IB Derivatives [253 marks]

A function f (x) has derivative f ′ (x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

1a. Find f (x). [6 marks]

 Markscheme
evidence of integration (M1)
eg ∫ f ′ (x)
correct integration (accept absence of C) (A1)(A1)
18 2
eg x3 + 2
x + C, x3 + 9x2
attempt to substitute x = −1 into their f = 0 (must have C) M1
3 2
eg (−1) + 9(−1) + C = 0, −1 + 9 + C = 0
Note: Award M0 if they substitute into original or differentiated function.
correct working (A1)
eg 8 + C = 0, C = −8
f (x) = x3 + 9x2 − 8 A1 N5
[6 marks]

1b. The graph of f has a point of inflexion at x = p. Find p. [4 marks]

 Markscheme
METHOD 1 (using 2nd derivative)
recognizing that f" = 0 (seen anywhere) M1
correct expression for f" (A1)
eg 6x + 18, 6p + 18
correct working (A1)
6p + 18 = 0
p = −3 A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed (M2)
eg − 2a
b
(must be clear this is for f′)

correct substitution (A1)

−18
eg 2×3

p = −3 A1 N3
[4 marks]

1c. Find the values of x for which the graph of f is concave-down. [3 marks]
 Markscheme
valid attempt to use f" (x) to determine concavity (M1)

eg f" (x) < 0, f" (−2), f" (−4), 6x + 18 ≤ 0

correct working (A1)

eg 6x + 18 < 0, f" (−2) = 6, f" (−4) = −6

f concave down for x < −3 (do not accept x ≤ −3) A1 N2

[3 marks]

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 π
cm3.

2a. Express h in terms of r. [2 marks]

Markscheme
correct equation for volume (A1)
eg πr2 h = 20π
20
h= r2
A1 N2

[2 marks]

The material for the base and top of the can costs 10 cents per cm 2 and the material for the
 The material for the base and top of the can costs 10 cents per cm 2 and the material for the
curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.

2b. Show that C = 20πr2 + 320π [4 marks]

r
.

Markscheme
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg 2πr2 × 10
correct expression for cost of curved side (seen anywhere) (A1)
eg 2πr × h × 8
correct expression for cost of curved side in terms of r A1
eg 8 × 2πr × 20 r2
, 320π
r2
320π
C = 20πr2 + r
AG N0

[4 marks]

Given that there is a minimum value for C, find this minimum value in terms of .
2c. Given that there is a minimum value for C, find this minimum value in terms of π. [9 marks]
 Markscheme
recognize C ′ = 0 at minimum (R1)
eg C ′ = 0, dCdr
=0
correct differentiation (may be seen in equation)
320π
C ′ = 40πr − r2
A1A1

correct equation A1
eg 40πr − 2 = 0, 40πr 320π
320π
2
r r
correct working (A1)
eg 40r3 = 320, r3 = 8
r = 2 (m) A1
attempt to substitute their value of r into C
eg 20π × 4 + 320 × π2 (M1)

correct working
eg 80π + 160π (A1)
240π (cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240 π is seen.
[9 marks]

Let ( ) = 12 cos − 5 sin , − ⩽ ⩽ 2 , be a periodic function with ( )= ( +2 )

 Let f (x) = 12 cos x − 5 sin x, −π ⩽ x ⩽ 2π, be a periodic function with f (x) = f (x + 2π)
The following diagram shows the graph of f .

There is a maximum point at A. The minimum value of f is −13 .

3a. Find the coordinates of A. [2 marks]

Markscheme
−0.394791,13
A(−0.395, 13) A1A1 N2
[2 marks]

3b. For the graph of f , write down the amplitude. [1 mark]

 Markscheme
13 A1 N1
[1 mark]

3c. For the graph of f , write down the period. [1 mark]

Markscheme
2π, 6.28 A1 N1
[1 mark]

3d. Hence, write f (x) in the form p cos (x + r). [3 marks]

 Markscheme
valid approach (M1)
eg recognizing that amplitude is p or shift is r
f (x) = 13 cos (x + 0.395) (accept p = 13, r = 0.395) A1A1 N3
Note: Accept any value of r of the form 0.395 + 2πk, k ∈ Z
[3 marks]

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so
that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by
d (t) = f (t) + 17, 0 ⩽ t ⩽ 5.

3e. Find the maximum speed of the ball. [3 marks]

 Markscheme
recognizing need for d ′(t) (M1)
eg −12 sin(t) − 5 cos( t)
correct approach (accept any variable for t) (A1)
eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32
maximum speed = 13 (cms −1) A1 N2
[3 marks]

−2.
3f. Find the first time when the ball’s speed is changing at a rate of 2 cm s [5 marks]
 Markscheme
recognizing that acceleration is needed (M1)
eg a(t), d "(t)
correct equation (accept any variable for t) (A1)
eg a (t) = −2, ∣∣ d (d ′ (t))∣∣ = 2, −12 cos (t) + 5 sin (t) = −2
dt

valid attempt to solve their equation (M1)

eg sketch, 1.33
1.02154
1.02 A2 N3
[5 marks]

A particle P moves along a straight line. The velocity v m s−1 of P after t seconds is given by v (t)
= 7 cos t − 5t cos t, for 0 ≤ t ≤ 7.
The following diagram shows the graph of v.

