IB Practice - Calculus - Differentiation Applications (V2 Legacy)

IB
Math – High Level Year 2 -‐ Calc Practice: Differentiation Applications Alei -‐ Desert Academy
IB Practice -‐ Calculus -‐ Differentiation Applications (V2 Legacy)

1. A particle moves along a straight line. When it is a distance s from a fixed point, where s > 1, the
(3s + 2)
velocity v is given by v = . Find the acceleration when s = 2.
(2s − 1)
(Total 4 marks)
2. Give exact answers in this part of the question.
The temperature g (t) at time t of a given point of a heated iron rod is given by
ln t
g (t) = , where t > 0.
t
(a) Find the interval where g′ (t) > 0.
(4)
(b) Find the interval where g″ (t) > 0 and the interval where g″ (t) < 0.
(5)
(c) Find the value of t where the graph of g (t) has a point of inflexion.
(3)
(d) Let t* be a value of t for which g′ (t*) = 0 and g″ (t*) < 0. Find t*.
(3)
(e) Find the point where the normal to the graph of g (t) at the point
(t*, g (t*)) meets the t-axis.
(3)
(Total 18 marks)
3. The acceleration, a(t) m s–2, of a fast train during the first 80 seconds of motion is given by
a(t) = – 1 t + 2
20
where t is the time in seconds. If the train starts from rest at t = 0, find the distance travelled by the
train in the first minute.
(Total 4 marks)
⎧ x 1n x − kx, x > 0
4. Consider the function fk (x) = ⎨ , where k ∈
⎩0, x=0
(a) Find the derivative of fk (x), x > 0.
(2)
(b) Find the interval over which f0 (x) is increasing.
The graph of the function fk (x) is shown below.
(2)
y
0 A x
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IB Math – High Level Year 2 -‐ Calc Practice: Differentiation Applications Alei -‐ Desert Academy
(c) (i) Show that the stationary point of fk (x) is at x = ek–1.

(ii) One x-intercept is at (0, 0). Find the coordinates of the other x-intercept.
(4)
(d) Find the area enclosed by the curve and the x-axis.
(5)
(e) Find the equation of the tangent to the curve at A.
(2)
(f) Show that the area of the triangular region created by the tangent and the
coordinate axes is twice the area enclosed by the curve and the x-axis.
(2)
(g) Show that the x-intercepts of fk (x) for consecutive values of k form a geometric sequence.
(3)
(Total 20 marks)
−t
5. The velocity, v, of an object, at a time t, is given by v = ke 2 , where t is in seconds and v is in m s–1.
Find the distance travelled between t = 0 and t = a.
Working:
Answer:
..................................................................
(Total 3 marks)
6. Find the coordinates of the point which is nearest to the origin on the line
L: x = 1 – λ, y = 2 – 3λ, z = 2.
Working:
Answer:
..................................................................
(Total 3 marks)
7. A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve
y = sin x, where 0 ≤ x ≤ n.
(a) Write down an expression for the area of the rectangle.
(b) Find the maximum area of the rectangle.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 3 marks)
8. Let f : x ! esin x.
(a) Find f ′ (x).
There is a point of inflexion on the graph of f, for 0 < x < 1.
(b) Write down, but do not solve, an equation in terms of x, that would allow you to find the value
of x at this point of inflexion.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 3 marks)

9. The diagram shows the graph of y = f′ (x).
y y = f’(x)
Indicate, and label clearly, on the graph

(a) the points where y = f (x) has minimum points;
(b) the points where y = f (x) has maximum points;
(c) the points where y = f (x) has points of inflexion.
Working:
(Total 3 marks)
10. Let f (x) = ln |x – 3x |, –0.5 < x < 2, x ≠ a, x ≠ b; (a, b are values of x for which f (x) is not defined).
5 2
(a) (i) Sketch the graph of f (x), indicating on your sketch the number of zeros of f (x). Show
also the position of any asymptotes.
(2)
(ii) Find all the zeros of f (x), (that is, solve f (x) = 0).
(3)
(b) Find the exact values of a and b.
(3)
(c) Find f (x), and indicate clearly where f′ (x) is not defined.
(3)
(d) Find the exact value of the x-coordinate of the local maximum of f (x), for 0 < x < 1.5. (You
may assume that there is no point of inflexion.)
(3)
(e) Write down the definite integral that represents the area of the region enclosed by f (x) and the
x-axis. (Do not evaluate the integral.)
(2)
(Total 16 marks)
11. (a) Sketch and label the curves
y = x2 for –2 ≤ x ≤ 2, and y = – 1 ln x for 0 < x ≤ 2.
2
(2)
(b) Find the x-coordinate of P, the point of intersection of the two curves.
(2)
(c) If the tangents to the curves at P meet the y-axis at Q and R, calculate the area of the triangle
PQR.
(6)
(d) Prove that the two tangents at the points where x = a, a > 0, on each curve are always
perpendicular.
(4)
(Total 14 marks)

12. (a) Let y = a + b sin x , where 0 < a < b.

b + a sin x
dy (b 2 + a 2 ) cos x
(i) Show that = .
dx (b + a sin x) 2
(4)
(ii) Find the maximum and minimum values of y.
(4)
(iii) Show that the graph of y = a + b sin x , 0 < a < b cannot have a vertical asymptote.
b + a sin x
(2)
(b) For the graph of y = 4 + 5 sin x for 0 ≤ x ≤ 2π,
5 + 4 sin x
(i) write down the y-intercept;
(ii) find the x-intercepts m and n, (where m < n) correct to four significant figures;
(iii) sketch the graph.
(5)
(c) The area enclosed by the graph of y = 4 + 5 sin x and the x-axis from x = 0 to x = n is denoted
5 + 4 sin x
by A. Write down, but do not evaluate, an expression for the area A.
(2)
(Total 17 marks)
13. The diagram shows a sketch of the graph of y = f′ (x) for a ≤ x ≤ b.
y
y = f’(x)
a b x
On the grid below, which has the same scale on the x-axis, draw a sketch of the graph of
y = f (x) for a ≤ x ≤ b, given that f (0) = 0 and f (x) ≥ 0 for all x. On your graph you should clearly
indicate any minimum or maximum points, or points of inflexion.
y
a b x
(Total 3 marks)

14. An astronaut on the moon throws a ball vertically upwards. The height, s metres, of the ball, after t
seconds, is given by the equation s = 40t + 0.5at2, where a is a constant. If the ball reaches its
maximum height when t = 25, find the value of a.
Working:
Answer:
..........................................................................
(Total 3 marks)
15. Let f ( x) = x⎛⎜ 3 ( x – 1) ⎞⎟, – 1.4 ≤ x ≤ 1.4
2 2
⎝ ⎠
(a) Sketch the graph of f (x). (An exact scale diagram is not required.)
On your graph indicate the approximate position of
(i) each zero;
(ii) each maximum point;
(iii) each minimum point.
(4)
(b) (i) Find f′ (x), clearly stating its domain.
(ii) Find the x-coordinates of the maximum and minimum points of f (x), for
–1 < x < 1.
(7)
(c) Find the x-coordinate of the point of inflexion of f (x), where x > 0, giving your answer correct
to four decimal places.
(2)
(Total 13 marks)
16. Consider the tangent to the curve y = x3 + 4x2 + x – 6.
(a) Find the equation of this tangent at the point where x = –1.
(b) Find the coordinates of the point where this tangent meets the curve again.
(Total 3 marks)
⎛ 1⎞
17. A point P(x, x2) lies on the curve y = x2. Calculate the minimum distance from the point A⎜ 2, – ⎟ to
⎝ 2⎠
the point P.
Working:
Answer:
..........................................................................
(Total 3 marks)
18. Let θ be the angle between the unit vectors a and b, where 0 < θ < π. Express |a – b| in terms of
1
sin θ .
2
(Total 3 marks)
19. A particle is moving along a straight line so that t seconds after passing through a fixed point O on
the line, its velocity v (t) m s–1 is given by
⎛π ⎞
v (t ) = t sin ⎜ t ⎟ .
⎝3 ⎠
(a) Find the values of t for which v(t) = 0, given that 0 ≤ t ≤ 6.
(3)
(b) (i) Write down a mathematical expression for the total distance travelled by the particle in
the first six seconds after passing through O.
(ii) Find this distance.
(4)
(Total 7 marks)

1
+
20. Consider the function f ( x) = x x , where x ∈ .
⎛ 1 – 1n x ⎞
(a) Show that the derivative f ′( x) = f ( x) ⎜ 2 ⎟.
⎝ x ⎠
(3)
(b) Sketch the function f (x), showing clearly the local maximum of the function and its horizontal
asymptote. You may use the fact that
1n x
lim = 0.
x→ ∞ x
(5)
(c) Find the Taylor expansion of f (x) about x = e, up to the second degree term, and show that this
polynomial has the same maximum value as f (x) itself.
(5)
(Total 13 marks)
21. A particle is projected along a straight line path. After t seconds, its velocity v metres per second is
1
given by v = .
2 + t2
(a) Find the distance travelled in the first second.
(b) Find an expression for the acceleration at time t.
(Total 6 marks)
3 2
22. A curve has equation xy + 2x y = 3. Find the equation of the tangent to this curve at the point (1, 1).
(Total 6 marks)
23. A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve
2
y = e – x . The area of this rectangle is denoted by A.
(a) Write down an expression for A in terms of x.
(b) Find the maximum value of A.
(Total 6 marks)
24. The diagram below shows the graph of y1 = f (x), 0 ″ x ″ 4.
y
0 1 2 3 4 x
x
On the axes below, sketch the graph of y2 = ∫ f (t )dt, marking clearly the points of inflexion.
0
y
0 1 2 3 4 x
(Total 6 marks)

25. A particle moves in a straight line with velocity v, in metres per second, at time t seconds, given by
v(t) = 6t2 – 6t, t ≥ 0
Calculate the total distance travelled by the particle in the first two seconds of motion.
Working:
Answer:
..........................................................................
(Total 6 marks)
x
26. Find the x-coordinate of the point of inflexion on the graph of y = xe , – 3 ≤ x ≤ 1.
Working:
Answer:
..........................................................................
(Total 6 marks)
3 3 –1
27. Air is pumped into a spherical ball which expands at a rate of 8 cm per second (8 cm s ). Find the
exact rate of increase of the radius of the ball when the radius is 2 cm.
Working:
Answer:
..........................................................................
(Total 6 marks)
2
28. The point B(a, b) is on the curve f (x) = x such that B is the point which is closest to A(6, 0).
Calculate the value of a.
Working:
Answer:
..........................................................................
(Total 6 marks)
29. (a) On the same axes sketch the graphs of the functions, f (x) and g (x), where
f (x) = 4 – (1 – x)2, for – 2 ≤ x ≤ 4,
g (x) = ln (x + 3) – 2, for – 3 ≤ x ≤ 5.
(2)
(b) (i) Write down the equation of any vertical asymptotes.
(ii) State the x-intercept and y-intercept of g (x).
(3)
(c) Find the values of x for which f (x) = g (x).
(2)
(d) Let A be the region where f (x) ≥ g (x) and x ≥ 0.
(i) On your graph shade the region A.
(ii) Write down an integral that represents the area of A.
(iii) Evaluate this integral.
(4)
(e) In the region A find the maximum vertical distance between f (x) and g (x).
(3)
(Total 14 marks)
30. A curve has equation x3 y2 = 8. Find the equation of the normal to the curve at the point (2, 1).
Working:
Answer:
.........................................................................
(Total 6 marks)
–1 – t
31. A particle moves in a straight line. Its velocity v m s after t seconds is given by v = e sin t. Find
the total distance travelled in the time interval [0, 2π].
(Total 6 marks)

