11. Straight line L is the tangent to the curve 12.

Straight line L is the normal to the curve

y = x2 – 4x + 1 at P(p , q) and L is parallel to y = ln x + x at R(r , s) and L is perpendicular to
the line L: 4x – y – 6 = 0. the line L: 2x – y + 4 = 0.
(a) Express the slope of L in terms of p. (a) Express the slope of L in terms of r.
(b) Find the coordinates of P. (b) Find the coordinates of R.
(c) Find the equation of L. (c) Find the equation of L.
Ex 5A: 22–26

Level Up Question
13. Find the equations of the tangent and the normal to the curve y = e5x + ln (x + 1) + 5 at the point where
the curve cuts the y-axis.

Find the point where

the curve cuts the
y-axis first.
M2A Lesson Worksheet 5.2C (Refer to Book M2A P.5.11)

Objective: To find the maximum and minimum points of a curve by the first derivative test.

First Derivative Test

Suppose a function f(x) is differentiable and x0 is a point in its domain.

(a) If f (x0) = 0 and the sign of f (x) changes (b) If f (x0) = 0 and the sign of f (x) changes
from positive to negative as x increases from negative to positive as x increases
through x0, f(x) has a relative maximum through x0, f(x) has a relative minimum
at x0. at x0.

In general, if f (x0) = 0, the point (x0 , f(x0)) is called a stationary point.

Instant Example 1 Instant Practice 1

Let f(x) = x2 – 6x + 4. Find the maximum or Let f(x) = –x2 + 4x + 5. Find the maximum or
minimum point of the graph of y = f(x). minimum point of the graph of y = f(x).
f (x) = 2x – 6
= 2(x – 3)
When f (x) = 0,
2(x – 3) = 0  Find the value of x for f (x) = 0.

x=3
f(3) = 32 – 6(3) + 4 = –5

x x<3 x=3 x>3

f (x) – 0 +

∵ The sign of f (x) changes from negative to

positive as x increases through 3.
∴ (3 , –5) is a minimum point.
In each of the following, find the maximum or minimum point of the graph of y = f(x). [Nos. 1–4]
1. f(x) = (x – 3)2 – 2x 2. f(x) = 4x – (x + 3)2 Ex 5B: 1–6

 3x 2
3. f(x) = 4. f(x) = e2(x – 1) – 2x + 4
x2  3
Instant Example 2 Instant Practice 2
Find the stationary point of the curve y = x3 – 1, and Find the stationary point of the curve y = 7 – (x – 2)5
determine whether it is a maximum or a minimum and determine whether it is a maximum or a
point. minimum point.
dy
= 3x2
dx
dy
When = 0,
dx
3x2 = 0
x=0
When x = 0, y = 03 – 1 = –1.

x x<0 x=0 x>0

dy
+ 0 +
dx
dy
∵ The sign of does not change as x increases
dx
through 0.
∴ The stationary point is (0 , –1) and it is neither
a maximum nor a minimum point.

In each of the following, find the stationary point(s) of the curve and determine whether each stationary
point is a maximum or a minimum point. [Nos. 5–6]
5. y = x3 – 3x2 + 3x + 5 6. y = 2x3 – 6x + 1 Ex 5B: 7–12
Find the absolute maximum and minimum of each of the following functions. [Nos. 7–8]
1
7. f(x) = 2
for –2  x  1 Remember to check 8. f(x) = ex – 1 + e1 – x for 0  x  3 Ex 5B: 13–17
x 1 the end-points.

If em = en,
then m = n.

Level Up Question
π π
9. Find the absolute maximum and minimum of f(x) = sin x – x for  x .
2 2
M2A Lesson Worksheet 5.2D(I) & (II) (Refer to Book M2A P.5.16)

Objective: To understand the concept of concavity of the graph of a function, and to find the maxima and
minima of a function by the second derivative test.

Second Derivative Test for Concavity

concave upward
Let f(x) be a twice differentiable function.
(a) If f (x) > 0 for all x in an interval I, the graph of y = f(x) is concave upward on I. concave downward
(b) If f (x) < 0 for all x in an interval I, the graph of y = f(x) is concave downward on I.