4a. Find the initial velocity of P. [2 marks]

 Markscheme
initial velocity when t = 0 (M1)
eg v(0)
v = 17 (m s −1) A1 N2
[2 marks]

4b. Find the maximum speed of P. [3 marks]

Markscheme
recognizing maximum speed when |v| is greatest (M1)
eg minimum, maximum, v' = 0
one correct coordinate for minimum (A1)
eg 6.37896, −24.6571
24.7 (ms−1) A1 N2
[3 marks]

Write down the number of times that the acceleration of P is 0 m s −2 .

 −2
4c. Write down the number of times that the acceleration of P is 0 m s . [3 marks]

Markscheme
recognizing a = v ′ (M1)
dv
eg a = dt
, correct derivative of first term

identifying when a = 0 (M1)

eg turning points of v, t-intercepts of v ′
3 A1 N3
[3 marks]

Find the acceleration of P when it changes direction.

4d. Find the acceleration of P when it changes direction. [4 marks]

Markscheme
recognizing P changes direction when v = 0 (M1)
t = 0.863851 (A1)
−9.24689
a = −9.25 (ms −2) A2 N3
[4 marks]

Find the total distance travelled by P.

4e. Find the total distance travelled by P. [3 marks]

Markscheme
correct substitution of limits or function into formula (A1)
7 0.8638 7 cos x
eg ∫0 | v | , ∫0 vdt − ∫0.8638 vdt, ∫ | 7 cos x − 5x | dx, 3.32 = 60.6
63.8874
63.9 (metres) A2 N3
[3 marks]

The following diagram shows the graph of ′, the derivative of .

 The following diagram shows the graph of f ′ , the derivative of f .

The graph of f ′ has a local minimum at A, a local maximum at B and passes through (4, − 2).

The point P(4, 3) lies on the graph of the function, f .

5a. Write down the gradient of the curve of f at P. [1 mark]

Markscheme
−2 A1 N1
[1 mark]

5b. Find the equation of the normal to the curve of f at P. [3 marks]

 Markscheme
1
gradient of normal = 2
(A1)

attempt to substitute their normal gradient and coordinates of P (in any order) (M1)
1
eg y − 4 = 2
(x − 3), 3 = 12 (4) + b, b = 1
y − 3 = 12 (x − 4), y = 12 x + 1, x − 2y + 2 = 0 A1 N3

[3 marks]

5c. Determine the concavity of the graph of f when 4 < x < 5 and justify your answer. [2 marks]
 Markscheme
correct answer and valid reasoning A2 N2
answer: eg graph of f is concave up, concavity is positive (between 4 < x < 5)
reason: eg slope of f ′ is positive, f ′ is increasing, f ′′ > 0,
sign chart (must clearly be for f ′′ and show A and B)

Note: The reason given must refer to a specific function/graph. Referring to “the graph”
or “it” is not sufficient.

[2 marks]

Let f(x) = cos x.

6a. (i) Find the first four derivatives of f(x). [4 marks]

(ii) Find f (19) (x).
 Markscheme
(i)
f ′(x) = − sin x, f ′′ (x) = − cos x, f (3) (x) = sin x, f (4) (x) = cos x A2 N2
(ii) valid approach (M1)
eg recognizing that 19 is one less than a multiple of 4, f (19) (x) = f (3) (x)

f (19) (x) = sin x A1 N2

[4 marks]

Let g(x) = xk , where k ∈ Z+ .

6b. (i) Find the first three derivatives of g(x). [5 marks]

(ii) Given that g (19) (x) = k!
(k−p)!
(xk−19 ), find p.
Markscheme
(i)
g ′(x) = kxk−1
g ′′ (x) = k(k − 1)xk−2 , g (3) (x) = k(k − 1)(k − 2)xk−3 A1A1 N2
(ii) METHOD 1
correct working that leads to the correct answer, involving the correct expression for the
19th derivative A2
(k−19)!
eg k(k − 1)(k − 2) … (k − 18) × , P
(k−19)! k 19

p = 19 (accept k!
(k−19)!
xk−19 ) A1 N1

METHOD 2
correct working involving recognizing patterns in coefficients of first three derivatives (may
be seen in part (b)(i)) leading to a general rule for 19th coefficient A2

eg g ′′ = 2! ( ) , k(k − 1)(k − 2) =
k k!
, g (3) (x)=k P3 (xk−3 )
2 (k−3)!

g (19) (x) = 19! ( ) , 19! ×

k k!
, P
(k−19)!×19! k 19
19
p = 19 (accept k!
(k−19)!
xk−19 ) A1 N1

[5 marks]

(19) (19)
 Let k = 21 and h(x) = (f (19) (x) × g (19) (x)).