32. The diagram below shows the graph of y1 = f (x).
y2
On the axes below, sketch the graph of y2 = f′ (x) .

y2
x
(Total 6 marks)
2
x
33. The function f is defined by f (x) = , for x > 0.
2x
(a) (i) Show that
2 x – x 2 ln 2
f′ (x) =
2x
(ii) Obtain an expression for f ″(x), simplifying your answer as far as possible.
(5)
(b) (i) Find the exact value of x satisfying the equation f ′(x) = 0
(ii) Show that this value gives a maximum value for f (x).
(4)
(c) Find the x-coordinates of the two points of inflexion on the graph of f.
(3)
(Total 12 marks)
34. An airplane is flying at a constant speed at a constant altitude of 3 km in a straight line that will take
1
it directly over an observer at ground level. At a given instant the observer notes that the angle θ is 3
1
π radians and is increasing at radians per second. Find the speed, in kilometres per hour, at which
60
the airplane is moving towards the observer.
Airplane
x
3 km
Observer
(Total 6 marks)

35. The following diagram shows an isosceles triangle ABC with AB = 10 cm and AC = BC. The vertex
C is moving in a direction perpendicular to (AB) with speed 2 cm per second.
A B
Calculate the rate of increase of the angle CÂB at the moment the triangle is equilateral.
Working:
Answer:
.........................................................................
(Total 6 marks)
36. Find the equation of the normal to the curve x3 + y3 – 9xy = 0 at the point (2, 4).
Working:
Answer:
.........................................................................
(Total 6 marks)
37. A closed cylindrical can has a volume of 500 cm3. The height of the can is h cm and the radius of the
base is r cm.
(a) Find an expression for the total surface area A of the can, in terms of r.
(b) Given that there is a minimum value of A for r > 0, find this value of r.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
38. The displacement s metres of a moving body B from a fixed point O at time t seconds is given by
s = 50t – 10t2 + 1000.
(a) Find the velocity of B in m s–1.
(b) Find its maximum displacement from O.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(Total 6 marks)
39. The line L is given by the parametric equations x = 1 – λ, y = 2 – 3λ, z = 2. Find the coordinates of the
point on L which is nearest to the origin.
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
(Total 6 marks)

ln x
40. The function f is defined on the domain x ≥ 1 by f (x) = .
x
(a) (i) Show, by considering the first and second derivatives of f, that there is one maximum
point on the graph of f.
(ii) State the exact coordinates of this point.
(9)
(iii) The graph of f has a point of inflexion at P. Find the x-coordinate of P.
(3)
Let R be the region enclosed by the graph of f, the x-axis and the line x = 5.
(c) Find the exact value of the area of R.
(6)
(d) The region R is rotated through an angle 2π about the x-axis. Find the volume of the solid of
revolution generated.
(3)
(Total 21 marks)
41. The normal to the curve y = k + ln x2, for x ≠ 0, k ∈ , at the point where x = 2, has equation
x
3x + 2y = b, where b ∈ . Find the exact value of k.
(Total 6 marks)
42. The function f is defined by f (x) = epx(x + 1), here p ∈ .
(a) (i) Show that f ′(x) = epx(p(x + 1) + 1).
(ii) Let f (n)(x) denote the result of differentiating f (x) with respect to x, n times.
Use mathematical induction to prove that
f (n)(x) = pn–1epx (p(x + 1) + n), n ∈ +.
(7)
(b) When p = 3 , there is a minimum point and a point of inflexion on the graph of f. Find the
exact value of the x-coordinate of
(i) the minimum point;
(ii) the point of inflexion.
(4)
(c) Let p = 1 . Let R be the region enclosed by the curve, the x-axis and the lines x = –2 and x = 2.
2
Find the area of R.
(2)
(Total 13 marks)
43. The diagram shows a trapezium OABC in which OA is parallel to CB. O is the centre of a circle
radius r cm. A, B and C are on its circumference. Angle OĈB = θ.
O r
A
C B

Let T denote the area of the trapezium OABC.
2
(a) Show that T = r (sin θ + sin 2θ).
2
(4)
For a fixed value of r, the value of T varies as the value of θ varies.
(b) Show that T takes its maximum value when θ satisfies the equation
4 cos2 θ + cos θ – 2 = 0, and verify that this value of T is a maximum.
(5)
(c) Given that the perimeter of the trapezium is 75 cm, find the maximum value of T.
(6)
(Total 15 marks)
x3
44. The curve y = – x2 – 3x + 4 has a local maximum point at P and a local minimum point at Q.
3
Determine the equation of the straight line passing through P and Q, in the form ax + by + c = 0,
where a, b, c ∈ .
(Total 6 marks)
45. The following diagram shows the points A and B on the circumference of a circle, centre O, and
radius 4 cm, where AÔB = θ. Points A and B are moving on the circumference so that θ is
increasing at a constant rate.
A B
Given that the rate of change of the length of the minor arc AB is numerically equal to the rate of
change of the area of the shaded segment, find the acute value of θ.
(Total 6 marks)
46. A man PF is standing on horizontal ground at F at a distance x from the bottom of a vertical wall GE.
He looks at the picture AB on the wall. The angle BPA is θ.
E
P D
x
F G
Let DA = a, DB = b, where angle PDˆ E is a right angle. Find the value of x for which tan θ is a
maximum, giving your answer in terms of a and b. Justify that this value of x does give a maximum
value of tan θ.
(Total 9 marks)
47. Let f (x) = 3x2 – x + 4. Find the values of m for which the line y = mx + 1 is a tangent to the graph of f.
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
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..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
(Total 6 marks)
48. Particle A moves in a straight line, starting from OA, such that its velocity in metres per second for 0
≤ t ≤ 9 is given by
1 3
vA = − t 2 + 3t + .
2 2
Particle B moves in a straight line, starting from OB, such that its velocity in metres per second for 0
≤ t ≤ 9 is given by
vB = e0.2t.
(a) Find the maximum value of vA, justifying that it is a maximum.
(5)
(b) Find the acceleration of B when t = 4.
(3)
The displacements of A and B from OA and OB respectively, at time t are sA metres and sB metres.
When t = 0, sA = 0, and sB = 5.
(c) Find an expression for sA and for sB, giving your answers in terms of t.
(7)
(d) (i) Sketch the curves of sA and sB on the same diagram.
(ii) Find the values of t at which sA = sB.
(8)
(Total 23 marks)
1
49. Let f be the function defined for x > − by f (x) = ln (3x + 1).
3
(a) Find f ′(x).
(b) Find the equation of the normal to the curve y = f (x) at the point where x = 2.
Give your answer in the form y = ax + b where a, b∈ .
..............................................................................................................................................
..............................................................................................................................................
(Total 6 marks)
50. The radius and height of a cylinder are both equal to x cm. The curved surface area of the cylinder is
increasing at a constant rate of 10 cm2/sec. When x = 2, find the rate of change of
(a) the radius of the cylinder;
(b) the volume of the cylinder.
..............................................................................................................................................
..............................................................................................................................................
..............................................................................................................................................
(Total 6 marks)

ln x
51. The function f is defined by f (x) = , x ≥1.
x3
(a) Find f ′(x) and f ′′(x), simplifying your answers.
(6)
(b) (i) Find the exact value of the x-coordinate of the maximum point and justify that this is a
maximum.
(ii) Solve f ′′(x) = 0, and show that at this value of x, there is a point of inflexion on the graph
of f.
(iii) Sketch the graph of f, indicating the maximum point and the point of inflexion.
(11)
The region enclosed by the x-axis, the graph of f and the line x = 3 is denoted by R.
(c) Find the volume of the solid of revolution obtained when R is rotated through 360° about the x-
axis.
(3)
1
(d) Show that the area of R is (4 – ln 3).
18
(6)
(Total 26 marks)
52. The volume of a solid is given by
4 3
V= πr + πr 2 h.
3
At the time when the radius is 3 cm, the volume is 1π cm3, the radius is changing at a rate of 2
cm/min and the volume is changing at a rate of 204π cm3/min. Find the rate of change of the height at
this time.
..............................................................................................................................................
(Total 6 marks)
53. A particle moves in a straight line. At time t seconds, its displacement from a fixed point O is s
metres, and its velocity, v metres per second, is given by v = 3t2 − 4t + 2, t ≥ 0. When t = 0, s = −3.
Find the value of t when the particle is at O.
..............................................................................................................................................
..............................................................................................................................................
(Total 6 marks)
−1
54. Car A is travelling on a straight east-west road in a westerly direction at 60 km h . Car B is
travelling on a straight north-south road in a northerly direction at 70 km h−1. The roads intersect at
the point O. When Car A is x km east of O, and Car B is y km south of O, the distance between the
cars is z km.
Find the rate of change of z when Car A is 0.8 km east of O and Car B is 0.6 km south of O.
(Total 6 marks)

55. A television screen, BC, of height one metre, is built into a wall. The bottom of the television screen
at B is one metre above an observer’s eye level. The angles of elevation ( AO ˆ C , AO
ˆ B ) from the
observer’s eye at O to the top and bottom of the television screen are α and β radians respectively.
The horizontal distance from the observer’s eye to the wall containing the television screen is x
ˆ C ) is θ radians, as shown below.
metres. The observer’s angle of vision ( BO
2 1
(a) (i) Show that θ = arctan – arctan .
x x
(ii) Hence, or otherwise, find the exact value of x for which θ is a maximum and justify that
this value of x gives the maximum value of θ.
(iii) Find the maximum value of θ.
(17)
(b) Find where the observer should stand so that the angle of vision is 15°.
(5)
(Total 22 marks)
2
56. Given that y = e − x find

2
(a) d y;
dx 2
(3)
−x 2
(b) the exact values of the x-coordinates of the points of inflexion on the graph of y = e ,
justifying that they are points of inflexion.
(3)
(Total 6 marks)
57. (a) A curve is defined by the implicit equation 2xy + 6x2 − 3y2 = 6.
⎛ 2⎞ 20
Show that the tangent at the point A with coordinates ⎜1, ⎟ has gradient .
⎝ 3⎠ 3
(6)
⎛ 2⎞
(b) The line x =1 cuts the curve at point A, with coordinates ⎜1, ⎟ , and at point B.
⎝ 3⎠
⎛a⎞ ⎛ c ⎞
Find, in the form r = ⎜⎜ ⎟⎟ + s⎜⎜ ⎟⎟
⎝b⎠ ⎝d ⎠
(i) the equation of the tangent at A;
(ii) the equation of the normal at B.
(10)
(c) Find the acute angle between the tangent at A and the normal at B.
(4)
(Total 20 marks)

58. The acceleration in m s–2 of a particle moving in a straight line at time t seconds, t > 0, is given by the
1
formula a = − 2
. When t =1, the velocity is 8 m s–1.
(1 + t )
(a) Find the velocity when t = 3.
(6)
(b) Find the limit of the velocity as t → ∞.
(1)
(c) Find the exact distance travelled between t =1 and t = 3.
(4)
(Total 11 marks)
2
59. If f (x) = x – 3x , x > 0, 3
(a) find the x-coordinate of the point P where f ′ (x) = 0;

(2)
(b) determine whether P is a maximum or minimum point.
(3)
(Total 5 marks)
60. The function f is defined by f (x) = x e2x.
It can be shown that f (n) (x) = (2n x + n 2n−1) e2x for all n∈ +, where f (n) (x) represents the nth
derivative of f (x).
(a) By considering f (n) (x) for n =1 and n = 2, show that there is one minimum point P on the
graph of f, and find the coordinates of P.
(7)
(b) Show that f has a point of inflexion Q at x = −1.
(5)
(c) Determine the intervals on the domain of f where f is
(i) concave up;
(ii) concave down.
(2)
(d) Sketch f, clearly showing any intercepts, asymptotes and the points P and Q.
(4)
(n) n n−1 2x +
(e) Use mathematical induction to prove that f (x) = (2 x + n2 ) e for all n∈ , where
f (n) (x) represents the nth derivative of f (x).
(9)
(Total 27 marks)
61. A gourmet chef is renowned for her spherical shaped soufflé. Once it is put in the oven, its volume
increases at a rate proportional to its radius.
(a) Show that the radius r cm of the soufflé, at time t minutes after it has been put in the oven,
dr k
satisfies the differential equation = , where k is a constant.
dt r
(5)
(b) Given that the radius of the soufflé is 8 cm when it goes in the oven, and 12 cm when it’s
cooked 30 minutes later, find, to the nearest cm, its radius after 15 minutes in the oven.
(8)
(Total 13 marks)
62. A normal to the graph of y = arctan (x − 1), for x > 0, has equation y = −2x + c, where c∈ .
Find the value of c.
(Total 6 marks)

63. André wants to get from point A located in the sea to point Y located on a straight stretch of beach. P
is the point on the beach nearest to A such that AP = 2 km and PY = 2 km. He does this by swimming
in a straight line to a point Q located on the beach and then running to Y.
When André swims he covers 1 km in 5 5 minutes. When he runs he covers 1 km in 5 minutes.

