 6 , a sin bt + 2 = 6
correct reasoning
e.g. asin bt = 4 and sin bt has amplitude of 1
a=4
AG
A1
N0
[2 marks]
Examiners report
Parts (a) and (b) were generally well done.
32b. The wheel turns at a rate of one rotation every 30 seconds.
Show that b =
π
15
[2 marks]
.
Markscheme
METHOD 1
(A1)
period = 30
b=
2π
30
A1
b=
π
15
AG
N0
METHOD 2
correct equation
(A1)
e.g. 2 = 4 sin 30b + 2 , sin 30b = 0
30b = 2π
b=
π
15
A1
AG
N0
[2 marks]
Examiners report
Parts (a) and (b) were generally well done, however there were several instances of candidates working backwards from the given
answer in part (b).
32c. In the first rotation, there are two values of t when the bucket is descending at a rate of 0.5 ms−1 .
Find these values of t .
[6 marks]
Markscheme
recognizing h′ (t) = −0.5 (seen anywhere)
R1
(M1)
attempting to solve
′
e.g. sketch of h , finding h′
correct work involving h′
A2
e.g. sketch of h′ showing intersection, −0.5 =
t = 10.6 , t = 19.4
A1A1
4π
π
cos ( 15
t)
15
N3
[6 marks]
Examiners report
Parts (c) and (d) proved to be quite challenging for a large proportion of candidates. Many did not attempt these parts. The most
common error was a misinterpretation of the word "descending" where numerous candidates took h′ (t) to be 0.5 instead of −0.5 but
incorrect derivatives for h were also widespread. The process required to solve for t from the equation −0.5 =
4π
π
cos ( 15
t)
15
overwhelmed those who attempted algebraic methods. Few could obtain both correct solutions, more had one correct while others
included unreasonable values including t < 0 .
32d. In the first rotation, there are two values of t when the bucket is descending at a rate of 0.5 ms−1 .
[4 marks]
Determine whether the bucket is underwater at the second value of t .
Markscheme
METHOD 1
valid reasoning for their conclusion (seen anywhere)
R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of substituting into h
e.g. h(19.4) , 4 sin
correct calculation
19.4π
15
(M1)
+2
A1
e.g. h(19.4) = −1.19
correct statement
A1
N0
e.g. the bucket is underwater, yes
METHOD 2
valid reasoning for their conclusion (seen anywhere)
R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of valid approach
(M1)
e.g. solving h(t) = 0 , graph showing region below x-axis
correct roots
A1
e.g. 17.5, 27.5
correct statement
A1
N0
e.g. the bucket is underwater, yes
[4 marks]
Examiners report
In part (d), not many understood that the condition for underwater was h(t) < 0 and had trouble interpreting the meaning of "second
value". Many candidates, however, did recover to gain some marks in follow through.
The following diagram shows the graph of f(x) = e−x .
2
The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.
33a. Identify the two points of inflexion.
[2 marks]
Markscheme
A1A1
B, D
N2
[2 marks]
Examiners report
Most candidates were able to recognize the points of inflexion in part (a).
33b. (i)
Find f ′ (x) .
[5 marks]
(ii) Show that f ′′ (x) = (4x2 − 2)e−x .
2
Markscheme
(i) f ′ (x) = −2xe−x
2
A1A1
Note: Award A1 for e
−x2
N2
and A1 for −2x .
(ii) finding the derivative of −2x , i.e. −2
evidence of choosing the product rule
(A1)
(M1)
e.g. −2e−x −2x × −2xe−x
2
2
−2e−x + 4x2 e−x
2
2
A1
f ′′ (x) = (4x2 − 2)e−x
2
AG
N0
[5 marks]
Examiners report
Most candidates were able to recognize the points of inflexion in part (a) and had little difficulty with the first and second derivatives
in part (b). A few did not recognize the application of the product rule in part (b).
33c. Find the x-coordinate of each point of inflexion.
[4 marks]
Markscheme
R1
valid reasoning
′′
e.g. f (x) = 0
attempting to solve the equation
(M1)
e.g. (4x2 − 2) = 0 , sketch of f ′′ (x)
p = 0.707 (=
1
)
√2
, q = −0.707 (= −
1
)
√2
A1A1
N3
[4 marks]
Examiners report
Obtaining the x-coordinates of the inflexion points in (c) usually did not cause many problems.
33d. Use the second derivative to show that one of these points is a point of inflexion.
[4 marks]
Markscheme
evidence of using second derivative to test values on either side of POI
M1
′′
e.g. finding values, reference to graph of f , sign table
correct working
A1A1
e.g. finding any two correct values either side of POI,
checking sign of f ′′ on either side of POI
reference to sign change of f ′′ (x)
R1
N0
[4 marks]
Examiners report
Only the better-prepared candidates understood how to set up a second derivative test in part (d). Many of those did not show, or
clearly 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 f ′′ (±
′′
1
)
√2
= 0 , completely missing the point of the question. The necessary condition for a point of
inflexion, i.e. f (x) = 0 and the change of sign for f ′′ (x) , seemed not to be known by the vast majority of candidates.
34. Let h(x) =
6x
cos x
. Find h′ (0) .
[6 marks]
Markscheme
METHOD 1 (quotient)
derivative of numerator is 6
(A1)
derivative of denominator is − sin x
attempt to substitute into quotient rule
(cos x)(6)−(6x)(− sin x)
(cos x)2
substituting x = 0
e.g.
(M1)
A1
correct substitution
e.g.
(A1)
(A1)
(cos 0)(6)−(6×0)(− sin 0)
h′ (0) = 6
(cos 0)2
A1
N2
METHOD 2 (product)
h(x) = 6x × (cos x)−1
(A1)
derivative of 6x is 6
derivative of (cos x)−1 is (−(cos x)−2 (− sin x))
attempt to substitute into product rule
(A1)
(M1)
A1
correct substitution
−2
e.g. (6x)(−(cos x) (− sin x)) + (6)(cos x)−1
substituting x = 0
(A1)
e.g. (6 × 0)(−(cos 0)−2 (− sin 0)) + (6)(cos 0)−1
h′ (0) = 6
A1
N2
[6 marks]
Examiners report
The 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 mistakes
in simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A few
candidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotient
rule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.
f(x) = 2
2
The following diagram shows part of the graph of the function f(x) = 2x2 .
The line T is the tangent to the graph of f at x = 1 .
35a. Show that the equation of T is y = 4x − 2 .
[5 marks]
Markscheme
f(1) = 2
(A1)
f ′ (x) = 4x
A1
evidence of finding the gradient of f at x = 1
M1
e.g. substituting x = 1 into f ′ (x)
finding gradient of f at x = 1
A1
′
e.g. f (1) = 4
evidence of finding equation of the line
M1
e.g. y − 2 = 4(x − 1) , 2 = 4(1) + b
y = 4x − 2
AG
N0
[5 marks]
Examiners report
The majority of candidates seemed to know what was meant by the tangent to the graph in part (a), but there were many who did not
fully show their work, which is of course necessary on a "show that" question. While many candidates knew they needed to find the
derivative of f , some failed to substitute the given value of x in order to find the gradient of the tangent.
35b. Find the x-intercept of T .
Markscheme
appropriate approach
(M1)
e.g. 4x − 2 = 0
x=
1
2
A1
N2
[2 marks]
Examiners report
Part (b), finding the x-intercept, was answered correctly by nearly every candidate.
[2 marks]
35c. The shaded region R is enclosed by the graph of f , the line T , and the x-axis.
(i)
[9 marks]
Write down an expression for the area of R .
(ii) Find the area of R .
Markscheme
(i) bottom limit x = 0 (seen anywhere)
(A1)
approach involving subtraction of integrals/areas
(M1)
e.g. ∫ f(x) − area of triangle , ∫ f − ∫ l
correct expression
A2
N4
1
1
e.g. ∫01 2x2 dx − ∫0.5
(4x − 2)dx , ∫01 f(x)dx − 12 , ∫00.5 2x2 dx + ∫0.5
(f(x) − (4x − 2))dx
(ii) METHOD 1 (using only integrals)
correct integration
∫ 2x2 dx =
2x3
3
, ∫ (4x − 2)dx =2x2 − 2x
substitution of limits
e.g.
1
12
area =
(A1)(A1)(A1)
(M1)
+ 23 − 2 + 2 − ( 121 − 12 + 1)
1
6
A1
N4
METHOD 2 (using integral and triangle)
area of triangle =
1
2
correct integration
∫ 2x2 dx =
(A1)
(A1)
2x3
3
substitution of limits
(M1)
e.g. 23 (1)3 − 23 (0)3 , 23 − 0
correct simplification
(A1)
e.g. 23 − 12
area =
1
6
A1
N4
[9 marks]
Examiners report
In part (c), most candidates struggled with writing an expression for the area of R . Many tried to use the difference of the two
functions over the entire interval 0– 1, not noticing that the area from 0– 0.5 only required the use of function f . Many of these
candidates were able to earn follow-through marks in the second part of (c) for their correct integration. There were a few candidates
who successfully found the area under the line as the area of a triangle.
The following diagram shows part of the graph of a quadratic function f .
The x-intercepts are at (−4, 0) and (6, 0) , and the y-intercept is at (0, 240) .
36a. Write down f(x) in the form f(x) = −10(x − p)(x − q) .
[2 marks]
Markscheme
f(x) = −10(x + 4)(x − 6)
A1A1
N2
[2 marks]
Examiners report
Parts (a) and (c) of this question were very well done by most candidates.
36b. Find another expression for f(x) in the form f(x) = −10(x − h)2 + k .
Markscheme
METHOD 1
attempting to find the x-coordinate of maximum point
(M1)
e.g. averaging the x-intercepts, sketch, y′ = 0 , axis of symmetry
attempting to find the y-coordinate of maximum point
e.g. k = −10(1 + 4)(1 − 6)
f(x) = −10(x − 1)2 + 250
A1A1
N4
METHOD 2
attempt to expand f(x)
(M1)
e.g. −10(x2 − 2x − 24)
attempt to complete the square
(M1)
e.g. −10((x − 1)2 − 1 − 24)
f(x) = −10(x − 1)2 + 250
[4 marks]
A1A1
N4
(M1)
[4 marks]
Examiners report
In part (b), many candidates attempted to use the method of completing the square, but were unsuccessful dealing with the coefficient
of −10. 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.
36c. Show that f(x) can also be written in the form f(x) = 240 + 20x − 10x2 .
[2 marks]
Markscheme
(M1)
attempt to simplify
e.g. distributive property, −10(x − 1)(x − 1) + 250
correct simplification
e.g.
−10(x2
A1
− 6x + 4x − 24) , −10(x2 − 2x + 1) + 250
f(x) = 240 + 20x − 10x2
AG
N0
[2 marks]
Examiners report
Parts (a) and (c) of this question were very well done by most candidates.
36d. A particle moves along a straight line so that its velocity, v ms−1 , at time t seconds is given by v = 240 + 20t − 10t2 , for
0≤t≤6.
(i)
[7 marks]
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, v′ (t) = 0
t=1
A1
N2
(ii) recognizing a(t) = v′ (t)
a(t) = 20 − 20t
(M1)
A1A1
speed is zero ⇒ t = 6
a(6) = −100 (ms−2 )
(A1)
A1
N3
[7 marks]
Examiners report
In part (d), it was clear that many candidates were not familiar with the relationship between velocity and acceleration, and did not
understand how those concepts were related to the graph which was given. A large number of candidates used time t = 1 in part b(ii),
rather than t = 6 . To find the acceleration, some candidates tried to integrate the velocity function, rather than taking the derivative of
velocity. Still others found the derivative in part b(i), but did not realize they needed to use it in part b(ii), as well.
37. A gradient function is given by
dy
dx
= 10e2x − 5 . When x = 0 , y = 8 . Find the value of y when x = 1 .