6c. (i) Find h′ (x). [7 marks]

−21! 2
(ii) Hence, show that h′ (π) = 2
π .
 Markscheme
(i) valid approach using product rule (M1)
eg uv′ + vu′, f (19) g (20) + f (20) g (19)
correct 20th derivatives (must be seen in product rule) (A1)(A1)
21!
eg g (20) (x) = (21−20)!
x, f (20) (x) = cos x

h′(x) = sin x(21!x) + cos x ( 21!

2
x2 ) (accept sin x ( 21!
1!
x) + cos x ( 21!
2!
x2 )) A1
N3
(ii) substituting x = π (seen anywhere) (A1)
eg f (19) (π)g (20) (π) + f (20) (π)g (19) (π), sin π 21!
1!
π + cos π 21!
2!
π2
evidence of one correct value for sin π or cos π (seen anywhere) (A1)
eg sin π = 0, cos π = −1
evidence of correct values substituted into h′ (π) A1

eg 21!(π) (0 − π
2!
), 21!(π) (− π2 ) , 0 + (−1) 21!
2
π2

Note: If candidates write only the first line followed by the answer, award A1A0A0.

−21! 2
2
π AG N0

[7 marks]

The following diagram shows the graph of f(x) = a sin bx + c, for 0 ⩽ x ⩽ 12.

The graph of f has a minimum point at (3, 5) and a maximum point at (9, 17).

7a. (i) Find the value of c. [6 marks]

(ii) Show that b = π
6
.

(iii) Find the value of a.

Markscheme
(i) valid approach (M1)
5+17
eg 2
c = 11 A1 N2
(ii) valid approach (M1)
2π
eg period is 12, per = b
, 9−3
2π
b= 12
A1

b= π
6
AG N0

(iii) METHOD 1
valid approach (M1)
eg 5 = a sin( π6 × 3) + 11, substitution of points

a = −6 A1 N2
METHOD 2
valid approach (M1)
17−5
eg 2
, amplitude is 6

a = −6 A1 N2
[6 marks]
 The graph of g is obtained from the graph of f by a translation of ( ). The maximum point on
k
0
the graph of g has coordinates (11.5, 17).

7b. (i) Write down the value of k. [3 marks]

(ii) Find g(x).

Markscheme
(i)
k = 2.5 A1 N1
(ii)
g(x) = −6 sin( π6 (x − 2.5)) + 11 A2 N2

[3 marks]

The graph of changes from concave-up to concave-down when = .

 The graph of g changes from concave-up to concave-down when x = w.

7c. (i) Find w. [6 marks]

(ii) Hence or otherwise, find the maximum positive rate of change of g.
Markscheme
(i) METHOD 1 Using g
recognizing that a point of inflexion is required M1
eg sketch, recognizing change in concavity
evidence of valid approach (M1)
eg g ′′ (x) = 0, sketch, coordinates of max/min on g ′
w = 8.5 (exact) A1 N2
METHOD 2 Using f
recognizing that a point of inflexion is required M1
eg sketch, recognizing change in concavity
evidence of valid approach involving translation (M1)
eg x = w − k, sketch, 6 + 2.5
w = 8.5 (exact) A1 N2
(ii) valid approach involving the derivative of g or f (seen anywhere) (M1)
eg g ′ (w), − π cos( π6 x), max on derivative, sketch of derivative

attempt to find max value on derivative M1

eg −π cos( π6 (8.5 − 2.5)), f ′ (6), dot on max of sketch

3.14159
max rate of change = π (exact), 3.14 A1 N2
[6 marks]

( )= 4 +5 ⩾ −1.25
 Let f(x) = √4x + 5, for x ⩾ −1.25.

8a. Find f ′ (1). [4 marks]

Markscheme
choosing chain rule (M1)
dy dy du
eg dx
= du
× dx
, u = 4x + 5, u′ = 4
correct derivative of f A2
1 − 12 2
eg 2
(4x + 5) × 4, f ′(x) =
√4x+5
2
f ′(1) = 3
A1 N2

[4 marks]

Consider another function . Let R be a point on the graph of . The -coordinate of R is 1. The
 Consider another function g. Let R be a point on the graph of g. The x-coordinate of R is 1. The
equation of the tangent to the graph at R is y = 3x + 6.

8b. Write down g ′ (1). [2 marks]

Markscheme
recognize that g ′ (x) is the gradient of the tangent (M1)
eg g ′ (x) = m
g ′(1) = 3 A1 N2
[2 marks]

8c. Find g(1). [2 marks]

Markscheme
recognize that R is on the tangent (M1)
eg g(1) = 3 × 1 + 6, sketch
g(1) = 9 A1 N2
[2 marks]

Let ( ) = ( ) × ( ). Find the equation of the tangent to the graph of at the

8d. Let h(x) = f(x) × g(x). Find the equation of the tangent to the graph of h at the [7 marks]
point where x = 1.
 Markscheme
f(1) = √4 + 5 (= 3) (seen anywhere) A1
h(1) = 3 × 9 (= 27) (seen anywhere) A1
choosing product rule to find h′ (x) (M1)
eg uv′ + u′ v
correct substitution to find h′ (1) (A1)
eg f(1) × g ′(1) + f ′(1) × g(1)
2
h′(1) = 3 × 3 + 3
× 9 (= 15) A1