(a) If PQ = x km, 0 ≤ x ≤ 2, find an expression for the time T minutes taken by André to reach
point Y.
(4)
dT 5 5x
(b) Show that = − 5.
dx x2 + 4
(3)
dT
(c) (i) Solve = 0.
dx
(ii) Use the value of x found in part (c) (i) to determine the time, T minutes, taken for André
to reach point Y.
d 2T 20 5
(iii) Show that = and hence show that the time found in part (c) (ii) is a
( )
2 3
dx 2
x +4 2
minimum.
(11)
(Total 18 marks)
64. A family of cubic functions is defined as fk (x) = k2x3 − kx2 + x, k∈ +.
(a) Express in terms of k
(i) f ′k (x) and f ′′k (x);
(ii) the coordinates of the points of inflexion Pk on the graphs of fk.
(6)
(b) Show that all Pk lie on a straight line and state its equation.
(2)
(c) Show that for all values of k, the tangents to the graphs of fk at Pk are parallel, and find the
equation of the tangent lines.
(5)
(Total 13 marks)
2
65. Consider the curve with equation f (x) = e −2 x for x < 0.
Find the coordinates of the point of inflexion and justify that it is a point of inflexion.
(Total 7 marks)

IB Math – High Level Year 2 -‐ Calc Practice: Differentiation Applications -‐ MarkScheme Alei -‐ Desert Academy
IB Practice -‐ Calculus -‐ Differentiation Applications (V2 Legacy) -‐ MarkScheme
(3 s + 2 )
1. Given v =
( 2 s − 1)
dv dv ds dv
then acceleration a = = × = ×v (M1)
dt ds dt ds
3(2 s − 1) − 2(3s + 2) (3s + 2)
therefore a = × (M1)
(2 s − 1) 2 (2 s − 1)
− 7 (3 s + 2 )
⇔a= (M1)
( 2 s − 1) 3
− 56
therefore when s = 2, a = (A1) (C4)
27
[4]
ln t 2 − ln t
2. (a) g (t) = . So g′(t) = . (M1)(A1)
t 2t 3 2
2
Hence, g′(t) > 0 when 2 > ln t or ln t < 2 or t < e . (M1)
2
Since the domain of g (t) is {t:t > 0}, g′(t) > 0 when 0 < t < e . (A1) 4
2 − ln t − 2 t − 3 t ( 2 − ln t )
(b) Since g′(t) = , g″(t) = (M2)
2t 32
4t 3
t [8 − 3 ln t ]
=– (A1)
4t 3
8/3
Hence g″(t) > 0 when 8 – 3 ln t < 0 ie t > e . (M1)
8/3
Similarly, g″(t) < 0 when 0 < t < e . (A1) 5
(c) g″(t) = 0 when t = 0 or 8 = 3 ln t.
8/3
Since, the domain of g is {t:t > 0}, g″(t) = 0 when t = e . (M1)
8/3 8/3
Since g″(t) > 0 when t > e and g″(t) < 0 when t < e , (M1)
⎛ 8 / 3 8 −4 / 3 ⎞ 8/3
⎜e , e ⎟ is the point of inflexion. The required value of t is e . (A1) 3
⎝ 3 ⎠
8/3
Note: Award (A1) for evaluating t as e .
2
(d) g′(t) = 0 when ln t = 2 or t = e . (M1)
2 2
2 e [8 − 3 ln e ] 1
Also g″(e ) = –
6
=− 5 <0 (M1)
4e 2e
2
Hence t* = e (A1) 3
(e) At (t*, g (t*)) the tangent is horizontal. (M1)
So the normal at the point (t*, g (t*)) is the line t = t*. (M1)
2
Thus, it meets the t axis at the point t = t* = e and hence the
2
point is (e , 0). (A1) 3
[18]
1
3. a(t) = – t+2
20
1 2
v(t) = – t + 2t + c (M1)
40
v = 0 when t = 0, and so c = 0
1 2 1
Thus, v(t) = – t + 2t = – t(t – 80). (A1)
40 40

60
Since v(t) ≥ 0 for 0 ≤ t ≤ 80, the distance travelled =
∫ 0
v(t )dt (M1)
60
⎡ 1 3 2⎤
= ⎢− t +t ⎥
⎣ 120 ⎦0
2⎛ 1⎞
= 60 ⎜1 − ⎟
⎝ 2⎠
= 1800 m. (A1) (C4)
[4]
4. (a) fk (x) = x ln x – kx
⇒ f k′ (x) = ln x + 1 – k (M1)(A1) 2
(b) f 0′ (x) = ln x + 1
f0 (x) increases where f 0′ (x) > 0 (M1)
1
⇒ ln x > –1 ⇒ x > (A1) 2
e
(c) (i) Stationary points happen where f k′ (x) = 0
⇒ ln x = k – 1 (M1)
k–1
⇒x=e (A1)
(ii) x intercepts are where fk (x) = 0
⇒ x ln x – kx = 0
⇒ x(ln x – k) = 0 (M1)
⇒ x = 0 or ln x = k
k
⇒ x =e
k
⇒ (e , 0) (A1) 4
ek ⎛ ek ⎞
(d) Area =
∫ 0
x ln x − kx dx = ⎜ =
⎝
∫ 0
(kx − x ln x)dx ⎟
⎠
(M1)
Integrate x ln x by parts.
x2 x
∫ x ln xdx =
2
ln x −
2
dx ∫ (M1)
x2 x2
= ln x − +C (A1)
2 4
ek
⎡ x2 x 2 kx 2 ⎤
⇒ Area = ⎢ ln x − − ⎥
⎣2 4 2 ⎦0
e 2k
= (A1) 5
4
Note: Given x ln x – kx = fk (x) ≈ 0 when x = 0.
k k k
(e) Gradient of the tangent at A(e , 0), m is f k′ (e ) = ln e + 1 – k (M1)
=1
k
Therefore, an equation of the tangent is y = x – e . (A1) 2
(f) The tangent forms a right angle triangle with the coordinate axes.
k
The perpendicular sides are each of length e . (M1)
1 1
Area of the triangle = × e k × e k = e 2 k (A1)
2 2
1 2k ⎛1 ⎞
e = 2⎜ e 2 k ⎟ ie The area of the triangle is twice the area
2 ⎝4 ⎠

enclosed by the curve and the x-axis. (AG) 2
k
(g) Since the x-intercepts are of the form xk = e , for k ∈ (M1)
k+l
then xk+1 = e
x k +1
and =e (A1)
xk
Therefore, the x-intercepts x0, x1, ...xk, ... form a geometric sequence
with x0 = 1 and a common ratio of e. (R1) 3
[20]
a
5. Total distance = k
∫ 0
e −t / 2 dt (M1)
= − 2k[e −t / 2 ]0a (M1)

–a/2 –a/2
= –2k(e – 1) metres (or equivalent eg 2k(1 – e )) (A1) (C3)
Note: Award (C2) if k is omitted in the final answer.
[3]
6. Let s be the distance from the origin to a point on the line, then
2 2 2
s = (1 – λ) + (2 – 3λ) + 4
2
= 10λ – 14λ + 9
d( s 2 )
= 20λ – 14 (M1)
dλ
d( s 2 ) 7
For minimum put = 0, ⇒ λ = (A1)
dλ 10
⎛ 3 −1 ⎞
The point is ⎜ , , 2⎟. (A1) (C3)
⎝ 10 10 ⎠
d( s 2 )
Note: At this point > 0, but this is not required.
dλ 2
OR
The position vector for the point nearest to the origin is perpendicular
to the direction of the line. At that point:
⎛ 1 − λ ⎞ ⎛ − 1⎞
⎜ ⎟⎜ ⎟
⎜ 2 − 3λ ⎟ . ⎜ − 3 ⎟ = 0 (M1)
⎜ 2 ⎟⎜ 0 ⎟
⎝ ⎠⎝ ⎠
Therefore, 10λ – 7 = 0
7
Therefore, λ = (A1)
10
⎛ 3 −1 ⎞
Therefore, the point is ⎜ , , 2⎟. (A1) (C3)
⎝ 10 10 ⎠
[3]
7.
1.0
0.8
0.6
0.4
0.2
0.0
x π
(a) Area = (π – 2x) sin x. (M1)(A1)

(C2)
2
(b) Maximum Area = 1.12 units (A1) (C1)
[3]
sin x
8. (a) Given f (x) = e
sin x
Then f ′ (x) = cos x × e (A1) (C1)
2 sinx sin x
(b) f ″(x) = cos x × e – sin x × e
sin x 2
=e (cos x – sin x) (M1)
For the point of inflexion, put f ″ (x) = 0
sin x 2
⇒e (cos x – sin x) = 0 (or equivalent) (A1) (C2)
Note: Award (C1) if the candidate only writes
f ″ (x) = 0
[3]
9. y y = f′ (x)
inf 10
min max min
–4 –3 –2 –1 1 2 3 x
inf
–10
–20
(a) Minimum points (A1) (C1)
(b) Maximum point (A1) (C1)
(c) Points of inflexion (A1) (C1)
Note: There is no scale on the question paper. For examiner reference the
scale has been added here and the numerical answers are minima at x = –3
and 2, maximum at x = 0 and points of inflexion at x = –1.79 and 1.12.
[3]
5 2
10. (a) (i) y = ln ⎜x – 3x ⎟
2.5
–0.5 0.5 1 1.5 2

–2.5
–5
–7.5
–10
–12.5
asymptote asymptote (G2)
Note: Award (G1) for correct shape, including three zeros, and (G1)
for both asymptotes
(ii) f (x) = 0 for x = 0.599, 1.35, 1.51 (G1)(G1)(G1) 5
(b) f (x) is undefined for
5 2
(x – 3x ) = 0 (M1)
2 3
x (x – 3) = 0
1/3
Therefore, x = 0 or x = 3 (A2) 3

5x 4 − 6 x ⎛ 5x 3 − 6 ⎞
(c) f ′(x) = ⎜ or ⎟ (M1)(A1)
x 5 − 3x 2 ⎜⎝ x 4 − 3x ⎟⎠
1/3
f ′(x) is undefined at x = 0 and x = 3 (A1) 3
(d) For the x-coordinate of the local maximum of f (x), where
0 < x < 1.5 put f ′(x) = 0 (R1)
3
5x – 6 = 0 (M1)
1
⎛6 ⎞3
x= ⎜ ⎟ (A1) 3
⎝5⎠
(e) The required area is
1.35
A=
∫ f ( x ) dx
0.599
(A2) 2
Note: Award (A1) for each correct limit.