[8 marks]
Markscheme
METHOD 1
(M1)
evidence of anti-differentiation
e.g. ∫ (10e2x − 5)dx
y = 5e2x − 5x + C
A2A1
Note: Award A2 for 5e2x , A1 for −5x . If “C” is omitted, award no further marks.
substituting (0, 8)
(M1)
e.g. 8 = 5 + C
C = 3 (y = 5e2x − 5x + 3)
substituting x = 1
y = 34.9 (5e2 − 2)
(A1)
(M1)
A1
N4
METHOD 2
evidence of definite integral function expression
e.g.
x
∫a f ′ (t)dt
=f(x) − f(a) ,
x
∫0 (10e2x
(M2)
− 5)
initial condition in definite integral function expression
e.g.
x
∫0 (10e2t
− 5)dt = y − 8 ,
x
∫0 (10e2x
(A2)
− 5)dx + 8
correct definite integral expression for y when x = 1
(A2)
e.g. ∫01 (10e2x − 5)dx + 8
y = 34.9 (5e2 − 2)
A1
N4
[8 marks]
Examiners report
Although a pleasing number of candidates recognized the requirement of integration, many did not correctly apply the reverse of the
chain rule to integration. While some candidates did not write the constant of integration, many did, earning additional follow-through
marks even with an incorrect integral. Weaker candidates sometimes substituted x = 1 into
line equation, earning no marks.
dy
dx
or attempted some work with a tangent
Let g(x) = 2x sin x .
38a. Find g ′ (x) .
[4 marks]
Markscheme
evidence of choosing the product rule
e.g.
uv′
correct derivatives cos x , 2
g′ (x)
(M1)
+ vu′
= 2x cos x + 2 sin x
(A1)(A1)
A1
N4
[4 marks]
Examiners report
Most candidates answered part (a) correctly, using the product rule to find the derivative, and earned full marks here. There were
some who did not know to use the product rule, and of course did not find the correct derivative.
38b. Find the gradient of the graph of g at x = π .
[3 marks]
Markscheme
attempt to substitute into gradient function
e.g.
(M1)
g′ (π)
(A1)
correct substitution
e.g. 2π cos π + 2 sin π
gradient = −2π
A1
N2
[3 marks]
Examiners report
In part (b), many candidates substituted correctly into their derivatives, but then used incorrect values for sin x and cos x , leading to
the wrong gradient in their final answers.
39. The graph of the function y = f(x) passes through the point ( 3 ,4) . The gradient function of f is given as f ′ (x) = sin(2x − 3) [6 marks]
2
. Find f(x) .
Markscheme
evidence of integration
e.g. f(x) = ∫ sin(2x − 3)dx
= − 12 cos(2x − 3) + C
(M1)
A1A1
substituting initial condition into their expression (even if C is missing)
M1
e.g. 4 = − 12 cos 0 + C
C = 4.5
(A1)
f(x) = − 12 cos(2x − 3) + 4.5
A1
N5
[6 marks]
Examiners report
While most candidates realized they needed to integrate in this question, many did so unsuccessfully. Many did not account for the
coefficient of x, and failed to multiply by 12 . Some of the candidates who substituted the initial condition into their integral were not
able to solve for "c", either because of arithmetic errors or because they did not know the correct value for cos 0 .
f(x) =
3
Let f(x) = x3 . The following diagram shows part of the graph of f .
The point P(a,f(a)) , where a > 0 , lies on the graph of f . The tangent at P crosses the x-axis at the point Q ( 23 ,0) . This tangent intersects
the graph of f at the point R(−2, −8) .
40a. (i)
Show that the gradient of [PQ] is
a3
a− 2
3
(ii) Find f ′ (a) .
(iii) Hence show that a = 1 .
.
[7 marks]
Markscheme
(i) substitute into gradient =
e.g.
y1 − y2
x1− x2
(M1)
f(a)−0
a− 2
3
substituting f(a) = a3
e.g.
a3−0
A1
a− 2
3
gradient
a3
AG
a− 2
N0
3
A1
(ii) correct answer
e.g.
3a2
′
N1
′
, f (a) = 3 , f (a) =
a3
a− 2
3
(iii) METHOD 1
evidence of approach
(M1)
′
e.g. f (a) = gradient , 3a2 =
a3
a− 2
3
A1
simplify
e.g. 3a2 (a − 23 ) = a3
A1
rearrange
e.g.
3a3
− 2a2 = a3
A1
evidence of solving
e.g. 2a3 − 2a2 = 2a2 (a − 1) = 0
a=1
AG
N0
METHOD 2
gradient RQ =
e.g.
−8
A1
3
A1
simplify
−8
−8
−2− 2
,3
3
evidence of approach
(M1)
e.g. f ′ (a) = gradient , 3a2 =
e.g.
a=1
,
3
a3
a− 2
3
=3
A1
simplify
3a2
−8
−2− 2
= 3 , a2 = 1
AG
N0
[7 marks]
Examiners report
Part (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.
The equation of the tangent at P is y = 3x − 2 . Let T be the region enclosed by the graph of f , the tangent [PR] and the line x = k , between
x = −2 and x = k where −2 < k < 1 . This is shown in the diagram below.
40b. Given that the area of T is 2k + 4 , show that k satisfies the equation k4 − 6k2 + 8 = 0 .
[9 marks]
Markscheme
approach to find area of T involving subtraction and integrals (M1)
k
k
e.g. ∫ f − (3x − 2)dx , ∫−2 (3x − 2) − ∫−2 x3 , ∫ (x3 − 3x + 2)
correct integration with correct signs
e.g.
1 4
x
4
−
3 2
x
2
+ 2x ,
3 2
x
2
− 2x −
A1A1A1
1 4
x
4
correct limits −2 and k (seen anywhere)
A1
k
e.g. ∫−2
(x3 − 3x + 2)dx , [ 14 x4 − 32 x2 + 2x]
attempt to substitute k and −2
k
−2
(M1)
correct substitution into their integral if 2 or more terms
A1
e.g. ( 14 k4 − 32 k2 + 2k) − (4 − 6 − 4)
setting their integral expression equal to 2k + 4 (seen anywhere)
simplifying
e.g.
1 4
k
4
−
(M1)
A1
3 2
k
2
+2 = 0
k4 − 6k2 + 8 = 0
AG
N0
[9 marks]
Examiners report
In part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtracting
the two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errors
which prevented candidates from correctly showing the desired final result.
−−−−−−−
f(x) = √16 − 4 2
−−−−−−−
The graph of f(x) = √16 − 4x2 , for −2 ≤ x ≤ 2 , is shown below.
41. The region enclosed by the curve of f and the x-axis is rotated 360∘ about the x-axis.
[6 marks]
Find the volume of the solid formed.
Markscheme
attempt to set up integral expression
M1
−−−−−−−2
−−−−−−−2
2
e.g. π∫ √16 − 4x2 dx , 2π ∫0 (16 − 4x2 ) , ∫ √16 − 4x2 dx
3
∫ 16dx = 16x , ∫ 4x2 dx = 4x3 (seen anywhere)
A1A1
evidence of substituting limits into the integrand
(M1)
e.g. (32 − 323 ) − (−32 + 323 ) , 64 − 643
volume = 128π
3
A2
N3
[6 marks]
Examiners report
Many candidates correctly integrated using f(x) , although some neglected to square the function and mired themselves in awkward
integration attempts. Upon substituting the limits, many were unable to carry the calculation to completion. Occasionally the π was
neglected in a final answer. Weaker candidates considered the solid formed to be a sphere and did not use integration.
f(x) =
1
3
−
2
− 3x
Let f(x) = 12 x3 − x2 − 3x . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
42. Write down the coordinates of
(i)
[6 marks]
the image of B after reflection in the y-axis;
(ii) the image of B after translation by the vector (
−2
);
5
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor 12 .
Markscheme
(i) (−3, − 9)
A1
(ii) (1, − 4)
A1A1
N1
N2
(iii) reflection gives (3, 9)
stretch gives ( 32 , 9)
(A1)
A1A1
N3
[6 marks]
Examiners report
Candidates were generally successful in finding images after single transformations in part (b). Common incorrect answers for (biii)
included ( 32 , 92 ) , (6, 9) and (6, 18) , demonstrating difficulty with images from horizontal stretches.
Let f(x) =
cos x
sin x
, for sin x ≠ 0 .
43a. Use the quotient rule to show that f ′ (x) =
−1
sin2x
.
[5 marks]
Markscheme
d
dx
sin x = cos x ,
d
dx
cos x = − sin x (seen anywhere)
M1
evidence of using the quotient rule
A1
correct substitution
e.g.
sin x(− sin x)−cos x(cos x)
sin2x
f ′ (x) =
−(sin2x+ cos 2x)
f ′ (x) =
−1
sin2x
(A1)(A1)
,
− sin2x− cos 2x
sin2x
A1
sin2x
AG
N0
[5 marks]
Examiners report
Many candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identity
achieves the numerator of the answer given. Whether changing to −(sin x)−2 , or applying the quotient rule a second time, most
candidates neglected the chain rule in finding the second derivative.
43b. Find f ′′ (x) .
[3 marks]
Markscheme
METHOD 1
(M1)
appropriate approach
−2
′
e.g. f (x) = −(sin x)
f ′′ (x) = 2(sin−3 x)(cos x) (= 2 cos3 x )
A1A1
sin x
N3
Note: Award A1 for 2sin−3 x , A1 for cos x .
METHOD 2
derivative of sin2 x = 2 sin x cos x (seen anywhere)
(M1)
evidence of choosing quotient rule
e.g. u = −1 , v = sin2 x , f ′′ =
f ′′ (x) =
2 sin x cos x
2
(sin2x)
(= 2 cos3 x )
sin x
A1
2
sin x×0−(−1)2 sin x cos x
2
(sin2x)
A1
N3
[3 marks]
Examiners report
Whether changing to −(sin x)−2 , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the
second derivative.
( )
( )
In the following table, f ′ ( π2 ) = p and f ′′ ( π2 ) = q . The table also gives approximate values of f ′ (x) and f ′′ (x) near x = π2 .
43c. Find the value of p and of q.
[3 marks]
Markscheme
evidence of substituting
e.g.
−1
sin2 π
,
2
π
2
M1
2 cos π
2
sin3 π
2
p = −1 , q = 0
A1A1
N1N1
[3 marks]
Examiners report
Those who knew the trigonometric ratios at π2 typically found the values of p and of q, sometimes in follow-through from an incorrect
f ′′ (x) .
44. Let f(x) = kx4 . The point P(1, k) lies on the curve of f . At P, the normal to the curve is parallel to y = − 1 x . Find the value [6 marks]
8
of k.
Markscheme
gradient of tangent = 8 (seen anywhere)
f ′ (x) = 4kx3 (seen anywhere)
(A1)
A1
recognizing the gradient of the tangent is the derivative
setting the derivative equal to 8
e.g.
4kx3
=8,
kx3
(A1)
=2
substituting x = 1 (seen anywhere)
k=2
A1
(M1)
(M1)
N4
[6 marks]
Examiners report
Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of
the 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 x = 1 , or no marks for working
that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the
derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the
candidates who demonstrated greater understanding, more used the gradient of the normal (the equation − 14 k = − 18 ) than the
gradient of the tangent (4k = 8 ) ; this led to more algebraic errors in obtaining the final answer of k = 2 . A number of unsuccessful
candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.
A function f is defined for −4 ≤ x ≤ 3 . The graph of f is given below.
The graph has a local maximum when x = 0 , and local minima when x = −3 , x = 2 .
45a. Write down the x-intercepts of the graph of the derivative function, f ′ .
[2 marks]
Markscheme
x-intercepts at −3, 0, 2
A2
N2
[2 marks]
Examiners report
Candidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left this
question blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often had
trouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Other
candidates listed individual x-values here rather than a range of values.