EITHER
attempt to substitute coordinates (in any order) into the equation of a straight line (M1)
eg y − 27 = h′ (1)(x − 1), y − 1 = 15(x − 27)
y − 27 = 15(x − 1) A1 N2
OR
attempt to substitute coordinates (in any order) to find the y-intercept (M1)
eg 27 = 15 × 1 + b, 1 = 15 × 27 + b
y = 15x + 12 A1 N2
[7 marks]

6−2x
Let f ′ (x) = 6x−x2
, for 0 < x < 6.

The graph of f has a maximum point at P.

9a. Find the x-coordinate of P. [3 marks]

 Markscheme
recognizing f ′ (x) = 0 (M1)
correct working (A1)
eg 6 − 2x = 0
x=3 A1 N2
[3 marks]

The y-coordinate of P is ln 27.

9b. Find f(x), expressing your answer as a single logarithm. [8 marks]

 Markscheme
evidence of integration (M1)
6−2x
eg ∫ f ′ , ∫ 6x−x2
dx
using substitution (A1)
1
eg ∫ u du where u = 6x − x2

correct integral A1
eg ln(u) + c, ln(6x − x2 )
substituting (3, ln 27) into their integrated expression (must have c) (M1)
eg ln(6 × 3 − 32 ) + c = ln 27, ln(18 − 9) + ln k = ln 27
correct working (A1)
eg c = ln 27 − ln 9
EITHER
c = ln 3 (A1)
attempt to substitute their value of c into f(x) (M1)
eg f(x) = ln(6x − x2 ) + ln 3 A1 N4
OR
attempt to substitute their value of c into f(x) (M1)
eg f(x) = ln(6x − x2 ) + ln 27 − ln 9
correct use of a log law (A1)

eg f(x) = ln(6x − x2 ) + ln( 27

9
), f(x) = ln(27(6x − x2 )) − ln 9

f(x) = ln(3(6x − x2 )) A1 N4
[8 marks]

9c. The graph of f is transformed by a vertical stretch with scale factor 1

ln 3
. The image of P under
this transformation has coordinates (a, b).
Find the value of a and of b, where a, b ∈ N.
 Markscheme
a=3 A1 N1
correct working A1
ln 27
eg ln 3

correct use of log law (A1)

3 ln 3
eg ln 3
, log3 27
b=3 A1 N2
[4 marks]

Fred makes an open metal container in the shape of a cuboid, as shown in the following
diagram.

The container has height x m, width x m and length y m. The volume is 36 m3 .

Let A(x) be the outside surface area of the container.

108
10a. Show that A(x) =
x + 2x2 . [4 marks]
 Markscheme
correct substitution into the formula for volume A1
eg 36 = y × x × x
valid approach to eliminate y (may be seen in formula/substitution) M1
36 36
eg y = x2
, xy = x

correct expression for surface area A1

eg xy + xy + xy + x2 + x2 , area = 3xy + 2x2
correct expression in terms of x only A1

eg 3x ( 362 ) + 2x2 , x2 + x2 + 36
x + 36
x + 36
x , 2x2 + 3 ( 36
x )
x
108
A(x) = x
+ 2x2 AG N0

[4 marks]

10b. Find A′ (x). [2 marks]

Markscheme
A′(x) = − 108
x2
+ 4x, 4x − 108x−2 A1A1 N2

Note: Award A1 for each term.

[2 marks]

Given that the outside surface area is a minimum, find the height of the container.
10c. Given that the outside surface area is a minimum, find the height of the container. [5 marks]

Markscheme
recognizing that minimum is when A′ (x) = 0 (M1)
correct equation (A1)
eg − 1082 + 4x = 0, 4x = 108
x x2
correct simplification (A1)
eg −108 + 4x3 = 0, 4x3 = 108
correct working (A1)
eg x3 = 27
height = 3 (m) (accept x = 3) A1 N2
[5 marks]

10 m2
10d. Fred paints the outside of the container. A tin of paint covers a surface area of 10 m2 [5 marks]
and costs $20. Find the total cost of the tins needed to paint the container.

Markscheme
attempt to find area using their height (M1)
108
eg 3
+ 2(3)2 , 9 + 9 + 12 + 12 + 12
minimum surface area = 54 m2 (may be seen in part (c)) A1
attempt to find the number of tins (M1)
54
eg 10
, 5.4
6 (tins) (A1)
$120 A1 N3
[5 marks]

1
Let f(x) = x−1 + 2, for x > 1.

11a. Write down the equation of the horizontal asymptote of the graph of f . [2 marks]

Markscheme
y = 2 (correct equation only) A2 N2
[2 marks]
11b. Find f ′ (x). [2 marks]

Markscheme
valid approach (M1)
0(x−1)−1
eg (x − 1)−1 + 2, f ′ (x) =
(x−1)2
−1
−(x − 1)−2 , f ′(x) = A1 N2
(x−1)2

[2 marks]

Let g(x) = ae−x + b, for x ⩾ 1. The graphs of f and g have the same horizontal asymptote.