[16]
11. (a)
y
4
y = x2
2
R P
x
–2 –1 Q O 1 2
–1
y = – –12 lnx (C2) 2
2 1
Note: Award (C1) for y = x , (C1) for y = – lnx. 2
2 1
(b) x + 2
ln x =0 when x = 0.548217.
Therefore, the x-coordinate of P is 0.548…. (G2) 2
2
(c) The tangent at P to y = x has equation y = 1.0964x – 0.30054, (G2)
and the tangent at P to y = – 12 ln x has equation y = –0.91205x + 0.80054. (G2)
1
Thus, the area of triangle PQR = 2 (0.30052 + 0.80054)(0.5482). (M1)
= 0.302 (3 sf) (A1)
OR
2 dy
y=x ⇒ = 2x (M1)
dx
2 2
Therefore, the tangent at (p, p ) has equation 2px – y = p . (C1)
dy 1
y = – 12 ln x ⇒ =− (M1)
dx 2x
2 3
Therefore, the tangent at (p, p ) has equation x + 2py = p + 2p . (C1)
2 2
Thus, Q = (0, –p ) and R = (0, p + 12 ).
Thus, the area of the triangle PQR
2
= 12 (2p + 12 )p (M1)
= 0.302 (3 sf) (A1) 6
2 dy
(d) y = x ⇒ when x = a, = 2a (C1)
dx

dy 1
y = – 12 ln x ⇒ when x = a, =− (a > 0) (C1)
dx 2a
⎛ 1 ⎞
Now, (2a) ⎜ − ⎟ = –1 for all a > 0. (M1)
⎝ 2a ⎠
Therefore, the tangents to the curve at x = a on each curve are
always perpendicular. (R1)(AG) 4
[14]
a + b sin x
12. (a) (i) y= ,0<a<b
b + a sin x
dy (b + a sin x)(b cos x) − (a + b sin x)(a cos x)
= (M1)(C1)
dx (b + a sin x) 2
b 2 cos x + ab sin x cos x − a 2 cos x − ab sin x cos x
= (M1)(C1)
(b + a sin x) 2
=
(b 2
)
– a 2 cos x
(AG) 4
(b + a sin x )2
dy 2 2
(ii) = 0 ⇒ cosx = 0 since b – a ≠ 0.
dx
π
This gives x = (+πk, k ∈ ) (M1)(C1)
2
π a+b
When x = , y = = 1,
2 b+a
3π a−b
and when x = ,y= = –1.
2 b−a
Therefore, maximum y = 1 and minimum y = –1. (A2) 4
(iii) A vertical asymptote at the point x exists if and only if
b + a sin x = 0. (R1)
b
Then, since 0 < a < b, sin x = – < –1 , which is impossible. (R1)
a
Therefore, no vertical asymptote exists. (AG) 2
(b) (i) y-intercept = 0.8 (A1)
4
(ii) For x-intercepts, sin x = – ⇒ x = 4.069, 5.356. (A2)
5
(iii)
y
1
0 π/2 π m 3π/2 n 2π x
–1
(C2) 5
4.069 4 + 5 sin x 5.356 4 + 5 sin x
(c) Area =
∫ 0 5 + 4 sin x
dx − ∫4.069 5 + 4 sin x
dx (M1)(C1)
OR
5.356 4 + 5 sin x
Area =
∫ 0 5 + 4 sin x
dx (M1)(C1) 2
[17]
13.
y
y = f’(x)
a b x
maximum
inflexion
inflexion
a minimum b x
(A1)(A1)(A1)
Note: Award (A1) for the sketch
(A1) for maximum and minimum points
(A1) for points of inflexion
[3]
2 ds
14. Given s = 40t + 0.5at , then the maximum height is reached when =0 (M1)
dt
at + 40 = 0 (M1)
− 40
a= = –1.6 (units not required) (A1) (C3)
25
[3]
15. (a) f (x) = x ⎛⎜ 3 (x − 1) ⎞⎟
2 2
⎝ ⎠
1.5
1
max
max 0.5
zero
–1 –0.5 zero
0.5 1
–0.5
min
min zero
–1
–1.5 (A4) 4
Notes: Award (A1) for the shape, including the two cusps (sharp points) at x
= ±1.
(i) Award (A1) for the zeros at x = ±1 and x = 0.
(ii) Award (A1) for the maximum at x = –1 and the minimum at
x = 1.
(iii) Award (A1) for the maximum at approx. x = 0.65 , and the minimum at
approx. x = –0.65
There are no asymptotes.

The candidates are not required to draw a scale.
( )
2
(b) (i) Let f (x) = x x 2 − 1 3
1 2
4 2 2 −
Then f ′(x) = x ( x − 1) 3 + ( x 2 − 1) 3 (M1)(A2)
3
1
− ⎡4 ⎤
f ′(x) = ( x 2 − 1) 3 ⎢ x 2 + ( x 2 − 1) ⎥
⎣3 ⎦
1
3⎛ ⎞
− 7 2
f ′(x) = ( x 2 − 1) ⎜ x − 1⎟ (or equivalent)
⎝3 ⎠
7x2 − 3
f ′(x) = 1
(or equivalent)
2
3( x − 1) 3
The domain is –1.4 ≤ x ≤ 1.4, x ≠ ±1 (accept –1.4 < x < 1.4, x ≠ ±1) (A1)
(ii) For the maximum or minimum points let f ′(x) = 0
2
ie (7x – 3) = 0 or use the graph. (M1)
Therefore, the x-coordinate of the maximum point is
3
x= (or 0.655) and (A1)
7
3
the x-coordinate of the minimum point is x = − (or –0.655). (A1) 7
7
Notes: Candidates may do this using a GDC, in that case award (M1)(G2).
(c) The x-coordinate of the point of inflexion is x = ±1.1339 (G2)
OR
4 x(7 x 2 − 9)
f ″(x) = , x ≠ ±1 (M1)
93 ( x 2 − 1) 4
For the points of inflexion let f ″(x) = 0 and use the graph,
9
ie x = = 1.1339. (A1) 2
7
Note: Candidates may do this by plotting f′ (x) and finding the x-coordinate
of the minimum point. There are other possible methods.
[13]
16. METHOD 1
(a) The equation of the tangent is y = –4x – 8. (G2) (C2)
(b) The point where the tangent meets the curve again is (–2, 0). (G1) (C1)
METHOD 2
dy 2
(a) y = –4 and = 3x + 8x + 1 = –4 at x = –1. (M1)
dx
Therefore, the tangent equation is y = –4x – 8. (A1) (C2)
3 2
(b) This tangent meets the curve when –4x – 8 = x + 4x + x – 6 which gives
3 2 2
x + 4x + 5x + 2 = 0 ⇒ (x + 1) (x + 2) = 0.
The required point of intersection is (–2, 0). (A1) (C1)
[3]
17. METHOD 1
2 2 2 1 2
Let S = AP = (x – 2) + (x + 2
) . (M1)
The graph of S is as follows:

41/4
x
0 1 2
The minimum value of S is 2.6686. (G1)
Therefore the minimum distance = 2.6686 = 1.63 (3 sf) (A1)
OR
The minimum point is (0.682, 1.63) (G1)
The minimum distance is 1.63 (3 sf) (G1) (C3)
METHOD 2
2 2 2 2
Let S = AP = (x – 2) + (x + 12 ) . (M1)
dS 2 1 3 2
= 2(x – 2) + 4x(x + 2
) = 4(x + x + 1)
dx
3
Solving x + x – 1 = 0 gives x = 0.68233 (G1)
Therefore the minimum distance = (0.68233 − 2) 2 + (0.682332 + 0.5) 2
= 1.63 (3 sf) (A1) (C3)
[3]
18. METHOD 1
b
a– b
1
a
⎪a – b⎪= 12 + 12 − 2(1)(1) cos θ (M1)

= 2(1 − cos θ ) (A1)
1
= 4 sin 2 θ
2
1
= 2sin θ. (A1) (C3)
2
METHOD 2
O
–12
a b
M
A a–b B AB=a–b

1
In ΔOAM, AM = OA sin θ. (M1)(A1)
2
1
Therefore, ⎪a – b⎪= 2 sin θ. (A1) (C3)
2
[3]
⎛π ⎞
19. (a) v(t) = t sin ⎜ t ⎟ = 0 when t = 0, t = 3 or t = 6 (C1)(C1)(C1) 3
⎝3 ⎠
3 ⎛π ⎞ 6 ⎛π ⎞
(b) (i) The required distance, d = t sin ⎜ t ⎟dt −
0 ⎝3 ⎠ ∫ ∫ t sin ⎜⎝ 3 t ⎟⎠dt
3
(M1)
(ii) d = 2.865 + 8.594 (G2)

= 11.459
= 11.5 m. (A1)
OR
6 ⎛π ⎞
(i) The required distance, d =
∫ 0
t sin ⎜ t ⎟ dt .
⎝3 ⎠
(M1)
(ii) d = 11.5 m. (G2)(A1)

OR
3 ⎛π ⎞ 6 ⎛π ⎞
(i) The required distance, d =
∫ t sin ⎜⎝ 3 t ⎟⎠dt − ∫ t sin ⎜⎝ 3 t ⎟⎠dt
0 3
(M1)
9 ⎧⎪ 3⎛ π ⎞ 2 ⎛π ⎞ 6⎛
π⎞
2
⎛ π ⎞ ⎫⎪
(ii) d=
π2 ∫
⎨ 0 ⎜ ⎟ t sin ⎜ t ⎟dt −
⎪⎩ ⎝ 3 ⎠ ⎝3 ⎠ ∫ ⎜ ⎟ t sin ⎜ t ⎟dt ⎬
3⎝ 3 ⎠ ⎝ 3 ⎠ ⎪⎭
(C1)
9 ⎧⎪⎡ ⎛ π ⎞ π ⎛ π ⎞⎤
3
⎡ ⎛π ⎞ π ⎛ π ⎞⎤ ⎫⎪
6
= ⎨⎢sin ⎜ t ⎟ − t cos⎜ t ⎟⎥ − ⎢sin ⎜ t ⎟ − t cos⎜ t ⎟⎥ ⎬
π2 ⎪⎩⎣ ⎝ 3 ⎠ 3 ⎝ 3 ⎠⎦ 0 ⎣ ⎝ 3 ⎠ 3 ⎝ 3 ⎠⎦ 3 ⎪⎭
[from (i)] (C1)
9
= (sin π – π cos π – sin 2π + 2π cos 2π + sin π – π cos π)
π2
36
= m (11.5 m). (A1) 4
π
18
Note: Award (A1) for ± (±5.73) which is obtained by integrating v
π
from 0 to 6.
[7]
20. (a) The derivative can be found by logarithmic differentiation. Let y = f (x).
1
1
y= xx ⇒ ln y =
ln x (M1)
x
y′ − 1 1 1 1 − ln x
⇒ = 2 ln x + × = (M1)(M1)
y x x x x2
⎛ 1 − ln x ⎞
⇒ y′ = y ⎜ ⎟
2
⎝ x ⎠
⎛ 1 − ln x ⎞
that is, f ′(x) = f (x) ⎜ 2
⎟ (AG) 3
⎝ x ⎠
(b) This function is defined for positive and real numbers only.
To find the exact value of the local maximum:
y′ = 0 ⇒ ln x = 1 ⇒ x = e (M1)
1
⇒y= ee (A1)
To find the horizontal asymptote:

1
ln x
lim ( y = x x ) ⇒ lim ln y = lim =0
x →∞ x →∞ x →∞ x
⇒ lim y = 1 (M1)(A1)
x→∞
y
1
(e,e (–e ))
y=1
x
(A1) 5
(c) By Taylor’s theorem we have
f ′′(e) 2
P2(x) = f (e) + f ′(e)(x – e) + (x – e) (A1)
2
⎛ 1 − ln x ⎞ ⎛ 2 ln x − 3 ⎞
f ″(x) = f ′(x) ⎜ 2 ⎟ + f ( x)⎜ 3 ⎟ (M1)
⎝ x ⎠ ⎝ x ⎠
1 1
⎛ 2 − 3⎞ e ⎛ − 1 ⎞ = −e e
−3
Also, f ′(e) = 0, and f ″(e) = 0 + f (e) ⎜ 3 ⎟ = e ⎜ 3 ⎟ (M1)(A1)
⎝ e ⎠ ⎝e ⎠
1
1 −3
ee 2
hence P2(x) = ee − (x – e) which is a parabola with vertex
2
1
at x = e and P2(e) = ee = f (e) (R1)(AG) 5
[13]
1 dt
21. (a) Distance = ∫ 2+t
0 2
(M1)(A1)
1
⎡ 1 ⎛ t ⎞⎤ ⎛ 1 1 ⎞
= ⎢ arctan⎜ ⎟⎥ ⎜ or tan –1 ⎟ (A1)
⎣ 2 ⎝ 2 ⎠⎦ 0 ⎝ 2 2⎠
= 0.435 (A1) (C4)
dv
(b) Acceleration = (M1)
dt
– 2t
= (A1) (C2)
(2 + t ) 2 2
[6]
3 2 dy 2 dy
22. y + 3xy + 4xy + 2x =0 (M1)(A1)
dx dx
⇒
dy
=
(
– y + 4 xy 3
) (A1)
dx 3xy 2 + 2 x 2
dy
At (1,1), = –1 (A1)
dx
Equation of tangent is y – 1 = –l(x – 1) or x + y = 2 (A2) (C6)
[6]
– x2 – x2
23. (a) A = 2x × e = 2x e (M1)(A1) (C2)

dA 2 2
(b) = 2 (1 – 2x ) e – x (A2)
dx
dA 1
= 0 when x = (A1)
dx 2
1
–
Amax = 2e 2 (or 0.858) (A1) (C4)
[6]
24. denotes pts of inflexion
y
0 1 2 3 4 x
(A2)(A2)(A2)
(C6)
Note: Award (A2) for the shape, and (A2) for each point of inflexion.
[6]
2
25. v(t) = 6t – 6t = 6t (t – 1)
+ – +
v(t)
0 1
METHOD 1
∫ (6t ) ∫ (6t )
1 2
2 2
Distance travelled = – – 6t dt + – 6t dt (M1)(M1)
0 1
= 1 + 5 (G1)(G1)
= 6 m. (G2) (C6)
METHOD 2
∫ (6t ) ∫ (6t )
1 2
2 2
Distance travelled = – – 6t dt + – 6t dt (M1)(M1)
0 1
3 2 1 3 2 2
= – [2t – 3t ] 0 + [2t – 3t ] 1 (A1)(A1)
= – (–1) + 2 (7) – 3(3)
= 6 m. (A2) (C6)
Note: Award (G1)(or (A1)) if the units are missing.
[6]
26. METHOD 1
x
y = xe
dy x x
= xe + e (M1)(A1)
dx
d2 y x x
= xe + 2e (M1)(A1)
dx 2
x
= e (x + 2) (A1)
Therefore the x-coordinate of the point of inflexion is x = –2 (A1) (C6)
METHOD 2
Sketching y = f '(x)

y
f ′(x)
–3 –2 –1 0 x
(G4)
f '(x) has a minimum when x = –2 (G1)
Thus, f (x) has point of inflexion when x = –2 (G1)
[6]
dV 4 3
27. = 8 (cm3s–1), V = πr
dt 3
dV 2
=> = 4πr (M1)(A1)
dr
dV dV dr dr ⎛ d V ⎞ ⎛ d V ⎞
= × => = ⎜ ⎟÷⎜ ⎟ (M1)
dt dr dt dt ⎝ d t ⎠ ⎝ d r ⎠
dr 2
When r = 2, = 8 ÷ (4π × 2 ) (M1)(A1)
dt
1 –1
= (cm s ) (do not accept 0.159) (A1) (C6)
2π
[6]
28.
y y= x 2
B(a, b)
0 A(6, 0) x
2
b=a
2 2 4
AB = (a – 6) + a (M1)(A1)
2 4
Minimum value of (x – 6) + x occurs at x = 1.33 (G3)
=> a = 1.33 (G1) (C6)
[6]
29. (a) (A1)(A1) 2
6
2 A
g(x)
–4 –3 –2 –1 1 2 3 4 5 6
–2
–4
–6 f(x)
Note: Award (A1) for showing the basic shape of f (x).
Award (A1) for showing both the vertical asymptote and
the basic shape of g (x).
(b) (i) x = –3 is the vertical asymptote. (A1)
2
(ii) x-intercept: x = 4.39 ( = e – 3) (G1)
y-intercept: y = –0.901 ( = ln 3 – 2) (G1) 3
(c) f (x) = g (x)
x = –1.34 or x = 3.05 (G1)(G1) 2
(d) (i) See graph
∫ (4 – (1 – x) ) – (ln (x + 3) – 2)dx
3.05 2
(ii) Area of A = (M1)(A1)
0
(iii) Area of A = 10.6 (G1) 4
(e) y = f (x) – g (x)
2
y = 5 + 2x – x – ln(x + 3)
dy 1
= 2 – 2x – (M1)
dx x+3
dy
Maximum occurs when =0
dx
1
2 – 2x =
x+3
2
5 – 4x – 2x = 0
x = 0.871 (A1)
y = 4.63 (A1)
OR
Vertical distance is the difference f (x) – g (x). (M1)
Maximum of f (x) – g (x) occurs at x = 0.871. (G1)
The maximum value is 4.63. (G1) 3
[14]
30. METHOD 1
2 2 3 dy
3x y + x 2y =0 (M1)(A1)
dx
dy
At (2, 1), 12 + 16 =0 (M1)
dx
dy 3
⇒ =– (A1)
dx 4
4
Gradient of normal = (A1)
3
4
Equation of normal is y – 1 = (x – 2) (A1) (C6)
3
METHOD 2
3
–
y=2 2x 2 (A1)
5
dy –
= –3 2x 2 (M1)(A1)
dx
3
= – when x = 2 (A1)
4
4
3
4
Equation of normal is y – 1 = (x – 2) (A1) (C6)
3
[6]

π 2π
∫ e– ∫
t
31. Distance travelled = sin tdt + e – t sin tdt
0 π
First integral = 0.620 (G2)
Second integral = (±)0.232 (G2)
Distance travelled = 0.852 (G2) (C6)
[6]
32.
y2
A B x
(C6)
Notes: Award (A1) for zero at A, (A1) for correct (concave) shape to left of
dotted line, (A1) for correct (concave) descent to right of dotted line,
(A1) for zero at B, (A1) for maximum at C, (A1) for asymptotic to x-axis as
x→ ∞ .
Please note that the first and fourth (A1) marks are given for the candidate’s
graph hitting the x-axis at A and B. No marks are given for the exact shape
of this graph at A and B since it is not possible to deduce from the given
graph whether or not the gradient of y2 is continuous at these points.
[6]
2 x ⋅ 2 – x 2 ln 2x 2 x
33. (a) (i) f ′(x) = (M1)(A1)
22 x
2 x – x 2 ln 2
= (AG)
2x
(ii) f "(x) =
2 x [2 – 2 x ln 2] – 2 x ln 2 2 x – x 2 ln 2 [ ] (M1)(A1)
22 x
x 2 (ln 2) – 4 x ln 2 + 2
2
= (A1) 5
2x
Note: Award the second (A1) for some form of simplification,
x ln 2(x ln 2 – 4) + 2
eg accept .
2x
2 2
(b) (i) 2x – x ln 2 = 0 giving x = (M1)(A1)
ln 2
Note: Award (M1)(A0) for x = 2.89.
(ii) With this value of x,
4–8+2
f′′ (x) = <0 (M1)(A1)
+ ve number
Therefore, a maximum. (AG) 4
(c) Points of inflexion satisfy f′′ (0) = 0, ie
2 2
x (ln 2) – 4x ln 2 + 2 = 0 (M1)
4 ln 2 ± 8(ln 2)
2
⇒x = (A1)
2(ln 2)
2

2± 2
= (= 0.845, 4.93) (A1)
ln 2
OR
x = 0.845, 4.93 (M1)(G1)(G1) 3
[12]
3
34. tan θ =
x
2 dθ − 3 dx
sec θ = (M1)
dt x 2 dt
π 2 2
when θ = , x = 3 and sec θ = 4 (A1)(A1)
3
dx − x 2 sec 2 θ dθ
= (M1)
dt 3 dt
dx − 3(4) ⎛ 1 ⎞
= ⎜ ⎟
dt 3 ⎝ 60 ⎠
dx 1 –1
=− km s
dt 15
dx –1
= –240 km h (A1)
dt
–1
The airplane is moving towards him at 240 km h (A1) (C6)
–1
Note: Award (C5) if the answer is given as –240 km h .
[6]
35. Let h = height of triangle and θ = CÂB . (A1)
Then, h = 5 tan θ (A1)
dh dθ
= 5 sec 2 θ × (M1)(A1)
dt dt
π
Put θ = .
3
dθ
2=5×4× (A1)
dt
dθ 1 ⎛ 18° ⎞
= rad per sec ⎜ Accept per second or 5.73° per second ⎟ (A1)(A1)(C6)
dt 10 ⎝ π ⎠
Note: Award (A1) for the correct value, and (A1) for the correct units.
[6]
3 3
36. x + y – 9xy = 0
Differentiating w.r.t. x
dy dy
⇒ 3x 2 + 3 y 2 – 9 y – 9x = 0 (A1)(A1)
dx dx
dy dy
Note: Award (A1)for 3 x 2 + 3 y 2 , and (A1) for – 9 y – 9 x .
dx dx
dy 9 y – 3x 2
⇒ = (A1)
dx 3 y 2 – 9 x
EITHER
at point (2, 4) gradient = 0.8. (A1)
⇒ Gradient of normal = –1.25 (A1)
OR

– 3y 2 + 9x
9 y – 3x 2
at point (2, 4), gradient is –1.25 (A1)
THEN
Equation of normal is given by
y – 4 = –1.25(x – 2) or y = –1.25x + 6.5 (A1) (C6)
[6]
3 2 500
37. (a) V = 500 cm ⇒ πr h = 500 ⇒ h = (A1)
πr 2
2 2 1000
Now S = 2πr + 2πrh ⇒ S = 2πr + (M1)(A1) (C3)
r
dS 1000
(b) = 4πr – 2 (A1)
dr r
dS
for min S we need =0 (A1)
dr
250
⇒r =3 (or r = 4.30) (A1) (C3)
π
[6]
2
38. (a) s = 50t = 10t + 1000
v = ds (M1)
dt
= 50 – 20t A1 (N2) 2
(b) Displacement is max when v = 0, M1
ie when t = 5 . A1
2
2
Substituting t = 5 , s = 50 × 5 – 10 × ⎛⎜ 5 ⎞⎟ + 1000 (M1)
2 2 ⎝2⎠
s = 1062.5 m A1 (N2) 4
[6]
39. EITHER
Let s be the distance from the origin to a point on the line, then
2 2 2
s = (l – λ) + (2 – 3λ) + 4 (M1)
2
= 10λ – 14λ + 9 A1
2
d(s )
= 20λ – 14 A1
dλ
d(s 2 )
For minimum = 0, ⇒ λ = 7 A1
dλ 10
OR
The position vector for the point nearest to the origin is perpendicular to
the direction of the line. At that point:
⎛ 1 − λ ⎞ ⎛ − 1⎞
⎜ ⎟ ⎜ ⎟
⎜ 2 − 3λ ⎟ • ⎜ − 3 ⎟ = 0 (M1)A1
⎜ 2 ⎟ ⎜ 0 ⎟
⎝ ⎠ ⎝ ⎠
Therefore, 10λ – 7 = 0 A1
Therefore, λ = 7 A1
10
THEN
x= 3 ,y=– 1 (A1)(A1)
10 10