45b. Write down all values of x for which f ′ (x) is positive.
[2 marks]
Markscheme
−3 < x < 0 , 2 < x < 3
A1A1
N2
[2 marks]
Examiners report
Candidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left this
question blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often had
trouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Other
candidates listed individual x-values here rather than a range of values.
45c. At point D on the graph of f , the x-coordinate is −0.5. Explain why f ′′ (x) < 0 at D.
Markscheme
correct reasoning
R2
e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing
therefore the second derivative is negative
[2 marks]
AG
[2 marks]
Examiners report
Many candidates had difficulty explaining why the second derivative is negative in part (c). A number claimed that since the point D
was “close” to a maximum value, the second derivative must be negative; this incorrect appeal to the second derivative test indicates a
lack of understanding of how the test works and the relative concept of closeness. Some candidates claimed D was a point of
inflexion, again demonstrating poor understanding of the second derivative. Among candidates who answered part (c) correctly, some
stated that f was concave down while others gave well-formed arguments for why the first derivative was decreasing. A few
candidates provided nicely sketched graphs of f ′ and f ′′ and used them in their explanations.
Consider the function f with second derivative f ′′ (x) = 3x − 1 . The graph of f has a minimum point at A(2, 4) and a maximum point at
B (− 43 , 358
).
27
46a. Use the second derivative to justify that B is a maximum.
[3 marks]
Markscheme
substituting into the second derivative
M1
e.g. 3 × (− 43 ) − 1
f ′′ (− 43 ) = −5
A1
since the second derivative is negative, B is a maximum
R1
N0
[3 marks]
Examiners report
Many candidates were successful with this question. In part (a), some candidates found f ′′ (− 43 ) and were unclear how to conclude,
but most demonstrated a good understanding of the second derivative test.
46b. Given that f ′ (x) = 3 x2 − x + p , show that p = −4 .
2
[4 marks]
Markscheme
setting f ′ (x) equal to zero
(M1)
evidence of substituting x = 2 (or x = − 43 )
(M1)
′
e.g. f (2)
correct substitution
A1
2
e.g. 32 (2)2 − 2 + p , 32 (− 43 ) − (− 43 ) + p
correct simplification
e.g. 6 − 2 + p = 0 , 83 + 43 + p = 0 , 4 + p = 0
p = −4
AG
A1
N0
[4 marks]
Examiners report
A large percentage of candidates were successful in showing that p = −4 but there were still some who worked backwards from the
answer. Others did not use the given information and worked from the second derivative, integrated, and then realized that p was the
constant of integration. Candidates who evaluated the derivative at x = 2 but set the result equal to 4 clearly did not understand the
concept being assessed. Few candidates used the point B with fractional coordinates.
f(x)
46c. Find f(x) .
[7 marks]
Markscheme
evidence of integration
f(x) =
1 3
x
2
−
1 2
x
2
(M1)
− 4x + c
A1A1A1
substituting (2, 4) or (− 43 , 358
) into their expression
27
A1
correct equation
1
2
3
(M1)
1
2
2
e.g. × 2 − × 2 − 4 × 2 + c = 4 , 12 × 8 − 12 × 4 − 4 × 2 + c = 4 , 4 − 2 − 8 + c = 4
f(x) = 12 x3 − 12 x2 − 4x + 10
A1
N4
[7 marks]
Examiners report
Candidates often did well on the first part of (c), knowing to integrate and successfully finding some or all terms. Some had trouble
with the fractions or made careless errors with the signs; others did not use the value of p = −4 and so could not find the third term
when integrating. It was very common for candidates to either forget the constant of integration or to leave it in without finding its
value.
Let f(x) = 6 + 6 sin x . 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.
47a. Solve for 0 ≤ x < 2π
(i)
[5 marks]
6 + 6 sin x = 6 ;
(ii) 6 + 6 sin x = 0 .
Markscheme
(i) sin x = 0
A1
x=0,x=π
A1A1
(ii) sin x = −1
A1
x=
3π
2
A1
[5 marks]
N1
N2
Examiners report
Many candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not show
sufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidates
worked in degrees; the majority worked in radians.
47b. Write down the exact value of the x-intercept of f , for 0 ≤ x < 2π .
[1 mark]
Markscheme
3π
2
A1
N1
[1 mark]
Examiners report
While some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), most
used their working from part (a)(ii) sometimes with follow-through on an incorrect answer.
47c. The area of the shaded region is k . Find the value of k , giving your answer in terms of π .
[6 marks]
Markscheme
evidence of using anti-differentiation
e.g. ∫0
3π
2
(M1)
(6 + 6 sin x)dx
correct integral 6x − 6 cos x (seen anywhere)
correct substitution
A1A1
(A1)
e.g. 6 ( 3π
) − 6 cos ( 3π
) − (−6 cos 0) , 9π − 0 + 6
2
2
k = 9π + 6
A1A1
N3
[6 marks]
Examiners report
Most 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; without
this working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with the
common trigonometric values.
47d. Let g(x) = 6 + 6 sin (x − π ) . The graph of f is transformed to the graph of g.
2
Give a full geometric description of this transformation.
Markscheme
translation of ( 2 )
0
π
[2 marks]
A1A1
N2
[2 marks]
Examiners report
While 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 π2 .
47e. Let g(x) = 6 + 6 sin (x − π ) . The graph of f is transformed to the graph of g.
2
p+ 3π
Given that ∫p
2
[3 marks]
g(x)dx = k and 0 ≤ p < 2π , write down the two values of p.
Markscheme
recognizing that the area under g is the same as the shaded region in f
p=
π
2
,p=0
A1A1
(M1)
N3
[3 marks]
Examiners report
Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority
tried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in the
wording of the question did not appear to have been noticed by candidates in this part either.
The velocity v ms−1 of an object after t seconds is given by v(t) = 15√t − 3t , for 0 ≤ t ≤ 25 .
48a. On the grid below, sketch the graph of v , clearly indicating the maximum point.
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for approximately correct shape, A1 for right endpoint at (25, 0) and A1 for maximum point in circle.
[3 marks]
Examiners report
The graph in part (a) was well done. It was pleasing to see many candidates considering the domain as they sketched their graph.
48b. (i)
Write down an expression for d .
[4 marks]
(ii) Hence, write down the value of d .
Markscheme
(i) recognizing that d is the area under the curve
(M1)
e.g. ∫ v(t)
correct expression in terms of t, with correct limits
9
A2
N3
9
e.g. d = ∫0 (15√t − 3t)dt , d = ∫0 vdt
(ii) d = 148.5 (m) (accept 149 to 3 sf)
A1
N1
[4 marks]
Examiners report
Part (b) (i) asked for an expression which bewildered a great many candidates. However, few had difficulty obtaining the correct
answer in (b) (ii).
Let f ′ (x) = −24x3 + 9x2 + 3x + 1 .
49a. There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.
[3 marks]
Markscheme
R1
valid approach
′′
e.g. f (x) = 0 , the max and min of f ′ gives the points of inflexion on f
−0.114, 0.364 (accept (−0.114, 0.811) and (0.364, 2.13))
A1A1
N1N1
[3 marks]
Examiners report
There were mixed results in part (a). Students were required to understand the relationships between a function and its derivative and
often obtained the correct solutions with incorrect or missing reasoning.
49b. Let g(x) = f ′′ (x) . Explain why the graph of g has no points of inflexion.
[2 marks]
Markscheme
METHOD 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
therefore no change in concavity
R1
R1
N1
N1
METHOD 3
g′′ (x) = −144
R1 N1
therefore no points of inflexion as g′′ (x) ≠ 0
R1
N1
[2 marks]
Examiners report
In part (b), the question was worth two marks and candidates were required to make two valid points in their explanation. There were
many approaches to take here and candidates often confused their reasoning or just kept writing hoping that somewhere along the
way they would say something correct to pick up the points. Many confused f ′ and g′ .
f(x) = x ln(4 −
2)
Let f(x) = x ln(4 − x2 ) , for −2 < x < 2 . The graph of f is shown below.
The graph of f crosses the x-axis at x = a , x = 0 and x = b .
50a. Find the value of a and of b .
[3 marks]
Markscheme
evidence of valid approach
(M1)
e.g. f(x) = 0 , graph
a = −1.73 , b = 1.73 (a = −√3, b = √3)
A1A1
N3
[3 marks]
Examiners report
This question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. For
example, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.
50b. The graph of f has a maximum value when x = c .
[2 marks]
Find the value of c .
Markscheme
attempt to find max
(M1)
′
e.g. setting f (x) = 0 , graph
c = 1.15 (accept (1.15, 1.13))
A1
N2
[2 marks]
Examiners report
This question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. For
example, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.
50c. The region under the graph of f from x = 0 to x = c is rotated 360∘ about the x-axis. Find the volume of the solid formed.
[3 marks]
Markscheme
attempt to substitute either limits or the function into formula
e.g. V =
c
π∫0 [f(x)]2 dx
V = 2.16 A2
2
, π∫ [x ln(4 − x2 )] ,
M1
1.149… 2
π ∫0
y dx
N2
[3 marks]
Examiners report
Most candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find the
correct answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) were
always futile and no marks were gained.
50d. Let R be the region enclosed by the curve, the x-axis and the line x = c , between x = a and x = c .
[4 marks]
Find the area of R .
Markscheme
valid approach recognizing 2 regions
(M1)
e.g. finding 2 areas
correct working
−1.73…
e.g. ∫0
(A1)
1.149…
f(x)dx + ∫0
area = 2.07 (accept 2.06)
A2
0
1.149…
f(x)dx , − ∫−1.73… f(x)dx + ∫0
f(x)dx
N3
[4 marks]
Examiners report
Most candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find the
correct answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) were
always futile and no marks were gained.
The diagram below shows a plan for a window in the shape of a trapezium.
Three sides of the window are 2 m long. The angle between the sloping sides of the window and the base is θ , where 0 < θ < π2 .
51a. Show that the area of the window is given by y = 4 sin θ + 2 sin 2θ .
[5 marks]
Markscheme
evidence of finding height, h
e.g. sin θ =
h
2
(A1)
, 2 sin θ
evidence of finding base of triangle, b
(A1)
e.g. cos θ = , 2 cos θ
b
2
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 θ )
A1
e.g. 2 ( 12 × 2 cos θ × 2 sin θ) + 2 × 2 sin θ , 12 (2 sin θ)(2 + 2 + 4 cos θ)
attempt to replace 2 sin θ cos θ by sin 2θ
M1
e.g. 4 sin θ + 2(2 sin θ cos θ)
y = 4 sin θ + 2 sin 2θ
AG
N0
[5 marks]
Examiners report
As the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) was
generally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in a
clear and organized manner. Some tried a "working backwards" approach, earning no marks.
51b. Zoe wants a window to have an area of 5 m2 . Find the two possible values of θ .
[4 marks]
Markscheme
correct equation
A1
e.g. y = 5 , 4 sin θ + 2 sin 2θ = 5
evidence of attempt to solve
(M1)
e.g. a sketch, 4 sin θ + 2 sin θ − 5 = 0
θ = 0.856 (49.0∘ ) , θ = 1.25 (71.4∘ )
A1A1
N3
[4 marks]
Examiners report
In part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and instead
attempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a second
solution outside the domain.
51c. John wants two windows which have the same area A but different values of θ .
Find all possible values for A .
[7 marks]
Markscheme
recognition that lower area value occurs at θ = π2
finding value of area at θ =
π
2
(M1)
(M1)
e.g. 4 sin ( π2 ) + 2 sin (2 × π2 ) , draw square
A=4
(A1)
(M1)
recognition that maximum value of y is needed
A = 5.19615 …
(A1)
4 < A < 5.20 (accept 4 < A < 5.19 )
A2
N5
[7 marks]
Examiners report
A pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or the
maximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident from
candidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken into
consideration for future paper writing.
p
Consider f(x) = x2 + x , x ≠ 0 , where p is a constant.