11c. Write down the value of b. [2 marks]

Markscheme
correct equation for the asymptote of g
eg y = b (A1)
b=2 A1 N2
[2 marks]

Given that ′ (1) = − , find the value of .

11d. Given that g ′ (1) = −e, find the value of a. [4 marks]

Markscheme
correct derivative of g (seen anywhere) (A2)
−x
eg g ′(x) = −ae
correct equation (A1)
eg −e = −ae−1
7.38905
a = e2 (exact), 7.39 A1 N2
[4 marks]

11e. There is a value of x, for 1 < x < 4, for which the graphs of f and g have the same [4 marks]
gradient. Find this gradient.
 Markscheme
attempt to equate their derivatives (M1)
−1
eg f ′ (x) = g ′ (x), = −ae−x
(x−1)2

valid attempt to solve their equation (M1)

eg correct value outside the domain of f such as 0.522 or 4.51,

correct solution (may be seen in sketch) (A1)

eg x = 2, (2, − 1)
gradient is −1 A1 N3
[4 marks]

Let = ( ), for −0.5 ≤ x ≤ 6.5. The following diagram shows the graph of ′, the derivative
 Let y = f(x), for −0.5 ≤ x ≤ 6.5. The following diagram shows the graph of f ′ , the derivative
of f .

The graph of f ′ has a local maximum when x = 2, a local minimum when x = 4, and it
crosses the
x-axis at the point (5, 0).

12a. Explain why the graph of f has a local minimum when x = 5. [2 marks]
 Markscheme
METHOD 1
f ′(5) = 0 (A1)
valid reasoning including reference to the graph of f ′ R1
eg f ′ changes sign from negative to positive at x = 5, labelled sign chart for f ′
so f has a local minimum at x = 5 AG N0

Note: It must be clear that any description is referring to the graph of f ′ , simply giving
the conditions for a minimum without relating them to f ′ does not gain the R1.

METHOD 2
f ′(5) = 0 A1
valid reasoning referring to second derivative R1
eg f ′′ (5) > 0
so f has a local minimum at x = 5 AG N0
[2 marks]

12b. Find the set of values of x for which the graph of f is concave down. [2 marks]

Markscheme
attempt to find relevant interval (M1)
eg f ′ is decreasing, gradient of f ′ is negative, f ′′ < 0
2 < x < 4 (accept “between 2 and 4”) A1 N2

Notes: If no other working shown, award M1A0 for incorrect inequalities such as 2 ≤ x
≤ 4, or “from 2 to 4”
[2 marks]

The following diagram shows the shaded regions , and .

12c. The following diagram shows the shaded regions A, B and C . [5 marks]

The regions are enclosed by the graph of f ′ , the x-axis, the y-axis, and the line x = 6.
The area of region A is 12, the area of region B is 6.75 and the area of region C is 6.75.
Given that f(0) = 14, find f(6).
 Markscheme
METHOD 1 (one integral)
correct application of Fundamental Theorem of Calculus (A1)
6 6
eg ∫0 f ′ (x)dx =f(6) − f(0), f(6) = 14 + ∫0 f ′ (x)dx
attempt to link definite integral with areas (M1)
6 6
eg ∫0 f ′ (x)dx = −12 − 6.75 + 6.75, ∫0 f ′ (x)dx = Area A + Area B + Area C
6
correct value for ∫0 f ′ (x)dx (A1)
6
eg ∫0 f ′ (x)dx = −12
correct working A1
eg f(6) − 14 = −12, f(6) = −12 + f(0)
f(6) = 2 A1 N3
METHOD 2 (more than one integral)
correct application of Fundamental Theorem of Calculus (A1)
2 2
eg ∫0 f ′ (x)dx = f(2) − f(0), f(2) = 14 + ∫0 f ′ (x)

attempt to link definite integrals with areas (M1)

2 5 6
eg ∫0 f ′(x)dx = 12, ∫2 f ′(x)dx = −6.75, ∫0 f ′(x) = 0
correct values for integrals (A1)
2 2
eg ∫0 f ′(x)dx = −12, ∫5 f ′(x)dx = 6.75, f(6) − f(2) = 0
one correct intermediate value A1
eg f(2) = 2, f(5) = −4.75
f(6) = 2 A1 N3
[5 marks]

The following diagram shows the shaded regions , and .