⎛ 3 −1 ⎞
The point is ⎜ , , 2⎟. N3
⎝ 10 10 ⎠
[6]
1
x – ln x × 1
40. (a) (i) Attempting to use quotient rule f ′(x) = x (M1)
x2
1 − ln x
f ′(x) = A1
x2
x 2 ⎛⎜ – 1 ⎞⎟ – (1 − ln x)2 x
f ″(x) =
⎝ x⎠ M1
x4
2ln x − 3
f ″(x) = A1
x3
Stationary point where f ′(x) = 0, M1
ie ln x = 1, (so x = e) A1
f ″(e) < 0 so maximum. R1AG N0
(ii) Exact coordinates x = e, y = 1 A1A1 N2 9
e
(b) Solving f ″(0) = 0 M1
1n x = 3 (A1)
2
3
x = e 2 (4.48) A1 N2 3
5 ln x
(c) Area =
∫ 1 x
dx A1
EITHER
Finding the integral by substitution/inspection
u = ln x, du = 1 dx (M1)
x
2 ⎛ (ln x) 2 ⎞
∫ udu = u ⎜=
⎜
⎟
⎟
M1A1
2 ⎝ 2 ⎠
5
⎡ (1n x) 2 ⎤
Area = ⎢
2
1 2
⎥ = 2 (1n 5) − (ln 1) (
2
) A1
⎣ ⎦1
Area = 1 (ln 5) (= 1.30)
2
A1 N2 6
2
OR
Finding the integral I by parts (M1)
u = ln x, dv = 1 ⇒ du = 1 , v = ln x
x x
I = uv – udv = (1n x) − ln x 1 dx = (1n x) 2 − I
∫ ∫
2
M1
x
2
2 (1n x)
⇒ 2I = (ln x) ⇒ I = A1
2
5
⎡ (1n x) 2 ⎤
⇒ area = ⎢
2
1 2
⎥ = 2 (1n 5) − (1n 1)
2
( ) A1
⎣ ⎦1
1 2
Area = (ln 5) (= 1.30) A1 N2 6
2
Note: Award N1 for 1.30 with no working.

b
(d) Using V =
∫ a
πy 2 dx (M1)
2
π ⎛⎜ 1n x ⎞⎟ dx
5
=
∫1 ⎝ x ⎠
A1
= 1.38 A1 N2 3
[21]
dy k 2
41. =− 2 + (M1)(A1)
dx x x
3
When x = 2 , gradient of normal = − (A1)
2
dy 2
⇒ = (A1)
dx 3
k 2 4
⇒ − +1 = ⇒ k= (M1)(A1) (C6)
4 3 3
[6]
42. (a) (i) f ′ ( x) = pe px ( x + 1) + e px (A1)
= e px ( p ( x + 1) + 1) (AG)
(ii) The result is true for n = 1 since
LHS = e px ( p ( x + 1) + 1)
and RHS = p1−1e px ( p ( x + 1) + 1) = e px ( p ( x + 1) + 1) . (M1)
Assume true for n = k : f (k ) k −1 px
( x) = p e ( p ( x + 1) + k ) (M1)
f ( k +1) ( x) = ( f ( k ) ( x) )′ = p k −1 pe px ( p ( x + 1) + k ) + p k −1e px p (M1)(A1)
= p k e px ( p ( x + 1) + k + 1) (A1)
Therefore, true for n = k ⇒ true for n = k + 1 and the
proposition is proved by induction. (R1) 7
(b) (i) f ′ ( x) = e 3x
( 3 ( x + 1) + 1 = 0 ) (M1)
1+ 3 ⎛ 3 + 3⎞
⇒x=− ⎜⎜ = − ⎟ (A1) N1
3 ⎝ 3 ⎟⎠
(ii) f ′′ ( x) = 3e 3x
( 3 ( x + 1) + 2 = 0 ) (M1)
2+ 3 ⎛ 2 3 + 3⎞
⇒x=− ⎜⎜ = − ⎟ (A1) N1 4
3 ⎝ 3 ⎟⎠
(c) f ( x) = e0.5 x ( x + 1)
EITHER
−1 2
area = − ∫ −2
f ( x)dx + ∫ f ( x)dx
−1
(M1)
= 8.08 (A1) N2
OR
2
area = ∫ −2
f ( x) dx (M1)
= 8.08 (A1) N2 2
[13]
43.

O r
A
C N B
(a) h = r sinθ CB = 2CN = 2r cosθ (A1)(A1)

h
Using T = (r + CB) (M1)
2
r2
T = (sinθ + 2sin θ cosθ ) (A1)
2
r2
= (sin θ + sin 2θ ) (AG) N0 4
2
dT r 2
(b) = (cosθ + 2cos2θ ) = 0 (for max) (M1)
dθ 2
⇒ cosθ + 2(2cos2 θ −1) = 4cos2 θ + cosθ − 2 = 0 (M1)(AG)
⇒ cosθ = 0.5931 (θ = 0.9359) (A1)
d 2T r 2
= (− sinθ − 4sin 2θ ) (M1)
dθ 2 2
d 2T
θ = 0.9359 ⇒ = −2.313r 2 < 0
dθ 2
⇒ there is a maximum (when θ = 0.9359 ) (R1) 5
θ
(c) In triangle AOB: AB = 2r sin (M1)(A1)
2
θ
Perimeter OABC = 2r + 2r cosθ + 2r sin = 75 (M1)
2
When θ = 0.9359 , r = 18.35 cm (A1)
r2 18.352
Area OABC = (sinθ + sin 2θ ) = (sin 0.9359 + sin 1.872) (M1)
2 2
= 296 cm2 (A1) N3 6
[15]
3
x
44. y= − x 2 − 3x + 4
3
dy
= x 2 − 2x − 3 (M1)
dx
dy
at = 0 , (x − 3)(x + 1) = 0 (M1)
dx
17
x = 3 , − 1; y = − 5 ,
3
⎛ 17 ⎞
So P(3, − 5) and Q ⎜ − 1 , ⎟ (A1)(A1)
⎝ 3⎠

y +5 x −3
Equation of (PQ) is = (M1)
⎛ 17 ⎞ − 1 − 3
⎜ + 5⎟
⎝ 3 ⎠
3 y + 15 x − 3
=
32 −4
3 y + 15 x − 3
=
8 −1
− 3y − 15 = 8x − 24
8 x + 3y − 9 = 0 (A1) (C6)
[6]
45.
A B
Arc length, s = 4θ (A1)

ds dθ
=4 (A1)
dt dt
1 2
Segment area, (4) (θ − sin θ )= 8 (θ − sin θ )
A= (A1)
2
dA dθ
= 8(1 − cos θ ) (A1)
dt dt
When numerically equal,
dθ dθ
4 = 8(1 − cos θ ) (M1)
dt dt
1
=1 − cos θ
2
1
= cos θ
2
π
θ = (Accept 60°) (A1) (C6)
3
[6]
46.
E
P D
x
F G

Let BP̂D = α and AP̂D = β then θ = α − β

tan α − tan β
tan θ = tan (α − β) = (M1)
1 + tan α tan β
b a
−
=
x x = (b − a ) x (A1)
ab x 2 + ab
1+ 2
x
( )
d (tan θ) x 2 + ab (b − a ) − (b − a ) 2 x 2
= (M1)
dx x 2 + ab
2
( )
(b − a )(ab − x 2
)
= (A1)
(x 2
+ ab )2
2
at maximum (ab − x ) = 0, b ≠ a
x= ab (A1)
d 2 (tan θ)
= (b − a )
⎡ x 2 + ab
⎢
( ) (− 2 x )− 4 x(ab − x )(x
2 2 2
+ ab ⎤
⎥
) (M1)
d x2 ⎢⎣ (x + ab )
2 4
⎥⎦
=
(b − a ) [− 2 x 3
− 2 xab − 4 xab + 4 x 3 ]
(x + ab
2
)3
=
(
(b − a ) 2 x 3 − 6 xab ) (A1)
(x 2
+ ab ) 3
at x =
d 2 (tan θ) (b − a ) − 4ab ab
ab , =
( )
dx 2 8a 3b 3
− (b − a ) ab
= (A1)
2a 2 b 2
d (tan θ)
2
since < 0 at x = ab this value is a maximum. (R1)
d x2
[9]
47. METHOD 1
2
Line and graph intersect when 3x − x + 4 = mx + 1 M1
2
ie 3x − (1 + m) x + 3 = 0 (A1)
y = mx + 1 tangent to graph ⇒
2
3x − (1 + m) x + 3 = 0 has equal roots (M1)
2
i.e b − 4ac = 0 (A1)
2
⇒ (1 + m) − 36 = 0 A1
⇒ m = 5, m = −7 A1 N3
METHOD 2
f ′(x) = 6x − 1 A1
2
3x − x + 4 = mx + 1 (M1)
Substitute m = 6x − 1 into above M1
2 2
3x −6x + 3 = 0 (A1)
⇒ x = ±1 A1
⇒ m=5, m = − 7 A1 N3
[6]

dv
48. (a) maximum when = 0 (or any correct method) (M1)
dt
t=3 (A2)
2
d vA
t = 3, = − 1⇒ maximum R1
dt 2
−1
⇒ maximum v A = 6 (m s ) A1 N3
dv
(b) Using acceleration = M1
dt
1 0 .2 t
= e (A1)
5
−2
when t = 4, a = 0.445 (m s ) A1 N2
(c) using s A = ∫ v A dt or s B = ∫ vB dt (M1)
1 3 3
sA = − t3 + t 2 + t + c A1
6 2 2
when t = 0, s A = 0 ⇒ c = 0 M1
∫
0.2t 0.2t
sB = e dt = 5e +d A1
when t = 0, s B = 5 ⇒ d = 0 (M1)
1 3 3
sA = − t3 + t 2 + t A1 N1
6 2 2
0.2t
s B = 5e A1 N1
(d) (i)
s
30
25
20
15
10
t
0 1 2 3 4 5 6 7 8 9 10 11
A1A1A1A1
Note: For each curve, award A1 for the s
intercept and A1 for the correct
shape.
(ii) t = 1.95 and t = 7.81 A2A2
[23]
1 ⎛ 3 ⎞
49. (a) f ′(x )= ×3 ⎜⎜ = ⎟⎟ M1A1 N2
3x + 1 ⎝ 3x + 1 ⎠
3
(b) Hence when x = 2, gradient of tangent = (A1)
7

7
⇒ gradient of normal is − (A1)
3
7
y − ln 7 = − (x − 2 ) M1
3
7 14
y = − x + + ln 7 A1 N4
3 3
(accept y = −2.33x + 6.61)
[6]
2
50. (a) A = 2πx (M1)
dA dx
= 4πx A1
dt dt
dx 10
= A1
d t 4π × 2
5
= (0.398) A1
4π
3
(b) V = πx
dV dx
= 3πx 2 M1
dt dt
5
= 3π × 2 2 ×
4π
= 15 A1
[6]
51. (a) Using quotient rule
1
x 3 × − 3x 2 ln x
f ′(x ) = x A1
x6
1 − 3 ln x
= A1 N2
x4
3
− × x 4 − 4 x 3 (1 − 3 ln x )
f ′ ′ (x ) = x M1A1
x8
− 7 + 12 ln x
= A1 N2
x5
(b) (i) For a maximum, f′ (x) = 0 giving (M1)
1
ln x=
3
1
x= e 3 A1 N2
EITHER
1
⎛ 13 ⎞ 12 × 3 − 7
f ′′⎜⎜ e ⎟⎟ = 5
<0 M1A1
⎝ ⎠ e3
∴ maximum AG N0
OR
1
for x < e , f ′(x )> 0
3