52a. Find f ′ (x) .
[2 marks]
Markscheme
f ′ (x) = 2x −
p
x2
A1A1
N2
Note: Award A1 for 2x , A1 for −
p
x2
.
[2 marks]
Examiners report
Candidates did well on (a).
52b. There is a minimum value of f(x) when x = −2 . Find the value of p .
Markscheme
evidence of equating derivative to 0 (seen anywhere)
′
evidence of finding f (−2) (seen anywhere)
correct equation
A1
p
4
e.g. −4 − = 0 , −16 − p = 0
p = −16
[4 marks]
A1
N3
(M1)
(M1)
[4 marks]
Examiners report
For (b), a great number of candidates substituted into the function instead of into the derivative.
p
The derivate of x2 was calculated without difficulties, but there were numerous problems regarding the derivative of x . There were
several candidates who considered both p and x as variables; some tried to use the quotient rule and had difficulties, others used
negative exponents and were not successful.
Let f(x) = 3 +
20
x2−4
, for x ≠ ±2 . The graph of f is given below.
The y-intercept is at the point A.
53a. (i)
Find the coordinates of A.
[7 marks]
(ii) Show that f ′ (x) = 0 at A.
Markscheme
(i) coordinates of A are (0, − 2)
(ii) derivative of
x2
A1A1
N2
− 4 = 2x (seen anywhere)
evidence of correct approach
(A1)
(M1)
e.g. quotient rule, chain rule
finding f ′ (x)
A2
e.g. f ′ (x) = 20 × (−1) × (x2 − 4)−2 × (2x) ,
(x2−4)(0)−(20)(2x)
(x2−4)2
substituting x = 0 into f ′ (x) (do not accept solving f ′ (x) = 0 )
′
at A f (x) = 0
AG
M1
N0
[7 marks]
Examiners report
Almost all candidates earned the first two marks in part (a) (i), although fewer were able to apply the quotient rule correctly.
40(3 2+4)
53b. The second derivative f ′′ (x) =
(i)
40(3x2+4)
(x2−4)3
[6 marks]
. Use this to
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 f ′ (x) = 0 (seen anywhere)
(R1)
reference to f ′′ (0) is negative (seen anywhere)
′′
evidence of substituting x = 0 into f (x)
finding f ′′ (0) =
40×4
(−4)3
(= − 52 )
M1
A1
AG
then the graph must have a local maximum
′′
(ii) reference to f (x) = 0 at point of inflexion
recognizing that the second derivative is never 0
e.g.
40(3x2
+ 4) ≠ 0 ,
3x2
R1
+4 ≠ 0 ,
x2
(R1)
A1
N2
4
3
≠ − , the numerator is always positive
Note: Do not accept the use of the first derivative in part (b).
[6 marks]
Examiners report
Many candidates were able to state how the second derivative can be used to identify maximum and inflection points, but fewer were
actually able to demonstrate this with the given function. For example, in (b)(ii) candidates often simply said "the second derivative
cannot equal 0" but did not justify or explain why this was true.
53c. Describe the behaviour of the graph of f for large |x| .
[1 mark]
Markscheme
correct (informal) statement, including reference to approaching y = 3
A1
N1
e.g. getting closer to the line y = 3 , horizontal asymptote at y = 3
[1 mark]
Examiners report
Not too many candidates could do part (c) correctly.
Let f(x) = √x . Line L is the normal to the graph of f at the point (4, 2) .
54a. Show that the equation of L is y = −4x + 18 .
[4 marks]
Markscheme
(A1)
finding derivative
1
2
e.g. f ′ (x) = 12 x , 2 1 x
√
correct value of derivative or its negative reciprocal (seen anywhere)
e.g.
1
2√4
,
gradient of normal =
e.g. −
A1
1
4
1
f ′(4)
1
gradient of tangent
(seen anywhere)
A1
= −4 , −2√x
substituting into equation of line (for normal)
M1
e.g. y − 2 = −4(x − 4)
y = −4x + 18
AG
N0
[4 marks]
Examiners report
Parts (a) and (b) were well done by most candidates.
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .
54b. Find an expression for the area of R .
[3 marks]
Markscheme
splitting into two appropriate parts (areas and/or integrals)
correct expression for area of R
4
A2
4.5
(M1)
N3
4
e.g. area of R = ∫0 √xdx + ∫4 (−4x + 18)dx , ∫0 √xdx + 12 × 0.5 × 2 (triangle)
Note: Award A1 if dx is missing.
[3 marks]
Examiners report
While quite a few candidates understood that both functions must be used to find the area in part (c), very few were actually able to
write a correct expression for this area and this was due to candidates not knowing that they needed to integrate from 0 to 4 and then
from 4 to 4.5.
54c. The region R is rotated 360∘ about the x-axis. Find the volume of the solid formed, giving your answer in terms of π .
[8 marks]
Markscheme
correct expression for the volume from x = 0 to x = 4
(A1)
e.g. V = ∫0 π [f(x)2 ]dx , ∫0 π√x2 dx , ∫0 πxdx
4
V = [ 12 πx2 ]
4
4
0
A1
V = π ( 12 × 16 − 12 × 0)
V = 8π
4
(A1)
A1
finding the volume from x = 4 to x = 4.5
EITHER
recognizing a cone
(M1)
e.g. V = 13 πr2 h
V = 13 π(2)2 × 12
=
2π
3
(A1)
A1
total volume is 8π + 23 π (=
26
π)
3
A1
N4
A1
N4
OR
4.5
V = π ∫4 (−4x + 18)2 dx
(M1)
4.5
= ∫4 π(16x2 − 144x + 324)dx
= π[ 163 x3 − 72x2 + 324x]
=
2π
3
4.5
4
A1
A1
total volume is 8π + 23 π (=
26
π)
3
[8 marks]
Examiners report
On part (d), some candidates were able to earn follow through marks by setting up a volume expression, but most of these
expressions 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 applied
distributive law of the exponent with respect to the addition or subtraction.
Let f(x) = cos 2x and g(x) = ln(3x − 5) .
55a. Find f ′ (x) .
[2 marks]
f ′ (x) = − sin 2x × 2(= −2 sin 2x)
− sin 2x
Examiners report
Almost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of the
chain rule, forgetting to account for the coefficient of x in their derivatives. A few did not know how to use the product rule, even
though it is in the information booklet.
55b. Find g ′ (x) .
[2 marks]
Markscheme
1
g′ (x) = 3 × 3x−5
(=
3
)
3x−5
Note: Award A1 for 3, A1 for
A1A1
1
3x−5
N2
.
[2 marks]
Examiners report
Almost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of the
chain rule, forgetting to account for the coefficient of x in their derivatives. A few did not know how to use the product rule, even
though it is in the information booklet.
55c. Let h(x) = f(x) × g(x) . Find h′ (x) .
[2 marks]
Markscheme
evidence of using product rule
(M1)
3
h′ (x) = (cos 2x) ( 3x−5
) + ln(3x − 5)(−2 sin 2x)
A1
N2
[2 marks]
Examiners report
Almost all candidates earned at least some of the marks on this question. Some weaker students showed partial knowledge of the
chain rule, forgetting to account for the coefficient of x in their derivatives. A few did not know how to use the product rule, even
though it is in the information booklet.
56. Consider the curve with equation f(x) = px2 + qx , where p and q are constants. The point A(1, 3) lies on the curve. The
tangent to the curve at A has gradient 8. Find the value of p and of q .
Markscheme
substituting x = 1 , y = 3 into f(x)
3 = p+q
A1
finding derivative
′
f (x) = 2px + q
(M1)
A1
correct substitution, 2p + q = 8
p = 5 , q = −2
[7 marks]
A1A1
N2N2
A1
(M1)
[7 marks]
Examiners report
A good number of candidates were able to obtain an equation by substituting the point 1, 3 into the function’s equation. Not as many
knew how to find the other equation by using the derivative. Some candidates thought they needed to find the equation of the tangent
line rather than recognising that the information about the tangent provided the gradient of the function at the point. While they were
usually able to find this equation correctly, it was irrelevant to the question asked.
A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m2, as shown below.
57. The fencing used for side AB costs $11 per metre. The fencing for the other three sides costs $3 per metre. The farmer creates [7 marks]
an enclosure so that the cost is a minimum. Find this minimum cost.
Markscheme
METHOD 1
correct expression for second side, using area = 525
e.g. let AB = x , AD =
(A1)
525
x
(M1)
attempt to set up cost function using $3 for three sides and $11 for one side
e.g. 3(AD + BC + CD) + 11AB
correct expression for cost
e.g.
525
x
×3+
525
x
A2
× 3 + 11x + 3x ,
525
AB
× 3 + 525
× 3 + 11AB + 3AB ,
AB
3150
x
+ 14x
EITHER
(M1)
sketch of cost function
(A1)
identifying minimum point
e.g. marking point on graph, x = 15
A1
minimum cost is 420 (dollars)
N4
OR
correct derivative (may be seen in equation below)
−1575
x2
e.g. C ′ (x) =
+ −1575
+ 14
2
x
setting their derivative equal to 0 (seen anywhere)
e.g.
−3150
x2
(A1)
(M1)
+ 14 = 0
A1
minimum cost is 420 (dollars)
N4
METHOD 2
correct expression for second side, using area = 525
e.g. let AD = x , AB =
(A1)
525
x
attempt to set up cost function using $3 for three sides and $11 for one side
e.g. 3(AD + BC + CD) + 11AB
correct expression for cost
A2
e.g. 3 (x + x + 525
) + 525
× 11 , 3 (AD + AD + 525
) + 525
× 11 , 6x + 7350
x
x
x
AD
AD
EITHER
sketch of cost function
(M1)
identifying minimum point
(A1)
e.g. marking point on graph, x = 35
minimum cost is 420 (dollars)
A1
N4
OR
correct derivative (may be seen in equation below)
(A1)
e.g. C ′ (x) = 6 − 7350
2
x
setting their derivative equal to 0 (seen anywhere)
e.g. 6 −
7350
x2
=0
minimum cost is 420 (dollars)
[7 marks]
A1
N4
(M1)
(M1)
Examiners report
Although 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 cost
function in two variables often could not use the area relationship to obtain a function in a single variable and so could make no
further progress. Of those few who created a correct cost function, most set the derivative to zero to find that the minimum cost
occurred at x = 15 , leading to $420. Although this is a correct approach earning full marks, candidates seem not to recognise that the
result can be obtained from the GDC, without the use of calculus.
Let f(x) = x cos x , for 0 ≤ x ≤ 6 .
58a. Find f ′ (x) .
[3 marks]
Markscheme
evidence of choosing the product rule
(M1)
e.g. x × (− sin x) + 1 × cos x
f ′ (x) = cos x − x sin x
A1A1
N3
[3 marks]
Examiners report
This problem was well done by most candidates. There were some candidates that struggled to apply the product rule in part (a) and
often wrote nonsense like −x sin x = − sin x2 .
58b. On the grid below, sketch the graph of y = f ′ (x) .
[4 marks]
Markscheme
A1A1A1A1
N4
Note: Award A1 for correct domain, 0 ≤ x ≤ 6 with endpoints in circles, A1 for approximately correct shape, A1 for local minimum
in circle, A1 for local maximum in circle.
[4 marks]
Examiners report
In part (b), few candidates were able to sketch the function within the required domain and a large number of candidates did not have
their calculator in the correct mode.
59. The acceleration, a ms−2 , of a particle at time t seconds is given by
a=
[7 marks]
1
+ 3 sin 2t, for t ≥ 1.
t
The particle is at rest when t = 1 .
Find the velocity of the particle when t = 5 .