Instant Example 1 Instant Practice 1

Let f(x) = x3 – 3x2 + 5. Find the range of values of x Let f(x) = –x3 + 5x – 4. Find the range of values of x
for which the graph of y = f(x) is concave upward for which the graph of y = f(x) is concave upward
and the range of values of x for which it is concave and the range of values of x for which it is concave
downward. downward.
f (x) = 3x – 6x and f (x) = 6x – 6
2

When f (x) = 0,
6x – 6 = 0
x=1
x x<1 x=1 x>1
f (x) – 0 +
∴ The graph is concave upward for x > 1.
The graph is concave downward for x < 1.

In each of the following, find the range of values of x for which the graph of y = f(x) is concave upward and
the range of values of x for which it is concave downward. [Nos. 1–2]
1. f(x) = 1 + 5x – 6x2 – x3 2. f(x) = ex – x2 Ex 5C: 1–5

Second Derivative Test for Relative Maxima and Minima

Suppose a function f(x) is twice differentiable and x0 is a point in its domain.
(a) If f (x0) = 0 and f (x0) < 0, f(x) has a relative maximum at x0.
(b) If f (x0) = 0 and f (x0) > 0, f(x) has a relative minimum at x0.
Instant Example 2 Instant Practice 2
Let f(x) = 2x3 + 6x2 – 7. Find the maximum and Let f(x) = x3 – 3x + 6. Find the maximum and
minimum points of the graph of y = f(x). minimum points of the graph of y = f(x).
f (x) = 6x2 + 12x
f (x) = 12x + 12
When f (x) = 0, 6x2 + 12x = 0
6x(x + 2) = 0
x = 0 or –2
3 2
f(0) = 2(0) + 6(0) – 7 = –7
f(–2) = 2(–2)3 + 6(–2)2 – 7 = 1
∵ f (0) = 12(0) + 12 = 12 > 0
∴ (0 , –7) is a minimum point.
∵ f (–2) = 12(–2) + 12 = –12 < 0
∴ (–2 , 1) is a maximum point.

In each of the following, find the maximum and minimum points of the graph of y = f(x). [Nos. 3–4]
3. f(x) = 11 – 9x2 – 2x3 4. f(x) = x3 – 3x2 – 9x + 8 Ex 5C: 6–8

Level Up Question
5. Let f(x) = x x  3 . Find the maximum/minimum point of the graph of y = f(x).
M2A Lesson Worksheet 5.2D(III) (Refer to Book M2A P.5.19)

Objective: To find points of inflexion.

Point of Inflexion
A point of inflexion on the graph of y = f(x) is a point where the concavity of the graph changes.

Steps for Finding the Point of Inflexion

Let f(x) be a twice differentiable function.
Step 1: Find f (x).
Step 2: Find the value of x0 such that f (x0) = 0 or f (x0) is undefined.
Step 3: Check whether the concavity changes at x0. If yes, then
(x0 , f(x0)) is a point of inflexion.

Instant Example 1 Instant Practice 1

Let f(x) = x5 – x. Find the point of inflexion of the Let f(x) = 7 – 6x2 – 2x3. Find the point of inflexion
graph of y = f(x). of the graph of y = f(x).
f (x) = 5x – 14

f (x) = 20x3  Step 1

When f (x) = 0,
20x3 = 0
x=0  Step 2
f(0) = 05 – 0 = 0
x x<0 x=0 x>0
 Step 3
f (x) – 0 +
∴ (0 , 0) is a point of inflexion.

In each of the following, find the point of inflexion of the graph of y = f(x). [Nos. 1–6]
1. f(x) = x3 + 2x + 5 2. f(x) = x3 – 3x2 – 4 Ex 5C: 11–16
 1 1
3. f(x) = 7x 7 f (x) is undefined 4. f(x) = ( x  1) 3
at x = .

5. f(x) = 2x2 – (x + 2)3 6. f(x) = (x + 1)ex

Level Up Question
7. Let f(x) = x2 ln x, where x > 0. Find the point of inflexion of the graph of y = f(x).
M2A Lesson Worksheet 5.3A & B (Refer to Book M2A P.5.23)

Objective: To determine whether a function is an even function or an odd function, and find horizontal,
vertical and oblique asymptotes.