12d. The following diagram shows the shaded regions A, B and C . [6 marks]

The regions are enclosed by the graph of f ′ , the x-axis, the y-axis, and the line x = 6.
The area of region A is 12, the area of region B is 6.75 and the area of region C is 6.75.
2
Let g(x) = (f(x)) . Given that f ′ (6) = 16, find the equation of the tangent to the graph of g at
the point where x = 6.
 Markscheme
correct calculation of g(6) (seen anywhere) A1
2
eg 2 , g(6) = 4
choosing chain rule or product rule (M1)
dy dy du
eg g ′ (f(x)) f ′ (x), dx
= du
× dx
, f(x)f ′(x) + f ′(x)f(x)
correct derivative (A1)
eg g ′(x) = 2f(x)f ′(x), f(x)f ′(x) + f ′(x)f(x)
correct calculation of g ′ (6) (seen anywhere) A1
eg 2(2)(16), g ′ (6) = 64
attempt to substitute their values of g ′ (6) and g(6) (in any order) into equation of a line
(M1)
eg 22 = (2 × 2 × 16)6 + b, y − 6 = 64(x − 4)
correct equation in any form A1 N2
eg y − 4 = 64(x − 6), y = 64x − 380
[6 marks]
[Total 15 marks]

A function f has its derivative given by f ′ (x) = 3x2 − 2kx − 9, where k is a constant.

13a. Find f ′′ (x). [2 marks]

Markscheme
f ′′ (x) = 6x − 2k A1A1 N2
[2 marks]

The graph of has a point of inflexion when = 1.

13b. The graph of f has a point of inflexion when x = 1. [3 marks]
Show that k = 3.

Markscheme
substituting x = 1 into f ′′ (M1)
eg f ′′ (1), 6(1) − 2k
recognizing f ′′ (x) = 0 (seen anywhere) M1
correct equation A1
eg 6 − 2k = 0
k=3 AG N0
[3 marks]

13c. Find f ′ (−2). [2 marks]

Markscheme
correct substitution into f ′ (x) (A1)
eg 3(−2)2 − 6(−2) − 9
f ′(−2) = 15 A1 N2
[2 marks]

Find the equation of the tangent to the curve of at (−2, 1), giving your answer in
13d. Find the equation of the tangent to the curve of f at (−2, 1), giving your answer in [4 marks]
the form y = ax + b.

Markscheme
recognizing gradient value (may be seen in equation) M1
eg a = 15, y = 15x + b
attempt to substitute (−2, 1) into equation of a straight line M1
eg 1 = 15(−2) + b, (y − 1) = m(x + 2), (y + 2) = 15(x − 1)
correct working (A1)
eg 31 = b, y = 15x + 30 + 1
y = 15x + 31 A1 N2
[4 marks]

13e. Given that f ′ (−1) = 0, explain why the graph of f has a local maximum when [3 marks]
x = −1.
 Markscheme
METHOD 1 (2nd derivative)
recognizing f ′′ < 0 (seen anywhere) R1
substituting x = −1 into f ′′ (M1)
eg f ′′ (−1), 6(−1) − 6
f ′′ (−1) = −12 A1
therefore the graph of f has a local maximum when x = −1 AG N0
METHOD 2 (1st derivative)
recognizing change of sign of f ′ (x) (seen anywhere) R1
eg sign chart
correct value of f ′ for −1 < x < 3 A1
eg f ′ (0) = −9
correct value of f ′ for x value to the left of −1 A1
eg f ′ (−2) = 15
therefore the graph of f has a local maximum when x = −1 AG N0
[3 marks]
Total [14 marks]

Consider a function , for 0 ≤ ≤ 10. The following diagram shows the graph of ′, the
 Consider a function f , for 0 ≤ x ≤ 10. The following diagram shows the graph of f ′ , the
derivative of f .

The graph of f ′ passes through (2, − 2) and (5, 1), and has x-intercepts at 0, 4 and 6.

14a. The graph of f has a local maximum point when x = p. State the value of p, and [3 marks]
justify your answer.

Markscheme
p=6 A1 N1
recognizing that turning points occur when f ′ (x) = 0 R1 N1
eg correct sign diagram
f ′ changes from positive to negative at x = 6 R1 N1
[3 marks]

14b. Write down f ′ (2). [1 mark]

 Markscheme
f ′(2) = −2 A1 N1
[1 mark]

14c. Let g(x) = ln(f(x)) and f(2) = 3. [4 marks]

Find g ′ (2).

Markscheme
attempt to apply chain rule (M1)
eg ln(x)′ × f ′ (x)
correct expression for g ′ (x) (A1)
1
eg g ′ (x) = f(x)
× f ′(x)
substituting x = 2 into their g ′ (M1)
f ′(2)
eg f(2)

−0.666667
g ′(2) = − 23 (exact), − 0.667 A1 N3

[4 marks]

a
14d. Verify that ln 3 + ∫2 g ′ (x)dx = g(a), where 0 ≤ a ≤ 10. [4 marks]
 Markscheme
evidence of integrating g ′ (x) (M1)

eg g(x)|a2 , g(x)|2a
applying the fundamental theorem of calculus (seen anywhere) R1
a
eg ∫2 g ′(x) = g(a) − g(2)
correct substitution into integral (A1)
eg ln 3 + g(a) − g(2), ln 3 + g(a) − ln(f(2))
ln 3 + g(a) − ln 3 A1
a
ln 3 + ∫2 g ′(x) = g(a) AG N0
[4 marks]