1
for x > e , f ′(x )< 0
3 M1A1
∴ maximum AG N0
7
(ii) f ′′(0 ) = 0 ⇒ ln (x ) = M1
12
7
x=e 12
(1.79) A1
f ″(1.5) = − 0.281 A1
f ″(2) = 0.0412 A1
Note: Accept any two sensible values either side
of 1.79.
∴ Change of sign ⇒ point of inflexion R1
(iii)
A1A1
Note: Award A1 for shape, A1 for marking
values which locate the maximum point
and point of inflexion correctly.
3
(c) Using V =
∫ 1
πy 2 dx (M1)
2
3 ⎛ ln x ⎞
=
∫ 1
π⎜ 3
⎝ x ⎠
⎟ dx (A1)
= 0.0458 A1 N2
3 ln x
(d) Area A =
∫1 x3
dx A1
Using integrating by parts (M1)

3
⎡ ln x ⎤ 1 3 1
A = ⎢− 2 ⎥ +
⎣ 2 x ⎦1 2
∫ 1 x3
dx A1
3
ln 3 1 ⎡ 1 ⎤
= − − A1A1
18 4 ⎢⎣ x 2 ⎥⎦ 1
ln 3 1 ⎛ 1 ⎞ ⎛ ln 3 2 ⎞
= − − ⎜ − 1⎟ ⎜ = − + ⎟ A1
18 4 ⎝ 9 ⎠ ⎝ 18 9 ⎠
1
= (4 − ln 3) AG N0
18
[26]
4
81π = π (3) + π (3) h
3 2
52. (M1)
3
81π = 36π + 9πh
h = 5 (cm) A1

4
V = πr 3 + πr 2 h
3
Attempt to differentiate with respect to time. M1
dV dr
= 4πr 2 + A1
dt dt
dr dh
2πrh + πr 2 A1
dt dt
dh
204π = 4π (3) (2) + 2π (3)(5)(2) + π (3)
2 2
dt
dh
204π = 72π + 60π + 9π
dt
dh
72π = 9π
dt
dh
= 8 (cm/min) A1 N2
dt
[6]
53. METHOD 1
(
s = ∫ 3t 2 − 4t + 2 dt ) A1
Attempting to integrate the RHS (M1)
s = t − 2t + 2t + C
3 2
A1
Note: The A1 above must include C.
Using s (0) = −3 gives C = − 3 ( s = t 3 − 2t 2 + 2t − 3) A1
Solving t − 2t 3 2
+ 2t − 3 = 0 for t (M1)
t = 1.81 (sec) A1 N3
METHOD 2
∫ (3t ) ∫
T 0
2
− 4t + 2 dt = ds (= 3) A1A1
0 −3
Solving for T (M1)
T = 1.81 (sec) A3 N3
[6]
2 2 2
54. z = x + y (or equivalent) (M1)
z = 0.8 + 0.6 2 2
(=1, initially) A1
Attempting to differentiate implicitly with respect to t M1
dz dx dy
2z = 2x + 2 y A1
dt dt dt
dz
= − (0.8 × 60 ) − (0.6 × 70 ) A1
dt
−1
Rate is −90 (km h ) A1 N0
[6]
55. (a) (i)

Attempting to express α and β in terms of arctan (M1)

α = arctan
2 and β = arctan 1 A1A1
x x
θ=α−β A1
θ = arctan
2 − arctan 1 AG N0
x x
(ii) METHOD 1
Attempting to differentiate either arctan ⎛⎜ 2 ⎞⎟ or arctan

⎝x⎠
⎛⎜ 1 ⎞⎟ . M1
⎝x⎠
2
− 2
d ⎛ ⎛ 2 ⎞⎞ x ⎛ −2 ⎞
⎜⎜ arctan ⎜ ⎟ ⎟⎟ = ⎜= 2 ⎟ A1A1
dx ⎝ ⎝ x ⎠⎠ ⎛ 2 ⎞
2 ⎜ x +4⎟
⎝ ⎠
⎜ ⎟ +1
⎝ x⎠
1
− 2
d ⎛ ⎛ 1 ⎞⎞ x ⎛ −1 ⎞
⎜⎜ arctan ⎜ ⎟ ⎟⎟ = ⎜= 2 ⎟ A1
dx ⎝ ⎝ x ⎠⎠ ⎛ 1 ⎞
2 ⎜ x +1 ⎟
⎝ ⎠
⎜ ⎟ +1
x
⎝ ⎠
dθ 2 1 ⎛ 2 − x2 ⎞
=− 2 + 2 ⎜= 2 ⎟
( )( )
Simplifying, A1
dx x + 4 x + 1 ⎜⎝ x + 4 x 2 + 1 ⎟
⎠
For a maximum
dθ = 0 (R1)
dx
2 2
x + 4 = 2(x + 1) (or equivalent) A1
x = 2 (as x > 0 ) A1 N1
EITHER
Justifying the maximum using appropriately chosen x-values M1
eg when x = 1, and when x = 2,
Correct gradients calculated for chosen values. A1
dθ 1 dθ 1
eg x = 1, = and x = 2 , = −
dx 10 dx 20
Showing that dθ > 0 for x < 2 and dθ < 0 for x > 2 R1
dx dx
⇒x= 2 is a maximum. AG N0
Note: Award M1A1R1 for a clearly labelled
sketch of either θ or
dθ against x.
dx
OR

Attempting to find a second derivative. M1
d 2θ
=
4x
−
2x ⎜= (
⎛ 2 x x 4 − 4 x 2 −14 ) ⎞⎟
dx 2
(
x2 + 4
2
) (
x 2 +1
2
) ⎝ ( )(
⎜ x 2 +1 2 x 2 + 4 2 ) ⎟
⎠
d 2θ 2
When x = 2 2
=− (= − 0.157 ). A1
dx 9
Since d 2θ < 0 for x = 2, R1
dx 2
then x = 2 is a maximum. AG N0
METHOD 2
Given that 0 < θ <
π , θ will be a maximum when
2
tanθ is a maximum. R1
tan α − tan β
Using tan θ = to get tanθ in terms of x. (M1)
1 + tan α tan β
2−1
⎛ ⎞
tan θ = x x 2 ⎜⎜ = 2 x ⎟⎟ A1
x +2 ⎠
1 + ⎛⎜ 2 ⎞⎟ ⎝
⎝x⎠
2
( )
d (tan θ ) = x + 2 − x(2 x ) ⎛⎜ = 2 − x ⎞⎟
2
M1A1
dx x2 + 2
2
( ⎜ x2 + 2 2 ⎟
⎝ ) ⎠ ( )
For a maximum d (tan θ ) = 0 (R1)
dx
2
2−x =0 A1
x = 2 (as x > 0) A1 N1
EITHER
Justifying the maximum using appropriately chosen x-values M1
eg when x = 1,
d (tan θ ) = 1 and when x = 2 , d (tan θ ) = − 1 A1
dx 9 dx 18
Since d (tan θ ) > 0 for x < 2 and d (tan θ ) < 0
dx dx
for x > 2 R1
Note: Award M1A1R1 for a clearly labelled
d
sketch of either tanθ or (tan θ ) against x.
dx
OR
Attempting to find a second derivative. M1
d 2
(tan θ ) = 2 x 2x − 63( 2
)
dx 2
x +2 ( )
2
2
When x = 2 d 2 (tan θ ) = − (= − 0.177) A1
dx 8
Since d 2 (tan θ )< 0 for x = 2 , R1
dx 2

(iii) METHOD 1
Substituting x = 2 into θ = arctan 2 − arctan 1 (M1)

x x
2 1
θ = arctan − arctan (= 0.340 radians ). A1 N2
2 2
METHOD 2
x
Substituting x = 2 into θ = arctan 2
(M1)
x +2
2
θ = arctan (= 0.340 radians). A1 N2
4
(b) METHOD 1
2 1 ⎛ π ⎞
Attempting to solve arctan − arctan = 15° ⎜ = ⎟ for x M1
x x ⎝ 12 ⎠
x = 0.649, 3.08 (m) A2A2 N4
METHOD 2
x
Attempting to solve 2
= tan 15° (or equivalent) for x M1
x +2
x = 0.649, 3.08 (m) A2A2 N4
[22]
− x2
56. (a) y =e
dy 2
⇒ = − 2 xe − x A1
dx
d2 y 2 2
⇒ 2
= − 2e − x + 4 x 2 e − x M1A1
dx
(b) 2e − x2
(
−1 + 2 x 2 = 0 ) M1
1
⇒ x=± A1
2
d2 y
when x = 0 , = − 2 (< 0 ).
dx 2
d2 y
when x = ± 1, 2
= 2e −1 (> 0 ).
dx
Hence the points are points of inflexion. R1
[6]
2 2
57. (a) 2xy + 6x – 3y = 6
dy dy
⇒ 2x + 2 y + 12 x − 6 y = 0 M1A1A1A1A1
dx dx
Note: Award A1 for each correctly differentiated term.
dy 4 dy
⇒ + + 12 − 4 = 0
2 A1
dx 3 dx
dy 40 dy 20
⇒ 2 = ⇒ = AG N0
dx 3 dx 3
⎛1⎞ ⎛ 3 ⎞
(b) (i) The equation of the tangent at A is r = ⎜ 2 ⎟ + s⎜ ⎟ (M1)A1A1N3
⎜ ⎟ ⎜⎝ 20 ⎟⎠
⎝3⎠
Note: Award M1A0A1 for omitting “r” or for writing “L”.
2
(ii) Substituting x = 1 2y + 6 –3y = 6 M1

2
⇒ 3y – 2y = 0
⇒ y (3y – 2) = 0
2
⇒ y = 0, A1
3
dy
Gradient at B satisfies 2 + 0 + 12 = 0 M1
dx
dy
⇒ =− 6 (A1)
dx
1
Gradient of the normal is A1
6
⎛1⎞ ⎛ 6⎞
Hence equation of the normal at B is r = ⎜⎜ ⎟⎟ + t ⎜⎜ ⎟⎟ M1A1 N0
⎝ 0⎠ ⎝1⎠
18 + 20
(c) cosθ = (= 0.309) M1A1A1
409 37
⇒ θ = 1.26 rads or 72.0° A1 N4
[20]
58. (a) ∫
v = a dt M1
dt
∫ (1 + t )
= −
2
A1
1
v= +k A1
1+ t
1
When t = 1, v = 8 ⇒ 8 = +k M1
2
15
⇒ k= A1
2
1 15 31 −1
When t =3, v = + = ms A1
4 2 4
15
(b) When t → ∞, v → m s −1 A1 N1
2
1 15
(c) v= +
1+ t 2
∫
s = v dt M1
3
⎡ 15 ⎤
s = ⎢ln 1 + t + t ⎥ A1A1
⎣ 2 ⎦1
45 15
= ln 4 + − ln 2 −
2 2
= (ln 2 + 15) metres A1
Note: Do not penalize if units missing.
[11]
2
59. (a) f ′(x ) =1 − 1
A1
x3

1
2
⇒1− 1
= 0⇒ x 3 = 2⇒ x =8 A1
x3
2
(b) f ′′(x ) = 4
A1
3x 3
f′′ (8) > 0 ⇒ at x = 8, f (x) has a minimum. M1A1
[5]
2x
60. (a) f′ (x) = (1 + 2x) e A1
f′ (x) = 0 M1
2x 1
⇒ (1 + 2x)e = 0 ⇒ x = − A1
2
2 2 − 1 2x 2x
f ′′(x) = (2 x + 2 × 2 )e = (4x + 4)e A1
⎛ 1⎞ 2
f ′′ ⎜ −⎟= A1
⎝ 2⎠ e
2 1
> 0 ⇒ at x = − , f (x) has a minimum. R1
e 2
⎛ 1 1 ⎞
P⎜− ,− ⎟ A1
⎝ 2 2e ⎠
(b) f′′(x) = 0 ⇒ 4x + 4 = 0 ⇒ x = −1 M1A1
nd ⎛ 1⎞ 2 4
Using the 2 derivative f′′ ⎜ − ⎟ = and f′′ (−2) = − 4 , M1A1
⎝ 2⎠ e e
the sign change indicates a point of inflexion. R1
(c) (i) f (x) is concave up for x > − 1. A1
(ii) f (x) is concave down for x < −1. A1
(d)
A1A1A1A1
Note: Award A1 for P and Q, with Q above P,
A1 for asymptote at y = 0,
A1 for (0, 0),
A1 for shape.
(e) Show true for n = 1 (M1)
2x 2x
f′ (x) = e + 2xe A1
2x 0 2x
= e (1 + 2x) = (2x + 2 ) e
(k) k k − 1 2x
Assume true for n = k, ie f x = (2 x + k × 2 )e ,k≥1 M1A1