Symmetry in the Graphs of Even and Odd Functions

Consider a function f(x).
(a) Even Function (b) Odd Function
If f(–x) = f(x) for all x in the domain of f(x), If f(–x) = –f(x) for all x in the domain of f(x),
f(x) is called an even function. Its graph is f(x) is called an odd function. Its graph is
symmetrical about the y-axis. symmetrical about the origin.
e.g. e.g.

In each of the following, determine whether the function f(x) is an even function or an odd function.
[Nos. 1–3]
1. The graph 2. 3. The graph
of an even of an odd
y y y
function has function has
reflectional rotational
symmetry. symmetry.
y = f(x) y = f(x)
y = f(x) x x
O O
x
O

Instant Example 1 Instant Practice 1

Determine whether f(x) = x – x3 is an even function Determine whether f(x) = cos x is an even function
or an odd function. or an odd function.
3 3 3
f(–x) = –x – (–x) = –x + x = –(x – x ) = –f(x)
∴ f(x) = x – x3 is an odd function.

In each of the following, determine whether the function f(x) is an even function or an odd function or neither.
[Nos. 4–7]
4. f(x) = 2x4 + 3 5. f(x) = x3 + x2 Ex 5D: 1–4
 1 x
6. f(x) = x  7. f(x) = 3
x x 1

Horizontal and Vertical Asymptotes of a Curve y = f(x)

(a) If lim f(x) = k or lim f(x) = k, then y = k is a horizontal asymptote of the curve y = f(x).
x x  

(b) If lim f(x) =  or lim f(x) = , then x = a is a vertical asymptote of the curve y = f(x).
xa xa

Instant Example 2 Instant Practice 2

Find the horizontal and vertical asymptotes of the Find the horizontal and vertical asymptotes of the
1 1
graph of the function y = 2 . graph of the function y =  .
x x2
1 1
∵ lim =0  or lim 2 = 0
x   x2 x   x

∴ y = 0 is a horizontal asymptote.
1 1
∵ lim =∞  or lim =∞
x  0 x2 x  0 x2

∴ x = 0 is a vertical asymptote.

In each of the following, find the horizontal and vertical asymptotes of the graph of the function. [Nos. 8–13]
4 2x
8. f(x) = 9. f(x) = Ex 5D: 5–12
3 x x

x x6
10. f(x) = 11. f(x) =
x5 x 1
 3x  1 4x2  1
12. f(x) = 13. f(x) =
x2 x2  4

x2 – 4
= (x + ____)(x – ____)

Oblique Asymptotes
g ( x)
For a rational function f(x) = , if h(x) is not a factor of g(x) and the degree of g(x) is larger than that
h( x)
of h(x) by 1, then the graph of y = f(x) has an oblique asymptote.

Instant Example 3 Instant Practice 3

Find the oblique asymptote of the graph of the Find the oblique asymptote of the graph of the
x2 x2 1
function y = . function y = .
x 1 x
x –1
x2 1
= x 1  x + 1 ) x2 + 0x + 0
x 1 x 1 x2 + x
 x 2
 –x + 0
1
lim   ( x  1) = lim –x – 1
x  x 1 x  x 1 1
 
=0
∴ y = x – 1 is an oblique asymptote.

In each of the following, find the oblique asymptote of the graph of the function. [Nos. 14–17] Ex 5D: 13, 14

x2 x2  3
14. y = 15. y =
x2 x 1
x – 2 ) x2 + 0x + 0 x + 1 ) x2 + 0x – 3
 x2  x x 2  3x  1
16. y = 17. y =
x3 x2
x – 3 ) x2 + x + 0 x + 2 ) x2 – 3x + 1

In each of the following, find all the asymptotes of the graph of the function. [Nos. 18–19]
2 x x2  7x  5
18. y = 19. y =
x2  6x x4
x + 4 ) x2 + 7x + 5

Level Up Question
 x3  x 2  2x  3
20. Find all the asymptotes of the graph of the function y = .
x2 1
M2A Lesson Worksheet 5.3C (Refer to Book M2A P.5.27)

Objective: To sketch the graphs of polynomial and rational functions.