Markscheme
evidence of integrating the acceleration function
(M1)
e.g. ∫ ( 1t + 3 sin 2t)dt
correct expression ln t − 32 cos 2t + c
evidence of substituting (1, 0)
A1A1
(M1)
e.g. 0 = ln 1 − 32 cos 2 + c
c = −0.624 (= 32 cos 2 − ln 1 or 32 cos 2)
(A1)
v = ln t − 32 cos 2t − 0.624 (= ln t − 32 cos 2t + 32 cos 2 or lnt − 32 cos 2t + 32 cos 2 − ln 1)
v(5) = 2.24 (accept the exact answer ln 5 − 1.5 cos 10 + 1.5 cos 2 ) A1
[7 marks]
N3
(A1)
Examiners report
This problem was not well done. A large number of students failed to recognize that they needed to integrate the acceleration
function. Even among those who integrated the function, there were many who integrated incorrectly. A great number of candidates
were not able to handle the given initial condition to find the integration constant but incorrectly substituted t = 5 directly into their
expression.
Let f(x) = Aekx + 3 . Part of the graph of f is shown below.
The y-intercept is at (0, 13) .
60a. Show that A = 10 .
[2 marks]
Markscheme
M1
substituting (0, 13) into function
e.g. 13 = Ae0 + 3
13 = A + 3
A = 10
A1
AG
N0
[2 marks]
Examiners report
This question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by most
centres. Parts (a) and (b) proved very accessible to many candidates.
60b. Given that f(15) = 3.49 (correct to 3 significant figures), find the value of k.
Markscheme
substituting into f(15) = 3.49
e.g. 3.49 =
10e15k
A1
+ 3 , 0.049 = e15k
evidence of solving equation
(M1)
e.g. sketch, using ln
k = −0.201 (accept ln 0.049
)
15
[3 marks]
A1
N2
[3 marks]
Examiners report
This question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by most
centres. Parts (a) and (b) proved very accessible to many candidates.
60c. (i)
Using your value of k , find f ′ (x) .
[5 marks]
(ii) Hence, explain why f is a decreasing function.
(iii) Write down the equation of the horizontal asymptote of the graph f .
Markscheme
(i) f(x) = 10e−0.201x + 3
f(x) = 10e−0.201x × −0.201 (= −2.01e−0.201x )
A1A1A1
N3
Note: Award A1 for 10e−0.201x , A1 for × − 0.201 , A1 for the derivative of 3 is zero.
(ii) valid reason with reference to derivative
R1
N1
′
e.g. f (x) < 0 , derivative always negative
(iii) y = 3
A1
N1
[5 marks]
Examiners report
The chain rule in part (c) was also carried out well. Few however, recognized the command term “hence” and that f ′ (x) < 0
guarantees a decreasing function. A common answer for the equation of the asymptote was to give y = 0 or x = 3 .
60d. Let g(x) = −x2 + 12x − 24 .
[6 marks]
Find the area enclosed by the graphs of f and g .
Markscheme
finding limits 3.8953 …, 8.6940 … (seen anywhere)
evidence of integrating and subtracting functions
correct expression
e.g.
8.69
∫3.90
A1A1
(M1)
A1
8.69
g(x) − f(x)dx , ∫3.90 [(−x2 + 12x − 24) − (10e−0.201x + 3)]dx
area = 19.5
A2
N4
[6 marks]
Examiners report
In part (d), it was again surprising and somewhat disappointing to see how few candidates were able to use their GDC effectively to
find the area between curves, often not finding correct limits, and often trying to evaluate the definite integral without the GDC,
which led nowhere.
f(x) =
x sin 2x + 10
Let f(x) = ex sin 2x + 10 , for 0 ≤ x ≤ 4 . 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 x = p and a local minimum point at N, where x = q .
61a. Write down the x-coordinate of A.
[1 mark]
Markscheme
2.31
A1
N1
[1 mark]
Examiners report
Parts (a) and (b) were generally well answered, the main problem being the accuracy.
61b. Find the value of
(i)
[2 marks]
p;
(ii) q .
Markscheme
(i) 1.02
A1
N1
(ii) 2.59
A1
N1
[2 marks]
Examiners report
Parts (a) and (b) were generally well answered, the main problem being the accuracy.
q
61c. Find ∫p f(x)dx . Explain why this is not the area of the shaded region.
[3 marks]
Markscheme
q
∫p f(x)dx = 9.96
A1
N1
split into two regions, make the area below the x-axis positive
[3 marks]
R1R1
N2
Examiners report
Many 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 was
the cause of the integral not giving the total area, but the explanations were not sufficiently clear. It was often stated that the area
below the axis was negative rather than the integral was negative.
Consider
f(x) = x ln(4 − x2 ) , for −2 < x < 2 . The graph of f is given below.
62a. 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)
[5 marks]
Find the x-coordinate of P and of Q.
(ii) Consider f(x) = k . Write down all values of k for which there are exactly two solutions.
Markscheme
(i) −1.15, 1.15
A1A1
N2
(ii) recognizing that it occurs at P and Q
(M1)
e.g. x = −1.15 , x = 1.15
k = −1.13 , k = 1.13
A1A1
N3
[5 marks]
Examiners report
Many candidates correctly found the x-coordinates of P and Q in (a)(i) with their GDC. In (a)(ii) some candidates incorrectly
interpreted the words “exactly two solutions” as an indication that the discriminant of a quadratic was required. Many failed to realise
that the values of k they were looking for in this question were the y-coordinates of the points found in (a)(i).
62b. Let g(x) = x3 ln(4 − x2 ) , for −2 < x < 2 .
4
Show that g′ (x) = −2x2 + 3x2 ln(4 − x2 ) .
4−x
[4 marks]
Markscheme
evidence of choosing the product rule
e.g.
uv′
derivative of x3 is 3x2
(A1)
derivative of ln(4 − x2 ) is
e.g.
×
−2x
4−x2
−2x
4−x2
(A1)
A1
correct substitution
x3
(M1)
+ vu′
+ ln(4 − x2 ) × 3x2
4
g′ (x) = −2x2 + 3x2 ln(4 − x2 )
4−x
AG
N0
[4 marks]
Examiners report
Many candidates were unclear in their application of the product formula in the verifying the given derivative of g. Showing that the
derivative was the given expression often received full marks though it was not easy to tell in some cases if that demonstration came
through understanding of the product and chain rules or from reasoning backwards from the given result.
62c. Let g(x) = x3 ln(4 − x2 ) , for −2 < x < 2 .
Sketch the graph of
g′
[2 marks]
.
Markscheme
A1A1
N2
[2 marks]
Examiners report
Some candidates drew their graphs of the derivative in (c) on their examination papers despite clear instructions to do their work on
separate sheets. Most who tried to plot the graph in (c) did so successfully.
62d. Let g(x) = x3 ln(4 − x2 ) , for −2 < x < 2 .
Consider g′ (x) = w . Write down all values of w for which there are exactly two solutions.
Markscheme
w = 2.69 , w < 0
[3 marks]
A1A2
N2
[3 marks]
Examiners report
Correct solutions to 10(d) were not often seen.
The diagram shows part of the graph of y = f ′ (x) . The x-intercepts are at points A and C. There is a minimum at B, and a maximum at D.
63a. (i)
Write down the value of f ′ (x) at C.
[3 marks]
(ii) Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.
Markscheme
(i) f ′ (x) = 0
A1
N1
(ii) METHOD 1
f ′ (x) < 0 to the left of C, f ′ (x) > 0 to the right of C
R1R1
N2
METHOD 2
f ′′ (x) > 0
R2
N2
[3 marks]
Examiners report
The 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 correctly
use the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking at
the graph of f ′ and not the graph of f .
63b. Which of the points A, B, D corresponds to a maximum on the graph of f ?
Markscheme
A
A1
[1 mark]
N1
[1 mark]
Examiners report
The 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 correctly
use the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking at
the graph of f ′ and not the graph of f .
63c. Show that B corresponds to a point of inflexion on the graph of f .
[3 marks]
Markscheme
METHOD 1
f ′′ (x) = 0
R2
discussion of sign change of f ′′ (x)
′′
R1
′′
e.g. f (x) < 0 to the left of B and f (x) > 0 to the right of B; f ′′ (x) 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 f ′
discussion of sign change of f ′′ (x)
′′
R2
R1
′′
e.g. f (x) < 0 to the left of B and f (x) > 0 to the right of B; f ′′ (x) changes sign either side of B
B is a point of inflexion
AG
N0
[3 marks]
Examiners report
In 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 information
should be given.
The acceleration, a ms−2 , of a particle at time t seconds is given by a = 2t + cos t .
64a. Find the acceleration of the particle at t = 0 .
[2 marks]
Markscheme
substituting t = 0
(M1)
e.g. a(0) = 0 + cos 0
a(0) = 1
A1
N2
[2 marks]
Examiners report
Parts (a) and (b) of this question were generally well done.
64b. Find the velocity, v, at time t, given that the initial velocity of the particle is ms−1 .
[5 marks]
Markscheme
(M1)
evidence of integrating the acceleration function
e.g. ∫ (2t + cos t)dt
correct expression t2 + sin t + c
A1A1
Note: If " +c" is omitted, award no further marks.
evidence of substituting (2,0) into indefinite integral
(M1)
e.g. 2 = 0 + sin 0 + c , c = 2
v(t) = t2 + sin t + 2
A1
N3
[5 marks]
Examiners report
Parts (a) and (b) of this question were generally well done.
64c. Find ∫ 3 vdt , giving your answer in the form p − q cos 3 .
0
[7 marks]
Markscheme
∫ (t2 + sin t + 2)dt =
t3
3
− cos t + 2t
A1A1A1
Note: Award A1 for each correct term.
evidence of using v(3) − v(0)
correct substitution
(M1)
A1
e.g. (9 − cos 3 + 6) − (0 − cos 0 + 0) , (15 − cos 3) − (−1)
16 − cos 3 (accept p = 16 , q = −1 )
A1A1
N3
[7 marks]
Examiners report
Problems arose in part (c) with many candidates not substituting s(3) − s(0) correctly, leading to only a partially correct final answer.
There were also a notable few who were not aware that cos 0 = 1 in both parts (a) and (c).
64d. What information does the answer to part (c) give about the motion of the particle?
[2 marks]
Markscheme
reference to motion, reference to first 3 seconds
R1R1
N2
e.g. displacement in 3 seconds, distance travelled in 3 seconds
[2 marks]
Examiners report
There 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
2
+ 5x − 8
Let f(x) = 5 cos π4 x and g(x) = −0.5x2 + 5x − 8 for 0 ≤ x ≤ 9 .
65. Let R be the region enclosed by the graphs of f and g . Find the area of R.
[5 marks]
Markscheme
METHOD 1
intersect when x = 2 and x = 6.79 (may be seen as limits of integration)
evidence of approach
(M1)
6.79
e.g. ∫ g − f , ∫ f(x)dx − ∫ g(x)dx , ∫2
area = 27.6
A2
A1A1
((−0.5x2 + 5x − 8) − (5 cos π4 x))
N3
METHOD 2
intersect when x = 2 and x = 6.79 (seen anywhere)
A1A1
evidence of approach using a sketch of g and f , or g − f .
(M1)
e.g. area = A + B − C , 12.7324 + 16.0938 − 1.18129 …
area = 27.6
A2
N3
[5 marks]
Examiners report
Part (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 find
points of intersection and the subsequent area.
66. Let f(x) = ex cos x . Find the gradient of the normal to the curve of f at x = π .
[6 marks]
Markscheme
evidence of choosing the product rule
′
f (x) =
ex
× (− sin x) + cos x × ex
substituting π
(M1)
(= ex cos x − ex sin x)
A1A1
(M1)
e.g. f ′ (π) = eπ cos π − eπ sin π , eπ (−1 − 0) , −eπ
taking negative reciprocal
e.g. −
(M1)
1
f ′(π)
gradient is
1
eπ
A1
N3
[6 marks]
Examiners report
Candidates familiar with the product rule easily found the correct derivative function. Many substituted π to find the tangent gradient,
but surprisingly few candidates correctly considered that the gradient of the normal is the negative reciprocal of this answer.