Consider n = k + 1, ie an attempt to find

d k
dx
(
f ( x) . ) M1
(k + 1) k 2x 2x k k−1
f (x) = 2 e + 2e (2 x + k × 2 ) A1
k k k −1 2x
= (2 + 2 (2 x + k × 2 )) e
k k k − 1 2x
= (2 × 2 x + 2 + k × 2 × 2 )e
k+1 k k 2x
= (2 x+2 +k×2 )e A1
k+1 k 2x
= (2 x + (k + 1) 2 ) e A1
P(n) is true for k ⇒ P(n) is true for k + 1, and since true
+
for n = 1, result proved by mathematical induction ∀ n ∈ R1
Note: Only award R1 if a reasonable attempt
th
is made to prove the (k + 1) step.
[27]
dV
61. (a) = cr A1
dt
4
V = πr 3
3
dV dr
= 4πr 2 M1A1
dt dt
dr
⇒ 4πr 2 = cr M1
dt
dr c
⇒ = A1
dt 4πr
k
= AG
r
dr k
(b) =
dt r
⇒ ∫ rdr = ∫ k dt M1
r2
= kt + d A1
2
An attempt to substitute either t = 0, r = 8 or t = 30, r = 12 M1
When t = 0, r = 8
⇒ d = 32 A1
2
r
⇒ = kt + 32
2
When t = 30, r = 12
12 2
⇒ = 30k + 32
2
4
⇒k = A1
3
2
r 4
∴ = t + 32
2 3
r2 4
When t = 15, = 15 + 32 M1
2 3
2
⇒ r = 104 A1

r ≈ 10 cm A1
Note: Award M0 to incorrect methods using
proportionality which give solution
r =10 cm.
[13]
d 1
62. (arctan (x − 1)) = (or equivalent) A1
1 + (x −1)
2
dx
1
mN = −2and so mT = (R1)
2
1 1
Attempting to solve = (or equivalent) for x M1
1 + (x −1)
2
2
x = 2 (as x > 0) A1
π
Substituting x = 2 and y = to find c M1
4
π
c=4+ A1 N1
4
[6]
2
63. (a) AQ = x + 4 (km) (A1)
QY = (2 − x) (km) (A1)
T = 5 5 AQ + 5QY (M1)
= 5 5 (x 2
)
+ 4 + 5(2 − x ) (mins) A1
(b) Attempting to use the chain rule on 5 5 (x + 4) 2

(M1)
d ⎛
(x ) 5 × (x + 4) × 2 x
1
1
+ 4 ⎞⎟ = 5
2 2 −
⎜5 5 2 A1
dx ⎝ ⎠ 2
⎛ 5 5x ⎞
⎜= ⎟
⎜ 2 ⎟
⎝ x +4 ⎠
d
(5(2 − x )) = − 5 A1
dx
dT 5 5x
= −5 AG N0
dx x2 + 4
(c) (i) 5 x = x 2 + 4 or equivalent A1
Squaring both sides and rearranging to
2 2
obtain 5x = x + 4 M1
x=1 A1 N1
Note: Do not award the final A1 for stating a
negative solution in final answer.
(ii) T = 5 5 1 + 4 + 5(2 −1) M1
= 30 (mins) A1 N1
Note: Allow FT on incorrect x value.
(iii) METHOD 1
Attempting to use the quotient rule M1
u = x ,v = x 2 + 4 ,
du
dx
=1 and
dv
dx
= x x2 + 4 ( )−1 / 2
(A1)

d 2T
⎡ 2 1 2
⎢ x +4 − 2 x +4
−1 / 2 ⎤
× 2x 2 ⎥ ( )
=5 5⎢
dx 2 ⎢ x2 + 4
⎥
⎥ ( ) A1
⎣⎢ ⎦⎥
Attempt to simplify (M1)
=
5 5
[x 2
]
+ 4 − x 2 or equivalent A1
(x 2
+4 )
3/ 2
20 5
= AG
(x 2
+4 )
3/ 2
20 5
When x = 1, > 0 and hence T = 30
(x 2
+4 )
3/ 2
is a minimum R1 N0
Note: Allow FT on incorrect x value, 0 ≤ x ≤ 2.
METHOD 2
Attempting to use the product rule M1
u = x , v = x2 + 4,
du
dx
=1 and
dv
dx
= x x2 + 4
−1 / 2
( ) (A1)
( ) ( )
2
d T −1 / 2 5 5x 2 −3 / 2
2
=5 5 x2 + 4 − x +4 × 2x A1
dx 2
⎛ 2 ⎞
⎜= 5 5 − 5 5 x ⎟
( )
⎜ x 2 + 4 1/ 2 x 2 + 4 3 / 2 ⎟
⎝ ⎠ ( )
Attempt to simplify (M1)
=
(
5 5 x2 + 4 −5 5 x2 ) (
⎛ 5 5 x2 + 4 − x2
⎜= )⎞⎟ A1
(x 2
+4 ) 3/ 2 ⎜
⎝ x2 + 4(3/ 2
) ⎟
⎠
20 5
= AG
(x 2
+4 ) 3/ 2
20 5
When x = 1, > 0 and hence T = 30 is a
(x 2
+4 )
3/ 2
minimum R1 N0
Note: Allow FT on incorrect x value, 0 ≤ x ≤ 2.
[18]
64. (a) (i) f k′ (x ) = 3k 2 x 2 − 2kx + 1 A1
f k′′ (x ) = 6k 2 x − 2k A1
(ii) Setting f ″(x) = 0 M1
2 1
⇒ 6k x − 2k = 0 ⇒ x = A1
3k
3 2
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
f ⎜ ⎟ = k 2 ⎜ ⎟ − k⎜ ⎟ + ⎜ ⎟ M1
⎝ 3k ⎠ ⎝ 3k ⎠ ⎝ 3k ⎠ ⎝ 3k ⎠
7
= A1
27k
⎛ 1 7 ⎞
Hence, Pk is ⎜ , ⎟
⎝ 3k 27k ⎠

7
(b) Equation of the straight line is y = x A1
9
As this equation is independent of k, all Pk lie on this straight line R1
(c) Gradient of tangent at Pk:
2
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 2
f ′(Pk ) = f ′⎜ ⎟ = 3k 2 ⎜ ⎟ − 2k ⎜ ⎟ + 1 = M1A1
⎝ 3k ⎠ ⎝ 3k ⎠ ⎝ 3k ⎠ 3
As the gradient is independent of k, the tangents are parallel. R1
7 2 1 1
= × +c⇒c = (A1)
27k 3 3k 27k
2 1
The equation is y = x + A1
3 27k
[13]
65. METHOD 1
EITHER
Using the graph of y = f ′(x) (M1)
A1
The maximum of f ′(x) occurs at x = −0.5. A1
OR
Using the graph of y = f ′(x). (M1)
A1
The zero of f ′′(x) occurs at x = − 0.5. A1
THEN
Note: Do not award this A1 for stating x = ± 0.5
as the final answer for x.
−0.5
f (−0.5) = 0.607 (= e ) A2
Note: Do not award this A1 for also stating
(0.5, 0.607) as a coordinate.
EITHER
Correctly labelled graph of f′ (x) for x < 0 denoting the maximum f′ (x) R1
(eg f′ (−0.6) = 1.17 and f′ (−0.4) = 1.16 stated) A1 N2
OR
Correctly labelled graph of f′′ (x) for x < 0 denoting the maximum f′ (x) R1
(eg f′′ (−0.6) = 0.857 and f′′ (−0.4) = −1.05 stated) A1 N2

OR
1
f′ (0.5) ≈ 1.21. f′ (x) < 1.21 just to the left of x=−
2
1
and f′ (x) < 1.21 just to the right of x=− R1
2
(eg f′ (−0.6) = 1.17 and f′ (−0.4) =1.16 stated) A1 N2
OR
1
f ′′(x) > 0 just to the left of x=− and f ′′ (x) < 0 just to the right
2
1
of x=− R1
2
(eg f ′′(−0.6) = 0.857 and f ′′(−0.4) = −1.05 stated) A1 N2
METHOD 2
− 2 x2
f ′(x) = −4x e A1
f ′(x) = −4 e
− 2x 2
+ 16x
2
e− 2x
2
(= (16 x − 4)e )
2 − 2 x2
A1
Attempting to solve f ′(x) = 0 (M1)
1
x=− A1
2
1
Note: Do not award this A1 for stating x=±
2
as the final answer for x.
⎛ 1⎞ 1
f ⎜− ⎟= (= 0.607) A1
⎝ 2⎠ e
Note: Do not award this A1 for also stating
⎛1 1 ⎞
⎜⎜ , ⎟⎟ as a coordinate.
⎝2 e⎠
EITHER
(eg f ′(−0.6) = 1.17 and f ′ (−0.4) = 1.16 stated) A1 N2
OR
(eg f′′ (−0.6) = 0.857 and f′′ (−0.4) = −1.05 stated) A1 N2
OR
1
f′ (0.5) ≈ 1.21. f′ (x) < 1.21 just to the left of x=−
2
1
and f′ (x) < 1.21 just to the right of x=− R1
2
(eg f′ (−0.6) = 1.17 and f′ (−0.4) =1.16 stated) A1 N2
OR
1
f ′′(x) > 0 just to the left of x=− and f ′′ (x) < 0 just to the right
2
1
of x=− R1
2
(eg f ′′(−0.6) = 0.857 and f ′′(−0.4) = −1.05 stated) A1 N2
[7]

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IB Practice -­‐ Calculus -­‐ Differentiation Applications (V2 Legacy)

Macintosh HD:Users:Shared:Dropbox:Desert:HL2:7Calculus:HL.CalcDiffAppsPractice.docx on 09/28/2016 at 9:24 AM Page 2 of 17

(c) (i) Show that the stationary point of fk (x) is at x = ek–1.

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Indicate, and label clearly, on the graph

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12. (a) Let y = a + b sin x , where 0 < a < b.

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On the axes below, sketch the graph of y2 = f′ (x) .

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56. Given that y = e − x find

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(a) find the x-coordinate of the point P where f ′ (x) = 0;

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When André swims he covers 1 km in 5 5 minutes. When he runs he covers 1 km in 5 minutes.

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= − 2k[e −t / 2 ]0a (M1)

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–0.5 0.5 1 1.5 2

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Note: Award (A1) for each correct limit.

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⎪a – b⎪= 12 + 12 − 2(1)(1) cos θ (M1)

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(ii) d = 2.865 + 8.594 (G2)

(ii) d = 11.5 m. (G2)(A1)

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f ( k +1) ( x) = ( f ( k ) ( x) )′ = p k −1 pe px ( p ( x + 1) + k ) + p k −1e px p (M1)(A1)

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(a) h = r sinθ CB = 2CN = 2r cosθ (A1)(A1)

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Arc length, s = 4θ (A1)

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Let BP̂D = α and AP̂D = β then θ = α − β

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Using integrating by parts (M1)

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IB Practice -‐ Calculus -‐ Differentiation Applications (V2 Legacy)