Graph Sketching
The following features can be considered when sketching the graph of a function:
(a) x- and y-intercepts
(b) Maximum and minimum points
(c) Concavity
(d) Points of inflexion
(e) Symmetry and asymptotes

Instant Example 1 Instant Practice 1

A polynomial f(x) has the following properties: A polynomial f(x) has the following properties:
(i) The graph of y = f(x) passes through (0 , 1). (i) The graph of y = f(x) passes through (4 , 0).
(ii) x x < –1 x = –1 –1 < x < 1 (ii) x x<2 x=2 2<x<6
f(x) / 3 / f(x) / –5 /
f (x) + 0 – f (x) – 0 +
x x=1 x>1 x x=6 x>6
f(x) –1 / f(x) 5 /
f (x) 0 + f (x) 0 –
(a) Find the maximum and minimum points of the (a) Find the maximum and minimum points of the
graph of y = f(x). graph of y = f(x).
(b) Sketch the graph of y = f(x). (b) Sketch the graph of y = f(x).

(a) From the table, the sign of f (x) changes from

positive to negative as x increases through –1.
∴ (–1 , 3) is a maximum point.
From the table, the sign of f (x) changes from
negative to positive as x increases through 1.
∴ (1 , –1) is a minimum point.

(b) y
(–1 , 3)

y = f(x)
(0 , 1)

x
O

(1 , –1)
1. A polynomial f(x) has the following properties:
(i) The graph of y = f(x) passes through (0 , 0).
(ii) x x < –3 x = –3 –3 < x < 3 x=3 x>3
f(x) / 2 / –2 /
f (x) + 0 – 0 +
(a) Find the maximum and minimum points of the graph of y = f(x).
(b) Sketch the graph of y = f(x). Ex 5D: 21, 22

2. A polynomial function f(x) has the following properties:

1 1 1
x x<0 x=0 0<x< x= <x<1 x=1 x>1
2 2 2
f(x) / 0 / 1 / 2 /
f (x) – 0 + + + 0 –
f (x) + + + 0 – – –
3
The x-intercepts of the graph of y = f(x) are 0 and .
2
(a) Find the maximum and minimum points and the point of inflexion of the graph of y = f(x).
(b) Sketch the graph of y = f(x).
3. Let f(x) = x4 – 2x2.
(a) Find the x- and y-intercepts of the graph of y = f(x).
(b) (i) Complete the following tables. Use ‘+’ and ‘–’ to denote ‘positive value’ and ‘negative
value’ respectively if necessary.
x x=0
f (x) 0

3
x x=–
3
f (x) 0
(ii) Hence, find the maximum and minimum points and the point(s) of inflexion of the graph of
y = f(x).
f(x) is an ( even / odd ) function.
(c) Sketch the graph of y = f(x) for –2 ≤ x ≤ 2. The graph of y = f(x) has
( reflectional / rotational ) symmetry.
(a)

(b) (i) f (x) = (c) f(–2) = Remember to find

f (x) = f(2) = the end-points.

(ii)
 12
4. Let f(x) = 2
.
x 3
(a) Find f (x).
72( x 2  1)
(b) It is given that f (x) = . Complete the following table. Use ‘+’ and ‘–’ to denote
( x 2  3) 3
‘positive value’ and ‘negative value’ respectively if necessary.
x x=0 x=1
f (x) 0
f (x)
(c) Find the maximum and/or minimum point(s) and the point(s) of inflexion of the graph of y = f(x).
(d) Find the asymptote of the graph of y = f(x).
If lim f ( x) = k or lim f ( x) = k, then
x x  
(e) Sketch the graph of y = f(x).
y = k is a ________________ asymptote.

Level Up Question
2
5. Let f(x) = 1  2 .
x
x x=0
(a) Find f (x) and f (x). Hence, complete the table.
f (x) undefined
(b) Find the asymptotes of the graph of y = f(x).
f (x)
(c) Sketch the graph of y = f(x).
(Use ‘+’ and ‘–’ to denote ‘positive value’ and
‘negative value’ respectively if necessary.)

If lim f ( x) =  or lim f ( x) = , then x = a is a __________ asymptote.

x  a x  a
M2A Lesson Worksheet 5.4A (Refer to Book M2A P.5.37)

Objective: To apply differentiation to solve problems relating to rate of change.