The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object as functions of time, t.
67a. Complete the following table by noting which graph A, B or C corresponds to each function.
Markscheme
A2A2
[4 marks]
N4
[4 marks]
Examiners report
Many candidates answered this question completely and correctly, showing a good understanding of the graphical relationship
between displacement, velocity and acceleration. When done incorrectly, many answered with the displacement as graph B and
acceleration as graph C.
67b. Write down the value of t when the velocity is greatest.
[2 marks]
Markscheme
t=3
A2
N2
[2 marks]
Examiners report
Many candidates found the value of t which gave a maximum in the remaining graph, and were awarded follow through marks.
68. The graph of y = √x between x = 0 and x = a is rotated 360∘ about the x-axis. The volume of the solid formed is 32π . Find [7 marks]
the value of a.
Markscheme
attempt to substitute into formula V = ∫ πy2 dx
A1
integral expression
a
π ∫0 (√x)2 dx
, π∫ x
correct integration
(A1)
e.g.
e.g. ∫ xdx =
(M1)
1 2
x
2
correct substitution V = π [ 12 a2 ]
(A1)
equating their expression to 32π
M1
e.g. π [ 12 a2 ] = 32π
a2 = 64
a=8
A2
N2
[7 marks]
Examiners report
Despite the “reverse” nature of this question, many candidates performed well with the integration. Some forgot to square the
function, while others did not discard the negative value of a. Some attempted to equate 32π to the formula for volume of a sphere,
which suggests this topic was not fully covered in some centres.
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 θ radians, where 0 ≤ θ ≤ π2 .
69. (i)
Find
dA
dθ
[8 marks]
.
(ii) Hence, find the exact value of θ which maximizes the area of the rectangle.
(iii) Use the second derivative to justify that this value of θ does give a maximum.
Markscheme
(i) dA
= 36 cos 2θ
dθ
A2
N2
(M1)
(ii) for setting derivative equal to 0
e.g. 36 cos 2θ = 0 ,
2θ = π2
θ = π4
dA
dθ
=0
(A1)
A1
N2
(iii) valid reason (seen anywhere)
e.g. at
π d2A
,
4 dθ2
R1
< 0 ; maximum when f ′′ (x) < 0
finding second derivative
evidence of substituting
π
4
d2A
dθ2
= −72 sin 2θ
A1
M1
e.g. −72 sin (2 × π4 ) , −72 sin ( π2 ) , −72
θ = π4 produces the maximum area
AG
N0
[8 marks]
Examiners report
As the area function was given in part (b), many candidates correctly found the derivative in (c) and knew to set this derivative to zero
for 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 a
maximum value. Simply calculating a negative value is not sufficient for a justification.
f(x) =
ax
Let f(x) =
ax
x2+1
, −8 ≤ x ≤ 8 , a ∈ R .The graph of f is shown below.
The region between x = 3 and x = 7 is shaded.
70a. Given that f ′′ (x) =
2ax(x2−3)
(x2+1)3
[7 marks]
, find the coordinates of all points of inflexion.
Markscheme
(M1)
evidence of appropriate approach
e.g. f ′′ (x) = 0
to set the numerator equal to 0
e.g.
2ax(x2
− 3) = 0 ;
(0, 0) , ( √3,
a√3
)
4
(x2
(A1)
− 3) = 0
, (−√3,−
a√3
)
4
(accept x = 0 , y = 0 etc)
A1A1A1A1A1
N5
[7 marks]
Examiners report
Although many recognized the requirement to set the second derivative to zero in (b), a majority neglected to give their answers as
ordered pairs, only writing the x-coordinates. Some did not consider the negative root.
70b. It is given that ∫ f(x)dx =
(i)
a
2
ln(x2 + 1) + C .
Find the area of the shaded region, giving your answer in the form p ln q .
(ii) Find the value of ∫48 2f(x − 1)dx .
[7 marks]
Markscheme
A2
(i) correct expression
7
e.g. [ a2 ln(x2 + 1)] , a2 ln 50 − a2 ln 10 , a2 (ln 50 − ln 10)
3
area = a2 ln 5
A1A1
N2
(ii) METHOD 1
recognizing the shift that does not change the area
e.g.
∫48 f(x − 1)dx
=
∫37 f(x)dx
(M1)
, ln 5
a
2
recognizing that the factor of 2 doubles the area
(M1)
e.g. ∫4 2f(x − 1)dx =2 ∫4 f(x − 1)dx (= 2 ∫3 f(x)dx)
8
8
7
8
∫4 2f(x − 1)dx = a ln 5 (i.e. 2 × their answer to (c)(i))
A1
N3
METHOD 2
changing variable
dw
dx
let w = x − 1 , so
=1
2a
ln(w2
2
2 ∫ f(w)dw =
+ 1) + c
(M1)
substituting correct limits
8
e.g. [a ln [ (x − 1)2 + 1]] , [aln(w2 + 1)]3 , a ln 50 − a ln 10
7
4
8
∫4 2f(x − 1)dx = a ln 5
A1
(M1)
N3
[7 marks]
Examiners report
For those who found a correct expression in (c)(i), many finished by calculating ln 50 − ln 10 = ln 40 . Few recognized that the
translation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).
Let f(x) = x3 − 4x + 1 .
f(x+h)−f(x)
71a. Use the formula f ′ (x) = lim
to show that the derivative of f(x) is 3x2 − 4 .
h→0
h
Markscheme
evidence of substituting x + h
A1
correct substitution
e.g. f ′ (x) = limh→0
simplifying
e.g.
(x+h)3−4(x+h)+1−(x3−4x+1)
h
A1
(x3+3x2h+3xh2+ h3−4x−4h+1− x3+4x−1)
h
factoring out h
e.g.
(M1)
A1
h(3x2+3xh+ h2−4)
h
f ′ (x) = 3x2 − 4
[4 marks]
AG
N0
[4 marks]
Examiners report
In part (b), it was clear that many candidates had difficulty with differentiation from first principles. Those that successfully set the
answer up, often got lost in the simplification.
71b. The tangent to the curve of f at the point P(1, − 2) is parallel to the tangent at a point Q. Find the coordinates of Q.
[4 marks]
Markscheme
f ′ (1) = −1
(A1)
M1
setting up an appropriate equation
e.g.
3x2
− 4 = −1
at Q, x = −1,y = 4 (Q is (−1, 4)) A1
A1
[4 marks]
Examiners report
Part (c) was poorly done with many candidates assuming that the tangents were horizontal and then incorrectly estimating the
maximum of f as the required point. Many candidates unnecessarily found the equation of the tangent and could not make any further
progress.
A function f has its first derivative given by f ′ (x) = (x − 3)3 .
72a. Find the second derivative.
[2 marks]
Markscheme
METHOD 1
f ′′ (x) = 3(x − 3)2
A2
N2
METHOD 2
attempt to expand (x − 3)3
′
e.g. f (x) =
x3
− 9x2
(M1)
+ 27x − 27
f ′′ (x) = 3x2 − 18x + 27
A1
N2
[2 marks]
Examiners report
Many candidates completed parts (a) and (b) successfully.
72b. Find f ′ (3) and f ′′ (3) .
[1 mark]
Markscheme
f ′ (3) = 0 , f ′′ (3) = 0
A1
N1
[1 mark]
Examiners report
Many candidates completed parts (a) and (b) successfully.
72c. The point P on the graph of f has x-coordinate 3. Explain why P is not a point of inflexion.
[2 marks]
Markscheme
METHOD 1
f ′′ does not change sign at P
evidence for this
R1
R1
N0
METHOD 2
f ′ changes sign at P so P is a maximum/minimum (i.e. not inflexion)
evidence for this
R1
R1
N0
METHOD 3
finding f(x) = 14 (x − 3)4 + c and sketching this function
indicating minimum at x = 3
R1
R1
N0
[2 marks]
Examiners report
A rare few earned any marks in part (c) - most justifying the point of inflexion with the zero answers in part (b), not thinking that
there is more to consider.
Let f(x) = e−3x and g(x) = sin (x − π3 ) .
73a. Write down
(i)
[2 marks]
′
f (x) ;
(ii) g′ (x) .
Markscheme
(i) −3e−3x
A1
(ii) cos (x − π3 )
N1
A1
N1
[4 marks]
Examiners report
A good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixed
success.
73b.
Let h(x) = e−3x sin (x − π3 ) . Find the exact value of h′ ( π3 ) .
[4 marks]
Markscheme
evidence of choosing product rule
e.g.
uv′
(M1)
+ vu′
correct expression
A1
e.g. −3e−3x sin (x − π3 ) + e−3x cos (x − π3 )
complete correct substitution of x = π3
−3 π
3
e.g. −3e
sin ( π3 − π3 ) + e
h′ ( π3 ) = e−π
−3 π
3
A1
(A1)
cos ( π3 − π3 )
N3
[4 marks]
Examiners report
Often the substitution of π3 was incomplete or not done at all.
In this question s represents displacement in metres and t represents time in seconds.
The velocity v m s–1 of a moving body is given by v = 40 − at where a is a non-zero constant.
74a. (i)
If s = 100 when t = 0 , find an expression for s in terms of a and t.
[6 marks]
(ii) If s = 0 when t = 0 , write down an expression for s in terms of a and t.
Markscheme
Note: In this question, do not penalize absence of units.
(i) s = ∫ (40 − at)dt
s = 40t − 12 at2 + c
(M1)
(A1)(A1)
substituting s = 100 when t = 0 (c = 100 )
s = 40t − 12 at2 + 100
(ii) s = 40t − 12 at2
A1
A1
(M1)
N5
N1
[6 marks]
Examiners report
Part (a) proved accessible for most.
Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by v = 40 − at , where
t = 0 at P. The station is 500 m from P.
74b. 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 a = 85 .
[6 marks]
Markscheme
(i) stops at station, so v = 0
t=
40
a
(seconds)
A1
(M1)
N2
(ii) evidence of choosing formula for s from (a) (ii)
substituting t =
40
a
(M1)
(M1)
2
e.g. 40 × 40
− 12 a × 40
a
a2
setting up equation
M1
2
e.g. 500 = s , 500 = 40 × 40
− 12 a × 40
, 500 = 1600
− 800
a
a
a
a2
evidence of simplification to an expression which obviously leads to a =
8
5
A1
e.g. 500a = 800 , 5 = a8 , 1000a = 3200 − 1600
a=
8
5
AG
N0
[6 marks]
Examiners report
Part (b), simple as it is, proved elusive as many candidates did not make the connection that v = 0 when the train stops. Instead, many
attempted to find the value of t using a = 85 .
74c. For a different train N, the value of a is 4.
Show that this train will stop before it reaches the station.
[5 marks]
Markscheme
METHOD 1
v = 40 − 4t , stops when v = 0
40 − 4t = 0
t = 10
(A1)
A1
M1
substituting into expression for s
s = 40 × 10 − 12 × 4 × 102
s = 200
A1
since 200 < 500 (allow FT on their s, if s < 500 )
train stops before the station
AG
R1
N0
METHOD 2
from (b) t =
40
4
= 10
A2
substituting into expression for s
e.g. s = 40 × 10 − 12 × 4 × 102
s = 200
M1
A1
since 200 < 500
R1
train stops before the station
AG
N0
METHOD 3
a is deceleration
4>
8
5
A2
A1
(A1)
so stops in shorter time
R1
so less distance travelled
so stops before station
AG
N0
[5 marks]
Examiners report
Few were successful in part (c).
Consider the curve y = ln(3x − 1) . Let P be the point on the curve where x = 2 .
75a. Write down the gradient of the curve at P.
[2 marks]
Markscheme
gradient is 0.6
A2
N2
[2 marks]
Examiners report
Although the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding the
derivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earned
follow through marks in part (b).