14e. The following diagram shows the graph of g ′ , the derivative of g. [4 marks]

The shaded region A is enclosed by the curve, the

x-axis and the line x = 2, and has area 0.66 units2 .
The shaded region B is enclosed by the curve, the x-axis and the line x = 5, and has area
0.21 units2 .
Find g(5).
 Markscheme
METHOD 1
substituting a = 5 into the formula for g(a) (M1)
5 5
eg ∫2 g ′ (x)dx, g(5) = ln 3 + ∫2 g ′ (x)dx (do not accept only g(5))
attempt to substitute areas (M1)
eg ln 3 + 0.66 − 0.21, ln 3 + 0.66 + 0.21
correct working
eg g(5) = ln 3 + (−0.66 + 0.21) (A1)
0.648612
g(5) = ln 3 − 0.45 (exact), 0.649 A1 N3
METHOD 2
attempt to set up an equation for one shaded region (M1)
5 4 5
eg ∫4 g ′ (x)dx = 0.21, ∫2 g ′ (x)dx = −0.66, ∫2 g ′ (x)dx = −0.45

two correct equations (A1)

eg g(5) − g(4) = 0.21, g(2) − g(4) = 0.66
combining equations to eliminate g(4) (M1)
eg g(5) − [ln 3 − 0.66] = 0.21
0.648612
g(5) = ln 3 − 0.45 (exact), 0.649 A1 N3
METHOD 3
attempt to set up a definite integral (M1)
5 5
eg ∫2 g ′(x)dx = −0.66 + 0.21, ∫2 g ′(x)dx = −0.45
correct working (A1)
eg g(5) − g(2) = −0.45
correct substitution (A1)
eg g(5) − ln 3 = −0.45
0.648612
g(5) = ln 3 − 0.45 (exact), 0.649 A1 N3
[4 marks]
Total [16 marks]

The following diagram shows the graph of a function . There is a local minimum point at ,
 The following diagram shows the graph of a function f . There is a local minimum point at A,
where x > 0.

The derivative of f is given by f ′ (x) = 3x2 − 8x − 3.

15a. Find the x-coordinate of A. [5 marks]

Markscheme
recognizing that the local minimum occurs when f ′ (x) = 0 (M1)
valid attempt to solve 3x2 − 8x − 3 = 0 (M1)
eg factorization, formula
correct working A1
8±√64+36
(3x + 1)(x − 3), x = 6
x=3 A2 N3

−1
Note: Award A1 if both values x = 3
, x = 3 are given.
[5 marks]

The -intercept of the graph is at ( 0, 6). Find an expression for ( ).

15b. The y-intercept of the graph is at ( 0, 6). Find an expression for f(x). [6 marks]
The graph of a function g is obtained by reflecting the graph of f in the y-axis, followed by a

translation of ( ).
m
n

Markscheme
valid approach (M1)
f(x) = ∫ f ′(x)dx
f(x) = x3 − 4x2 − 3x + c (do not penalize for missing “ +c”) A1A1A1
c=6 (A1)
f(x) = x3 − 4x2 − 3x + 6 A1 N6
[6 marks]

The following diagram shows part of the graph of

 The following diagram shows part of the graph of
y = f(x).

The graph has a local maximum at

A, where
x = −2, and a local minimum at
B, where
x = 6.

16a. On the following axes, sketch the graph of y = f ′ (x). [4 marks]

Markscheme

A1A1A1A1 N4

Note: Award A1 for x-intercept in circle at −2, A1 for x-intercept in circle at 6.

Award A1 for approximately correct shape.
Only if this A1 is awarded, award A1 for a negative y-intercept.

[4 marks]

(0), ′ (6), ′′ (−2)

16b. Write down the following in order from least to greatest: f(0), f ′ (6), f ′′ (−2). [2 marks]

Markscheme
f ′′ (−2), f ′(6), f(0) A2 N2
[2 marks]

Consider the graph of the semicircle given by

f(x) = √6x − x2 , for
0 ⩽ x ⩽ 6. A rectangle
PQRS is drawn with upper vertices
R and
S on the graph of
f , and
PQ on the
x-axis, as shown in the following diagram.

17a. Let OP = x.
(i) Find PQ, giving your answer in terms of x.
(ii) Hence, write down an expression for the area of the rectangle, giving your answer in terms
of x.

Markscheme
(i) valid approach (may be seen on diagram) (M1)
eg Q to 6 is x
PQ = 6 − 2x A1 N2
(ii) A = (6 − 2x)√6x − x2 A1 N1
[3 marks]

17b. Find the rate of change of area when x = 2. [2 marks]

 Markscheme
dA
recognising dx
at x = 2 needed (must be the derivative of area) (M1)
dA 7√2
dx
=− 2
, − 4.95 A1 N2

[2 marks]

17c. The area is decreasing for a < x < b. Find the value of a and of b. [2 marks]

Markscheme
a = 0.879 b = 3 A1A1 N2
[4 marks]

Let
6x
f(x) = x+1
, for
x>0.