Rate of Change
dy
If y is a quantity varied with time t, then is the rate of change of y with respect to t.
dt

[In this worksheet, give the answers correct to 3 significant figures if necessary.]

Instant Example 1 Instant Practice 1

A particle is moving along a straight line and its A particle is moving along a straight line and its
displacement s m from a fixed point after t seconds displacement s m from a fixed point after t seconds
is given by s = 2 + 7t – 3t . Find the velocity and the is given by s = t2 – 3t – 8. Find the velocity and the
2

acceleration of the particle at t = 1. acceleration of the particle at t = 4.

Let v m/s and a m/s2 be the velocity and the

acceleration respectively at time t.
ds Velocity is the rate of change of
v = = 7 – 6t displacement with respect to time.
dt
dv Acceleration is the rate of change
a = = –6
dt of velocity with respect to time.
When t = 1, v = 7 – 6(1) = 1
a = –6
∴ The velocity is 1 m/s and the acceleration
is –6 m/s2.

1. An object is moving along a straight line 2. An object is moving along a straight line in a
such that its displacement s m from a fixed way that its displacement x m from a fixed
point after t seconds is given by point varies with the time t s as follows:
s = 13 – 8t – 5t2. Find the velocity and the x = 4t3 + 2t2 – 6t + 11
acceleration of the object at t = 2. Find the velocity and the acceleration of the
object at t = 3. Ex 5E: 1
Instant Example 2 Instant Practice 2
Oil is dropped on a table and it covers a region Air is pumped into a spherical balloon with initial
which is circular in shape. The initial area of the volume 16 cm3 at a constant rate of 4 cm3/s. Let
region is 10 cm2. The area of the region is V cm3 and r cm be the volume and the radius of
increasing at a constant rate of 2 cm2/s. Let A cm2 the balloon respectively after t seconds.
and r cm be the area and the radius of the region (a) Express V and r in terms of t.
respectively after t seconds. (b) Find the rate of change of the radius of the
(a) Express A and r in terms of t. balloon at t = 2.
(b) Find the rate of change of the radius of the Volume of
sphere
region at t = 5.
4 3
= r
3
(a) A is increased by 2t at time t.
∴ A = 10 + 2t
∵ A = r2
A
∴ r=
π
10  2t
r
π
1 1
 
dr 1  10  2t  2 2 1  10  2t  2
(b) =    =  
dt 2  π  π π π 
When t = 5,
1

dr 1 10  2(5)  2
=
dt π  π 

= 0.126, cor. to 3 sig. fig.
∴ The radius of the region is increasing
at a rate of 0.126 cm/s.

3. The initial volume of a right circular cylinder is 12 m3. The volume of the cylinder is decreasing at a
constant rate of 0.1 m3/s. Let V m3, r m and 2r m be the volume, the base radius and the height of the
cylinder respectively after t seconds.
(a) Express V and r in terms of t. Volume of (b) Find the rate of change of the base radius
cylinder
= r2h
of the cylinder at t = 6. Ex 5E: 3, 4

The radius is
decreasing.
dr
is ( positive /
dt
negative ).
4. The radius of a hemisphere is decreasing at a 5. An empty container is in the form of an
constant rate of 2 cm/s. When the radius of the inverted right circular cone with base radius
hemisphere is 12 cm, find the rate of decrease of 10 cm and height 20 cm. Water is filled into
its curved surface area. Ex 5E: 5 the container at a constant rate of 3 cm3/s.
Curved surface Find the rate of increase of the water level
area of hemisphere
1 when the water is 1 cm deep. Ex 5E: 7
= (4r2)
2 Volume of
cone
1 2
= r h
3

6. Lisa is standing below a helicopter which is flying in the air at 800 m above her. Then, the helicopter
flies horizontally along a straight path at a constant speed of 45 m/s. Find the rate of change of the
angle of elevation of the helicopter from Lisa when the helicopter travels 200 m.