75b. The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.
[5 marks]
Markscheme
at R, y = 0 (seen anywhere)
A1
at x = 2 , y = ln 5 (= 1.609 …)
(A1)
gradient of normal = −1.6666 …
(A1)
evidence of finding correct equation of normal
e.g. y = ln 5 = −
5
(x − 2)
3
x = 2.97 (accept 2.96)
A1
, y = −1.67x + c
A1
coordinates of R are (2.97,0)
N3
[5 marks]
Examiners report
Although the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding the
derivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earned
follow through marks in part (b).
Let f(x) = x(x − 5)2 , for 0 ≤ x ≤ 6 . The following diagram shows the graph of f .
Let R be the region enclosed by the x-axis and the curve of f .
76a. Find the area of R.
[3 marks]
Markscheme
finding the limits x = 0 , x = 5
integral expression
e.g.
(A1)
A1
5
∫0 f(x)dx
area = 52.1
A1
N2
[3 marks]
Examiners report
Many candidates set up a completely correct equation for the area enclosed by the x-axis and the curve. Also, many of them tried an
analytic approach which sometimes returned incorrect answers. Using the wrong limits 0 and 6 was a common error.
76b. Find the volume of the solid formed when R is rotated through 360∘ about the x-axis.
[4 marks]
Markscheme
evidence of using formula v = ∫ πy2 dx
correct expression
e.g. volume =
(M1)
A1
π ∫05 x2 (x − 5)4 dx
volume = 2340
A2
N2
[4 marks]
Examiners report
The formula for the volume of revolution given in the data booklet was seen many times in part (b). Some candidates wrote the
integrand incorrectly, either missing the π or not squaring. A good number of students could write a completely correct integral
expression for the volume of revolution but fewer could evaluate it correctly as many started an analytical approach instead of using
their 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 the
volume of revolution. This probably caused a shortage of time for later questions.
Let f(x) = 3 sin x + 4 cos x , for −2π ≤ x ≤ 2π .
77. Write down one value of x such that f ′ (x) = 0 .
[2 marks]
Markscheme
evidence of correct approach
(M1)
e.g. max/min, sketch of f ′ (x) indicating roots
one 3 s.f. value which rounds to one of −5.6, −2.5, 0.64, 3.8
A1
N2
[2 marks]
Examiners report
The most common approach in part (d) was to differentiate and set f ′ (x) = 0 . Fewer students found the values of x given by the
maximum or minimum values on their graphs.
Let f(x) = x cos(x − sin x) , 0 ≤ x ≤ 3 .
78. The graph of f is revolved 360∘ about the x-axis from x = 0 to x = a . Find the volume of the solid formed.
[4 marks]
Markscheme
evidence of using V = π∫ [f(x)]2 dx
A2
fully correct integral expression
2.31
e.g. V = π∫0
(M1)
2.31
[x cos(x − sin x)]2 dx , V = π∫0
[f(x)]2 dx
A1
N2
V = 5.90
[4 marks]
Examiners report
A good number of candidates could set up the correct integral expression for volume, but surprisingly few were able to use their
GDC to find the correct value. Some attempted to analytically integrate the square of this unusual function, expending valuable time
in this effort. A small but significant number of candidates wrote a final answer as 1.88π , which accrued the accuracy penalty.
Let f(x) = e2x cos x , −1 ≤ x ≤ 2 .
79a. Show that f ′ (x) = e2x (2 cos x − sin x) .
[3 marks]
Markscheme
correctly finding the derivative of e2x , i.e. 2e2x
A1
correctly finding the derivative of cos x , i.e. − sin x
evidence of using the product rule, seen anywhere
A1
M1
′
e.g. f (x) = 2e2x cos x − e2x sin x
f ′ (x) = 2e2x (2 cos x − sin x)
AG
N0
[3 marks]
Examiners report
A good number of candidates demonstrated the ability to apply the product and chain rules to obtain the given derivative.
79b. Let the line L be the normal to the curve of f at x = 0 .
Find the equation of L .
Markscheme
evidence of finding f(0) = 1 , seen anywhere
attempt to find the gradient of f
A1
(M1)
′
e.g. substituting x = 0 into f (x)
value of the gradient of f
A1
e.g. f ′ (0) = 2 , equation of tangent is y = 2x + 1
gradient of normal = − 12
(A1)
y − 1 = − 12 x (y = − 12 x + 1)
[5 marks]
A1
N3
[5 marks]
Examiners report
Where candidates recognized that the gradient of the tangent is the derivative, many went on to correctly find the equation of the
normal.
79c. The graph of f and the line L intersect at the point (0, 1) and at a second point P.
(i)
[6 marks]
Find the x-coordinate of P.
(ii) Find the area of the region enclosed by the graph of f and the line L .
Markscheme
M1
(i) evidence of equating correct functions
e.g.
e2x cos x
x = 1.56
=−
A1
1
x+1
2
, sketch showing intersection of graphs
N1
(ii) evidence of approach involving subtraction of integrals/areas
(M1)
e.g. ∫ [f(x) − g(x)]dx , ∫ f(x)dx − area under trapezium
fully correct integral expression
1.56
e.g. ∫0
A2
[ e2x cos x − (− 12 x + 1)]dx , ∫0
1.56 2x
e cos xdx − 0.951 …
area = 3.28
A1
N2
[6 marks]
Examiners report
Few candidates showed the setup of the equation in part (c) before writing their answer from the GDC. Although a good number of
candidates correctly expressed the integral to find the area between the curves, surprisingly few found a correct answer. Although this
is a GDC paper, some candidates attempted to integrate this function analytically.
80a. Find ∫
1
dx
2x+3
.
[2 marks]
Markscheme
∫
1
dx
2x+3
= 12 ln(2x + 3) + C (accept 12 ln |(2x + 3)| + C ) A1A1
N2
[2 marks]
Examiners report
Many candidates were unable to correctly integrate but did recognize that the integral involved the natural log function; they most
often missed the factor 12 or replaced it with 2.
80b. Given that ∫ 3
0
1
dx
2x+3
−
−
= ln √P , find the value of P.
[4 marks]
Markscheme
1
∫03 2x+3
dx = [ 12 ln(2x + 3)]
3
0
(M1)
evidence of substitution of limits
e.g. 12 ln 9 − 12 ln 3
evidence of correctly using ln a − ln b = ln ab (seen anywhere)
(A1)
e.g. 12 ln 3
evidence of correctly using aln b = ln ba (seen anywhere)
−
−
e.g. ln √ 93
P = 3 (accept ln √3 )
A1
(A1)
N2
[4 marks]
Examiners report
Part (b) proved difficult as many were unable to use the basic rules of logarithms.
81. A particle moves along a straight line so that its velocity, v ms−1 at time t seconds is given by v = 6e3t + 4 . When t = 0 , the
[7 marks]
displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.
Markscheme
evidence of anti-differentiation
e.g. s = ∫
(6e3x
+ 4)dx
s = 2e3t + 4t + C
A2A1
substituting t = 0 ,
(M1)
7 = 2+C
(M1)
A1
C=5
s = 2e3t + 4t + 5
A1
N3
[7 marks]
Examiners report
There were a number of completely correct solutions to this question. However, there were many who did not know the relationship
between velocity and position. Many students differentiated rather than integrated and those who did integrate often had difficulty
with the term involving e. Many who integrated correctly neglected the C or made C = 7 .
( )=
1
3
+2
2
−5
Consider f(x) = 13 x3 + 2x2 − 5x . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.
82a. Find f ′ (x) .
[3 marks]
Markscheme
f ′ (x) = x2 + 4x − 5
A1A1A1
N3
[3 marks]
Examiners report
This question was very well done with most candidates showing their work in an orderly manner. There were a number of
candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression
rather than an equation.
82b. Find the x-coordinate of M.
[4 marks]
Markscheme
evidence of attempting to solve f ′ (x) = 0
evidence of correct working
e.g. (x + 5)(x − 1) ,
x = −5, x = 1
so x = −5
A1
(M1)
A1
−4± √16+20
2
, sketch
(A1)
N2
[4 marks]
Examiners report
This question was very well done with most candidates showing their work in an orderly manner. There were a number of
candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression
rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a
maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was
unnecessary and used up valuable time on the exam.
82c. Find the x-coordinate of N.
[3 marks]
Markscheme
METHOD 1
f ′′ (x) = 2x + 4 (may be seen later)
A1
evidence of setting second derivative = 0
(M1)
e.g. 2x + 4 = 0
x = −2
A1
N2
METHOD 2
evidence of use of symmetry
(M1)
e.g. midpoint of max/min, reference to shape of cubic
correct calculation
e.g.
A1
−5+1
2
x = −2
A1
N2
[3 marks]
Examiners report
This question was very well done with most candidates showing their work in an orderly manner. There were a number of
candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression
rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a
maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was
unnecessary and used up valuable time on the exam.
82d. The line L is the tangent to the curve of f at (3, 12). Find the equation of L in the form y = ax + b .
[4 marks]
Markscheme
attempting to find the value of the derivative when x = 3
′
f (3) = 16
(M1)
A1
valid approach to finding the equation of a line
M1
e.g. y − 12 = 16(x − 3) , 12 = 16 × 3 + b
y = 16x − 36
A1
N2
[4 marks]
Examiners report
This question was very well done with most candidates showing their work in an orderly manner. There were a number of
candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression
rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a
maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was
unnecessary and used up valuable time on the exam.
y=
′
(x)
The diagram below shows part of the graph of the gradient function, y = f ′ (x) .
83a. On the grid below, sketch a graph of y = f ′′ (x) , clearly indicating the x-intercept.
[2 marks]
Markscheme
A1A1
N2
Note: Award A1 for negative gradient throughout, A1 for x-intercept of q. It need not be linear.
[2 marks]
Examiners report
Several candidates had a correct sketch in part (a).
83b. Complete the table, for the graph of y = f(x) .
[2 marks]
Markscheme
A1A1
N1N1
Examiners report
The majority of the errors occurred in parts (b) and (c). In part (b), some seemed to just guess while others left it blank.
83c. Justify your answer to part (b) (ii).
[2 marks]
Markscheme
METHOD 1
R1R1
Second derivative is zero, second derivative changes sign.
N2
METHOD 2
There is a maximum on the graph of the first derivative.
R2
N2
Examiners report
In part (c), justification lacked completeness. For example, many stated that the second derivative must equal zero but said nothing of
its change in sign.
The following diagram shows the graphs of f(x) = ln(3x − 2) + 1 and g(x) = −4 cos(0.5x) + 2 , for 1 ≤ x ≤ 10 .
84a. 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.
[6 marks]
Markscheme
(i) intersection points x = 3.77 , x = 8.30 (may be seen as the limits)
approach involving subtraction and integrals
fully correct expression
(A1)(A1)
(M1)
A2
8.30
8.30
8.30
e.g. ∫3.77
((−4 cos(0.5x) + 2) − (ln(3x − 2) + 1))dx , ∫3.77
g(x)dx − ∫3.77
f(x)dx
(ii) A = 6.46
A1
N1
[6 marks]
Examiners report
Many candidates did not make good use of the GDC in this problem. Most had the correct expression but incorrect limits. Some tried
to integrate to find the area without using their GDC. This became extremely complicated and time consuming.
84b. (i)
Find f ′ (x) .
(ii) Find
g′ (x)
[4 marks]
.
Markscheme
(i) f ′ (x) =
3
3x−2
A1A1
N2
Note: Award A1 for numerator (3), A1 for denominator (3x − 2) , but penalize 1 mark for additional terms.
(ii) g′ (x) = 2 sin(0.5x)
A1A1
N2
Note: Award A1 for 2, A1 for sin(0.5x) , but penalize 1 mark for additional terms.
[4 marks]
Examiners report
In part (b), the chain rule was not used by some.