18a. Find f ′ (x) . [5 marks]

 Markscheme
METHOD 1
evidence of choosing quotient rule (M1)
u′v−uv′
e.g. v2

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

d d
e.g. dx
(6x) =6, dx
(x + 1) = 1
correct substitution into quotient rule A1
(x+1)6−6x 6x+6−6x
e.g. ,
(x+1)2 (x+1)2
6
f ′(x) = A1 N4
(x+1)2

[5 marks]

METHOD 2
evidence of choosing product rule (M1)
e.g. 6x(x + 1)−1 , uv′ + vu′
evidence of correct differentiation (must be seen in product rule) (A1)(A1)
d d
e.g. dx
(6x) =6, dx
(x + 1)−1 = −1(x + 1)−2 × 1
correct working A1
−6x+6(x+1)
e.g. 6x × −(x + 1)−2 + (x + 1)−1 × 6 ,
(x+1)2
6
f ′(x) = A1 N4
(x+1)2

[5 marks]

18b. Let g(x) = ln( 6x ) , for x > 0 . [4 marks]

x+1
1
Show that g ′ (x) = x(x+1)
.
 Markscheme
METHOD 1
evidence of choosing chain rule (M1)

e.g. formula, 1
× ( x+1
6x
)
( x+1
6x
)

1 x+1
correct reciprocal of is 6x
(seen anywhere) A1
( 6x
x+1
)

correct substitution into chain rule A1

e.g. 1
× 6
,( 6
) ( x+1
6x
)
( 6x
) (x+1)2 (x+1)2
x+1

working that clearly leads to the answer A1

e.g. ( (x+1) ) ( 6x ),( ) ( x+1

x )
6 1 1 6(x+1)
,
(x+1)2 6x(x+1)2

1
g ′(x) = x(x+1)
AG 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) A1A1
6 1 1 1
e.g. 6x
− x+1 , x − x+1

working that clearly leads to the answer A1

x+1−x 6x+6−6x 6(x+1−x)
e.g. x(x+1)
, 6x(x+1)
, 6x(x+1)
1
g ′(x) = x(x+1)
AG N0

[4 marks]

1
18c. Let h(x) = . The area enclosed by the graph of h , the x-axis and the lines [7 marks]
x(x+1)
1 1
x= 5
and x = k is ln 4 . Given that k > 5
, find the value of k .
 Markscheme
valid method using integral of h(x) (accept missing/incorrect limits or missing dx ) (M1)

e.g. area = ∫ 1 h(x)dx , ∫ ( x(x+1)

1
)
k
5

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

e.g. ∫ ( x(x+1)
1
)dx = ln( x+1
6x
)

correct substitution and subtraction A1

) , ln( k+1
6× 15
e.g. ln( k+1 ) − ln( ) − ln(1)
6k 6k
1
5
+1

setting their expression equal to ln 4 (M1)

6k
e.g. ln( k+1 ) − ln(1) = ln 4 , ln( k+1
6k
) = ln 4 , ∫ 1 h(x)dx = ln 4
k
5

correct equation without logs A1

6k
e.g. k+1 = 4 , 6k = 4(k + 1)
correct working (A1)
e.g. 6k = 4k + 4 , 2k = 4
k=2 A1 N4
[7 marks]

Let
f(x) = cos(ex) , for
−2 ≤ x ≤ 2 .

19a. Find f ′ (x) . [2 marks]

Markscheme
f ′(x) = −ex sin(ex ) A1A1 N2
[2 marks]

On the grid below, sketch the graph of ′( ).

19b. 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]

The following diagram shows the graph of

 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) .

20a. Use the graph to write down the value of [3 marks]

(i) a;
(ii) c;
(iii) d .

Markscheme
(i) a = 8 A1 N1
(ii) c = 2 A1 N1
(iii) d = 4 A1 N1
[3 marks]

20b. Show that b = π

4
. [2 marks]
 Markscheme
METHOD 1
recognizing that period = 8 (A1)
correct working A1
2π 2π
e.g. 8 = b
,b= 8
b= π
4
AG N0

METHOD 2
attempt to substitute M1
e.g. 12 = 8 sin(b(4 − 2)) + 4
correct working A1
e.g. sin 2b = 1
b= π
4
AG N0

[2 marks]

20c. Find f ′ (x) . [3 marks]

Markscheme
evidence of attempt to differentiate or choosing chain rule (M1)
e.g. cos π
4
(x − 2) , π
4
×8
f ′(x) = 2π cos( π4 (x − 2)) (accept 2π cos π4 (x − 2) ) A2 N3

[3 marks]

20d. 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
correct equation A1
e.g. −2π = 2π cos( π4 (x − 2)) , −1 = cos( π4 (x − 2))

correct working (A1)

e.g. cos−1 (−1) = π
4
(x − 2)
using cos−1 (−1) = π (seen anywhere) (A1)
e.g. π = π
4
(x − 2)
simplifying (A1)
e.g. 4 = (x − 2)
x=6 A1 N4
[6 marks]

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