800 m
7. In the figure, boat X is sailing due east of pier P at a constant speed of 15 km/h and boat Y is sailing
due south of P at a constant speed of 20 km/h. At a moment, X and Y are 100 km and 80 km away
from P respectively. Find the rate of increase of the distance between X and Y at that moment.
15 km/h
P boat X

boat Y
N
20 km/h

Level Up Question
8. A piece of string of 50 cm long is bent to form a right-angled triangle with base x cm and height y cm.
It is given that xy – 50x – 50y + 1 250 = 0. If x increases at a constant rate of 0.5/s, find the rate of
change of y when x = 5.
M2A Lesson Worksheet 5.4B (Refer to Book M2A P.5.44)

Objective: To apply differentiation to solve problems relating to maximum and minimum.

General Steps in Solving a Problem on Maximization and Minimization

Step 1: Define the variables (independent and dependent variables).
Step 2: Express the quantity to be maximized or minimized as a function of an independent variable.
Step 3: By using differentiation, find the absolute maximum or minimum of the function as required by
the problem.
Step 4: Write down the answer in words.

Instant Example 1 Instant Practice 1

A wire of length 100 cm is bent to form a rectangle A paper sector is folded into a right circular cone
with length x cm and width y cm. Let A cm2 be the with base radius r cm and slant height  cm. The
area of the rectangle. perimeter of the sector is 12 cm. Let A cm2 be the
(a) Show that A = 50x – x2. curved surface area of the cone.
(b) Hence, find the length of the rectangle so that (a) Show that A = 2(6r – r2).
its area is a maximum. (b) Hence, find the base radius of the cone so that
its curved surface area is a maximum.
(a) 2(x + y) = 100
Express  in terms of r.
x + y = 50 Then express A in
terms of r.
 y = 50 – x
∵ A = xy
∴ A = x(50 – x)
A = 50x – x2  Step 2: Express A in
terms of one variable x.
dA
(b) = 50 – 2x  Step 3: Use the first/
dx second derivative test.
dA
When = 0, 50 – 2x = 0
dx
x = 25
When x = 25, A = 50(25) – 252 = 625.

x x < 25 x = 25 x > 25
dA
+ 0 –
dx
∵ A has only one relative maximum but no
relative minimum on the interval.
∴ A has the absolute maximum at x = 25.
∴ The length of the rectangle is 25 cm so
that its area is a maximum.  Step 4
1. An open cuboid container is made from metal 2. An open right cylindrical can is made from
3
sheet to hold 500 cm of water. The base of the 108 cm2 of iron sheet. Let r cm and C cm3
container is a square of side x cm. Let A cm2 be the base radius and the capacity of the
be the total area of the material used. can respectively.
2 000 π
(a) Show that A = x2 + . (a) Show that C = (108r – r3).
x 2
(b) Find the minimum total area of the (b) Find the maximum capacity of the can
material used and the corresponding and the corresponding base radius.
length of the base of the container. Ex 5F: 11, 12
3. In the figure, a straight line L cuts the x-axis 4. Find the coordinates of the point on the
and the y-axis at (p , 0) and (0 , q) parabola y = x2 that is closest to (3 , 0).
respectively, where p > 0 and q > 0. L also Hence, find that shortest distance.
passes through (4 , 6). Ex 5E: 13, 14
y

(0 , q)
(4 , 6)
x
O (p , 0) L
(a) Express p in terms of q.
(b) Find the minimum area of the region
bounded by L, the x-axis and the y-axis.
For (a), consider
two different
ways to express
the slope of L.
5. The owner of a house is building a rectangular backyard PQRS of area 324 m2. The three sides of the
backyard will be rebuilt into straight paths and the remaining part will become a small farm as shown
in the figure. Each of the paths on the left and right sides is 2 m wide. The path on the remaining side
is 1 m wide. Find the dimensions of the small farm such that its area is a maximum.
P Q

Backyard 2m Small Farm 2m

1m
S R
First find a relation between the dimensions of
the farm and the dimensions of the backyard.

Level Up Question
6. In the figure, Andrew is 16 km due north of a point X and Bolton is 15 km due east of X. They start to
cycle at the same time. Andrew cycles due south at a constant speed of 18 km/h. Bolton cycles due
west at a constant speed of 12 km/h.
(a) Show that the distance between Andrew and Bolton after t hours is 468t 2  936t  481 km.
(b) Can the distance between Andrew and Bolton be less than 3.5 km? Explain your answer.
Andrew N
18 km/h

16 km
12 km/h
Bolton
X 15 km