Let ∫15 3f(x)dx = 12 .
85a. Show that ∫ 1 f(x)dx = −4 .
5
[2 marks]
Markscheme
A1
evidence of factorising 3/division by 3
5
5
e.g. ∫1 3f(x)dx = 3 ∫1 f(x)dx ,
12
3
5 3f(x)dx
3
, ∫1
(do not accept 4 as this is show that)
evidence of stating that reversing the limits changes the sign
e.g.
1
∫5 f(x)dx
A1
5
= − ∫1 f(x)dx
∫51 f(x)dx = − 4
AG
N0
[2 marks]
Examiners report
This question was very poorly done. Very few candidates provided proper justification for part (a), a common error being to write
∫15 f(x)dx =f(5) − f(1) . What was being looked for was that ∫15 3f(x)dx =3 ∫15 f(x)dx and ∫51 f(x)dx = − ∫15 f(x)dx .
85b. Find the value of ∫ 2 (x + f(x))dx + ∫ 5 (x + f(x))dx .
1
2
[5 marks]
Markscheme
(A1)
evidence of correctly combining the integrals (seen anywhere)
e.g. I =
2
5
∫1 (x + f(x))dx + ∫2 (x + f(x))dx
5
= ∫1 (x + f(x))dx
evidence of correctly splitting the integrals (seen anywhere)
e.g. I =
(A1)
∫15 xdx + ∫15 f(x)dx
2
∫ xdx = x2 (seen anywhere)
5
∫15 xdx =[ x2 ] =
2
1
I = 16
A1
25
2
− 12 (=
A1
24
,12)
2
A1
N3
[5 marks]
Examiners report
Part (b) had similar problems with neither the combining of limits nor the splitting of integrals being done very often. A common error
5
5
was to treat f(x) as 1 in order to make ∫1 f(x)dx = 4 and then write ∫1 (x + f(x))dx = [x + 1]51 .
Let f : x ↦ sin3 x .
86a. Find f ′ (x) , giving your answer in the form asin p xcosq x where a, p, q ∈ Z .
[2 marks]
Markscheme
f ′ (x) = 3sin2 x cos x
A2
N2
[2 marks]
Examiners report
This question was not done well by most candidates.
1
86b. Let g(x) = √3 sin x(cos x) 2 for 0 ≤ x ≤
x-axis.
π
2
. Find the volume generated when the curve of g is revolved through 2π about the [7 marks]
b
V = ∫a πy2 dx
π
1
V = ∫0 2 π(√3 sin xcos 2 x)2 dx
π
= π ∫0 2 3sin2 x cos xdx
π
V = π[ sin3 x] 2 (= π ( sin3 ( π2 ) − sin3 0))
0
sin π2 = 1
π (1 − 0)
V =π
sin 0 = 0
Examiners report
This question was not done well by most candidates. No more than one-third of them could correctly give the range of f(x) = sin3 x
and few could provide adequate justification for there being exactly one solution to f(x) = 1 in the interval [0, 2π] . Finding the
derivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of the
formula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the
1
necessary function and limits into π ∫ab y2 dx and fewer still could square √3 sin xcos 2 x correctly. From those who did square
correctly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.
Let S be the total area of the two segments shaded in the diagram below.
87a. Find the value of θ when S is a local minimum, justifying that it is a minimum.
Markscheme
METHOD 1
attempt to differentiate
e.g.
dS
dθ
(M1)
= −4 cos θ
setting derivative equal to 0
correct equation
(M1)
A1
e.g. −4 cos θ = 0 , cos θ = 0 , 4 cos θ = 0
θ = π2
A1
N3
EITHER
(M1)
evidence of using second derivative
′′
S (θ) = 4 sin θ A1
S′′ ( π2 ) = 4
A1
it is a minimum because S′′ ( π2 ) > 0
R1
N0
OR
evidence of using first derivative
(M1)
′
for θ < , S (θ) < 0 (may use diagram)
π
2
A1
for θ > π2 , S′ (θ) > 0 (may use diagram) A1
it is a minimum since the derivative goes from negative to positive
METHOD 2
2π − 4 sin θ is minimum when 4 sin θ is a maximum
4 sin θ is a maximum when sin θ = 1
θ=
π
2
A3
[8 marks]
N3
(A2)
R3
R1
N0
[8 marks]
Examiners report
Only a small number of candidates recognized the fact S would be minimum when sin was maximum, leading to a simple noncalculus solution. Those who chose the calculus route often had difficulty finding the derivative of S, failing in a significant number of
cases to recognize that the derivative of a constant is 0, and also going through painstaking application of the product rule to find the
simple derivative. When it came to justify a minimum, there was evidence in some cases of using some form of valid test, but
explanation of the test being used was generally poor.
87b. Find a value of θ for which S has its greatest value.
[2 marks]
Markscheme
S is greatest when 4 sin θ is smallest (or equivalent)
θ = 0 (or π )
A1
(R1)
N2
[2 marks]
Examiners report
Candidates who answered part (d) correctly generally did well in part (e) as well, though answers outside the domain of θ were
frequently seen.
Let f(x) = ex (1 − x2 ) .
88a. Show that f ′ (x) = ex (1 − 2x − x2 ) .
[3 marks]
Markscheme
M1
evidence of using the product rule
f ′ (x) = ex (1 − x2 ) + ex (−2x)
Note: Award A1 for
ex (1 − x2 )
f ′ (x) = ex (1 − 2x − x2 )
AG
A1A1
, A1 for ex (−2x) .
N0
[3 marks]
Examiners report
Many candidates clearly applied the product rule to correctly show the given derivative. Some candidates missed the multiplicative
nature of the function and attempted to apply a chain rule instead.
y = f(x)
Part of the graph of y = f(x), for −6 ≤ x ≤ 2 , is shown below. The x-coordinates of the local minimum and maximum points are r and s
respectively.
88b. Write down the value of r and of s.
[4 marks]
Markscheme
at the local maximum or minimum point
f ′ (x) = 0 (ex (1 − 2x − x2 ) = 0)
⇒ 1 − 2x − x2 = 0
(M1)
(M1)
r = −2.41 s = 0.414
A1A1
N2N2
[4 marks]
Examiners report
Although part (c) was a “write down” question where no working is required, a good number of candidates used an algebraic method
of solving for r and s which sometimes returned incorrect answers. Those who used their GDC usually found correct values, although
not always to three significant figures.
88c. Let L be the normal to the curve of f at P(0, 1) . Show that L has equation x + y = 1 .
[4 marks]
Markscheme
f ′ (0) = 1
A1
gradient of the normal = −1
A1
evidence of substituting into an equation for a straight line
correct substitution
(M1)
A1
e.g. y − 1 = −1(x − 0) , y − 1 = −x , y = −x + 1
x+y = 1
AG
N0
[4 marks]
Examiners report
In part (d), many candidates showed some skill showing the equation of a normal, although some tried to work with the gradient of
the tangent.
y = f(x)
88d. Let R be the region enclosed by the curve y = f(x) and the line L.
(i)
[5 marks]
Find an expression for the area of R.
(ii) Calculate the area of R.
Markscheme
(i) intersection points at x = 0 and x = 1 (may be seen as the limits)
approach involving subtraction and integrals
fully correct expression
A2
(A1)
(M1)
N4
1
1
1
e.g. ∫0 (ex (1 − x2 ) − (1 − x))dx , ∫0 f(x)dx − ∫0 (1 − x)dx
(ii) area R = 0.5
A1
N1
[5 marks]
Examiners report
Surprisingly few candidates set up a completely correct expression for the area between curves that considered both integration and
the correct subtraction of functions. Using limits of −6 and 2 was a common error, as was integrating on f(x) alone. Where
candidates did write a correct expression, many attempted to perform analytic techniques to calculate the area instead of using their
GDC.
A toy car travels with velocity v ms−1 for six seconds. This is shown in the graph below.
89a. Write down the car’s velocity at t = 3 .
[1 mark]
Markscheme
4 (m s−1 )
A1
N1
[1 mark]
Examiners report
[N/A]
89b. Find the car’s acceleration at t = 1.5 .
[2 marks]
Markscheme
recognizing that acceleration is the gradient
e.g. a(1.5) =
M1
4−0
2−0
a = 2 (m s−2 )
A1
N1
[2 marks]
Examiners report
[N/A]
89c. Find the total distance travelled.
[3 marks]
Markscheme
recognizing area under curve
M1
e.g. trapezium, triangles, integration
A1
correct substitution
e.g. 12 (3 + 6)4 , ∫06 |v(t)|dt
distance ​18 (m)
A1
N2
[3 marks]
Examiners report
[N/A]
90. Find ∫
ex
dx
1+ex
.
[3 marks]
Markscheme
attempt to use substitution or inspection
e.g. u = 1 + ex so
du
dx
correct working
A1
e.g. ∫
du
u
M1
= ex
= ln u
ln(1 + ex ) + C
A1
N3
[3 marks]
Examiners report
[N/A]
Given that f(x) = x1 , answer the following.
91a. Find the first four derivatives of f(x) .
[4 marks]
Markscheme
f ′ (x) = −x−2 (or −
1
x2
f ′′ (x) = 2x−3 (or
)
2
x3
A1
f ′′′ (x) = −6x−4 (or −
6
x4
f (4) (x) = 24x−5 (or
)
24
x5
A1
)
)
N1
N1
A1
A1
N1
N1
[4 marks]
Examiners report
[N/A]
91b. Write an expression for f (n) (x) in terms of x and n .
[3 marks]
Markscheme
f (n) (x) =
(−1)nn!
xn+1
or (−1)n n!(x−(n+1) )
A1A1A1
N3
[3 marks]
Examiners report
[N/A]
Let f(x) = cos x + √3 sin x , 0 ≤ x ≤ 2π . The following diagram shows the graph of f .
The y-intercept is at (0, 1) , there is a minimum point at A (p, q) and a maximum point at B.
92a. Find f ′ (x) .
[2 marks]
Markscheme
f ′ (x) = − sin x + √3 cos x
A1A1
[2 marks]
Examiners report
[N/A]
N2
92b. Hence
(i)
[10 marks]
show that q = −2 ;
(ii) verify that A is a minimum point.
Markscheme
(i) at A, f ′ (x) = 0
R1
A1
correct working
e.g. sin x = √3 cos x
tan x = √3
x=
π
3
,
4π
3
A1
A1
attempt to substitute their x into f(x)
M1
e.g. cos ( 4π
) + √3 sin ( 4π
)
3
3
correct substitution
e.g. − 12 + √3 (−
A1
√3
)
2
correct working that clearly leads to −2
​
A1
e.g. − 12 − 32
​ q = −2
AG
N0
(ii) correct calculations to find f ′ (x) either side of x =
4π
3
A1A1
e.g. f ′ (π) = 0 − √3 , ​ f ′ (2π) = 0 + √3​ ​ ​ ​
f ′ (x) changes sign from negative to positive
so A is a minimum
AG
R1
N0
[10 marks]
Examiners report
[N/A]
Let f(x) = 4x − ex−2 − 3 , for 0 ≤ x ≤ 5 .
93. Write down the gradient of the graph of f at x = 3 .
[1 mark]
Markscheme
gradient is 1.28
A1
N1
[1 mark]
Examiners report
[N/A]
Let h(x) =
2x−1
x+1
, x ≠ −1 .
94. Let R be the region in the first quadrant enclosed by the graph of h , the x-axis and the line x = 3.
(i)
Find the area of R.
(ii) Write down an expression for the volume obtained when R is revolved through 360∘ about the x-axis.
[5 marks]
Markscheme
(i) area = 2.06
A2
N2
b
(ii) attempt to substitute into volume formula (do not accept π ∫a y2 dx )
2
volume = π∫ 13 ( 2x−1
) dx
x+1
2
A2
M1
N3
[5 marks]
Examiners report
[N/A]
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